url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://stats.stackexchange.com/questions/467265/distributions-on-an-ellipsoid | 1,620,979,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00626.warc.gz | 545,438,680 | 38,248 | # Distributions on an ellipsoid?
This is my first question. I hope somebody can help me with this. I am after the name of a distribution and references would be incredibly useful. I am looking for a generalisation of the von-Mises distribution (on the unit circle) or von-Mises Fisher distribution (on the unit sphere) to an ellipsoid. I assume this has been done before, but I have not been successful in finding any relevant literature.
• In what way do you wish to make the analogue on an ellipse? What properties should the function have and how will they be defined? The reason that I am asking is because for circular distribution I am always thinking about a parameter with a uniform metric like the angle, ranging from 0 to 2pi. But what is it for an ellipse? Also an angle ranging from 0 to 2pi, but with some different metric? – Sextus Empiricus May 19 '20 at 7:59
• Hi, thank you for responding. I am interested in distributions akin to the ones above, but on an ellipsoid not an ellipse. Although, if you happen to know generalisations to an ellipse (for vM), than that would also be useful. What I am looking for is a symmetric unimodal distribution that is like the normal distribution in Euclidean space, but on an ellipisoid. I did try searching, but couldn't find any distributions defined on an ellipisoid or even an ellipse. – dataMonkey May 19 '20 at 8:21
• "that is like the normal distribution in Euclidean space" What do you mean by 'is like'? – Sextus Empiricus May 19 '20 at 8:37
• Because an ellipsoid is a sphere when the coordinate axes are aligned with the ellipsoid's axes and suitably scaled, there's nothing to do: all spherical distributions can be applied to ellipsoids without modification. That might explain why it's hard to find literature specific to ellipsoids. – whuber May 19 '20 at 11:34
• @whuber: do you want to post your comment(s) as an answer? Better to have a short answer than no answer at all. Anyone who has a better answer can post it. – Stephan Kolassa Jul 24 '20 at 6:49 | 494 | 2,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.928539 |
https://www.examrace.com/ACET/ACET-Model-Questions/Quantitative-Analysis-Questions/Mathematics-Exponents.html | 1,643,132,571,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304859.70/warc/CC-MAIN-20220125160159-20220125190159-00578.warc.gz | 789,277,825 | 5,325 | # Exponents: Practice Questions ACET
Get top class preparation for ACET right from your home: get questions, notes, tests, video lectures and more- for all subjects of ACET.
1. 0.00875 equals:
1. 8.75 × 10 − 2
2. 8.75 × 10 − 3
3. 8.75 × 10 − 3
4. 87.5 × 10 − 4
5. 87.513 × 104
2. 104 is not equal to which of the following?
1. 100,000
2. 1.0 × 10
3. 10 × 10 × 10 × 10
4. 10 × 10
5. 10,000
3. Multiply 104 by 102
1. 10
2. 10
3. 10
4. 10
5. 101
4. Divide x5 by x2
1. x
2. x
3. x
4. x
5. x2.5
5. Find 8.23 × 109
1. 0.00000000823
2. 0.000000823
3. 8.23
4. 8230000000
5. 823000000000 | 289 | 584 | {"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} | 3.375 | 3 | CC-MAIN-2022-05 | latest | en | 0.569703 |
https://nrich.maths.org/187 | 1,623,705,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00462.warc.gz | 407,117,170 | 4,602 | ### Homes
Six new homes are being built! They can be detached, semi-detached or terraced houses. How many different combinations of these can you find?
### Stairs
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Train Carriages
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
# Teddy Bear Line-up
##### Age 5 to 7Challenge Level
Lachlan was playing with his bear counters.
He had four blue, then four red followed by four yellow and finally four green bears.
"What are you doing now?" Jenni asked.
"I want to arrange them so that no two bears of the same colour are next to each other", he said.
"That's easy!" cried Jenni.
"Ah, but you have to do it in the least number of moves possible", replied Lachlan.
What's the least number of moves you can take to rearrange the bears? | 233 | 986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.973886 |
http://www.cs.tau.ac.il/~efif/courses/Software1_Spring_03/ | 1,603,381,820,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00549.warc.gz | 132,827,238 | 3,739 | 0368-2157-12 0368-2157-09
# Software 1 Spring 2003
## News
• A review practice will be held at Orenstein 102 on Thu. the 19th. at 16:00.
• I'm offering a review practice on Thu. the 19th, before the final. If you are interested, send an email message to efif, with the title "Software 1". Feel free to ask for spesific topics you would like me to cover. If I get at least 10 requests, I'll reserve a class room and announce the location and exact time in this page.
No. of requests: 10
• Instructors
1. Prof. Dan Halperin, 0368-2157-12, course page
2. Iris G., 0368-2157-09
• The time of the practice
CourseDayTimeLocation
0368215713Sunday10:00 - 11:00Kaplun 118
0368215710Sunday11:00 - 12:00Kaplun 118
0368215711Sunday12:00 - 13:00Kaplun 118
0368215714Tuesday15:00 - 16:00Orenstein 102
• Office hours: Sunday, 15:00-16:00. Location: Computational Geometry Lab. Schreiber basement 18.
• By popular demand, the due date of assignment 6 has moved 2 days back. The new deadline is Tue. Jun. 10th midnight.
• Assignment No. 6 has been posted. You need to complete it by the end of the semester. Its due data is June 8th, and there are no special allowances. Take a look at assign6 for a list of standard C functions I used in my implementation, and more.
• If there are at least 3 students interested, I'll extend the practice on Tue. May the 13th by ~30 minutes to make up for last week.
• The due date of assignment No. 4 has been postponed to Sun. May 4th. Nevertheless, assignment No. 5 has been posted. I suggest for those that don't need the extra days, to start working on the new one.
• The formula for the cubic B-spline originally presented in the assignment has an error. The correct formula is:
`R(x) = 1/6 ( P(x+2)3 - 4P(x+1)3 + 6P(x)3 - 4P(x-1)3 )`
Notice the coefficients of the 2nd and 3rd terms. Thanks for Omri Iluz for pointing it out me.
• A make up practice is going to be held on Sun. Apr. the 13th at Orenstein 102 between 14:00-15:00. I'll concentrate on the photo program (needed for assignment 4). You are all invited.
• There is going to be a make up practice at Orenstein 102, 15:00-16:00 on Thu. the 10th, instead of the Tue. practice that didn't take place (everyone is invited). For those of you that cannot make it, I may conduct another practice on Sun. the 13th. If you are interested in a practice on Sunday, send a message to me and choose the time convenient to you, either 14:00-15:00, 15:00-16:00, or 16:00-17:00. I'll reply back and post the final time if at all.
• Assignment No. 4 has been posted.
• See ~tochna1/03b/assign3 for examples for the exercises of assignment 3
• The 2nd round of the checking of assignment 1 has been completed, and a response has been sent to all registered students via email. If you haven't received an email, please contact me.
• Assignment No. 3 has been posted.
• All the mistakes in the examples for the exercises of assignment-2 have been fixed. Make sure to test your programs and compare the results.
• See ~tochna1/03b/assign2 for examples for the exercises of assignment 2
• There won't be any practice on Tue. Mar. 19th due to Purim. I suggest those student enroled to this group, to join one of the practices on Sun. If there is a demand, I'll offer a make up practice the week after.
• No office hours this Sun. Mar. the 16th.
• Assignment No. 2 has been posted.
• See ~tochna1/03b/assign1 for examples for the exercises of assignment 1
```/* .....
* Full Name: you-full-name
* Id No: your-id
* User Name: your-user-name
* Group No: your-group-number
* Assignment No: the-assignment-number
* .....
*/
```
• Correction to exercise 1 of Assignment 1 (d2u, u2d). In DOS format the '\n' follows the '\r' to form "\r\n", and not vise verse.
• Assignment No. 1 has been posted.
• On the second week of the semester Yaniv, one of the system administrators, will give a presentation about all kind of system related issues relevant to this course. There will be three identical presentations back to back on Sun. the 23rd 10:00 - 11:00, 11:00 - 12:00, and 12:00 - 13:00 in Kaplun 118. There won't be any other practices during this week. Students registered to the Tuesday practice are invited to attend one of the three presentations on Sunday instead. | 1,221 | 4,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | latest | en | 0.91113 |
https://www.kgdcgymandcheer.com/answers-to-questions-about-gymnastics/faq-10-gymnastics-how-calculated.html | 1,632,064,054,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00090.warc.gz | 857,099,058 | 23,210 | # FAQ: 10 Gymnastics How Calculated?
## How are gymnastics scores calculated?
Score for each gymnast is determined by adding together the score for the routine’s content (Difficulty Score ) and execution (Execution Score ). Each routine was given a start value, and the actual score was the total of credit given for the routine minus deductions for execution.
## How old are Level 10 gymnasts?
There are three optional only levels: 8,9, 10. The minimum age for level 8 is 8 years old, while for levels 9 and 10, it is 9 years of age. Level 9 is the second level of optional competition. Its difficulty requirements and expectations are accordingly more difficult than at level 8.
## How do you get a perfect 10 in gymnastics?
Execution scores are still out of 10, so the theoretical possibility exists for a gymnast to get a “perfect 10 ” for execution in addition to whatever number they get for difficulty, but no such score has been awarded in decades.
You might be interested: Quick Answer: What Is A Layout In Gymnastics?
## What are gymnastics scores out of?
Individually, the scoring system is pretty straight-forward. Every gymnast starts from a 10.0. Throughout the routine, the judges deduct points, tenths, and even hundredths for mistakes in execution. Once the routine is over, the final score is tallied and the result is posted.
## What do gymnastic judges look for?
Judges are looking for things like amplitude, expression, maturity, confidence, and personal style on these events in the way that the gymnast performs her routine. Specific deductions are used when a gymnast fails to demonstrate these qualities.
## What is the max score in gymnastics?
4 Perfect 10s in Gymnastics History Traditionally, the highest attainable score in artistic gymnastics was a perfect 10. This is because, up until 2006, a competitive gymnast began a routine with a start value of 10 and then had points deducted over the course of their routine.
## What level is Simone Biles?
Simone Biles
Discipline Women’s artistic gymnastics
Level Senior international elite
Years on national team 2012–2016, 2018–present (US)
Gym World Champions Centre (current) Bannon’s Gymnastix Inc. (2003–2014)
12
## Is 12 too old to start gymnastics?
You can begin gymnastics at almost any age you develop an interest, but you may want to stick with recreational gymnastics if you start older than 12. Starting later than 12 years old may not give you enough time to develop the skills you need to go up against people who have been at it since they were toddlers.
## What do Level 10 gymnasts do?
Level 10 is the highest level in the USA Gymnastics Junior Olympics Program. The level is open to women’s artistic, men’s artistic, trampoline, acrobatic and rhythmic gymnasts.
You might be interested: Question: When Did The Shape Of The Gymnastics Vault Change?
## Has any gymnast gotten a 10?
Until the 1976 Summer Olympics no -one had ever achieved a perfect score in gymnastics. Indeed, the scoreboard only had three digits, to display scores such as 9.50, or 9.85. Nadia Comaneci, a 14-year-old Romanian gymnast competing at her first Olympics was to change that.
## Did Simone Biles get a perfect 10?
Well, don’t weep for the perfect 10. Its destruction created this glorious and implausible athlete called Simone Biles. Let’s talk about the perfect 10 for a moment. I was just 9 years old when Romania’s Nadia Comaneci scored the first 10 in Olympic gymnastics history, but I remember it.
## Why did Katelyn Ohashi quit?
Ohashi, now 22, was once an Olympic hopeful—she beat Simone Biles in competition in 2013 with an extraordinarily difficult balance beam routine. But after a series of injuries and a battle with body shaming, Ohashi decided to quit elite gymnastics and enroll in UCLA.
## What is the hardest skill in gymnastics?
The Produnova It takes a daredevil to perform a Produnova, the hardest Vault in Women’s Gymnastics.
## What level are most college gymnasts?
Level 10 and elites usually become college gymnasts. It is important to be a level 10 or elite while looking for a college to do gymnastics at because NCAA Gymnastics competes at a level 10 difficulty.
## Who is the best gymnast in the world?
World Artistic Gymnastics Championships
Rank Gymnastics Nation
1 Simone Biles United States
2 Svetlana Khorkina Russia
3 Gina Gogean Romania
4 Larisa Latynina Soviet Union
18 | 1,034 | 4,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.94274 |
https://brainly.in/question/28736 | 1,487,687,784,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170741.49/warc/CC-MAIN-20170219104610-00162-ip-10-171-10-108.ec2.internal.warc.gz | 721,523,138 | 9,703 | # One kilometre is equal to how many miles ?
2
by pravin3103
• Brainly User
2014-07-27T13:47:37+05:30
One kilometer =1/3 miles
hope it helped u
2014-07-27T13:48:53+05:30
1km=.33333.....miles | 81 | 192 | {"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-2017-09 | latest | en | 0.617373 |
http://mathhelpforum.com/calculus/140337-help-me-find-limit-integral-problem-i-m-lost.html | 1,480,812,314,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00044-ip-10-31-129-80.ec2.internal.warc.gz | 178,298,380 | 11,006 | # Thread: Help me find the limit? Is this an integral problem? I'm lost...
1. ## Help me find the limit? Is this an integral problem? I'm lost...
Compute the limit
limit as n approaches infinity of [1+(x/n)]^(n)
exactly, for each fixed x and use your calculator to write a decimal approximation of the limit when x = 1.
Basically, I don't know what the question is asking or how to approach the problem. If somebody could break it down for me that would be great.
We have a test tomorrow on integration and he wants us to use what we learned from that chapter.
Also, if somebody could provide a link that might benefit me I'd appreciate that too... the biggest problem I'm having is that I really don't know WHAT to google in this case. lol
Thanks guys.
2. Test it out.
$\left(1 + \frac{x}{1}\right)^{1}\;=\;1 + x$
$\left(1 + \frac{x}{2}\right)^{2}\;=\;1 + x + \frac{x^{2}}{4}$
$\left(1 + \frac{x}{4}\right)^{4}\;=\;1 + x + \frac{3x^{2}}{8}+ \frac{x^{3}}{16}+ \frac{x^{4}}{256}$
Are we getting anywhere? If we've done nothing else, it is hoped that you might be discouraged from determining the limit to be unuty, (1). It's more than 1+x, whatever it is.
We could also try the Binomial Theorem on it. Is that in your bag of tricks?
OR, you could find the limit of the logarithm and see where that leads.
$log[(1+x/n)^{n}] = n\cdot log(1+x/n) = \frac{log(1+x/n)}{\frac{1}{n}}$
Your task is to determine why on earth I would write it in that last form.
3. I would imagine that he wants us to use the last method you provided.
And the reason you would do so would be to relocate the nth power to the front of the equation, right?
4. Also, using the first method... I don't know the binomial theorem well enough to use it, but taking the limit of the equation to the nth power... through using l'Hopitals Rule, would the solution be x/2?
The last bit of the equation would be limit (x^(n))/(n^(2)) =LH= limit nx/2n = x/2?
5. It seems like you are on the edge of understanding. The purpose of the last form is to demonstrate the applicability of l'Hopital.
Rewrite the argument of the logarithm, and then try l'Hopital.
$
\frac{log(\frac{n+x}{n})}{\frac{1}{n}}
$ | 621 | 2,181 | {"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": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-50 | longest | en | 0.930953 |
https://plainmath.net/98021/if-the-function-f-x-log-2x-x-2-is-giv | 1,670,083,873,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00749.warc.gz | 478,816,685 | 12,131 | # If the function f(x)=log_(2x) x^2 is given, what is f′(4)?
If the function $f\left(x\right)={\mathrm{log}}_{2x}{x}^{2}$ is given, what is ${f}^{\prime }\left(4\right)$?
I tried to use the formula for derivative of logarithm but here the base is 2x, so it made me confused.
Note that the answer is $1/\left(18\mathrm{ln}2\right)$
You can still ask an expert for help
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Faith Wise
Hint:
Note that:
${\mathrm{log}}_{b}a=\frac{\mathrm{ln}a}{\mathrm{ln}b}$
Now you can use the formula for the derivative of a fraction. | 216 | 687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "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} | 3.515625 | 4 | CC-MAIN-2022-49 | latest | en | 0.875026 |
http://convertit.com/Go/SalvageSale/Measurement/Converter.ASP?From=chain | 1,632,652,466,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00201.warc.gz | 15,571,661 | 3,430 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```surveyors chain = 20.1168 meter (length) ``` Related Measurements: Try converting from "chain" to bolt (of cloth), caliber (gun barrel caliber), cloth finger, engineers chain, fermi, foot, football field, gradus (Roman gradus), Greek fathom, inch, Israeli cubit, ken (Japanese ken), m (meter), point (typography point), ri (Japanese ri), Roman cubit, sazhen (Russian sazhen), shaku (Japanese shaku), sun (Japanese sun), vara (Mexican vara), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: chain = 201,168,000,000 angstrom, 2,376 barleycorn, 44 Biblical cubit, .55 bolt (of cloth), .09166667 cable length, 79,200 caliber (gun barrel caliber), .22 city block (informal), 88 cloth quarter, 17.6 ell, 2.01E+16 fermi, 60,350.4 French, 198 hand, 792 inch, .0125 mile, .73333333 naval shot, 4,752 pica (typography pica), .01359423 Roman mile, 66 survey foot, .01249998 UK mile (British mile), 24.01 vara (Mexican vara).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 443 | 1,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-39 | longest | en | 0.676492 |
https://brainmass.com/economics/utility/utility-functions-498455 | 1,713,834,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00858.warc.gz | 129,916,297 | 7,273 | Purchase Solution
Utility Functions
Not what you're looking for?
Attached are 2 utility functions that I'm having trouble with.
Problem 3
For each of the following utility functions, draw indifference curves for utility level 12 and 16. Indicate three bundles on each of these indifference curves by specifying the coordinates of each bundle. Shade the weakly preferred set of bundles you indicated on the indifference curves of utility 16. For each of the utility functions, determine and explain whether the preferences they represent are convex, concave, or non-convex preferences.
a) U_3 (x_1,x_2 )=?x_1?^2-?2x?_2;
b) U_4 (x_1,x_2 )=Min{4x_1-4x_2,2x_1-x_2}.
Solution Summary
The solution includes drawings of the requested utility function, as well as a full justification of why they are drawn that way. Additionally, example bundles are given for each of the functions.
Solution Preview
The solution to this problem is attached. Please let me know if you have any questions.
I hope this helps!
-------
Text included for search purposes:
Problem 3
For each of the following utility functions, draw indifference curves for utility level 12 and 16. Indicate three bundles on each of these indifference curves by specifying the coordinates of each bundle. Shade the weakly preferred set of bundles you indicated on the indifference curves of utility 16. For each of the utility functions, determine and explain whether the preferences they represent are convex, concave, or non-convex preferences.
a) U_3 (x_1,x_2 )=?x_1?^2-?2x?_2;
b) U_4 (x_1,x_2 )=Min{4x_1-4x_2,2x_1-x_2}.
In order to complete this problem, the first step is to obtain x2 as a function of x1. This will allow us to draw the indifference curves. This can be done as follows. For the utility function in (a):
Where U is any given utility level (in this particular case, the utility levels of interest are 12 and 16). Note that the final equation above ...
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Lecture 26
# EPS 7 Lecture Notes - Lecture 26: Sea Level Rise, Tacticity, Thermal Expansion
1 Page
96 Views
Fall 2017
Department
Earth And Planetary Science
Course Code
EPS 7
Professor
David Romps
Lecture
26
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Lecture 25 10/25/17
Ice and sea level: Where to invest in property
Paleo sea level data exhibit no discernable change in sea level over the past 3000
years
- Salt marshes and tide gauges
- Tide gauges
- Satellite altimetry
- Everything together
Observations show a sea level rise of 0.3 m (a foot) so far
With sudden warming, sea level rises at an initial rate of roughly 3mm/year/K
- T = +1K, 3mm/ year
- T = +2K, 6mm/year
- T = +3K, 9mm/year
Climate has warmed (on average) +1K = 0.3m
In the U.S. and most of Europe, sea level rise is around the global average of
3mm/year
In parts of Canada and northern Europe, sea level is falling due to isotactic rebound
➢ During a glacial vs. during an interglacial
Isotactic rebound in mm/year (blue is positive)
For a sustained amount of global warming, the committed sea-level rise is the
eventual amount of rise you will get once the ocean and ice sheets have equilibrated
to the new global temperature
To understand sea-level rise, you must understand history
Forecast year 2100, cant tell
With warming, the committed sea-level rise is about 6m/K (what is coming to you)
- T = +1K, 6m
- T = +2K, 12m
- T = +3K, 18m
- If the mean global warming hangs out at +3K for thousands of years, roughly what
is the final sea-level rise?
20m (very uncertain)
- What is the timescale for adjustment to the new sea level?
We have two numbers
3mm/year/K and 6m/K
3mm/year/K = 3mm/K/year
6m/K = (6 m/K) X (1000 mm/m) = 6000 mm/K
(6000 mm/K) / (3mm/K/year) = 2000 years
Timescale for ice sheets to respond to warming is about 2000 years
PCP 8.5, sea level will have risen about 1m in 2100
-What contributes to sea level rise?
- Thermal expansion is currently contributing about half of the sea-level rise
- How much will melting Arctic sea ice contribute to sea-level rise?
Practically zero
- Why?
Archimedes principle says that a floating objects displaces its weight in water
When floating ice melts, it simply fills out the space it takes
Only the ice out of the water contributes to sea level rise
When an ice sheet melts, it pours into the ocean and raises the sea level
GRACE observations and Antarctic Ice Mass Change
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Description
Lecture 25 102517 Ice and sea level: Where to invest in property Paleo sea level data exhibit no discernable change in sea level over the past 3000 years Salt marshes and tide gauges Tide gauges Satellite altimetry Everything together Observations show a sea level rise of 0.3 m (a foot) so far With sudden warming, sea level rises at an initial rate of roughly 3mmyearK T = +1K, 3mm year T = +2K, 6mmyear T = +3K, 9mmyear Climate has warmed (on average) +1K = 0.3m In the U.S. and most of Europe, sea level rise is around the global average of 3mmyear In parts of Canada and northern Europe, sea level is falling due to isotactic rebound During a glacial vs. during an interglacial Isotactic rebound in mmyear (blue is positive) For a sustained amount of global warming, the committed sealevel rise is the eventual amount of rise you will get once the ocean and ice sheets have equilibrated to the new global temperature To understand sealevel rise, you must understand history Forecast year 2100, cant tell With warming, the committed sealevel rise is about 6mK (what is coming to you) T = +1K, 6m T = +2K, 12m T = +3K, 18m If the mean global warming hangs out at +3K for thousands of years, roughly what is the final sealevel rise? 20m (very uncertain) What is the timescale for adjustment to the new sea level? We have two numbers 3mmyearK and 6mK 3mmyearK = 3mmKyear 6mK = (6 mK) X (1000 mmm) = 6000 mmK (6000 mmK) (3mmKyear) = 2000 years Timescale for ice sheets to respond to warming is about 2000 years PCP 8.5, sea level will have risen about 1m in 2100 What contributes to sea level rise? Thermal expansion is currently contributing about half of the sealevel rise How much will melting Arctic sea ice contribute to sealevel rise? Practically zero Why? Archimedes principle says that a floating objects displaces its weight in water When floating ice melts, it simply fills out the space it takes Only the ice out of the water contributes to sea level rise When an ice sheet melts, it pours into the ocean and raises the sea level GRACE observations and Antarctic Ice Mass Change
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So we can recommend you notes for your school. | 1,653 | 6,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-43 | longest | en | 0.858064 |
https://man.linuxreviews.org/man3/cspmv.f.3.html | 1,721,532,768,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00635.warc.gz | 321,486,214 | 2,804 | # cspmv.f
Section: LAPACK (3)
Updated: Tue Nov 14 2017
Page Index
cspmv.f
## SYNOPSIS
### Functions/Subroutines
subroutine cspmv (UPLO, N, ALPHA, AP, X, INCX, BETA, Y, INCY)
CSPMV computes a matrix-vector product for complex vectors using a complex symmetric packed matrix
## Function/Subroutine Documentation
### subroutine cspmv (character UPLO, integer N, complex ALPHA, complex, dimension( * ) AP, complex, dimension( * ) X, integer INCX, complex BETA, complex, dimension( * ) Y, integer INCY)
CSPMV computes a matrix-vector product for complex vectors using a complex symmetric packed matrix
Purpose:
``` CSPMV performs the matrix-vector operation
y := alpha*A*x + beta*y,
where alpha and beta are scalars, x and y are n element vectors and
A is an n by n symmetric matrix, supplied in packed form.
```
Parameters:
UPLO
``` UPLO is CHARACTER*1
On entry, UPLO specifies whether the upper or lower
triangular part of the matrix A is supplied in the packed
array AP as follows:
UPLO = 'U' or 'u' The upper triangular part of A is
supplied in AP.
UPLO = 'L' or 'l' The lower triangular part of A is
supplied in AP.
Unchanged on exit.
```
N
``` N is INTEGER
On entry, N specifies the order of the matrix A.
N must be at least zero.
Unchanged on exit.
```
ALPHA
``` ALPHA is COMPLEX
On entry, ALPHA specifies the scalar alpha.
Unchanged on exit.
```
AP
``` AP is COMPLEX array, dimension at least
( ( N*( N + 1 ) )/2 ).
Before entry, with UPLO = 'U' or 'u', the array AP must
contain the upper triangular part of the symmetric matrix
packed sequentially, column by column, so that AP( 1 )
contains a( 1, 1 ), AP( 2 ) and AP( 3 ) contain a( 1, 2 )
and a( 2, 2 ) respectively, and so on.
Before entry, with UPLO = 'L' or 'l', the array AP must
contain the lower triangular part of the symmetric matrix
packed sequentially, column by column, so that AP( 1 )
contains a( 1, 1 ), AP( 2 ) and AP( 3 ) contain a( 2, 1 )
and a( 3, 1 ) respectively, and so on.
Unchanged on exit.
```
X
``` X is COMPLEX array, dimension at least
( 1 + ( N - 1 )*abs( INCX ) ).
Before entry, the incremented array X must contain the N-
element vector x.
Unchanged on exit.
```
INCX
``` INCX is INTEGER
On entry, INCX specifies the increment for the elements of
X. INCX must not be zero.
Unchanged on exit.
```
BETA
``` BETA is COMPLEX
On entry, BETA specifies the scalar beta. When BETA is
supplied as zero then Y need not be set on input.
Unchanged on exit.
```
Y
``` Y is COMPLEX array, dimension at least
( 1 + ( N - 1 )*abs( INCY ) ).
Before entry, the incremented array Y must contain the n
element vector y. On exit, Y is overwritten by the updated
vector y.
```
INCY
``` INCY is INTEGER
On entry, INCY specifies the increment for the elements of
Y. INCY must not be zero.
Unchanged on exit.
```
Author:
Univ. of Tennessee
Univ. of California Berkeley | 837 | 2,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.781351 |
https://manoptjl.org/stable/solvers/convex_bundle_method/ | 1,721,435,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00168.warc.gz | 329,748,295 | 7,928 | Convex bundle method
Manopt.convex_bundle_methodFunction
convex_bundle_method(M, f, ∂f, p)
perform a convex bundle method $p_{j+1} = \mathrm{retr}(p_k, -g_k)$, where $\mathrm{retr}$ is a retraction and
$$$g_k = \sum_{j\in J_k} λ_j^k \mathrm{P}_{p_k←q_j}X_{q_j},$$$
$p_k$ is the last serious iterate, $X_{q_j} ∈ ∂f(q_j)$, and the $λ_j^k$ are solutions to the quadratic subproblem provided by the convex_bundle_method_subsolver.
Though the subdifferential might be set valued, the argument ∂f should always return one element from the subdifferential, but not necessarily deterministic.
For more details, see [BHJ24].
Input
• M: a manifold $\mathcal M$
• f: a cost function $f:\mathcal M→ℝ$ to minimize
• ∂f: the subgradient $∂f: \mathcal M → T\mathcal M$ of f restricted to always only returning one value/element from the subdifferential. This function can be passed as an allocation function (M, p) -> X or a mutating function (M, X, p) -> X, see evaluation.
• p: (rand(M)) an initial value $p_0 ∈ \mathcal M$
Optional
• atol_λ: (eps()) tolerance parameter for the convex coefficients in λ.
• atol_errors: (eps()) tolerance parameter for the linearization errors.
• m: (1e-3) the parameter to test the decrease of the cost: $f(q_{k+1}) \le f(p_k) + m \xi$.
• diameter: (50.0) estimate for the diameter of the level set of the objective function at the starting point.
• domain: ((M, p) -> isfinite(f(M, p))) a function to that evaluates to true when the current candidate is in the domain of the objective f, and false otherwise, for example domain = (M, p) -> p ∈ dom f(M, p) ? true : false.
• k_max: upper bound on the sectional curvature of the manifold.
• evaluation: (AllocatingEvaluation) specify whether the subgradient works by allocation (default) form ∂f(M, q) or InplaceEvaluation in place, that is of the form ∂f!(M, X, p).
• inverse_retraction_method: (default_inverse_retraction_method(M, typeof(p))) an inverse retraction method to use
• retraction_method: (default_retraction_method(M, typeof(p))) a retraction(M, p, X) to use.
• stopping_criterion: (StopWhenLagrangeMultiplierLess(1e-8; names=["-ξ"])) a functor, see StoppingCriterion, indicating when to stop
• vector_transport_method: (default_vector_transport_method(M, typeof(p))) a vector transport method to use
• sub_problem: a function evaluating with new allocations that solves the sub problem on M given the last serious iterate p_last_serious, the linearization errors linearization_errors, and the transported subgradients transported_subgradients
Output
the obtained (approximate) minimizer $p^*$, see get_solver_return for details
source
Manopt.convex_bundle_method!Function
convex_bundle_method!(M, f, ∂f, p)
perform a bundle method $p_{j+1} = \mathrm{retr}(p_k, -g_k)$ in place of p.
Input
• M: a manifold $\mathcal M$
• f: a cost function $f:\mathcal M→ℝ$ to minimize
• ∂f: the (sub)gradient $∂f:\mathcal M→ T\mathcal M$ of F restricted to always only returning one value/element from the subdifferential. This function can be passed as an allocation function (M, p) -> X or a mutating function (M, X, p) -> X, see evaluation.
• p: an initial value $p_0=p ∈ \mathcal M$
for more details and all optional parameters, see convex_bundle_method.
source
State
Manopt.ConvexBundleMethodStateType
ConvexBundleMethodState <: AbstractManoptSolverState
Stores option values for a convex_bundle_method solver.
Fields
• atol_λ: (eps()) tolerance parameter for the convex coefficients in λ
• atol_errors: (eps()) tolerance parameter for the linearization errors
• bundle: bundle that collects each iterate with the computed subgradient at the iterate
• bundle_cap: (25) the maximal number of elements the bundle is allowed to remember
• diameter: (50.0) estimate for the diameter of the level set of the objective function at the starting point
• domain: ((M, p) -> isfinite(f(M, p))) a function to that evaluates to true when the current candidate is in the domain of the objective f, and false otherwise, for example domain = (M, p) -> p ∈ dom f(M, p) ? true : false
• g: descent direction
• inverse_retraction_method: the inverse retraction to use within
• k_max: upper bound on the sectional curvature of the manifold
• linearization_errors: linearization errors at the last serious step
• m: (1e-3) the parameter to test the decrease of the cost: $f(q_{k+1}) \le f(p_k) + m \xi$.
• p: current candidate point
• p_last_serious: last serious iterate
• retraction_method: the retraction to use within
• stop: a StoppingCriterion
• transported_subgradients: subgradients of the bundle that are transported to p_last_serious
• vector_transport_method: the vector transport method to use within
• X: (zero_vector(M, p)) the current element from the possible subgradients at p that was last evaluated.
• stepsize: (ConstantStepsize(M)) a Stepsize
• ε: convex combination of the linearization errors
• λ: convex coefficients that solve the subproblem
• ξ: the stopping parameter given by $ξ = -\lvert g\rvert^2 – ε$
• sub_problem: ([convex_bundle_method_subsolver]) a function that solves the sub problem on M given the last serious iterate p_last_serious, the linearization errors linearization_errors, and the transported subgradients transported_subgradients,
• sub_state: an AbstractEvaluationType indicating whether sub_problem works in-place of λ or allocates a solution
Constructor
ConvexBundleMethodState(M::AbstractManifold, p; kwargs...)
with keywords for all fields with defaults besides p_last_serious which obtains the same type as p. You can use for example X= to specify the type of tangent vector to use
source
Stopping criteria
Manopt.StopWhenLagrangeMultiplierLessType
StopWhenLagrangeMultiplierLess <: StoppingCriterion
Stopping Criteria for Lagrange multipliers.
Currently these are meant for the convex_bundle_method and proximal_bundle_method, where based on the Lagrange multipliers an approximate (sub)gradient $g$ and an error estimate $ε$ is computed.
The mode=:both requires that both $ε$ and $\lvert g \rvert$ are smaller than their tolerances for the convex_bundle_method, and that $c$ and $\lvert d \rvert$ are smaller than their tolerances for the proximal_bundle_method.
The mode=:estimate requires that, for the convex_bundle_method $-ξ = \lvert g \rvert^2 + ε$ is less than a given tolerance. For the proximal_bundle_method, the equation reads $-ν = μ \lvert d \rvert^2 + c$.
Constructors
StopWhenLagrangeMultiplierLess(tolerance=1e-6; mode::Symbol=:estimate, names=nothing)
Create the stopping criterion for one of the modes mentioned. Note that tolerance can be a single number for the :estimate case, but a vector of two values is required for the :both mode. Here the first entry specifies the tolerance for $ε$ ($c$), the second the tolerance for $\lvert g \rvert$ ($\lvert d \rvert$), respectively.
source
Debug functions
Manopt.DebugWarnIfLagrangeMultiplierIncreasesType
DebugWarnIfLagrangeMultiplierIncreases <: DebugAction
print a warning if the Lagrange parameter based value $-ξ$ of the bundle method increases.
Constructor
DebugWarnIfLagrangeMultiplierIncreases(warn=:Once; tol=1e2)
Initialize the warning to warning level (:Once) and introduce a tolerance for the test of 1e2.
The warn level can be set to :Once to only warn the first time the cost increases, to :Always to report an increase every time it happens, and it can be set to :No to deactivate the warning, then this DebugAction is inactive. All other symbols are handled as if they were :Always:
source
Helpers and internal functions
Manopt.convex_bundle_method_subsolverFunction
λ = convex_bundle_method_subsolver(M, p_last_serious, linearization_errors, transported_subgradients)
convex_bundle_method_subsolver!(M, λ, p_last_serious, linearization_errors, transported_subgradients)
solver for the subproblem of the convex bundle method at the last serious iterate $p_k$ given the current linearization errors $c_j^k$, and transported subgradients $\mathrm{P}_{p_k←q_j} X_{q_j}$.
The computation can also be done in-place of λ.
The subproblem for the convex bundle method is
\begin{align*} \operatorname*{arg\,min}_{λ ∈ ℝ^{\lvert J_k\rvert}}& \frac{1}{2} \Bigl\lVert \sum_{j ∈ J_k} λ_j \mathrm{P}_{p_k←q_j} X_{q_j} \Bigr\rVert^2 + \sum_{j ∈ J_k} λ_j \, c_j^k \\ \text{s. t.}\quad & \sum_{j ∈ J_k} λ_j = 1, \quad λ_j ≥ 0 \quad \text{for all } j ∈ J_k, \end{align*}
where $J_k = \{j ∈ J_{k-1} \ | \ λ_j > 0\} \cup \{k\}$. See [BHJ24] for more details
Tip
A default subsolver based on RipQP.jl and QuadraticModels is available if these two packages are loaded.
source
Manopt.DomainBackTrackingStepsizeType
DomainBackTrackingStepsize <: Stepsize
Implement a backtrack as long as we are $q = \operatorname{retr}_p(X)$ yields a point closer to $p$ than $\lVert X \rVert_p$ or $q$ is not on the domain. For the domain this step size requires a ConvexBundleMethodState
source
Literature
[BHJ24]
R. Bergmann, R. Herzog and H. Jasa. The Riemannian Convex Bundle Method, preprint (2024), arXiv:2402.13670. | 2,422 | 9,033 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.561295 |
http://mathoverflow.net/revisions/70489/list | 1,369,058,645,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699036375/warc/CC-MAIN-20130516101036-00017-ip-10-60-113-184.ec2.internal.warc.gz | 160,623,286 | 6,969 | 6 added 18 characters in body
Let $G/k$ be a finite group scheme over a field $k$ and $X$ be $k$-scheme of finite type. An action of $G$ on $X$ is a $k$-morphism $\mu : G \times_k X \rightarrow X$ satisfying the usual conditions. In SGA3-V-4 and 5, it states that the quotient $X/G$ exists if $\mu$ is a finite flat morphism with other conditions. But in somewhere, I saw that the quotient $X/G$ exists if $G$ is a finite group scheme over $k$, but it doesn't require that the action morphism $\mu$ to be finite or flat. My question is that if we need $\mu$ to be finite flat to ensure the existence of the quotient $X/G$.
Now I will show that under the assumption that $X$ is geometrically reduced, an action of a finite group scheme over $k$ on $X$ is always finite flat. So the above question reduces to if we really need the geometrically reduced assumption. This assumption enables us to get a scheme morphism from a "morphism" on its closed points. Also, even under this assumption, is there another way to show that the finiteness of $\mu$ without working on the base change to the algebraically closure of $k$ first.
Notice that a finite $k$-scheme $G$ is a finite disjoint union of $U_g := \mathrm{Spec}(A_g)$ with each $A_g$ being a finite $k$-algebra ( $\mathrm{dim}(A_g) = 0$ ) such that $U_g$ is a one-point set and each $U_g$ is both open and closed in $G$. For each point $g \in G$, we denote the one point set $g$ by $U_g$, or simply by $g$, and $(U_g)_\mathrm{red}$ by $\overline{g}$.
We also denote the point in $G_{\mathrm{red}}$ which corresponds to the point $g$ in $G$ by $\overline{g}$.
Let first assume that $k$ is algebraically closed. In this case, each point $g$ of $G$ is a $k$-rational point and that $\overline{g}$ is an open affine subset of $G_{\mathrm{red}}$, which is $k$-isomorphic to $\mathrm{Spec}(k)$. For each $\overline{g}$, we consider the natural morphism $\overline{g} \times_k X = (U_g)_\mathrm{red} \times_k X \rightarrow X$. From the conditions of a group scheme acting on a scheme, we know that each point $\overline{g}$ gives an isomorphism from $X(k)$ to $X(k)$. The assumption on geometrically reduced tells us that it in fact gives an $k$-isomorphism of $X$. Hence the composition $\mu_{\overline{g}} : {\overline{g}} \times_k X \rightarrow g \times_k X \rightarrow X$ is a $k$-isomorphism. Using the fact that the natural morphism $i : Y_{\mathrm{red}} \rightarrow Y$ has a property that $i(U)$ is an open affine subset of $Y$ for any open affine subset $U$ of $Y_{\mathrm{red}}$ ( for any noetherian scheme $Y$ ), one sees that $\mu_{\overline{g}}$ is an affine morphism. For any open affine subset $\mathrm{Spec}(A)$ of $X$, let $\mathrm{Spec}(B)$ be its inverse image under $\mu_g : g \times_k X \rightarrow X$. We have that the composition of $A \leftarrow B \leftarrow A$ is $\mathrm{id}_A$. Also notice that $A \leftarrow B$ is the quotient of $B$ by a nilpotent ideal. This implies that for each $b \in b$, there exists $a \in A$ such that $b-a$ is nilpotent. Hence $B \leftarrow A$ is an integral homomorphism. It's easy to see that $\mu_g$ is of finite type. Hence $B \leftarrow A$ is in fact finite. This proves that $\mu_g : g \times_k X \rightarrow X$ is finite flat. Since $\mu : G \times_k X \rightarrow X$ is the finite disjoint union of $\mu_g$, we know that $\mu$ is finite flat.
Now for $k$ not necessary being algebraically close, we take the base change to $\overline{k}$ and apply the above result. So the question is that, if the base change $\overline{\mu}$ of $\mu : G \times_k X \rightarrow X$ to $\overline{k}$ is finite flat, then could we conclude that $\mu$ is finite flat. Notice that $\mu$ is of finite type. One can show that $\mu$ is separated and universally closed from the separateness of $\overline{\mu}$ which is at the same time universally closed ( since $\overline{\mu}$ is finite). So one concludes that $\mu$ is proper. It's easy to show that $\mu$ is quasi-finite hence $\mu$ is finite. Finally, it's easy to show that $\mu$ is flat.
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Let $G/k$ be a finite group scheme over a field $k$ and $X$ be $k$-scheme of finite type. An action of $G$ on $X$ is a $k$-morphism $\mu : G \times_k X \rightarrow X$ satisfying the usual conditions. In SGA3-V-4 and 5, it states that the quotient $X/G$ exists if $\mu$ is a finite flat morphism with other conditions. But in somewhere, I saw that the quotient $X/G$ exists if $G$ is a finite group scheme over $k$, but it doesn't require that the action morphism $\mu$ to be finite or flat. My question is that if we need $\mu$ to be finite flat to ensure the existence of the quotient $X/G$.
Now I will show that under the assumption that $X$ is geometrically reduced, an action of a finite group scheme over $k$ on $X$ is always finite flat. So the above question reduces to if we really need the geometrically reduced assumption. This assumption enables us to get a scheme morphism from a "morphism" on its closed points. Also, even under this assumption, is there another way to show that the finiteness of $\mu$ without working on the base change to the algebraically closure of $k$ first.
Notice that a finite $k$-scheme $G$ is a finite disjoint union of $U_g := \mathrm{Spec}(A_g)$ with each $A_g$ being a finite $k$-algebra ( $\mathrm{dim}(A_g) = 0$ ) such that $U_g$ is a one-point set and each $U_g$ is both open and closed in $G$. For each point $g \in G$G$, we denote$ g $g$ by $U_g$, U_g$, or simply by$g$g$, and $(U_g)\mathrm{red}$ (U_g)_\mathrm{red}$by$\overline{g}$. We also denote the point in$G{\mathrm{red}}$G_{\mathrm{red}}$ which corresponds to the point $g$ in $G$ by $\overline{g}$.
Let first assume that $k$ is algebraically closed. In this case, each point $g$ of $G$ is a $k$-rational point and that $\overline{g}$ is an open affine subset of $G_{\mathrm{red}}$, which is $k$-isomorphic to $\mathrm{Spec}(k)$. For each $\overline{g}$, we consider the natural morphism $\overline{g} \times_k X = (U_g)_\mathrm{red} \times_k X \rightarrow X$. From the conditions of a group scheme acting on a scheme, we know that each point $\overline{g}$ gives an isomorphism from $X(k)$ to $X(k)$. The assumption on geometrically reduced tells us that it in fact gives an $k$-isomorphism of $X$. Hence the composition $\mu_{\overline{g}} : {\overline{g}} \times_k X \rightarrow g \times_k X \rightarrow X$ is a $k$-isomorphism. Using the fact that the natural morphism $i : Y_{\mathrm{red}} \rightarrow Y$ has a property that $i(U)$ is an open affine subset of $Y$ for any open affine subset $U$ of $Y_{\mathrm{red}}$ ( for any noetherian scheme $Y$ ), one sees that $\mu_{\overline{g}}$ is an affine morphism. For any open affine subset $\mathrm{Spec}(A)$ of $X$, let $\mathrm{Spec}(B)$ be its inverse image under $\mu_g : g \times_k X \rightarrow X$. We have that the composition of $A \leftarrow B \leftarrow A$ is $\mathrm{id}_A$. Also notice that $A \leftarrow B$ is the quotient of $B$ by a nilpotent ideal. This implies that for each $b \in b$, there exists $a \in A$ such that $b-a$ is nilpotent. Hence $B \leftarrow A$ is an integral homomorphism. It's easy to see that $\mu_g$ is of finite type. Hence $B \leftarrow A$ is in fact finite. This proves that $\mu_g : g \times_k X \rightarrow X$ is finite flat. Since $\mu : G \times_k X \rightarrow X$ is the finite disjoint union of $\mu_g$, we know that $\mu$ is finite flat.
Now for $k$ not necessary being algebraically close, we take the base change to $\overline{k}$ and apply the above result. So the question is that, if the base change $\overline{\mu}$ of $\mu : G \times_k X \rightarrow X$ to $\overline{k}$ is finite flat, then could we conclude that $\mu$ is finite flat. Notice that $\mu$ is of finite type. One can show that $\mu$ is separated and universally closed from the separateness of $\overline{\mu}$ which is at the same time universally closed ( since $\overline{\mu}$ is finite). So one concludes that $\mu$ is proper. It's easy to show that $\mu$ is quasi-finite hence $\mu$ is finite. Finally, it's easy to show that $\mu$ is flat.
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1 | 2,411 | 8,344 | {"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} | 3.09375 | 3 | CC-MAIN-2013-20 | latest | en | 0.890715 |
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Question From class 12 Chapter DEFAULT
locus of the middle point of the chord which subtend angle at the centre of the circle
Locus of the middle points of chords of the circle which subtend a right angle at the centre is
1:45
Locus of the point, the chord of contact of which subtends an angle at the centre of circle is
5:21
Locus of midpoint of chord of circle which subtend right angle at is
Locus of midpoint of chord of circle which subtend right angle at is
Find the condition that the chord of contact of tangents from the point to the circle should subtend a right angle at the centre. Hence find the locus of .
6:50
Find the locus of the midpoint of the chords of the circle which subtend a right angle at the point
3:00
Find the locus of the mid-point of the chords of the circle which subtend an angle of at the centre of the circle.
5:18
The locus of the mid point of a chord of the circle which subtends a right angle at the origin is
3:49
Locus of mid point of chord AB of a circle which subtends angle (1,0)
9:56
The locus of the middle point of the chord of the circle such that the segment of the chord on the parabola subtends a right angle at the origin, is a circle whose centre and radius respectively are
3:59
Then locus of the point of intersection of the tangents at the extremities of a chord of the circle which touches the circle passes through the point,
Find the locus of the middle points of the chords of the parabola which subtend a right angle at the vertex of the parabola.
5:02
The locus of the mid-point of the chords of the circle which subtend a right angle at the point(h, k), is
7:40
The equation of the locus of the middle point of a chord of the circle such that the pair of lines joining the origin to the point of intersection of the chord and the circle are equally inclined to the x-axis is (b) (d) none of these
5:19
The locus of the mid points of a chord the circle which subtends right angle at the origin is
2:22
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http://www.gabormelli.com/RKB/Categorical_distribution | 1,718,756,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861796.49/warc/CC-MAIN-20240618234039-20240619024039-00531.warc.gz | 39,798,009 | 22,297 | Discrete Probability Function Structure
(Redirected from Categorical distribution)
• … Note that, in some fields, such as natural language processing, the categorical and multinomial distributions are conflated, and it is common to speak of a "multinomial distribution" when a categorical distribution is actually meant. This stems from the fact that it is sometimes convenient to express the outcome of a categorical distribution as a "1-of-K" vector (a vector with one element containing a 1 and all other elements containing a 0) rather than as an integer in the range $\displaystyle{ 1 \dots K }$; in this form, a categorical distribution is equivalent to a multinomial distribution over a single trial. | 146 | 708 | {"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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.896206 |
http://www.crossfitmph.com/2009/08/20/wod-210809/ | 1,600,645,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00252.warc.gz | 173,760,552 | 13,304 | ## Friday, August 21, 2009
### Friday, August 21, 2009
for time:
5x shoulder press @ body weight
10x overhead squat @ body weight
15x front squat @ body weight
If performing this workout below prescription, it requires use of the general formula below—yes, math:
• [(body weight x number of reps in exercise) / (weight you plan to use in this exercise)] = repetitions to be completed
In other words, you need to shoulder press the heaviest possible load as many repetitions as necessary to equal 5x your body weight, overhead squat the heaviest possible load as many repetitions as necessary to equal 10x your body weight and front squat the heaviest possible load as many repetitions as necessary to equal 15x your body weight.
For example, if Jack Athlete weighs 200 pounds, cannot shoulder press his weight, but can shoulder press 100 pounds, he would press 100 pounds 10 times (200 x 5 = 100 x 10). If he cannot overhead squat his body weight, but can overhead squat 125 pounds, he would overhead squat 125 pounds 16 times (200 x 10 = 125 x 16). If he can front squat his body weight, he changes nothing, and front squats 200 pounds 15 times (200 x 15 = 200 x 15).
You must complete this calculation for each of the three lifts after the warm-up, but before the workout begins. A different weight may be used for each lift. Transition quickly if you plan on switching weights. Scores will be calculated as the total time of workout, the weight of each exercise (as a percentage of your body weight—the heavier the better) and the number of repetitions completed for each exercise.
This workout is scored by the time, load(s) and scheme.
Compare results to April 20, 2009. | 394 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-40 | latest | en | 0.936227 |
https://www.shaalaa.com/question-bank-solutions/a-wire-whose-cross-sectional-area-increasing-linearly-its-one-end-other-connected-across-battery-v-volts-which-following-quantities-remain-constant-wire-a-drift-speed-b-current-density-c-electric-drift-electrons-origin-resistivity_16487 | 1,653,360,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00215.warc.gz | 1,124,839,713 | 10,645 | Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire? (a) drift speed (b) current density (c) electric current (d) electric field - Physics
A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts.
Which of the following quantities remain constant in the wire?
(a) drift speed
(b) current density
(c) electric current
(d) electric field
Solution
The electric current will remain constant in a wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts.
Because current is the only quantity that does not depend on the area of cross- sections of the wire.
I=(dq)/(dt), that is the rate of flow of charge, where as drift speed, current density and electric field are depends on the increasing area of cross-section with the following relations:
Drift speed : ν_d=I/(An""e)
Current density = I/A
Electric field = J/σ
Concept: Drift of Electrons and the Origin of Resistivity
Is there an error in this question or solution? | 282 | 1,285 | {"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} | 3.203125 | 3 | CC-MAIN-2022-21 | latest | en | 0.931267 |
https://yanabeea.net/family-fun-11-disney-themed-activities-for-kids/ | 1,685,939,850,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00131.warc.gz | 1,189,207,431 | 20,972 | Family Fun: 11 Disney-Themed Activities for Kids
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In this house, we do Disney. Whether it’s in our favorite movies or in our homes, the magic of Disney is a feeling that never goes away. We love to bring the joy of Disney into our lives and share it with all members of the family. In my new blog post I am sharing 11 fun activities for kids that are themed around Disney!
-Go on a family scavenger hunt in your house or neighborhood looking for things related to Disney. When you find one, take a picture and share it with the group!
-Play hide and seek using toy figurines of characters from any movie. Remember to all use character names when calling out who’s “it”!
-Create an art project that incorporates at least three different types of popular Disney movies together, such as The Lion King + Toy Story (my favorite!) = Pride Rock inspired by Andy’s Room – Assemble parts of Tinker Bell from Peter Pan into the air while everyone else is blindfolded (or just turn off the lights) and see who can identify the Tinker Bell parts in the shortest amount of time.
-Play Disney trivia and test your knowledge!
-Practice math by counting how many times an animal appears on a card, then tally up your score with everyone else’s to find out which team has done better (think Go Fish or Old Maid). The first person/team to reach 100 points wins! This is great for practicing addition facts too if you use cards that have numbers as well as animals.
-Be creative and draw something inspired by Disney – try drawing a favorite character from any movie, painting some scenery from their world, or even sketching one of those iconic circles we always hear about!
-Go on YouTube and search for Disney songs. Sing along!
-Find a coloring page for your favorite song from the movie and color it in, or find an image that inspires you and paint on top of it to create your own unique piece.
-Play a word game where each player gets four cards with words drawn on them (again, try using categories such as Disney lyrics) and put them in order by first letter so they read like sentences: “I love Sally’s birthday cake” would be A I L S B C K D E – this is similar to Hangman but instead of guessing letters to figure out what someone has guessed wrong, we’re figuring out which word follows another one alphabetically. The person who wins can start the next round if they want.
-Find a word that rhymes with something in your favorite Disney movie (for example, “snow” could rhyme with the title of Aladdin). Write out as many versions of this word as you can think up and see which one is funniest to read aloud.
-Name an animal or insect native to a place in the world like Africa for instance; then find five different words that are spelled similarly but have completely different meanings – such as giraffe, gharial, garifuna, geriatric etcetera) and put them all on index cards laid face down in front of each player. Have everyone pick a card at random while someone else flips it over so they can’t tell what’s on the other side. The player who picks a word that is spelled in the same way as their animal or insect gets to keep it and swap another card with someone else if they want.
-Pretend you’re going on a Disney cruise, listing all of your things for packing (for example: “I need my princess hat, my Captain Hook costume, sunscreen”).
-Choose two categories that have nothing in common except one word in each category – for instance: Saltwater Animals vs. Native American Tribes; then give five different answers to fill out both columns. For example: Whale Sharks vs Carnival Indians)
-Find something from around the house like an apple or a can of beans and describe how if feels before giving it to someone else
-Find a person in the room and make up an imaginary conversation with them about what they would want their last meal to be.
-Pretend you’re going on your first date with another player, listing all of your things for packing (for example: “I need my princess hat, my Captain Hook costume, sunscreen”).
-Get up from where you are sitting and look around the house or yard – then tell everyone one thing that makes it special to you.
-Draw two different animals without lifting pencil from paper between drawings; name both animals. (eagle & lion)
-Name as many Disney movies as possible in 60 seconds
In this house we do disney!
-We love to play in the Disney store
-It’s always a magical trip. | 966 | 4,460 | {"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-2023-23 | latest | en | 0.936733 |
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## Convert long ton per (square micrometer) to long ton per (square inch)
### (long tn/μm² to long tn/in²)
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Mil, also known as a thou is equal to one thousandth of an inch (0.001 inch). The mil or thou is still used in the United States to measure thickness of various materials. The mil should not be confused with a millimeter (0.001 meter), a multiple of a base metric unit of length.
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Viscosity is a measure of how resistive a fluid is to flow. Viscosity describes how thick a fluid is, e.g. honey is more resistive to flow than water, and so it is said, that honey is thick or more viscous, i.e. having a higher viscosity, than water. Viscosity in fluids is analogues to friction in solids. Both, viscosity and friction, are mechanisms by which kinetic energy of moving solid objects (friction), or molecules in fluids (viscosity), is converted to thermal energy. The metric (SI) measurement unit of dynamic viscosity is pascal-second (Pa·s) or N × s ÷ m2 or kilogram per meter per second (kg/m/s). | 419 | 1,766 | {"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} | 3.15625 | 3 | CC-MAIN-2013-20 | latest | en | 0.866506 |
https://bearsemath.com/2018/03/06/popcorn-day/ | 1,686,103,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00560.warc.gz | 142,644,902 | 24,121 | We popped corn today. We started off by brainstorming all the things we wondered about, and all the questions we could make and solve regarding popcorn.
We measured volumes, and volume changes, we timed how long it took to pop
We made containers out of one cheed of paper and tried to maximize the volume.
We noticed that the buckets aren’t perfect cylinders, so We figured out how to approximate the volume, and then checked it using our known cylinders.
We did a lot of math and explored so many different things! It is hard to imagine how many explorations came from a bag of popcorn. | 127 | 593 | {"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-2023-23 | latest | en | 0.959192 |
https://convertoctopus.com/166-milliliters-to-tablespoons | 1,627,742,274,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00347.warc.gz | 195,246,303 | 7,977 | ## Conversion formula
The conversion factor from milliliters to tablespoons is 0.06762804511761, which means that 1 milliliter is equal to 0.06762804511761 tablespoons:
1 ml = 0.06762804511761 tbsp
To convert 166 milliliters into tablespoons we have to multiply 166 by the conversion factor in order to get the volume amount from milliliters to tablespoons. We can also form a simple proportion to calculate the result:
1 ml → 0.06762804511761 tbsp
166 ml → V(tbsp)
Solve the above proportion to obtain the volume V in tablespoons:
V(tbsp) = 166 ml × 0.06762804511761 tbsp
V(tbsp) = 11.226255489523 tbsp
The final result is:
166 ml → 11.226255489523 tbsp
We conclude that 166 milliliters is equivalent to 11.226255489523 tablespoons:
166 milliliters = 11.226255489523 tablespoons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 tablespoon is equal to 0.089076896649398 × 166 milliliters.
Another way is saying that 166 milliliters is equal to 1 ÷ 0.089076896649398 tablespoons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred sixty-six milliliters is approximately eleven point two two six tablespoons:
166 ml ≅ 11.226 tbsp
An alternative is also that one tablespoon is approximately zero point zero eight nine times one hundred sixty-six milliliters.
## Conversion table
### milliliters to tablespoons chart
For quick reference purposes, below is the conversion table you can use to convert from milliliters to tablespoons
milliliters (ml) tablespoons (tbsp)
167 milliliters 11.294 tablespoons
168 milliliters 11.362 tablespoons
169 milliliters 11.429 tablespoons
170 milliliters 11.497 tablespoons
171 milliliters 11.564 tablespoons
172 milliliters 11.632 tablespoons
173 milliliters 11.7 tablespoons
174 milliliters 11.767 tablespoons
175 milliliters 11.835 tablespoons
176 milliliters 11.903 tablespoons | 506 | 1,990 | {"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} | 4.15625 | 4 | CC-MAIN-2021-31 | latest | en | 0.753503 |
https://builditsolar.com/Projects/PoolHeating/HotTubCalc/HotTubCalc.htm | 1,582,392,163,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00488.warc.gz | 315,018,543 | 5,272 | # Solar Collector Size to Heat a Hot Tub
This is a pretty good method to determine how large a solar collector is needed to heat your hot tub. The size of the collector needed to heat a hot tub depends on a lot of things: volume of the tub, how well its insulated, how good the cover is, what your climate is like, how hot you keep it, ... Its not easy to take all that into account to get a good idea what size collector you need. But, I ran across the method detailed below in "More Other Homes and Garbage" that uses the temperature drop of the hot tub over 24 hours to determine the daily heat loss for your hot tub in your climate and for the way you use it -- it seems like a good approach to me. By the way "More Other Homes and Garbage" is a terrific book written the 80's, but full of good ideas, good analysis, and written in a style that won't put you asleep -- and, its only \$8 used!
## Using the Method
Here is the step by ste:
1. Heat the tub up to the temperature you want to use it at (or maybe a touch more) -- record the starting temperature.
2. Turn the hot tub heater off so that the tub gets no heat for the period of the test.
3. Let the tub cool down for 24 hours.
Duplicate your normal usage pattern for the test period -- that is, use the tub as usual, then put the cover on and let it sit.
4. After 24 hours, note the temperature that the tub has cooled to.
5. If you don't know the size of hour hot tub in gallons, estimate what it is (see note below).
6. Use the plot below to get an estimate of solar collector area.
For example, if you have a 300 gallon tub and it cools by 10F in 24 hours, then you need about 25 sqft of solar collector to heat it.
If you can, its best to do this test in the winter when the heat loss is higher.
If the collector area comes out quite low (say 6 sqft), I'd build a larger collector and just have some reserve in hand. If you are buying a commercial collector, you will need to just pick a size that is a reasonable match to the estimated size -- maybe the next larger size. If you are building the collector, you will want to consider standard sizes of materials -- that is, there is no point in trying to build a 26 sqft collector when standard sheet and glazing sizes would make it easier to build a 32 sqft (4 by 8) collector.
### Notes:
Ideally, you would do the test in the coldest part of the year in which you use the hot tub, so that the collector size is up to matching the tub heat loss in the coldest time period you use the tub. But, I would not pick an extreme cold day -- just a day that is typical of the coldest time of year you use the tub.
If you can't do the temperature drop test in the coldest part of the year in which you use the tub, then you should probably adjust the collector area upward based on the temperature difference between the hot tub and the outside air. For example, if you do the test in the spring when the hot tub temp is 104F, and the average outdoor temperature over 24 hrs is 50F, then the temperature difference is 104F - 50F = 54F. If in the winter the average outside temperature is 30F, then the temperature difference is 104F - 30F = 74F. So, you would probably want to ratio the collector area up by 74F/54F = 1.4 to get a collector area that would heat the tub on an average winter day. When estimating these temperatures, you want to use averages not extremes -- you want the 24 hour average temperature on an average day for the season you are looking -- e.g. where I am, we can get down to -40F on an extreme winter night, but our 24 hour average for the same season might be more like 25F. WeatherSpark.com is a good place to get averages.
If you don't like where you end up on the plot above (i.e. you have a large 24 hour temperature drop), consider beefing up the insulation, getting a better cover, or placing the the hot tub in a protective environment (e.g. a sunspace). Hot tubs use a lot of energy (probably more than anything else in your house), so it worth taking all the steps you can to improve their efficiency.
### How the method works:
The method is from "More Other Homes and Garbage.
The hot tub heat loss can be calculated from the temperature drop of the water -- so, if the temperature drops 10F in a 300 gallon tub, the heat loss is:
Heat loss 24 hours = (300 gal)*(8.34 lb/gal)*(10F)*(1 BTU/lb-F) = 25,000 BTU
To estimate the collector area from the heat loss, the method assumes that each sqft of collector will produce 1000 BTU on a sunny day. This seems about right to me, and matches the SRCC test data for glazed flat plate collectors pretty well. For example, a Heliodyne 4 by 8 collector produces 32K BTU on a sunny day with a temperature difference of 36F...
The thing I like about this method is that by using the actual temperature drop for your hot tub where you are, it takes into account most of the many variables that would go into a fully analytical approach.
Bear in mind that in sizing the collector, a sunny day is assumed. On part cloudy or cloudy days, that may not be enough to heat the tub fully depending on the season.
Gary May 12, 2012 | 1,250 | 5,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-10 | latest | en | 0.930325 |
epbank.ir | 1,524,637,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00477.warc.gz | 94,706,084 | 21,314 | # The Falling Value of the US Dollar
what does it mean that the U.S. dollar is «falling»? Fall- ing from where?
No, dollar bills are not dropping out of the sky — although that would be pretty cool. It means that any money you have is fall- ing in value compared with the money used by other countries. A bit of math might help explain this. Let›s use Canada as an ex- ample, since lots of Americans go on vacation there. A year ago, if you traded 10 U.S. dollars for Canadian money, you would have received 14 Canadian dollars in return. Today, 10 U.S. dol- lars would get you about 10 Canadian dollars.
Who decides what the dollar is worth? How do they do it? Dollars are traded on markets the same way that football cards are swapped on the playground. Most children who collect football cards want at least one of Ronaldinho because the Brazilian is one of the world›s most famous footballers. But what if Ronaldinho started to play badly and couldn›t score goals any more? The other kids at school might feel that his card was worth less and want to trade it for one they thought was more valuable.
That is what›s happening to the dollar. People are worried that
the U.S. economy is in a slump. So they don›t want U.S. dollars as much as they once did. They are trading them for other countries› money, and that makes the dollar less valuable.
Just how bad is it for the dollar? It›s pretty bad. The value of the U.S. dollar has hit an all-time low compared with the euro, the official money of 13 European countries. And, for the first time in 31 years, Canada›s dollar is about equal to the U.S. dollar.
Is the falling dollar bad for Americans economy? Yes and no. Staying with Canada as our example, the falling U.S. dollar means that if you go there on vacation and want to buy a souvenir — a beaded belt, perhaps — it might cost you 10 U.S. dollars instead of \$7 if you had bought it a year ago. In that way, the falling U.S. dollar hurts American travellers and people who buy goods from other countries. Toys made overseas and sold here could cost more this holiday season.
But people coming from Europe or other countries will find that American goods are now less expensive for them to buy. More of them will visit Disney World and other U.S. tourist attractions , spending money in American hotels and restaurants. U.S. farmers will sell more food to other countries and American factories will sell more cars and other goods overseas.
• آسان پرداخت بانک سپه (24)
• اسان پرداخت بانک سپه (12)
• شعبه ی آسان پرداخت کرج (1) | 588 | 2,539 | {"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-2018-17 | latest | en | 0.973935 |
https://studyres.com/doc/135389/chapter-4 | 1,709,580,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00794.warc.gz | 529,848,688 | 14,418 | Survey
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Transcript
```Elasticity: A Measure
of Responsiveness
PREPARED BY:
FERNANDO QUIJANO, YVONN QUIJANO,
KYLE THIEL & APARNA SUBRAMANIAN
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
• price elasticity of demand (Ed)
A measure of the responsiveness of
the quantity demanded to changes in
price; equal to the absolute value of
the percentage change in quantity
demanded divided by the percentage
change in price.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
2 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Computing Percentage Changes and Elasticities
Table 4.1
Data
Computation
with Initialvalue method
COMPUTING PRICE ELASTICITY WITH INITIAL VALUES AND
MIDPOINTS
Price
Quantity
Initial
\$20
100
New
22
80
Price
Quantity
Percentage change
Price elasticity of demand
10%
\$2
x 100
\$20
2.0
Percentage change
Price elasticity of demand
9.52%
2.33
20
x 100
100
- 20%
10%
Price
Computation
with Midpoint
method
20%
\$2
x 100
\$21
Quantity
22.22%
20
x 100
90
- 22.22%
9.52%
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
3 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Price Elasticity and the Demand Curve
Figure 4.1 shows five different demand curves, each with a different elasticity. We
can divide products into five types, depending on their price elasticities of demand.
► FIGURE 4.1
Elasticity and Demand Curves
• elastic demand
The price elasticity of demand is
greater than one.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
4 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Price Elasticity and the Demand Curve
► FIGURE 4.1 (cont’d.)
Elasticity and Demand Curves
• inelastic demand
The price elasticity of demand is
less than one.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
5 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Price Elasticity and the Demand Curve
► FIGURE 4.1 (cont’d.)
Elasticity and Demand Curves
• unit elastic demand
The price elasticity of demand is
one.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
6 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Price Elasticity and the Demand Curve
► FIGURE 4.1 (cont’d.)
Elasticity and Demand Curves
• perfectly inelastic demand
The price elasticity of demand is
zero.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
7 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Price Elasticity and the Demand Curve
► FIGURE 4.1 (cont’d.)
Elasticity and Demand Curves
• perfectly elastic demand
The price elasticity of demand is
infinite.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
8 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Elasticity and the Availability of Substitutes
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez
9 of 37
chapter
4.1
THE PRICE ELASTICITY OF DEMAND
Other Determinants of the Price Elasticity of Demand
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 10 of 37
chapter
4.2
USING PRICE ELASTICITY TO PREDICT
CHANGES IN QUANTITY
If we have values for two of the three variables in the elasticity formula, we
can compute the value of the third. The three variables are:
(1) the price elasticity of demand itself,
(2) the percentage change in quantity, and
(3) the percentage change in price.
Specifically, we can rearrange the elasticity formula:
percentage change in quantity demanded = percentage change in price × Ed
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 11 of 37
chapter
Extra Application 8
WHY YOU’LL PAY MORE IN RENT THIS YEAR
The current rental market is extremely hot with the average rent reaching \$940 per month
for the last quarter of 2005. Some areas such as New York City reported rents averaging
\$2,400 per month. In spite of these numbers, many analysts believe rent is still low in
several regions. In some areas such as Ft. Lauderdale Florida, rent has climbed almost
12% in the recent months as occupancy rates approached 100% and is expected to go
higher. Much of the reason for higher rents is supply related due to many apartment units
being converted to condominiums. Other regions, particularly along the Gulf Coast, can
thank Hurricane Katrina for the rental unit shortage. With apartments being difficult to find
vacant, many owners are using the hot market to increase rents a little. Rising home prices
and increasing rates also alter the picture and push a number of people back to the rental
market. Houses are getting too expensive for some people to own.
A decrease in the supply of rental units will
automatically push prices higher. Owners, looking to
make a profit, tend to react very quickly to high
occupancy rates. However, new units will no doubt
soon be constructed to lessen the shortage.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 12 of 37
chapter
BEER TAXES AND HIGHWAY DEATHS
APPLYING THE CONCEPTS #1: How can we use the price elasticity
of demand to predict the effects of taxes?
We can use the concept of price elasticity to predict the
effects of a change in the price of beer on drinking and
• The price elasticity of demand for beer among young adults is about 1.30.
• If a state imposes a beer tax that increases the price of beer by 10 percent, how will
the price hike affect beer consumption among young adults?
Using the elasticity formula, we predict that beer consumption will decrease by 13 percent:
percentage change in quantity demanded = percentage change in price × Ed
= 10% × 1.30 = 13%
• The number of highway deaths among young adults is roughly proportional to their
beer consumption, so the number of deaths will also decrease by 13 percent.
• Larger taxes would decrease beer consumption and highway deaths by larger
amounts.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 13 of 37
chapter
SUBSIDIZED MEDICAL CARE IN CÔTE D’IVOIRE AND PERU
APPLYING THE CONCEPTS #2: Does the responsiveness of
consumers to changes in price vary by income?
Many developing nations subsidize medical care, charging consumers a small
fraction of the cost of providing the services. If a nation were to cut its subsidies
and thus increase the price of medical care for consumers, how would the higher
price affect its poor and wealthy households?
• In Côte d’Ivoire in Africa, the price elasticity of demand for hospital services is 0.47 for
poor households and 0.29 for wealthy households. A 10-percent increase in the price of
hospital services would cause poor households to cut back their hospital care by:
percentage change in quantity demanded = 10% × 0.47 = 4.7%
In contrast, wealthy households would cut back by:
percentage change in quantity demanded = 10% × 0.29 = 2.9%
• In Peru, the price elasticity is 0.67 for poor households but only 0.03 for wealthy
households. How would the higher price affect its poor and wealthy households?
• The poor are much more sensitive to price, so when prices increase, they experience
much larger reductions in medical care.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 14 of 37
chapter
4.3
PRICE ELASTICITY AND TOTAL REVENUE
• total revenue
The money a firm generates from
selling its product.
total revenue = price per unit × quantity sold
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 15 of 37
chapter
HOW TO CUT TEEN SMOKING BY 60 PERCENT
APPLYING THE CONCEPTS #3: How can we use the price elasticity of
demand to predict the effects of public policies?
Under the 1997 federal tobacco settlement, if smoking by teenagers does not
decline by 60 percent by the year 2007, cigarette makers will be fined \$2 billion.
The settlement increased cigarette prices by about 62 cents per pack, a
percentage increase of about 25 percent. Will the price hike be large enough to
meet the target reduction of 60 percent?
• The demand for cigarettes by teenagers is elastic, with an elasticity of 1.3. Therefore, a
25-percent price hike will reduce teen smoking by only 32.5 percent, far short of the
target reduction:
percentage change in quantity demanded = 25% × 1.30 = 32.5%
• About half of the decrease in consumption occurs because fewer teenagers will become
smokers, and the other half occurs because each teenage smoker will smoke fewer
cigarettes. To meet the target reduction of teenage smoking, the price of cigarette prices
must increase by about 46 percent:
percentage change in quantity demanded = 46% × 1.3 = 60%
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 16 of 37
chapter
4.3
PRICE ELASTICITY AND TOTAL REVENUE
Elastic Versus Inelastic Demand
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 17 of 37
chapter
4.3
PRICE ELASTICITY AND TOTAL REVENUE
Elastic Versus Inelastic Demand
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 18 of 37
chapter
A BUMPER CROP IS BAD NEWS FOR FARMERS
APPLYING THE CONCEPTS #4: If demand is inelastic, how does an increase
in supply affect total expenditures?
Suppose that favorable weather generates a “bumper crop” for soybeans
that is 30 percent larger than last year’s harvest.
The good news is that farmers will sell more bushels of soybeans. The bad
news is that the increase in supply will decrease the equilibrium price of
soybeans, so they will get less money per bushel. Which will be larger, the
increase in quantity or the decrease in price?
• The demand for soybeans and many other agricultural products is inelastic.
• To increase the quantity demanded by 30 percent to meet the higher supply,
the price must decrease by more than 30 percent.
• Consumers need a large price reduction to buy more of the product.
If the price elasticity of demand is 0.75, the price must decrease by 40 percent to increase
the quantity demanded by 30 percent. To show this, we can rearrange the elasticity formula:
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 19 of 37
chapter
4.4
ELASTICITY AND TOTAL REVENUE FOR
A LINEAR DEMAND CURVE
Price Elasticity Along a Linear Demand Curve
► FIGURE 4.2
Elasticity and Total Revenue
Along a Linear Demand Curve
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 20 of 37
chapter
4.4
ELASTICITY AND TOTAL REVENUE FOR
A LINEAR DEMAND CURVE
Price Elasticity Along a Linear Demand Curve
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 21 of 37
chapter
4.5
OTHER ELASTICITIES OF DEMAND
Income Elasticity of Demand
• income elasticity of demand
A measure of the responsiveness of
demand to changes in consumer income;
equal to the percentage change in the
quantity demanded divided by the
percentage change in income.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 22 of 37
chapter
4.5
OTHER ELASTICITIES OF DEMAND
Cross-Price Elasticity of Demand
• cross-price elasticity of demand
A measure of the responsiveness of demand to changes
in the price of another good; equal to the percentage
change in the quantity demanded of one good (X) divided
by the percentage change in the price of another good
(Y).
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 23 of 37
chapter
Extra Application 9
HIGH FUEL PRICES IMPACT VACATION PLANS
Memorial Day vacation travel by automobile will be up by only 0.7 percent this year, the
smallest increase in several years. A 34 percent year-to-year increase in the price of
gasoline is the culprit. The AAA Travel survey also indicated that many vacationers would
alter their plans to take advantage of cheaper motels, closer destinations, and/or shorter
stays.
• Air, bus, and train travel all expect similar slight increases in people traveling, or
stable numbers.
• Airplane ticket prices are up about 10 percent and hotels have increased rates by
The substitution effect helps explain why the demand
curve is downward sloping and to the right. As the price of
gasoline increases people will alter their behavior by
buying less gasoline. Part of this behavior can be
explained by consumers substituting a portion of gasoline
purchases. Vacation and entertainment travel happens to
be a substitutable component of gasoline consumption.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 24 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
• price elasticity of supply
A measure of the responsiveness of the
quantity supplied to changes in price;
equal to the percentage change in
quantity supplied divided by the
percentage change in price.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 25 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
▼ FIGURE 4.3
The Slope of the Supply Curve and Supply Elasticity
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 26 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
What Determines the Price Elasticity of Supply?
The price elasticity of supply is determined by how rapidly
production costs increase as the total output of the
industry increases. If the marginal cost increases rapidly,
the supply curve is relatively steep and the price elasticity
is relatively low.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 27 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
The Role of Time: Short-Run Versus Long-Run Supply
Elasticity
Time is an important factor in determining the price elasticity of supply for a product.
The market supply curve is positively sloped because of two responses to an
increase in price:
• Short run. A higher price encourages existing firms to increase their output by
purchasing more materials and hiring more workers.
• Long run. New firms enter the market and existing firms expand their production
facilities to produce more output.
The short-run response is limited because of the principle of diminishing returns.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 28 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
Extreme Cases: Perfectly Inelastic Supply and Perfectly
Elastic Supply
▼ FIGURE 4.4
Perfectly Inelastic Supply and Perfectly Elastic Supply
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 29 of 37
chapter
Extra Application 11
THE MARKET FOR RARE GUITARS
The price of rare guitars is escalating faster than most corporate stocks. A sunburst Les
Paul Standard that originally sold for \$265 in 1960 can now command \$300,000 in pristine
condition. Les Paul’s are not alone in this market. Pre-war Martin acoustic guitars and pre1966 Fender Stratocasters have seen similar run-ups. Much of the recent price
appreciation has been driven by non-musician investors attempting to cash in on the gains.
However, scarcity is also to blame. For example, there were only approximately 1,700
sunburst Les Pauls ever manufactured. Many experts caution buyer beware since the
astronomical prices have stimulated a heavy trade in fakes. The current estimated count of
2,200 sunburst Les Pauls illustrates this point. Where did the other 500 come from?
The known supply of rare guitars is virtually fixed.
Therefore, increasing demand by collectors pushes the
price rapidly higher. Even a very small shift in demand due
to an increase in the number of collectors/investors results
in a substantial change in the price. The opposite is also
true. If investor/collectors suddenly became concerned
decrease in the number of potential buyers) the price could
fall just as rapidly.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 30 of 37
chapter
4.6
THE PRICE ELASTICITY OF SUPPLY
Extreme Cases: Perfectly Inelastic Supply and Perfectly
Elastic Supply
• perfectly inelastic supply
The price elasticity of supply equals
zero.
• perfectly elastic supply
The price elasticity of supply is equal
to infinity.
Predicting Changes in Quantity Supplied
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 31 of 37
chapter
Extra Application 10
OPEC’S OIL STRANGLEHOLD
The Organization of Petroleum Exporting Countries (OPEC) is considering production
cutbacks to halt the falling world price of oil. However, OPEC members are also cognizant
of the fact that high oil prices spur development of alternative fuel sources. Many analysts
point out that this fact forces OPEC to allow prices to fall periodically so that alternative fuel
projects do not become viable. However, as oil demand continues to increase, OPEC may
not be able to periodically lower prices.
Since oil has a relatively inelastic short run demand, a
cutback in supply results in only a very small reduction
in the number of units sold but a substantial increase
in price. Since the production and distribution costs
will remain constant the price increase means a
significant increase in profits for oil producers.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 32 of 37
chapter
4.7
USING ELASTICITIES TO PREDICT
CHANGES IN EQUILIBRIUM PRICE
The Price Effects of a Change in Demand
► FIGURE 4.5
An Increase in Demand
Increases the Equilibrium
Price
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 33 of 37
chapter
4.7
USING ELASTICITIES TO PREDICT
CHANGES IN EQUILIBRIUM PRICE
The Price Effects of a Change in Demand
Under what conditions will an increase in demand
cause a relatively small increase in price?
• Small increase in demand.
• Highly elastic demand.
• Highly elastic supply.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 34 of 37
chapter
4.7
USING ELASTICITIES TO PREDICT
CHANGES IN EQUILIBRIUM PRICE
The Price Effects of a Change in Supply
► FIGURE 4.6
A Decrease in Supply
Increases the Equilibrium
Price
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 35 of 37
chapter
4.7
USING ELASTICITIES TO PREDICT
CHANGES IN EQUILIBRIUM PRICE
The Price Effects of a Change in Supply
Under what conditions will a decrease in supply
cause a relatively small increase in price?
• Small decrease in supply.
• Highly elastic demand.
• Highly elastic supply.
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 36 of 37
chapter
AN IMPORT BAN AND SHOE PRICES
APPLYING THE CONCEPTS #7: How do import restrictions
affect prices?
We can use the supply version of the price-change formula
to predict the effects of import restrictions on equilibrium prices.
Consider a nation that limits shoe imports.
• Suppose the import restrictions decrease the supply of shoes by 30 percent.
To use the price-change formula, we need the price elasticities of supply and demand.
• Suppose the supply elasticity is 2.3 and, as shown in Table 20.2, the demand elasticity
is 0.70.
• Plugging these numbers into the price-change formula, we predict a 10-percent
increase in price:
© 2007 Pearson/Prentice Hall, Survey of Economics: Principles, Applications & Tools, 3e, O’Sullivan • Sheffrin • Perez 37 of 37
```
Related documents | 5,403 | 20,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-10 | latest | en | 0.70748 |
https://www.stat.math.ethz.ch/pipermail/r-help/2016-May/438695.html | 1,653,419,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00571.warc.gz | 1,177,644,488 | 5,090 | # [R] projectRaster function no values
Ben Tupper btupper at bigelow.org
Wed May 18 15:05:17 CEST 2016
```Hi Adrienne,
You'll always get great help for these kinds of questions if you subscribe and post to the R-SIG-Geo mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-geo I highly recommend that you join!
I have encountered this before, and usually it is because I have mistakenly assumed that source and destination data are roughly coincident. I'm not sure if that is true in your case. I have tried to replicate your steps. I transformed the coordinates of your source and destination rasters into SpatialPoints objects, and then I reprojected your source coordinates to the projection of your destination coordinates. Unless I have messed up a step, you can see in the plot generated that there is a significant difference in the extent of the source and destination rasters. Perhaps your destination coordinates are amiss?
Cheers,
Ben
### Start
library(sp)
library(raster)
nc <- 234
nr <- 229
m <- matrix(runif(nc*nr), ncol = nc, nrow = nr)
r1 <- raster(m, xmn = -1747.5, xmx = 1762.5, ymn = -1710, ymx = 1725,
crs = CRS('+proj=lcc +lon_0=-77 +lat_0=38 +lat_1=30 +lat_2=60'))
newlon <- c(-102.97288, -74.05399)
newlat <- c(25.13511, 40.27023)
newnr <- 113
newnc <- 215
template <- raster(nrows = newnr, ncol = newnc,
xmn = newlon[1], xmx = newlon[2],
ymn = newlat[1], ymx = newlat[2],
crs = CRS("+proj=longlat +datum=WGS84"))
r2 <- setValues(template, runif(ncell(template)))
xy_r1 <- SpatialPoints(coordinates(r1),
proj4string = CRS('+proj=lcc +lon_0=-77 +lat_0=38 +lat_1=30 +lat_2=60'))
xy_r1_tr <- spTransform(xy_r1, CRS("+proj=longlat +datum=WGS84"))
xy_r2 <- SpatialPoints(coordinates(r2),
proj4string = CRS("+proj=longlat +datum=WGS84"))
plot(xy_r2, pch = '.', axes = TRUE)
points(xy_r1_tr, pch = 1, col = 'orange')
### END
> On May 17, 2016, at 1:02 PM, Adrienne Wootten <amwootte at ncsu.edu> wrote:
>
> All,
>
> Greetings! Any help with this problem is appreciated!
>
> I'm working to get a netcdf file that has a Lambert Conformal Conic
> projection into geographic, but also a smaller area. Here's the issue I'm
> having - essentially it looks like projectRaster is working, but the
> resulting raster has no values.
>
> The data itself is massive so I can't include that, but here's what's going
> on.
>
>
>> testvar2 = raster("SE/test.nc",band=t,varname="TAMAX") # pulling first
> time slice of my netcdf
>> testvar2
>
> class : RasterLayer
> band : 1 (of 4 bands)
> dimensions : 229, 234, 53586 (nrow, ncol, ncell)
> resolution : 15, 15 (x, y)
> extent : -1747.5, 1762.5, -1710, 1725 (xmin, xmax, ymin, ymax)
> coord. ref. : +proj=lcc +lon_0=-77 +lat_0=38 +lat_1=30 +lat_2=60
> +ellps=WGS84
> data source : /projdata/dcerp/DSdata/regcmdata/SE/test.nc
> names : Avg.Max.Aneom.Temperature
> z-value : 1960-01-01
> zvar : TAMAX
>
>> summary(testvar2) # yes is does have values
> Avg.Max.Aneom.Temperature
> Min. -23.347107
> 1st Qu. -4.706635
> Median 4.347733
> 3rd Qu. 16.109032
> Max. 24.556152
> NA's 0.000000
>
>> newlocs <- raster(ncols=length(newlon), nrows=length(newlat)) # dummy
> raster with new grid I want
>> projection(newlocs)=CRS("+proj=longlat +datum=WGS84")
>
>> newlat=c(25.13511, 25.27025, 25.40538, 25.54052, 25.67565, 25.81079,
> 25.94592, 26.08106, 26.21619, 26.35133, 26.48646, 26.62160, 26.75673,
> 26.89187, 27.02700, 27.16214, 27.29727, 27.43241, 27.56754, 27.70268,
> 27.83781, 27.97295, 28.10808, 28.24322, 28.37835, 28.51349, 28.64862,
> 28.78376, 28.91889, 29.05403, 29.18916, 29.32430, 29.45943, 29.59457,
> 29.72970, 29.86484, 29.99997, 30.13511, 30.27024, 30.40538, 30.54051,
> 30.67565, 30.81078, 30.94592, 31.08105, 31.21619, 31.35132, 31.48646,
> 31.62159, 31.75673, 31.89186, 32.02700, 32.16213, 32.29727, 32.43240,
> 32.56754, 32.70267, 32.83781, 32.97294, 33.10808, 33.24321, 33.37835,
> 33.51348, 33.64862, 33.78375, 33.91889, 34.05402, 34.18916, 34.32429,
> 34.45943, 34.59456, 34.72970, 34.86483, 34.99997, 35.13510, 35.27024,
> 35.40537, 35.54051, 35.67564, 35.81078, 35.94591, 36.08105, 36.21618,
> 36.35132, 36.48645, 36.62159, 36.75672, 36.89186, 37.02699, 37.16213,
> 37.29726, 37.43240, 37.56753, 37.70267, 37.83780, 37.97294, 38.10807,
> 38.24321, 38.37834, 38.51348, 38.64861, 38.78375, 38.91888, 39.05402,
> 39.18915, 39.32429, 39.45942, 39.59456, 39.72969, 39.86483, 39.99996,
> 40.13510, 40.27023) # the new regular grid in geographic I'd like to work
> with
>
>> newlon=c(-102.97288, -102.83774, -102.70261, -102.56747, -102.43234,
> -102.29720, -102.16207, -102.02693, -101.89180, -101.75666, -101.62153,
> -101.48639, -101.35126, -101.21612, -101.08099, -100.94585, -100.81072,
> -100.67558, -100.54045, -100.40531, -100.27018, -100.13504, -99.99991,
> -99.86477, -99.72964, -99.59450, -99.45937, -99.32423, -99.18910,
> -99.05396, -98.91883, -98.78369, -98.64856, -98.51342, -98.37829,
> -98.24315, -98.10802, -97.97288, -97.83775, -97.70261, -97.56748,
> -97.43234, -97.29721, -97.16207, -97.02694, -96.89180, -96.75667,
> -96.62153, -96.48640, -96.35126, -96.21613, -96.08099, -95.94586,
> -95.81072, -95.67559, -95.54045, -95.40532, -95.27018, -95.13505,
> -94.99991, -94.86478, -94.72964, -94.59451, -94.45937, -94.32424,
> -94.18910, -94.05397, -93.91883, -93.78370, -93.64856, -93.51343,
> -93.37829, -93.24316, -93.10802, -92.97289, -92.83775, -92.70262,
> -92.56748, -92.43235, -92.29721, -92.16208, -92.02694, -91.89181,
> -91.75667, -91.62154, -91.48640, -91.35127, -91.21613, -91.08100,
> -90.94586, -90.81073, -90.67559, -90.54046, -90.40532, -90.27019,
> -90.13505, -89.99992, -89.86478, -89.72965, -89.59451, -89.45938,
> -89.32424, -89.18911, -89.05397, -88.91884, -88.78370, -88.64857,
> -88.51343, -88.37830, -88.24316, -88.10803, -87.97289, -87.83776,
> -87.70262, -87.56749, -87.43235, -87.29722, -87.16208, -87.02695,
> -86.89181, -86.75668, -86.62154, -86.48641, -86.35127, -86.21614,
> -86.08100, -85.94587, -85.81073, -85.67560, -85.54046, -85.40533,
> -85.27019, -85.13506, -84.99992, -84.86479, -84.72965, -84.59452,
> -84.45938, -84.32425, -84.18911, -84.05398, -83.91884, -83.78371,
> -83.64857, -83.51344, -83.37830, -83.24317, -83.10803, -82.97290,
> -82.83776, -82.70263, -82.56749, -82.43236, -82.29722, -82.16209,
> -82.02695, -81.89182, -81.75668, -81.62155, -81.48641, -81.35128,
> -81.21614, -81.08101, -80.94587, -80.81074, -80.67560, -80.54047,
> -80.40533, -80.27020, -80.13506, -79.99993, -79.86479, -79.72966,
> -79.59452, -79.45939, -79.32425, -79.18912, -79.05398, -78.91885,
> -78.78371, -78.64858, -78.51344, -78.37831, -78.24317, -78.10804,
> -77.97290, -77.83777, -77.70263, -77.56750, -77.43236, -77.29723,
> -77.16209, -77.02696, -76.89182, -76.75669, -76.62155, -76.48642,
> -76.35128, -76.21615, -76.08101, -75.94588, -75.81074, -75.67561,
> -75.54047, -75.40534, -75.27020, -75.13507, -74.99993, -74.86480,
> -74.72966, -74.59453, -74.45939, -74.32426, -74.18912, -74.05399)
>
>> extent(newlocs) = c(min(newlon),max(newlon),min(newlat),max(newlat)) #
> dummy raster put to the right extent
>
>> testproj2 = projectRaster(from=testvar2,to=newlocs,method="bilinear") #
> project raster itself
>
>> testproj2 # literally no values.
> class : RasterLayer
> dimensions : 113, 215, 24295 (nrow, ncol, ncell)
> resolution : 0.1345065, 0.1339391 (x, y)
> extent : -102.9729, -74.05399, 25.13511, 40.27023 (xmin, xmax, ymin,
> ymax)
> coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0
> data source : in memory
> names : Avg.Max.Aneom.Temperature
> values : NA, NA (min, max)
>
>
> I'm quite perplexed with this one, I feel like I'm doing everything right
> so I'm not sure what's failing. The R version is R 3.2.3 in a Linux/Unix
> environment.
>
> Many thanks for your help!
>
> --
> Ph.D Candidate / Graduate Research Assistant
> State Climate Office of North Carolina
> Department of Marine, Earth and Atmospheric Sciences
> North Carolina State University
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> and provide commented, minimal, self-contained, reproducible code.
Ben Tupper
Bigelow Laboratory for Ocean Sciences
60 Bigelow Drive, P.O. Box 380
East Boothbay, Maine 04544
http://www.bigelow.org
``` | 3,547 | 8,521 | {"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-2022-21 | latest | en | 0.673505 |
https://www.helpwithassignment.com/Solution-Library/calculation-of-present-value-and-internal-rate-of-return | 1,553,087,921,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202347.13/warc/CC-MAIN-20190320125919-20190320151919-00202.warc.gz | 791,400,137 | 15,688 | Calculation Of Present Value And Internal Rate Of Return
Question
1. What is the present value (PV) of an investment that will pay \$4000 in one year’s time and \$400 every year after that, when the interest rate is 5% per year?
A. \$2400
B. \$3600
C. \$7200
D. \$8000
2. You are considering investing in a zero-coupon bond that will pay you its face value of \$1000 in ten years. If the bond is currently selling for \$485.20, then the internal rate of return (IRR) for investing in this bond is closest to:
A. 12%
B. 8.0%
C. 7.5%
D. 10%
Summary
These short questions belong to Finance. The 1st question is about calculating the present value of an investment. The 2nd question is about calculating the internal rate of return for a zero-coupon bond.
Total Word Count 17
• Rasha
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• maani
I have 50 questions for the same test your page is showing only 28
• joeanne
• joeanne
hi can anyone help or guide me to my assignments. thanks
• Monik
• Cristina
This solution is perfect ...thanks
• Janete
Hello Allison,I love the 2nd image that you did! I also, had never heard of SumoPaint, is something that I will have to exolpre a bit! I understand completely the 52 (or so) youtube videos that you probably watched. Sometimes they have what you want, sometimes they don't! However, it is always satisfying when you are able to produce something that you have taught yourself. Great job!Debra 0 likes
• Sandeep
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• Oxana
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• Paul Brandon-Fritzius
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• tina Johnson
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• Giuseppe
works fine. | 442 | 1,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-13 | latest | en | 0.946701 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-11-section-11-3-geometric-sequences-and-series-concept-and-vocabulary-check-page-853/1 | 1,576,002,413,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528490.48/warc/CC-MAIN-20191210180555-20191210204555-00206.warc.gz | 727,141,657 | 11,904 | ## Intermediate Algebra for College Students (7th Edition)
RECALL: A geometric sequence has a common ratio $r$ such that the next term is equal to the product of the previous term and the common ratio $r$. Thus, the missing expressions in the given statements are geometric and common ratios, respectively. | 63 | 307 | {"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} | 3.078125 | 3 | CC-MAIN-2019-51 | latest | en | 0.947269 |
https://cs.stackexchange.com/questions/92519/generate-all-sub-paths | 1,571,409,152,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00009.warc.gz | 434,505,833 | 33,982 | # Generate all sub paths
I have a path from node 1 to node n, which I can represent as a set: S = {1, 2, ..., n-1, n}. I want to efficiently generate the set of all subpaths from 1 to n. For instance, for n=5, we have S={1,2,3,4,5}.
So all the subpaths are as follows:
{{1,2},{2,3},{3,4},{4,5}, {1,2},{2,3},{3,4,5}, {1,2},{2,3,4},{4,5}, {1,2},{2,3,4,5}, {1,2,3},{3,4},{4,5}, {1,2,3},{3,4,5}, {1,2,3,4},{4,5}, {1,2,3,4,5}}
• Can you define what a subpath means for you? And, what approaches have you considered? – D.W. May 31 '18 at 0:59
• I was considering a depth first enumeration, but I was looking for a different conceptualization. I admit it was late in the day and I was tired. There's a better way to visualize it, as presented further below in this thread. – Bruno Repetto May 31 '18 at 17:11
• A path is the enumeration of all nodes in it, from 1 to n. Edges are always of the form $(k,k+1)$. A subpath $\{i,j\}$, for $i<j$ is a collection of edges such that we have a path from i to j where $(i, i+1), \cdots,(j-1,j)$. A collection of subpaths from 1 to n is therefore of the form: $\{a_{1k},a_{2k}\}$ for $k=1,\cdots,p$ where $a_{11}=1$ and $a_{2p}=n$ and $a_{2(k-1)} = a_{1k}$. – Bruno Repetto May 31 '18 at 17:13
• Rather than putting additional information or clarifications in the comments, please edit the question so it stands on its own and so people can understand your question without having ot read the comments; incorporate that information into the question. – D.W. May 31 '18 at 18:14
Suppose that the path consists of edges $e_1,\ldots,e_\ell$. The following answer assumes that by "subpath" you mean a partition of $e_1,\ldots,e_\ell$ into intervals. Go over all possible values of $b_1,\ldots,b_{\ell-1} \in \{T,F\}$. If $b_i = T$ then the edge $e_i$ ends an interval. Otherwise it doesn't. Here is how this looks in your case, $\ell=4$: $$\begin{array}{ccc|c} b_1 & b_2 & b_3 & \text{subpath} \\\hline F & F & F & e_1e_2e_3e_4 \\ F & F & T & e_1e_2e_3;e_4 \\ F & T & F & e_1e_2;e_3e_4 \\ F & T & T & e_1e_2;e_3;e_4 \\ T & F & F & e_1;e_2e_3e_4 \\ T & F & T & e_1;e_2e_3;e_4 \\ T & T & F & e_1;e_2;e_3e_4 \\ T & T & T & e_1;e_2;e_3;e_4 \\ \end{array}$$
• Viewed this way it becomes obvious! I'll test this out with the longer paths, but I think it should work correctly. Thanks @Yuval! – Bruno Repetto May 31 '18 at 16:53
• This works beatifully! Thanks again @Yuval! – Bruno Repetto May 31 '18 at 21:45
I am storing the nodes in an array and using recursion to generate the combinations of possible subpaths.
The method generateSubPaths(int[] arr, int index) takes an array and an index as input. The array arr[] contains the nodes and the index represents the current node.
We call this method recursively by updating index to index - 1. This will return us an array of strings. Each element of the returned array represents a combination of subpaths up to node index - 1. The number of combinations gets double if we increase the number of nodes by 1.
For n = 2 : Paths =>
(12)
For n = 3 : Paths =>
(12)(23)
(123)
For n = 4 : Paths =>
(12)(23)(34)
(123)(34)
(1234)
(12)(234)
For generating combinations after adding an extra node, repeat these steps for all existing combinations:
1. Create a new path between the previous node and current node.
2. Append new node to the last subpath in the combination.
Example :-
For n = 3 : One of the combinations is (12)(23).
For n = 4 :
1. Create a new path between node 3 and node 4 i.e. (34). Therefore, the new path is
(12)(23)(34).
2. Append node 4 to the last subpath in the combination. Therefore, the new path is
(12)(234).
Source Code :-
import java.util.Scanner;
public class SubPaths {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// Input number of nodes (>1)
System.out.println("Enter number of nodes (>1) :");
int n = sc.nextInt();
int[] arr = new int[n];
// Initialize the array of nodes
for (int i = 1; i <= n; ++i) {
arr[i - 1] = i;
}
String[] result = generateSubPaths(arr, n);
// Print each combination in a separate line
for (String str : result) {
System.out.println(str);
}
sc.close();
}
/*
* This method will return an array of Strings , where each String
represents a
* possible combination of subpaths
*/
private static String[] generateSubPaths(int[] arr, int index) {
/*
* If there are only two nodes, only 1 combination is posssible.
*/
if (index == 2) {
String[] result = new String[1];
// Join node 1 and node 2
result[0] = "(" + arr[0] + arr[1] + ") ";
return result;
}
/*
* smallResult[] stores all the combinations of subpaths where
* maximum node is represented by 'index'
*/
String[] smallResult = generateSubPaths(arr, index - 1);
int x = smallResult.length;
/*
* If one new node is added i.e. the current node, the possible
* combinations will be double of the previous ones.
*/
String[] result = new String[x * 2];
String temp;
for (int i = 0; i < x; ++i) {
temp = smallResult[i];
// Create a new path between current node and previous node.
result[i] = temp + "(" + temp.charAt(temp.length() - 2) +
arr[index - 1] + ")";
// Append current node to the last subpath in the combination
result[x + i] = temp.substring(0, temp.length() - 1) +
arr[index - 1] + ") ";
}
return result;
}
}
• This is not a programming site. There is no reason to include code in an answer. – Yuval Filmus May 31 '18 at 13:36
• @YuvalFilmus I do not know much about how to answer a question. Should I remove this code block? – user89160 Jun 1 '18 at 5:34
• You can replace it with succinct pseudocode. – Yuval Filmus Jun 1 '18 at 5:35 | 1,786 | 5,614 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-43 | latest | en | 0.900134 |
https://www.wyzant.com/resources/answers/485435/function_word_problem | 1,618,331,958,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00081.warc.gz | 1,165,861,041 | 14,120 | Nikki C.
# Function word problem
Twitter monthly worldwide active users increased approximately linearly from 42 million users in 2010 to 218 million users in 2013. Let n be the number(in millions) of worldwide active users of twitter at t years since 2000. Find an equation of a linear model to describe the data.
## 2 Answers By Expert Tutors
By:
Tutor
4.9 (27)
Math/Physics Tutor
Tutor
4.9 (99)
Andy C.
I respectfully disagree.
At time t=0, the enrollment will be 2000
At time t=10 years, the enrollment will be much WORSE.
The independent variable is the number of years since 2000
Intercept B = y - mx where m is the slope and (x,y) is a point on the line.
slope m=176/3 ; Using (2010-2000,42) --> X=10 and y=42
Then the intercept is B = y - mx = 42 - 10*176/3 = -544 and 2/3
Report
05/09/18
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https://www.jiskha.com/questions/459168/a-tree-casts-a-shadow-10-feet-long-a-6-ft-man-casts-a-shadow-4-ft-long-the-triangle | 1,579,415,173,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594209.12/warc/CC-MAIN-20200119035851-20200119063851-00262.warc.gz | 944,839,005 | 4,452 | # Math
A tree casts a shadow 10 feet long. A 6-ft. man casts a shadow 4 ft. long. The triangle formed by the tree and its shadow is similar to the triangle formed by the man and his shadow. How tall is the tree?
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## Similar Questions
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2. ### prealgebra
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7. ### ALGERBRA
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9. ### math
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More Similar Questions | 720 | 2,403 | {"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} | 3.484375 | 3 | CC-MAIN-2020-05 | latest | en | 0.939894 |
https://www.mapleprimes.com/users/sarra/questions?page=4 | 1,670,373,047,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00805.warc.gz | 927,906,885 | 21,260 | ## 270 Reputation
8 years, 289 days
## Plot of solution of ODE...
Maple
Dear all;
Thanks in advance for helping me to plot the solution of this second order ode.
with(plots):
ode := diff(y(x), x, x) = x*y(x)+x^(17/12);
ics := y(1000) = 0, y(1001) = 1;
dsolve({ics,ode}):
How can I plot the solution obtained in the range (1000, 1001).
Thanks
## Plot of function in prescribed interval...
Maple
Dear all,
I have the function
y:=x->-9.8455282400*10^9142*exp(-(2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+3.3889331940*10^(-9169)*exp((2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+(16/153)*x^(7/6)*sqrt(Pi)*exp((2/3)*x^(3/2))+Pi*((1/2)*exp(-(2/3)*x^(3/2))*(-1+exp((2/3)*x^(2/3)))/(x^(1/4)*Pi)-(16/153)*x^(7/6)*exp((2/3)*x^(3/2))/sqrt(Pi));
how can I plot y versus x with x in the interval (1000, 1001).
First, it's simple to verify that y(1000)=0; y(1001)=1; So (1000, 0) and (1001,1) belong to our graph.
I tried plot( y, x=1000..1001); but there is no curves.
Thank you in advance to help me to plot the graph of this function.
with brest regards,
## Bisection method, roots, polynom...
Dear all,
I need you help to finish some steps of this idea to approximate the roots of a given equation (polynom). Thanks in advance for your help.
I have a sturm sequence, I would like to use Bisection method to approximation the roots using Sturm decomposition of my polynom. For example, my polynom is P=x^6-4*x^3+x-2
s := sturmseq(x^6-4*x^3+x-2,x);
sturm(s,x,-2,2); # The number of roots in the interval (-2,2)
Here, i would like to find the roots in (-M,M) :
Bounding all roots in [-M,M] where M = max{1, sum^(n-1) |ai|/an}.
f0 = f, f1 = f', then use -remainder,
I know that sturm(s,x,-M,M); gives the number of roots in (-M,M) but is it possible to use the variation of sign like :
gives a Sturm sequence for f.
variation of sign, varsign(a0,a1,...,ar).
Thm: (Sturm) varsign(f0(alpha),...,fr(alpha)) - varsign(f0(beta),..., fr(beta))
is the number of distinct roots of f in [alpha,beta].
then i would like Isolating roots of rational polynomials
Method: reduce, remove rational roots, divide and conquer in [-M,M],
then use bisection in disjoint closed intervals ctg one root each
Bisection method :
`Bisection`
` Setup: f(a) < 0, f(b) > 0 (or conversely).`
` Repeated subdivision of [a,b] guaranteed to get close to a root.`
Error analysis: for error eps, solve (b-a)/ 2^(n+1) < tol for n. where tol is the tolerance
Thanks
## Sturm sequence and roots of equation...
Dear all;
I need you help for solving this problem, and thanks in advantage for your help.
I have a polynom like P =x^6-4*x^3+x-2; and i would like to find an approximate value of the roots in some interval [a,b] =[-2,2] using sturm sequence.
The method is based on:
1) first construct the sturm sequence:
For given polynom P =x^6-4*x^3+x-2;
Let S0=P;
S1=diff(p,x);
let s:=quo(S0,S1,x);
S2:=-rem(S0,S1,x);
.... S[k+1-rem(S[k-1],S[k]);
S[k] is the sturm sequence.
2) let f(a)= number of change of sign in the sturm sequence and f(b) the same . so f(b)-f(a) give the number of roots in the interval [a,b].
3) If f(b)-f(a) =0 so there are no roots
and if f(a)-f(b)=1 one can find the root
4) if f(a) -f(b) >2 :
given toterance tol=0.001; for example
if the abs(a-b)<2*epsilon we display a message that there are k roots at (b+a)/2
with our error tolerance
5) otherwise if c=(b+a)/2 is not a root of P_k(x) for any k, ( where p_k is an element of the sturm sequence )
we divide the interval into equal halves [a,c] and [x,b] and we run step 2 on each interval
else if c is a root of one of these p_k(x) add any time account to c so that c lies close the middle of [a,b] and not a root
6) Give all the roots ( approximate the rrots with small error epsilon).
## ode, dsolve, ics...
Dear all;
Thank you very much for helping me to understand this problem.
I need your help for this question, it's seem for correct but when I run the code there is no dispaly of the solution with this command dsolve({ode,ics}) ;
ode:=x*diff(y(x),x)+y(x)=x*exp(-x^2);
ics:=y(0)=1;
dsolve(ode);
dsolve({ode,ics}) ;
2 3 4 5 6 7 8 Last Page 4 of 21
| 1,378 | 4,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-49 | latest | en | 0.781783 |
http://stackoverflow.com/questions/19328095/calculating-the-sum-of-values-in-2-linked-list | 1,386,851,058,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164583115/warc/CC-MAIN-20131204134303-00051-ip-10-33-133-15.ec2.internal.warc.gz | 167,012,207 | 14,815 | # Calculating the Sum of values in 2 linked list [duplicate]
So I got a programming question at an interview recently.
There are 2 linked lists, each node's store a value from 1 through 9 (indicating one index of the number). Hence 123 would be a linked list 1->2->3
The task was to create a function:
`static LinkedListNode getSum(LinkedListNode a, LinkedListNode b)`
that would return the sum of the values in the 2 linked list arguements.
If the array a is: 1->2->3->4
And the array b is: 5->6->7->8
Here is my algorithm:
Go through each node in a and b, get the values as an integer and add them. Create a new linked list with the those values.
Here is the code: It runs with a complexity of O(3n) roughly I assume. Once through each of the array inputs and once to create the output array.
Any improvements? Better algorithms... or code improvements
``````public class LinkedListNode {
int value;
this.value = value;
this.next = null;
}
int value = node.value;
while (node.next != null) {
node = node.next;
value = value * 10 + node.value;
}
return value;
}
int aval = getValue(a);
int bval = getValue(b);
int result = aval + bval;
while (result > 0) {
int len = (int) Math.pow((double) 10,
(double) String.valueOf(result).length() - 1);
int val = result / len;
ans = ans.next;
result = result - val*len;
}
}
}
``````
-
It looks like the task was to add numbers where the decimal digits are represented as a linked list. You should add that to the question, as it is this is very hard to understand. – Kilian Foth Oct 11 at 14:27
Just to let everyone know, this question already has an answer here. This copy was originally cross-posted at Programmers, and migrated here instead of closed. – Generic Holiday Name Oct 12 at 2:30
## migrated from programmers.stackexchange.comOct 11 at 22:15
This question came from our site for professional programmers interested in conceptual questions about software development.
## marked as duplicate by Generic Holiday Name, Mena, Luc M, nwellnhof, madth3Oct 12 at 17:59
let me give it a shot...
``````static LinkedListNode getSum(LinkedListNode a, LinkedListNode b) {
//some checks first if any computation will be needed at all
if(a == null) {
if(b == null)
return null;
else
return b;
} else if (b == null)
return a;
//initialize the variables
//move the contents of a & b into stacka & stackb respectively at the same time
//best case is when a & b are of equal size
//worst case is when the size of a & b are worlds apart.
while(a != null || b != null){
if(a != null) {
if(stacka == null){
} else {
temp.next = stacka;
stacka = temp;
}
}
if(b != null) {
if(stackb == null){
} else {
temp.next = stackb;
stackb = temp;
}
}
if(a != null) a = a.next;
if(b != null) b = b.next;
}
int remainder = 0;
//just pop off the stack then merge! also, don't forget the remainder~
while(stacka != null || stackb != null){
//pop from the top of the stack
int i = ((stacka == null) ? 0 : stacka.value) + ((stackb == null) ? 0 : stackb.value) + remainder;
//set the value of the remainder if any as well as the value of i
remainder = i / 10;
i %= 10;
if(ans == null) {
ans = temp;
} else {
temp.next = ans;
ans = temp;
}
if(stacka != null) stacka = stacka.next;
if(stackb != null) stackb = stackb.next;
}
return ans;
}
``````
Since I didn't use the getValue() function, this should be around O(2n) at best case. What I did here was use the LinkedListNode as a stack to temporarily store the nodes while I invert them, then pop the values off one at a time to populate the output LinkedListNode.
Then again, In the end, both algorithms still fall under O(n) so the difference can be negligible.
I'll try to make a recursive version later if I have time.
P.S. Sorry if i didn't add curly braces to some of my if else statements, its hard to tab them using the answer form
-
It can be optimized by constructing the resulting linked list from back-to-front:
``````int aval = getValue(a);
int bval = getValue(b);
int result = aval + bval;
while (result > 0) {
int val = result % 10;
result /= 10;
}
// Assuming you want to return 0 rather than null if the sum is 0
}
``````
This avoids the repeated Math.pow calls.
I think the overall algorithm you used should be fastest. One alternative that comes to mind is doing some kind of add-with-carry operation on each pair of digits (i.e. doing the addition "manually"), but that would very likely be slower.
-
Doing the carries manually would likely be faster, since you're only adding single digits and 9+9 = 18, meaning you'd only have to test whether it was > 10, and if so subtract 10 and do the carry. But since we're using a linked list that stores single digits, I don't think it would make a real difference in speed, because it's almost certainly swamped with all those pointer dereferences. – Michael Shaw Oct 11 at 19:18
Well, the main issue performance-wise is that you would need to traverse each list twice, once to get to the least-significant digit, then back again as we add each pair of digits. – Cyanfish Oct 11 at 19:34
If we altered the problem by having the linked lists store numbers from smallest digit to largest digit, that problem would go away. That representation would actually make more sense in general as well, at least for a singly linked list. – Michael Shaw Oct 11 at 20:58
Usually in these types of exercises one is expected to perform the operation without converting into a more common intermediate form first (like an integer). The next question I would expect to get is, "What if the numbers are 100 digits long?" Try to solve it using only linked lists, although you probably have to reverse the direction of the operands in order to provide a reasonable running time.
-
First, run through both lists, flipping the directions of the arrows and ending with the final nodes in memory. This is linear time and constant space.
Now you have a pair of linked lists that represent the numbers from their lowest to highest digit. Run through the lists again, creating your new linked list and flipping the arrows back as you go. This is linear time and linear space (for the new list).
-
The original question is in Java, but here's a very simple Scala solution. It left pads the lists with 0's so that they're the same length. Then, it zips the lists together so that we have a single list of pairs. Finally, it adds the pairs right to left passing along a carry value. (The same way you learned how to add numbers in first grade.) It shows how we can solve problems quickly and with small amounts of code using functional techniques:
``````def add(nums1: List[Int], nums2: List[Int]): List[Int] = {
val nums1Size = nums1.size
val nums2Size = nums2.size
val maxSize = nums1Size max nums2Size
val nums1Padded = List.fill(maxSize - nums1Size)(0) ++ nums1
val nums2Padded = List.fill(maxSize - nums2Size)(0) ++ nums2 | 1,732 | 6,870 | {"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} | 3.59375 | 4 | CC-MAIN-2013-48 | latest | en | 0.847795 |
http://www.mathskey.com/question2answer/11764/prove-trigonometric-identities | 1,726,879,730,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00533.warc.gz | 51,785,465 | 10,679 | # prove trigonometric identities
secxtanx=cscxcotx
asked Apr 16, 2014
Sec x tan x = csc x cot x
Left hand side identity = Sec x tan x
= 1 / cos x * ( sin x / cos x
= sin x / cos 2 x
Left hand side identity = sin x / cos 2 x
Right hand side identity = csc x cot x
= 1 / sin x * (cos x / sin x
= cos x / sin 2 x
Right hand side identity = cos x / sin 2 x
Left hand side identity = Right hand side identity
sin x /cos 2 x = cos x / sin 2 x
sin 3 x = cos 3
Let us assume that x = π / 4
sin 3 x = sin 3 ( π / 4 ) = ( 1/ √ 2)3 = 0 . 3535
cos 3 x = cos 3 (π / 4) = ( 1/ √ 2)3 = 0 . 3535
sin 3 x = cos 3 x
Since period of a sinusoidal function is 2π.
So. it is also true for all x = π / 4 + 2nπ where n is integer.
For all x values except x = π / 4 + 2nπ the identity is not true.
Therefore Sec x tan x ≠ csc x cot x .
answered Apr 17, 2014 | 341 | 865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.467342 |
http://anabolicminds.com/forum/politics/60717-free-tax-lessons-post747251.html | 1,461,935,165,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111324.43/warc/CC-MAIN-20160428161511-00183-ip-10-239-7-51.ec2.internal.warc.gz | 9,087,577 | 14,615 | 1. Free Tax Lessons For Everyone!!!
Let's put tax cuts in terms everyone can understand.
Suppose that every day, ten men go out for beer and the bill for all ten
comes to \$100. If they paid their bill the way we pay our taxes, it would go something
like this:
The first four men (the poorest) would pay nothing.
The fifth would pay \$1.
The sixth would pay \$3.
The seventh would pay \$7.
The eighth would pay \$12.
The ninth would pay \$18.
The tenth man (the richest) would pay \$59.
So, that's what they decided to do.
The ten men drank in the bar every day and seemed quite happy
with the arrangement, until on day, the owner threw them a curve.
"Since you are all such good customers," he said, "I'm going to reduce
the cost of your daily beer by \$20."Drinks for the ten now cost just \$80.
The group still wanted to pay their bill the way we pay our taxes so the
first four men were unaffected. They would still drink for free. But
what about the other six men - the paying customers? How could they divide
the \$20 windfall so that everyone would get his 'fair share?'
They realized that \$20 divided by six is \$3.33. But if they subtracted
that from everybody's share, then the fifth man and the sixth man would
each end up bei ng paid to drink his beer.
So, the bar owner suggested that it would be fair to reduce each
man's bill by roughly the same amount, and he proceeded to work out the
amounts each should pay.
And so:
The fifth man, like the first four, now paid nothing (100% savings).
The sixth now paid \$2 instead of \$3 (33%savings).
The seventh now pay \$5 instead of \$7 (28%savings).
The eighth now paid \$9 instead of \$12 (25% savings).
The ninth now paid \$14 instead of \$18 (22% savings).
The tenth now paid \$49 instead of \$59 (16% savings).
Each of the six was better off than before. And the first four continued
to drink for free. But once outside the restaurant, the men began to
compare their savings.
"I only got a dollar out of the \$20,"declared the sixth man. He pointed
to the tenth man," but he go t \$10!"
"Yeah, that's right," exclaimed the fifth man. "I only saved a dollar,
too. It's unfair that he got ten times more than I!"
"That's true!!" shouted the seventh man. "Why should he get \$10 back
when I got only two? The wealthy get all the breaks!"
"Wait a minute," yelled the first four men in unison. "We didn't get
anything at all. The system exploits the poor!"
The nine men surrounded the tenth and beat him up.
The next night the tenth man didn't show up for drinks, so the nine
sat down and had beers without him. But when it came time to pay the
bill, they discovered something important. They didn't have enough money between
all of them for even half of the bill!
And that, boys and girls, journalists and college professors, is how our
tax system work s. The people who pay the highest taxes get the most
benefit from a tax reduction. Tax them too much, attack them for being
wealthy, and they just may not show up anymore. In fact, they might
start drinking overseas where the atmosphere is somewhat friendlier.
David R. Kamerschen, Ph.D.
Professor of Economics
University of Georgia
For those who understand, no explanation is needed. For those who do
not understand, no explanation is possible.
2. Once Obama is elected taxes will be a thing of the past!!!
•
3. Originally Posted by klugman
Once Obama is elected taxes will be a thing of the past!!!
Too much hype around Obama.. If he gets in a position where he could potentially take the polls he will be Though, I don't think he'll ever get that far before 08.
4. Originally Posted by anabolicrhino
Let's put tax cuts in terms everyone can understand.
Suppose that every day, ten men go out for beer and the bill for all ten
comes to \$100. If they paid their bill the way we pay our taxes, it would go something
like this:
The first four men (the poorest) would pay nothing.
The fifth would pay \$1.
The sixth would pay \$3.
The seventh would pay \$7.
The eighth would pay \$12.
The ninth would pay \$18.
The tenth man (the richest) would pay \$59.
So, that's what they decided to do.
The ten men drank in the bar every day and seemed quite happy
with the arrangement, until on day, the owner threw them a curve.
"Since you are all such good customers," he said, "I'm going to reduce
the cost of your daily beer by \$20."Drinks for the ten now cost just \$80.
The group still wanted to pay their bill the way we pay our taxes so the
first four men were unaffected. They would still drink for free. But
what about the other six men - the paying customers? How could they divide
the \$20 windfall so that everyone would get his 'fair share?'
They realized that \$20 divided by six is \$3.33. But if they subtracted
that from everybody's share, then the fifth man and the sixth man would
each end up bei ng paid to drink his beer.
So, the bar owner suggested that it would be fair to reduce each
man's bill by roughly the same amount, and he proceeded to work out the
amounts each should pay.
And so:
The fifth man, like the first four, now paid nothing (100% savings).
The sixth now paid \$2 instead of \$3 (33%savings).
The seventh now pay \$5 instead of \$7 (28%savings).
The eighth now paid \$9 instead of \$12 (25% savings).
The ninth now paid \$14 instead of \$18 (22% savings).
The tenth now paid \$49 instead of \$59 (16% savings).
Each of the six was better off than before. And the first four continued
to drink for free. But once outside the restaurant, the men began to
compare their savings.
"I only got a dollar out of the \$20,"declared the sixth man. He pointed
to the tenth man," but he go t \$10!"
"Yeah, that's right," exclaimed the fifth man. "I only saved a dollar,
too. It's unfair that he got ten times more than I!"
"That's true!!" shouted the seventh man. "Why should he get \$10 back
when I got only two? The wealthy get all the breaks!"
"Wait a minute," yelled the first four men in unison. "We didn't get
anything at all. The system exploits the poor!"
The nine men surrounded the tenth and beat him up.
The next night the tenth man didn't show up for drinks, so the nine
sat down and had beers without him. But when it came time to pay the
bill, they discovered something important. They didn't have enough money between
all of them for even half of the bill!
And that, boys and girls, journalists and college professors, is how our
tax system work s. The people who pay the highest taxes get the most
benefit from a tax reduction. Tax them too much, attack them for being
wealthy, and they just may not show up anymore. In fact, they might
start drinking overseas where the atmosphere is somewhat friendlier.
David R. Kamerschen, Ph.D.
Professor of Economics
University of Georgia
For those who understand, no explanation is needed. For those who do
not understand, no explanation is possible.
Great post, you put it in terms that anyone can understand.
5. good post.
•
6. Tax system has some definate flaws...What would be ways to fix it?
7. Originally Posted by Jayhawkk
Tax system has some definate flaws...What would be ways to fix it?
I would just eliminate the federal tax system all together.
The money collected from the private citizens is basically used to pay off the interest on our national debt(~\$900,000,000,000)
The money collected from corporations is about \$400,000,000,000
which is oddly enough the same amount as our defense budget!
This is a great documentary on the subject, it has lots of valuable information | 1,817 | 7,523 | {"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} | 3.71875 | 4 | CC-MAIN-2016-18 | latest | en | 0.980783 |
https://jeremycote.me/estimation-modeling-accuracy | 1,558,942,059,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262029.97/warc/CC-MAIN-20190527065651-20190527091651-00027.warc.gz | 515,995,465 | 5,602 | I’m currently studying both mathematics and physics in university, and I have to admit that it can be difficult to straddle the line between the two. Both are similar, yet demand different mindsets in terms of how to think about tackling a problem and actually coming up with a solution. In mathematics, not only is the right answer desirable. Every step along the way should be rigourously justified. That’s because the conclusion that one wants to get to rests on the arguments that come beforehand. Without those arguments, you don’t have anything. This is why mathematics classes require students to create proofs that carefully apply definitions. I’m not saying that there isn’t any playfulness involved, but when it comes down to making an argument, the clearer the supporting propositions, the easier it is for others to become convinced of the truth of your claim.
In physics, I’ve found that the situation is quite different. Being mathematically coherent is of course necessary within developing a theory, but the truth is that physicists are much “looser” with their mathematics, for lack of a better word. In physics, it’s often taken for granted that certain complications “are so small that they won’t make a difference”, which allows them to drop the complications. This is something that absolutely would not be allowed when proving statements in mathematics, because any weak argument is the first thing that gets attacked when someone critiques a proof. Many people think that π+e is transcendental, but since we don’t have a proof of this, it’s an unjustified belief.
The difference in physics (and science in general) is the fact that we often know what the answer should be. This makes a huge difference in terms of the way that we work through theory to get to a result. It’s a lot easier to say “these other contributions won’t have a large effect” when we know that continuing in this manner will give the observed result. Of course, it probably is true that certain contributions aren’t as important (and one can show this mathematically), but that extra work is often hand-waved away. Because of this, I’ve observed that we often will simplify matters a lot more than what I would have thought appropriate, because it gives the correct answer.
I’ve had mixed feelings about this, particularly because I’ve been on the other side in my mathematics classes, where it was necessary to go through the steps, even if something seemed obvious or didn’t make a huge difference. I often thought it was annoying (and still do, at times) when the mathematics were “simplified” in the sense that rigour was sacrificed for brevity and the final result. I wished we would rigourously justify each and every step, in order to make things mathematically correct. I also didn’t like the fact that sometimes we would “guess” results, in the sense that the best way to solve an equation was to try a solution and see what came out of it. This all seemed far removed from my studies in mathematics.
Recently though, I’ve not had a change of heart, but rather I’ve understood more of the rationale behind a lot of these decisions. As I’ve wrote about before, science is about making models of the world that both explain and predict the various features we see around us. However, in order to be mathematically tractable, simplifications and approximations are necessary. Furthermore, they aren’t fundamentally a bad thing, as long as one keeps in mind the simplifications throughout. This was the key I was missing. It’s not that we’re deliberately ignoring thorny issues, it’s that we are making a first model, which can always be refined and improved. It’s unrealistic to expect to have hyper-realistic models when first learning a subject, so these toy models with their approximations will have to do. Even if I don’t like the fact that we approximate irregular shapes as spheres, it’s done so that the problem is tractable and it doesn’t change the end result drastically.
My shift in mindset has come after really digging into some of the work of Tadashi Tokieda, who has some interesting resources from an old course available here. He is an applied mathematician who is also a great communicator. If you look at the website I linked to, you will see that he is very good at explaining things, and I particularly like how he characterizes the kind of work an applied mathematician should do. He says that an applied mathematician should be trying to do a back-of-the-envelope every day in order to increase one’s skills. The goal here isn’t to be analytically exact. Instead, it’s about probing the relationships between the items of interest. It’s about using mathematics to get to a result, without being overly worried about the formalism. That can wait for later. This has inspired me to start doing the same.
I’ve begun working on asking myself questions that delve into this sort of thing, where it’s unclear how to exactly begin, but by making approximations, a reasonable estimate can be found. It’s not easy, but it has gotten me to be more open with estimation. As the author of a book I’m reading on the subject writes, “It’s okay to say that 2 3=10.” The point isn’t to be precise. It’s to make a calculation tractable.
The more that I think about it, the more that I realize that we quickly discourage students from doing this at school. We say, “Don’t guess. Find the exact answer.” The truth is that estimation is important, and should be more frequently used. We should be able to take any kind of statement with a number attached and make sense of it. This ability is crippled when everything has to be exact. As such, I think we should be encouraging more estimation and less accuracy in order to get a foothold into a problem. Only then should we move onto refining and making a model more accurate. After all, that’s what we often do in science. We start with something we can handle, and make it more and more sophisticated. | 1,242 | 5,966 | {"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} | 2.546875 | 3 | CC-MAIN-2019-22 | latest | en | 0.955539 |
http://mathoverflow.net/questions/66299/ramanujans-incorrect-formula/66317 | 1,467,133,656,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396949.33/warc/CC-MAIN-20160624154956-00125-ip-10-164-35-72.ec2.internal.warc.gz | 203,099,834 | 15,519 | # Ramanujan's Incorrect formula
I actually looked at one of my Questions (posted at MATH.SE) again and found a formula which actually Ramanujan had discovered.
Ramanujan: If $\alpha$ and $\beta$ are positive numbers such that $\alpha \cdot \beta = \pi^{2}$ then, $$\alpha \cdot \sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\alpha} -1} + \beta \cdot\sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\beta}-1} = \frac{\alpha+\beta}{24} -\frac{1}{4}$$
I actually heard that this result is not true. I would like to know where the mistake is and whether something can be rectified in this proof so that, my above problem can be summed by using this result.
• I would also like to know the Intuitive idea behind discovering such mysterious formulas.
-
Is there a typo somewhere? The left side has three instances of $\alpha$ and only one of $\beta$, whereas one expects something more symmetric, by consideration of what happens by interchanging $\alpha$ and $\beta$. (Or, maybe Ramanujan had a typo? :-D) – Todd Trimble May 28 '11 at 18:18
Unfortunately, your bulleted question has no answer... :) – David Hansen May 29 '11 at 0:31
@Todd: The second sum actually has a $\beta$ which i had actually typed as $\alpha$ – S.C. May 29 '11 at 3:16
@David: Yeah, It's ok. I actually have a solution at the given link :) Was just looking for another method. – S.C. May 29 '11 at 3:48
I believe this formula is true, provided the $\alpha$ in the second sum is changed to a $\beta$, as suggested by Todd Trimble's comment. Let $$P(x) = \prod_{n=1}^\infty \frac{1}{1-x^n}$$ be the generating function for the number of partitions of a non-negative integer $n$. Dedekind proved that $P$ satisfies the transformation formula $$\log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr) + \frac{1}{2} \log t$$ for $t > 0$. Differentiating this formula with respect to $t$ gives $$-\sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n t}-1} - \frac{1}{t^2} \sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n/t} -1} = \frac{\pi}{12} \Bigl( -\frac{1}{t^2} - 1\Bigr) + \frac{1}{2t}$$ Now multiply through by $-t/2$ and substitute $\alpha = \pi t$, $\beta = \pi /t$ to get $$\sum_{n=1}^\infty \frac{\alpha n}{e^{2n\alpha}-1} + \sum_{n=1}^\infty \frac{\beta n}{e^{2n\beta}-1} = \frac{1}{24}(\beta+\alpha) - \frac{1}{4}$$ which is Ramanujan's formula.
The transformation formula for $P$ is related to the theory of modular forms, of which the Eisenstein series mentioned in Derek Jennings' answer to your question on math.stackexchange are important examples. Briefly, if we define $$\eta(\tau) = \frac{e^{2\pi i \tau/24}}{P(e^{2\pi i \tau})} = e^{2\pi i \tau/24} \prod_{n=1}^\infty (1-e^{2\pi i n \tau}),$$ then $\eta(\tau)^{24}$ is a modular form of weight $12$. As such, $\eta$ satisfies the identity $$\eta(-1/\tau) = \sqrt{-i \tau}\; \eta(\tau).$$ The transformation formula for $P$ follows by setting $\tau = it$ and taking logs. | 962 | 2,913 | {"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} | 3.703125 | 4 | CC-MAIN-2016-26 | latest | en | 0.867719 |
https://gyanipandit.com/programming/java-program-to-find-largest-number-in-an-array/ | 1,718,862,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00017.warc.gz | 249,392,382 | 27,316 | # Java Program to Find Largest Number in an Array
In this article, we are going to understand and perform a simple task in Java, in which, we are given an array, and we need to find the largest number in the array. We are required to write a Java program for the same. So, we will first understand the logic behind the program, and then we would try to implement the program.
## Java Program to Find Largest Number in an Array
In order to completely understand the program, one needs to be familiar with some different concepts –
• Array in Java
• Loops in Java
Have a look at the below input-output examples, which would give us a clear idea about our objective.
Example Input Output –
Input array: {34, 23, 5, 4, 34, 45, 78, 55}
Output: The largest number in the array is: 78
Input array: {5,4, 23, 1, 77, 65, 99}
Output: The largest number in the array is: 99
So, from the input-output, the objective becomes more clear. There can be different methods through which, you can achieve your goal, and here, we are going to understand some simple methods of finding the largest number in the array in Java.
#### Way 1 – Sorting the array
We have an array, and we need to find the largest number in the array. So, in this simple method, all we have to do is to sort the array in ascending order, and the largest number in the array is going to be the last element in the sorted array. To sort the array, we can use the sort method from the Arrays class in Java.
Have a look at a simple program, in which, we try to sort the array in ascending order, and then get the last element of the sorted array, as the largest number in the array.
import java.util.Arrays;
public class Main
{
public static void main(String[] args) {
int[] arr = new int[] {5,4, 23, 1, 77, 65, 99};
// sorting the array.
Arrays.sort(arr);
// the last element in the array is the largest number.
System.out.println(“The largest number in the array is:” + arr[arr.length – 1]);
}
}
As you can see in the above program, we are given an array, and we just need to get the largest number in the array. For that, we are just sorting the array in ascending order, and in the sorted array, the largest number is the last element in the array. You can try executing the above program on your own and try changing the values in the array to get some different outputs. Let’s try to have a look at the output of the above program now –
Output
The largest number in the array is:99
As you can see, we were able to get the largest number in the array, by easily sorting the array in ascending order. We have used the sort method from the Arrays class here, but you can also sort it manually.
#### Way 2 – Using an iterative approach
In this approach, we are going to iterate through the array, finding the largest number in the array. So, here is the set of steps that we are going to follow, to achieve our goal using this approach –
Assume that the first element in the array is the largest number in the given array
Iterate through the array, through each element, and if the current element in the array is greater than the assumed largest number, then we need to update our largest number since we found a number larger than the largest.
After we are done with the iteration, we just need to print the largest element as our output.
So, now let’s have a look at the Java program, which tries to demonstrate the same thing.
import java.util.Arrays;
public class Main
{
public static void main(String[] args) {
int[] arr = new int[] {34, 23, 5, 4, 34, 45, 78, 55};
// assuming that the first element is the largest.
int largest = arr[0];
for (int i=0;i<arr.length ;i++)
{
if(arr[i] > largest)
largest = arr[i];
}
System.out.println(“The largest element in the array is:” + largest);
}
}
As you can see in the above program, we are having an array, and then we are assuming that the first element in the array is the largest number in the array(also we assume that the array is non-empty). After that, we are looping through the array, and check if the current element in the array is larger than the assumed largest element, and if the condition is true, then we are just updating our largest number.
After we are done looping, we have the largest number in the array. So, we just need to show the largest number as our output for the program.
Let’s have a look at the output for the above program –
Output –
The largest element in the array is:78
As you can see, we have got the largest number in the array. You can try executing the above program and try playing with the numbers in the array, to get some different outputs.
#### Way 3 – Using the enhanced for a loop –
This approach is quite similar to the previous one, since here as well, we are going to loop. But here, we are going to make use of the enhanced for loop(also referred to as for-each loop). As mentioned earlier, this approach is much similar to the previous one, so here is the set of steps that we are going to follow, to achieve our goal –
import java.util.Arrays;
public class Main
{
public static void main(String[] args) {
int[] arr = new int[] {34, 23, 5, 4, 34, 45, 78, 55};
// assuming that the first element is the largest.
int largest = arr[0];
for(int number : arr)
{
if (number>largest)
largest = number;
}
System.out.println(“The largest element in the array is:” + largest);
}
}
As you can see in the above program, we have an array, and then we are assuming that the first element in the array is the largest number in the array(also we are assuming that the array is non-empty). After that, we are looping, and check if the current number in the array is greater than the assumed largest and if yes, we are updating our largest number.
After we are out of the loop, we have our largest number stored in the variable largest, and we just need to show the largest number as our output.
Let’s have a look at the output of the above program –
Output –
The largest element in the array is:78
As you can see, using the enhanced for loop, we were able to get the largest number in the array.
#### Conclusion
In this article, we have seen some different methods to get the largest number in the array, and we performed Java programs for the same. There can be some other ways to achieve the same goal, but here, we have seen some simple methods, through which, we can get the largest number in the array, in Java.
### Q: What is a simple method to find the largest number in the array?
Ans: To get the largest number in the array, you can simply sort the array in ascending order, and you would find the largest number as the last element in the array.
### Q: What is a simple method to find the smallest number in the array?
Ans: To get the smallest element in the array, we just need to sort the array in ascending order, and we can find that the smallest number in the array is the first element in the array. | 1,633 | 6,896 | {"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-2024-26 | latest | en | 0.880038 |
http://physics.stackexchange.com/questions/tagged/thermodynamics+energy-conservation | 1,411,345,119,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657136494.66/warc/CC-MAIN-20140914011216-00314-ip-10-234-18-248.ec2.internal.warc.gz | 218,237,161 | 19,028 | # Tagged Questions
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I don't know how to resolve a problem, but I don't want the answer since I'm almost going to have it resolved. What the problem says is we have 85 liters of water at 7ºC in an iron pot of 29kg. We ... | 1,428 | 6,399 | {"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": 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} | 2.671875 | 3 | CC-MAIN-2014-41 | latest | en | 0.926922 |
http://ebs.cu.edu.tr/?upage=fak&page=drs&f=3&b=140&ch=1&dpage=all&yil=2017&lang=en&dpage=tnm&InKod=6352&dpage=all | 1,607,105,376,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00450.warc.gz | 33,924,791 | 18,360 | •
• Information on Degree Programmes
COURSE INFORMATON
Course Title Code Semester L+P Hour Credits ECTS
Statistical Reliability Analysis * ISB 492 8 3 3 5
Prerequisites and co-requisites Yok Recommended Optional Programme Components None
Language of Instruction Turkish
Course Level First Cycle Programmes (Bachelor's Degree)
Course Type
Course Coordinator Prof. Dr. Güzin YÜKSEL
Instructors
Prof. Dr. GÜZİN YÜKSEL 1. Öğretim Grup:A Prof. Dr. GÜZİN YÜKSEL 2. Öğretim Grup:A
Assistants
Goals
The aim of this course is to introduce reliability concept to students.Reliability estimation for different systems, reliability growth models will be given.
Content
Reliability estimation for different systems,Hazard rate estimation,Estimation based on life distributions,Exponential distribution,Weibull distribution,Gamma distribution,Testing based on life distributions,Probability plotting techniques.
Learning Outcomes
-
Course's Contribution To Program
NoProgram Learning OutcomesContribution
12345
1
Explain the essence fundamentals and concepts in the field of Probability, Statistics and Mathematics
X
2
Emphasize the importance of Statistics in life
X
3
Define basic principles and concepts in the field of Law and Economics
4
Produce numeric and statistical solutions in order to overcome the problems
X
5
Use proper methods and techniques to gather and/or to arrange the data
X
6
Utilize computer systems and softwares
X
7
Construct the model, solve and interpret the results by using mathematical and statistical tehniques for the problems that include random events
X
8
Apply the statistical analyze methods
X
9
Make statistical inference(estimation, hypothesis tests etc.)
X
10
Generate solutions for the problems in other disciplines by using statistical techniques
X
11
Discover the visual, database and web programming techniques and posses the ability of writing programme
X
12
Construct a model and analyze it by using statistical packages
X
13
Distinguish the difference between the statistical methods
X
14
Be aware of the interaction between the disciplines related to statistics
X
15
Make oral and visual presentation for the results of statistical methods
X
16
Have capability on effective and productive work in a group and individually
X
17
Professional development in accordance with their interests and abilities, as well as the scientific, cultural, artistic and social fields, constantly improve themselves by identifying training needs
X
18
Develop scientific and ethical values in the fields of statistics-and scientific data collection
X
Course Content
WeekTopicsStudy Materials _ocw_rs_drs_yontem
1 Reliability estimation for different systems Reading related books
2 Hazard rate estimation Reading related books
3 Estimation based on life distributions Reading related books
4 Exponential distribution Reading related books
5 Weibull distribution Reading related books
6 Gamma distribution Reading related books | 625 | 2,960 | {"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-2020-50 | latest | en | 0.784708 |
https://skeptic78240.wordpress.com/2016/03/28/quiz-question-42/ | 1,511,373,943,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806620.63/warc/CC-MAIN-20171122175744-20171122195744-00472.warc.gz | 729,528,590 | 17,246 | # Quiz Question
### One of a continuing series
There once was in Carrollton, Texas, at the intersection of Belt Line Road and Webb Chapel Road a concern called Otis Engineering. I was by there for a visit. They made “down-line” equipment for oil wells. This was stuff that went down into oil wells, sometimes miles deep, to make measurements and such.
They had a pressure chamber for testing their stuff, and this chamber was well-armored, to withstand those deep pressures. They filled it with water and applied the pressure. They took great care that there was no air in the chamber. Why was that. What would be the problem if there were some air in the pressure chamber?
This one is some basic science and I expect some quick answers. Provide your response in the comments section below.
## Update and answer
Greg has answered this one, but I do not see evidence he has posted a comment. The reason is this:
A body of water under pressure represents an amount of stored energy to the extent it is compressed. Air is multiply-compressible over water, and at great pressures can story a dangerous amount of energy. The potential energy of a compressed fluid is proportional to the pressure and the square of the compressed volume. Since water compresses very little under great pressure, compressed water does not represent so much potential energy. Any gas in the chamber, however, would be compressed tremendously and would represent a large amount of potential energy.
Greg is correct in that it would take more energy to raise the pressure of the chamber if had some air in it. However, the real concern is the danger of releasing the potential energy accidentally. | 335 | 1,678 | {"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-2017-47 | longest | en | 0.984431 |
https://www.lacontracorrent.cat/how-to-determine-limiting-reactant-from-mole-ratio/ | 1,653,291,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00252.warc.gz | 985,201,073 | 16,205 | # How To Determine Limiting Reactant From Mole Ratio 2021
How To Determine Limiting Reactant From Mole Ratio. 1 calculate moles of each reactant: 10.0 g n 2 x 1 mole n 2 /28.0 g n 2 = 0.357 moles n 2 have 10.0 g h 2 x 1 mole h 2 /2.02 g h 2 = 4.95 moles h 2 have step 3:
3) the volume of reactants limits the precision of the reaction by keeping all of the data within a comparable range. 3lmol 2l(coefficient) = 1.5 −.
### High School Chemistry Formula Sheet Chemistry Reference
4) naclo is the limiting reagent in the reaction up until a ratio of 41 ml naclo to 9 ml naoh. A third method identifies the limiting reactant by calculating the moles of each reactant present, then comparing those values to the required ratio of reactants as indicated by the balanced equation coefficients (for an example, see zumdahl and zumdahl, 2014, pp.
### How To Determine Limiting Reactant From Mole Ratio
Begin with a balanced chemical equation and starting amounts for each reactant.Calculate the number of moles used for each reactant.Convert mass of each starting reactants to moles.Determine the amount (in grams) of excess reactant that remains after the reaction is complete.
Determine the whole number mole ratio of the two reactants.Draw two best fit straight lines, and determine where they intersect.Find the stoichiometric mole ration of reactants from the line of intersection on the graph.Finding the limiting reactant is an important step in finding the percentage yield of the reaction.
Formula of limiting reactant = amount of excess reactant remaining =From the equation we know that the equation mole ratio is 1:2.However, we need the actual mole ratio to find out the limiting reactant.Identify limiting reactants (mole ratio method).
Identify the limiting reactant in the reaction of nitrogen and oxygen to form no, if 8.35 g of n 2 and 5.02 g of o 2 are combined.If at any point do not fall close to the lines, repeat these measurements.If equal moles of reactants are present, then both the reactants will be completely consumed, and none of the reactants will act as a limiting reactant or as an excess reactant.In order to determine the limiting reactant, we need to determine which of the reactants will give less product.
Is this necessary, or even important, for the success of the experiment?Learning targets determine mole ratios from balanced chemical equationsLimiting reactant also determine how long the reaction will last for.Mg + 2hcl = mgcl2 + h2 the balanced equatio
n is needed to determine the mole ratio between the two reactants.
Mole ratios are used as conversion factors between products and reactants in many chemistry problems.Moles = mass/mr moles cl2 = 125 / 71 = 1.76 moles moles c = 125 / 12 = 10.4 moles 2 check the required ratio compared to the actual ratio.Notice also how the limiting reactant isn’t necessarily the reactant.Remember, this is determined based on the mole ratio of h 2 and h 2 o, which is 2:2 (the coefficients) in front of each molecule.
Require 2moles cl2 / 1mole c available 1.76moles cl2 /10.4 moles c = 0.17 cl2 / 1 mole c cl2 is the limiting reagent 6.The amount of the limiting reagent controls how much the two substances can react.The key is to keep the same reactant on top as the step above.The limiting reactant is that whose value is smallest after dividing the mole number by their coefficient:
The limiting reactant is the species that is consumed first based on available quantities and stoichiometric relations.The masses of two reactants cannot be compared directly.The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is.The method of continuous variations allows scientists to determine the mole ratio of two reactants in a chemical reaction.
The molarities of the reactant solutions were equal in this experiment.The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.The percentage yield of a reaction is the ratio of its actual yield to its theoretical yield times 100.Theoretical yield is the yield predicted by stoichiometric calculations, assuming the limiting reactant reacts completely.
This gives a ratio in which no number is less than 1.This gives you a molar ratio of #al# to #i_2# of #0.04448/0.009456# usually, you divide each number in the fraction by the smaller number of moles.To calculate the molar ratios, you put the moles of one reactant over the moles of the other reactant.To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation.
Typically the mole ratios of reactants and products can be found from a balanced chemical equation, however when the formulas of the products are unknown, experiments can be conducted to discover this ratio.Use the balanced chemical equation to determine the mole ratio (stoichiometric ratio) of the reactants in the chemical reaction compare the available moles of each reactant to the moles required for complete reaction using the mole ratio (i) the limiting reagent is the reactant that will be completely used up during the chemical reaction.We see that the number for hydrogen is the lower value, so hydrogen is the limiting reagent.What we need to do is determine an amount of one product (either moles or mass) assuming all of each reactant reacts.
Which solution was the limiting reactant in each trial?Whichever reactant gives the lesser amount of product is the limiting reactant.Ø ø ø chapter 9:
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How To Get Wrinkles Out Of Polyester Spandex. A dry cleaner told me this trick as i wanted to remove the creases from drapery fabric i had… | 1,586 | 6,735 | {"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-2022-21 | latest | en | 0.847757 |
https://priorart.ip.com/IPCOM/000095053 | 1,529,776,307,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865145.53/warc/CC-MAIN-20180623171526-20180623191526-00589.warc.gz | 676,221,751 | 7,669 | Browse Prior Art Database
# Resistance Controlled Oscillator
IP.com Disclosure Number: IPCOM000095053D
Original Publication Date: 1965-Aug-01
Included in the Prior Art Database: 2005-Mar-06
Document File: 2 page(s) / 39K
IBM
## Related People
Fant, WJ: AUTHOR [+2]
## Abstract
This oscillator provides an output signal having a frequency which varies as a function of the resistance of one arm in a bridge circuit. The oscillator consists of five sections.
This text was extracted from a PDF file.
At least one non-text object (such as an image or picture) has been suppressed.
This is the abbreviated version, containing approximately 71% of the total text.
Page 1 of 2
Resistance Controlled Oscillator
This oscillator provides an output signal having a frequency which varies as a function of the resistance of one arm in a bridge circuit. The oscillator consists of five sections.
In the bridge and mixer, either R1 or R2 can be the sensor. The sensor resistance changes with the measured variable. R1 and R2 are equal for one value of the measured variable. R3 and C1 form a phase-shifting network called the quadrature network. This network provides a voltage having a phase shift from 1 to 180 degrees of lag with respect to Vo/2. as frequency goes from zero to m. R6 and R7 are equal and form a reference source for Vo/2. R5 attenuates the output of the quadrature network. Q1 prevents R5 from loading the quadrature.
If the frequency of oscillation when R1 = R2 is assumed to be omega = 1/RC, the phase contribution of the quadrature network is -90 degrees. There is no contribution from the sensor network.
If R1 differs from R2, some in-phase voltage, positive or negative, appears and, when summed with the quadrature voltage, the resultant is no longer 90 degrees.
The differential amplifier subtracts the quadrature voltage from the in-phase sensor voltage, producing the resultant described above. It introduces no additional phase shift.
The AGC section can be a variable resistance device, a clipper, or a combination of both. For best results, it is referenced to the output voltage of the system. It prevents saturation of amplifier stages... | 484 | 2,176 | {"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-2018-26 | longest | en | 0.89719 |
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# Powerscore CR Notes-Hope that someone finds it kudosofiable! v2.0 OUT
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Powerscore CR Notes-Hope that someone finds it kudosofiable! v2.0 OUT [#permalink]
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19 Jul 2014, 08:54
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It is a consolidated revision notes for those who have studied using powerscore CR.
Handy for a quick look through to refresh the concepts from the book.
If you find it useful please kudosify !
Topics Covered:
1.Must be true/Most Supported
2. Main point
3. Assumption
4. Strenghten/ Support
6. Weaken
7. Method of reasoning
8. Flaw in the reasoning
9. Parallel reasoning
10.Evaluate the argument
Attachments
Power-Score GMAT Notes-CR V2.0.xlsx [3.44 MiB]
File comment: It a consolidated revision notes for those who have studied using powerscore CR. If you find it useful please kudosify !
Power-Score GMAT Notes-CR.xlsx [36.6 KiB]
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Last edited by royQV on 01 Dec 2014, 00:10, edited 2 times in total.
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Re: Powerscore CR Notes-Hope that someone finds it kudosofiable! v2.0 OUT [#permalink]
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20 Jul 2014, 10:18
royQV wrote:
It a consolidated revision notes for those who have studied using powerscore CR. If you find it useful please kudosify !
Great Job!!
I have completed reading major part of this book and liked the way you summed up the concepts...
Thanks again for sharing the excel.
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### John's Train Is on Time
##### Age 11 to 14 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### Overlap
##### Age 11 to 14 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Robotic Rotations
##### Age 11 to 16 Challenge Level:
How did the the rotation robot make these patterns?
### Turning Triangles
##### Age 11 to 14 Challenge Level:
A triangle ABC resting on a horizontal line is "rolled" along the line. Describe the paths of each of the vertices and the relationships between them and the original triangle.
### Rolling Triangle
##### Age 11 to 14 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Weighty Problem
##### Age 11 to 14 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### In a Spin
##### Age 14 to 16 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010.
### Turning Tangles
##### Age 11 to 14 Challenge Level:
Look carefully at the video of a tangle and explain what's happening.
### Rollin' Rollin' Rollin'
##### Age 11 to 14 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### Simplifying Transformations
##### Age 11 to 14 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Get Cross
##### Age 14 to 16 Challenge Level:
A white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Coke Machine
##### Age 14 to 16 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Napoleon's Theorem
##### Age 14 to 18 Challenge Level:
Triangle ABC has equilateral triangles drawn on its edges. Points P, Q and R are the centres of the equilateral triangles. What can you prove about the triangle PQR?
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### Illusion
##### Age 11 to 16 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Notes on a Triangle
##### Age 11 to 14 Challenge Level:
Can you describe what happens in this film?
### Arrow Arithmetic 1
##### Age 14 to 16 Challenge Level:
The first part of an investigation into how to represent numbers using geometric transformations that ultimately leads us to discover numbers not on the number line.
### Symmetric Trace
##### Age 14 to 16 Challenge Level:
Points off a rolling wheel make traces. What makes those traces have symmetry?
### Shaping up with Tessellations
##### Age 7 to 14
This article describes the scope for practical exploration of tessellations both in and out of the classroom. It seems a golden opportunity to link art with maths, allowing the creative side of your. . . .
### A Roll of Patterned Paper
##### Age 14 to 16 Challenge Level:
A design is repeated endlessly along a line - rather like a stream of paper coming off a roll. Make a strip that matches itself after rotation, or after reflection
### Rotations Are Not Single Round Here
##### Age 14 to 16 Challenge Level:
I noticed this about streamers that have rotation symmetry : if there was one centre of rotation there always seems to be a second centre that also worked. Can you find a design that has only. . . .
### Attractive Rotations
##### Age 11 to 14 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach. | 1,524 | 6,772 | {"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} | 3.859375 | 4 | CC-MAIN-2019-35 | latest | en | 0.920793 |
https://math.stackexchange.com/questions/2646016/how-do-i-prove-that-these-two-set-of-sets-have-this-specific-relationship | 1,563,499,596,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00536.warc.gz | 469,753,411 | 36,825 | # How do I prove that these two set of sets have this specific relationship?
Suppose for some $F\subseteq \mathcal{P}(A)$, we have $U = \{X \subseteq A: \forall S \in F . S \subseteq X\}.$ We want to show that $\bigcup F = \bigcap U$, and then subsequently generalize and show that for some $L \subseteq \mathcal{P}(A)$, $\bigcap F = \bigcup L$.
So there's essentially 2 things to prove, and one is just the opposite of the other. I'm not sure why $U$ is defined so abstractly, but it seems to me that $U$ is just a subset of $F$, where each element is a subset of A as well. Can we simplify $U = \{S\subseteq A: S \in F\}$, or just $U \subseteq F$?
I have no idea how I could begin proving the union of sets equals the intersection of sets and vice versa. If $U\subseteq F$, how could we have $\bigcup F = \bigcap U$? My thoughts:
1. Maybe I need to show that $\bigcup F \subseteq \bigcap U$ and $\bigcup U \subseteq \bigcap F$ to show the first one. It might be able to show that for any set $X \in F$, we have $X \in F \lor X \in U$. But since $X\in U$ implies $X \in \bigcap U$, we have our relationship shown immediately.
Can I prove along this line of argument?
But the second part doesn't have $L$ related to $F$ by anything so it is even harder to prove it.
To prove this, you have to show two things: $\bigcap U \subseteq \bigcup F$ and $\bigcap U \supseteq \bigcup F$. This proves equality of sets. The idea behind proving something like "$A\subseteq B$" is to assume $a\in A$ and then show $a\in B$. This question can be approached like this, but here are a few hints.
Hints:
• To show $\bigcap U \subseteq \bigcup F$: Does $\bigcup F \in U$?
• To show $\bigcup F \subseteq \bigcap U$: If $x\in\bigcup F$, then for each $X\in U$, is $x\in X$?
Edit: Note that $F$ is just a set of sets. $U$ is the set of sets $X$ for which any set $S\in F$ is a subset of $X$. That is, every set in $U$ contains every set in $F$.
Concerning the set $L$, you are wanting to just make an analogous set to $U$, except this time so that we have $\bigcap F = \bigcup L$. Two hints for this:
• Try writing $L$ in a similar form to $U$. That is, in the form of: $$L = \{X\subseteq A \mid \forall S\in F.\underline{\text{blank}}\},$$ where you fill in the $\underline{\text{blank}}$.
• If you still have difficulties, think about what $U$ was. $U$ was the set of sets containing $\bigcup F$. We took the intersection of $U$ to get only $F$. In this way, $\bigcap U$ is the "smallest set containing $\bigcup F$". In the case of $L$, because you are taking its union, you will end up with the "largest set containing $\bigcap F$".
• Thanks for your answer. Can you clarify if the $x$ and $X$ are the same thing? Also, do you have a hint on how the second part of the proof could be solved since we don't know $F \subseteq L$ or vice versa? – oldselflearner1959 Feb 11 '18 at 16:11
• Also, if we can show $X \in U$, how do I then show that $X \in \bigcap U$? aren't they different? – oldselflearner1959 Feb 11 '18 at 16:16
• Here, $x$ is an element of $\bigcup F\subseteq A$, while $X\in U$ is a subset of $A$. So no they are not the same thing. I have edited the mistake I made typing up the question originally to address your last comment, and I will add another edit to address $L$ in a bit. – anakhro Feb 11 '18 at 16:19
• For the first part, I think it's possible to get from $x \in \bigcup F \implies \forall x \in \forall X \in U$ but I cannot get $\bigcap U$ still. – oldselflearner1959 Feb 11 '18 at 16:58
• Sorry, which inclusion are you having difficulties with? $\bigcup F\subseteq\bigcap U$ or $\bigcap U\subseteq\bigcup F$? – anakhro Feb 11 '18 at 17:01 | 1,119 | 3,663 | {"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} | 4.09375 | 4 | CC-MAIN-2019-30 | latest | en | 0.911012 |
https://shnaton.huji.ac.il/index.php/NewSyl/80629/2/2022/ | 1,652,984,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00323.warc.gz | 586,428,885 | 3,232 | Syllabus High dimensional statistics - 80629
עברית
Students needing academic accommodations based on a disability should contact the Center for Diagnosis and Support of Students with Learning Disabilities, or the Office for Students with Disabilities, as early as possible, to discuss and coordinate accommodations, based on relevant documentation. For further information, please visit the site of the Dean of Students Office. Print close PDF version Last update 02-09-2021 HU Credits: 3 Degree/Cycle: 2nd degree (Master) Responsible Department: Mathematics Semester: 1st Semester Teaching Languages: Hebrew Campus: E. Safra Course/Module Coordinator: Zemer Kosloff Coordinator Email: zemer.kosloff@mail.huji.ac.il Coordinator Office Hours: Teaching Staff: Prof Zemer Kosloff Course/Module description: The course will serve as an introduction to the methods of analyzing high dimensional statistical models. The first part will deal with the basic of tail and concentration bounds, sub-Gaussian random variables, entropic methods and uniform laws of large numbers. After that we aim to apply these methods to some problems such as covariance estimators, sparse-linear regression and the Lasso algorithm. Course/Module aims: Learning outcomes - On successful completion of this module, students should be able to: Be familiar with the mathematical foundations and methods underlying modern research in the rapidly evolving field of high-dimensional statistics. Attendance requirements(%): 0 Teaching arrangement and method of instruction: Lectures Course/Module Content: 0) Introduction and some nice examples. 1) Basic tail bounds (Chernoff, Hoeffelding inequalities and martingale difference methods). 2) SubGaussian random variables, equivalent definitions and the sub-Gaussian norm. 3) Uniform laws of large numbers, Rademacher complexity and Vapnik-Chernovakis dimension. 4) Metric entropy and its uses: Covering, Packing, chainning and Dudley's integral. 5) Random matrices and covariance estimation. 6) Sparse linear regression. Required Reading: none Additional Reading Material: a) M.J. Wainwright. High-Dimensional Statistics, A Non-Asymptotic Viewpoint. Cambridge university press. b) R. Vershynin, Introduction to the non-asymptotic analysis of random matrices. Cambridge University Press, c) A New Look at Independence – Special Invited Paper, by M. Talagrand, the Annals of Applied Probability, 24(1),1–34, 1996. Course/Module evaluation: End of year written/oral examination 0 % Presentation 50 % Participation in Tutorials 0 % Project work 0 % Assignments 50 % Reports 0 % Research project 0 % Quizzes 0 % Other 0 % Additional information: The interested students must have completed the course "introduction to probability theory and statistics. Knowledge in measure theory and continuous probability is an advantage. Print | 601 | 2,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-21 | latest | en | 0.833593 |
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#### Resources tagged with Factors and multiples similar to Train Routes:
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### There are 112 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Nineteen Hexagons
##### Age 5 to 7 Challenge Level:
In this maze of hexagons, you start in the centre at 0. The next hexagon must be a multiple of 2 and the next a multiple of 5. What are the possible paths you could take?
### One of Thirty-six
##### Age 5 to 7 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Same Length Trains
##### Age 5 to 7 Challenge Level:
How many trains can you make which are the same length as Matt's, using rods that are identical?
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Mystery Matrix
##### Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Lots of Lollies
##### Age 5 to 7 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Are You Well Balanced?
##### Age 5 to 7 Challenge Level:
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Growing Garlic
##### Age 5 to 7 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### A Square Deal
##### Age 7 to 11 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Multiplication Squares
##### Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Grouping Goodies
##### Age 5 to 7 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Constant Counting
##### Age 5 to 7 Challenge Level:
You can make a calculator count for you by any number you choose. You can count by ones to reach 24. You can count by twos to reach 24. What else can you count by to reach 24?
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Biscuit Decorations
##### Age 5 to 7 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### A Mixed-up Clock
##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Skip Counting
##### Age 5 to 7 Challenge Level:
Find the squares that Froggie skips onto to get to the pumpkin patch. She starts on 3 and finishes on 30, but she lands only on a square that has a number 3 more than the square she skips from.
### Tables Teaser
##### Age 5 to 7 Challenge Level:
How will you work out which numbers have been used to create this multiplication square?
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Fractions in a Box
##### Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
### Scoring with Dice
##### Age 7 to 11 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### Mrs Trimmer's String
##### Age 5 to 7 Challenge Level:
Can you help the children in Mrs Trimmer's class make different shapes out of a loop of string?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Little Squares
##### Age 5 to 7 Challenge Level:
Look at the squares in this problem. What does the next square look like? I draw a square with 81 little squares inside it. How long and how wide is my square?
### What's Left?
##### Age 5 to 7 Challenge Level:
Use this grid to shade the numbers in the way described. Which numbers do you have left? Do you know what they are called?
### The Set of Numbers
##### Age 5 to 7 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Doubling Fives
##### Age 5 to 7 Challenge Level:
This activity focuses on doubling multiples of five.
### Sorting Numbers
##### Age 5 to 7 Challenge Level:
Can you sort numbers into sets? Can you give each set a name?
### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
### Two Primes Make One Square
##### Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Being Collaborative - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level to work on with others.
### Domino Pick
##### Age 5 to 7 Challenge Level:
Are these domino games fair? Can you explain why or why not?
### Making Pathways
##### Age 7 to 11 Challenge Level:
Can you find different ways of creating paths using these paving slabs?
### Share Bears
##### Age 5 to 7 Challenge Level:
Yasmin and Zach have some bears to share. Which numbers of bears can they share so that there are none left over?
### Money Measure
##### Age 7 to 11 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Crossings
##### Age 7 to 11 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Multiplication Series: Number Arrays
##### Age 5 to 11
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Number Tracks
##### Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see? | 2,154 | 9,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-39 | latest | en | 0.907714 |
https://itprospt.com/qa/348835/question-1-15-points-do-farmers-in-poor-countries | 1,686,393,835,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00722.warc.gz | 367,948,820 | 11,386 | 1
# Question 1 (15 points) Do farmers in poor countries respond to economic incentives? A well-known argument...
## Question
###### Question 1 (15 points) Do farmers in poor countries respond to economic incentives? A well-known argument...
Question 1 (15 points) Do farmers in poor countries respond to economic incentives? A well-known argument is that farmers might be poor but they are efficient and do respond to incentives in a normal way. An investigator, interested in this question, examines the supply response of Bangladeshi farmers in the cultivation of sugar-cane to the price of sugar-cane as well as to the price of jute. Both sugarcane and jute are crops that compete for the same land (that is, are substitutes in production), which is a relatively scarce resource in the country. The researcher's model is: where AREA- area planted to sugar-cane (thousands of hectares) and PS price of sugar-cane (\$ per ton) and Pj-price of jute (S per ton). The prices are in Canadian dollars converted from the local currency (Taka) what would you expect β2 and β3 to be if farmers did not respond to price incentives? what would you expect the signs of these parameters to be if they responded to such incentives in a normal way? (2 points) A. Using annual time series data, the researcher obtained the following regression: B. AREAA 134.425.19PS 6.97PJ se (46.2) (3.59) (2.66) N 34, R-squared 0.62 Sample means: AREA 125.53, PS 4.98, PJ -19.00 1. Use t tests to determine whether Bangladeshi farmers respond to incentives? Set α to 0.05. (3 points) 2. Suppose PS and PJ were to rise simultaneously by \$1 per ton and \$3 per ton respectively a. write down the expression for the change in area planted ΔΕ(AREA), using the population regression (the first b. Obtain a point estimate of λ using the estimated model. Construct a 95% confidence interval for λ, given that c. Test the hypothesis that these simultaneous price changes have no impact on the area planted to sugarcane equation). This is a linear combination of parameters. Call this linear combination A. (2 points) cov(b2,b3)-4.1201. (4 points) Use the 5 percent significance level. (4 points)
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You run a trading strategy on a constant and the return on the market. The intercept and slope estimates are 0.541 and-0.243 respectively. The t-ratio associated with the slope is 0.99, what is the standard error of the slope? 2. a) 0.245 b) c) 0.020 d) There is not enough information to answer this...
##### Activity 2.a - Analyze the Transactions Review the transactions and determine the accounts, the account types,...
Activity 2.a - Analyze the Transactions Review the transactions and determine the accounts, the account types, if they increase/decrease and if they are DR/CR. List accounts in the order they would be in the journal entry. Purchased office supplies on account Account #1 Account Type Increase/Decreas...
##### 7. (10 points) Given three pure tones with the following frequencies and intensity levels: 100 Hz...
7. (10 points) Given three pure tones with the following frequencies and intensity levels: 100 Hz at 60 d8, 500 Hz at 70 dB, and 1000 Hz at 80 dB. (a) Compute the total loudness in sones of these three pure tones. (b) what is the combined intensity level of the three, and (c) Find the intensity leve... | 1,505 | 6,164 | {"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-2023-23 | latest | en | 0.891998 |
https://www.codingbroz.com/concave-polygon-hackerrank-solution/ | 1,723,364,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00500.warc.gz | 558,709,524 | 66,408 | # Concave Polygon – HackerRank Solution
In this post, we will solve Concave Polygon HackerRank Solution. This problem (Concave Polygon) is a part of HackerRank Functional Programming series.
You are given the cartesian coordinates of a set of points in a 2D plane (in no particular order). Each of these points is a corner point of some Polygon, P, which is not self-intersecting in nature. Can you determine whether or not P is a concave polygon?
## Input Format
The first line contains an integer, N, denoting the number of points.
The N subsequent lines each contain 2 space-separated integers denoting the respective x and y coordinates of a point.
## Constraints
• 3 <= N <= 1000
• 0 <= x, y <= 1000
## Output Format
Print YES if P is a concave polygon; otherwise, print NO.
Sample Input
``````4
0 0
0 1
1 1
1 0``````
Sample Output
``NO``
Explanation
The given polygon is a square, and each of its 4 internal angles are 90. As none of these are over 180, the polygon is not concave and we print NO.
Scoring
The percentage score awarded for your submission will be:
`` 100 - 2*(percentage of tests which you solve incorrectly) ``
If this value is negative, the percentage score for your submission will be 0.
So if you get half or more of the tests incorrect, your score will be a zero.
## Solution – Concave Polygon – HackerRank Solution
Scala
```import java.util.Scanner
import scala.collection.mutable
object Orientation extends Enumeration {
type Orientation = Value
val ColLinear, Clockwise, Counterclockwise = Value
}
object Solution {
import Orientation.Orientation
def main(args: Array[String]): Unit = {
val sc = new Scanner(System.in)
val n = sc.nextInt
case class Point(x: Int, y: Int)
val initialPoints = (0 until n).map(_ => Point(sc.nextInt, sc.nextInt))
val bottommostIndex = initialPoints.indices.reduce((accIndex, pIndex) => {
val acc = initialPoints(accIndex)
val p = initialPoints(pIndex)
if (acc.y < p.y || acc.y == p.y && acc.x < p.x) accIndex else pIndex
})
val basePoint = initialPoints(bottommostIndex)
def polarAngle(point: Point) = math.atan2(point.y - basePoint.y, point.x - basePoint.x)
def distance2(p0: Point, p1: Point) = {
val dx = p1.x - p0.x
val dy = p1.y - p0.y
dx * dx + dy * dy
}
val points = initialPoints.indices.filter(_ != bottommostIndex).map(initialPoints(_))
.sortWith((p0, p1) => {
val polar0 = polarAngle(p0)
val polar1 = polarAngle(p1)
polar0 < polar1 || polar0 == polar1 && distance2(p0, basePoint) < distance2(p1, basePoint)
})
.toList
val filteredPoints = basePoint :: points
val orderedCount = 3
val stack = mutable.Stack[Point]()
stack.pushAll(filteredPoints.take(orderedCount))
def orientation(p0: Point, p1: Point, p2: Point): Orientation = {
val cross = (p1.y - p0.y) * (p2.x - p1.x) - (p1.x - p0.x) * (p2.y - p1.y)
if (cross == 0) Orientation.ColLinear else if (cross > 0) Orientation.Clockwise else Orientation.Counterclockwise
}
filteredPoints.drop(orderedCount)
.foreach(p => { | 845 | 3,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.669884 |
https://www.slideserve.com/evelyn/math-metaphors | 1,511,286,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00345.warc.gz | 871,823,560 | 12,673 | 1 / 10
# Math Metaphors - PowerPoint PPT Presentation
Math Metaphors. Atif Chaudhry Malvin Hiew Ruth Limberg Thi Nguyen. Abstract Math: Images and Metaphors http://www.abstractmath.org/MM/MMImagesMetaphors.htm. This website contains mathematical metaphors. Math metaphors argue that people use metaphors and images to better understand math.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Math Metaphors' - evelyn
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Presentation Transcript
• Atif Chaudhry
• Malvin Hiew
• Ruth Limberg
• Thi Nguyen
Abstract Math: Images and Metaphorshttp://www.abstractmath.org/MM/MMImagesMetaphors.htm
This website contains mathematical metaphors. Math metaphors argue that people use metaphors and images to better understand math.
• ¨ “Continuous functions don’t have gaps in the graph”. As a student, this helps you to visualize and understand continuity of functions.
• ¨ You may think of the real numbers as lying along a straight line (the real line) that extends infinitely far in both directions. This is both a visual and a metaphor (a real number “is” a place on the real line).
Understanding without Proof argue that people use metaphors and images to better understand math.http://www.rattlesnake.com/notions/math-metaphor.html
People use math metaphors in our day to day life from toddlers to adults.
• There are actually four ‘grounding’ metaphors (metaphors based on experiences many of us had as a child).:
• Adding and taking away objects from a collection (playing with pebbles);
• Construction of a larger whole from smaller objects (playing with blocks);
• Measuring the width or height of something (by stretching our hands to the ends of the object or standing up to see how high it is);
• Moving from one place to another (by crawling or walking).
• These experiences provide us with four metaphors that work with arithmetic. Four inference-preserving cross-domain mapping mechanisms that work consistently with each other and the world.
Understanding Integrated Mathematics Using Living Metaphors toddlers to adults.http://www.nctm.org/resources/contnet.aspx?id=1674
Grow rather than build toddlers to adults.
• This is an article written by F.Merlino, project director of the Greater Philadelphia Secondary Mathematics Project at La Salle University. In this article, Merlino is disputing the learning methods of math that he grew up with and embracing the new mathematical era; which involves metaphors and analogies rather than repetition and “laying a good foundation.” He has proved that mathematical metaphors have been outdated and should be done away with.
Grow rather than build con’t toddlers to adults.
Some quotes from Merlino’s article include:
• “Metaphors can also limit how we think, can prevent us from "thinking outside the box," can constrain our imaginations, and can veil our eyes from seeing things as they really are.”
• Integrated curricula can be better understood if we view the act of thinking as a living system, like a growing tree, rather than like a structure made of bricks and mortar. Using "living" metaphors has enormous implications for how we think about mathematics education. | 814 | 3,693 | {"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} | 3.484375 | 3 | CC-MAIN-2017-47 | latest | en | 0.841733 |
https://www.sawaal.com/numbers-questions-and-answers/-nbspif-n-is-even-which-of-the-following-cannot-be-odd_4403 | 1,716,409,631,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00059.warc.gz | 864,170,995 | 13,702 | 6
Q:
# If n is even, which of the following cannot be odd?
A) n+3 B) 2n C) n+4 D) 3n
Explanation:
In case B, odd times even will give even
Subject: Numbers
Exam Prep: GRE
Q:
Which number will fit in '?' place in the following series?
12, 23, 45, 89, ?
A) 175 B) 176 C) 177 D) 178
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
7 1832
Q:
Assuming that the numbers in each of the following figures follow a similar pattern, select the option that can replace the question mark (?) in figure C.
A) 61 B) 21 C) 16 D) 81
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
1 438
Q:
The equation of a line is given by 4y-6x+9=0. Find the slope of other line which is perpendicular to it.
A) -2/3 B) 2/3 C) 3/2 D) -3/2
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
2 550
Q:
What will be the 20th term in the given sequence?
-50, -47, -44, _
A) -10 B) 10 C) -7 D) 7
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
9 2524
Q:
In the given series of numbers how many 7's are there which are immediately followed by 9 and immediately not preceeded by 5 ?
A) 6 B) 2 C) 1 D) 4
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
3 743
Q:
The value of $1+22+23+24+...+29$ is
A) 255 B) 511 C) 1021 D) 2047
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams , CAT
9 1396
Q:
In an arithmetic progression if 13 is the 3rd term, -47 is the 13th term, then -17 is which term?
A) 9 B) 10 C) 7 D) 8
Explanation:
Filed Under: Numbers
Exam Prep: Bank Exams
5 1180
Q:
What is the value of (81 + 82 + 83 + ......... +130)?
A) 5275 B) 10550 C) 15825 D) 21100 | 567 | 1,625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.793816 |
https://www.adventuresinmachinelearning.com/efficient-ways-to-create-and-append-lists-in-python/ | 1,726,183,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00618.warc.gz | 595,937,842 | 16,971 | # Efficient Ways to Create and Append Lists in Python
## Creating Lists with Repeated Values
When working with lists in Python, you sometimes need to create a list with repeated values, either the same value or a set of values. There are different ways to achieve this, as outlined below.
## Using the Multiplication Operator
The multiplication operator, denoted by the asterisk symbol (*), allows you to repeat a list or a string a specified number of times. Here’s an example:
``````my_list = [0] * 5
print(my_list)
``````
### Output:
``````[0, 0, 0, 0, 0]
``````
In this case, we created a list of five zeros by multiplying the integer 0 by 5. You can also use this method to repeat a string by using quotes.
Here’s an example:
``````my_string = "hello " * 3
print(my_string)
``````
### Output:
``````hello hello hello
``````
The multiplication operator works well when you need to create a list with repeated values of the same type.
## Using a List Comprehension
A list comprehension is a concise way to create lists in Python. You can also use it to create a list with repeated values.
Here’s an example:
``````my_list = [1 for i in range(5)]
print(my_list)
``````
### Output:
``````[1, 1, 1, 1, 1]
``````
In this case, we used a list comprehension to create a list of five ones by iterating over a range of 5. You can also use nested lists to create more complex patterns.
Here’s an example:
``````my_list = [[i]*2 for i in range(3)]
print(my_list)
``````
### Output:
``````[[0, 0], [1, 1], [2, 2]]
``````
In this case, we created a nested list with two values repeated for each iteration of the outer loop. Using itertools.repeat()
The itertools module is a built-in Python library that contains functions for creating and manipulating iterables.
The repeat() function allows you to repeat a specific value a specified number of times. Here’s an example:
``````import itertools
my_list = list(itertools.repeat('hello', 3))
print(my_list)
``````
### Output:
``````['hello', 'hello', 'hello']
``````
In this case, we used the repeat() function to create a list with the string ‘hello’ repeated three times.
## Creating a List with Multiple Values Repeated
Sometimes you may need to create a list with multiple values repeated a specified number of times. There are different ways to achieve this, as outlined below.
## Using a for Loop
One way to create a list with multiple values repeated is to use a for loop. Here’s an example:
``````my_list = []
for i in range(3):
my_list.extend([1, 2, 3])
print(my_list)
``````
### Output:
``````[1, 2, 3, 1, 2, 3, 1, 2, 3]
``````
In this case, we used a for loop to iterate over a range of 3 and extended the list with the values 1, 2, and 3. Using itertools.repeat() with List Comprehension
Another way to create a list with multiple values repeated is to use the itertools.repeat() function with a list comprehension.
This method works well when you need to create mutable objects like lists or dictionaries. Here’s an example:
``````import itertools
my_list = [[i]*3 for i in itertools.repeat(['a', 'b', 'c'], 2)]
print(my_list)
``````
### Output:
``````[['a', 'b', 'c', 'a', 'b', 'c'], ['a', 'b', 'c', 'a', 'b', 'c']]
``````
In this case, we created a list of two nested lists, each containing the values ‘a’, ‘b’, and ‘c’ repeated three times. We used the repeat() function to repeat the list [‘a’, ‘b’, ‘c’] two times.
## Conclusion
In this article, we explored different ways to create lists with repeated values in Python. We looked at using the multiplication operator, list comprehension, and itertools.repeat() functions.
We also looked at creating a list with multiple values repeated using a for loop and itertools.repeat() with list comprehension. Using these methods, you can easily create lists with the values you need, whether you are working with immutable or mutable objects.
## Appending an Item to a List Multiple Times
When working with lists in Python, you may need to append an item to a list multiple times. The append() method only adds one item at a time, so using it multiple times can become tedious.
Fortunately, there are different ways to append an item to a list multiple times in a more efficient way. In this article, we will explore one method that uses a generator expression and the list.extend() method.
## Using a Generator Expression and list.extend()
A generator expression is similar to a list comprehension, but it returns a generator object rather than a list. A generator object is an iterable that returns the values on-the-fly rather than creating a list all at once.
The list.extend() method is used to add the items from the generator object to the list. Here’s an example of how to append an item to a list multiple times using a generator expression and list.extend():
``````my_list = ['a', 'b', 'c']
my_list.extend('d' for i in range(3))
print(my_list)
``````
### Output:
``````['a', 'b', 'c', 'd', 'd', 'd']
``````
In this code, we have created a list containing the values ‘a’, ‘b’, and ‘c’. We then used the extend() method to add the value ‘d’ to the list three times.
The generator expression ‘d’ for i in range(3) returns the value ‘d’ three times. Using the generator expression to repeat a value multiple times is more memory-efficient than creating a list with the same value repeated multiple times using the multiplication operator or a list comprehension.
This is because the generator expression generates the values on-the-fly without creating a list, which can be useful if you have limited available memory or are working with a large dataset. You can also use this method to append other types of objects to a list, such as lists or tuples.
Here’s an example:
``````my_list = ['a', 'b', 'c']
my_list.extend([1, 2] for i in range(2))
print(my_list)
``````
### Output:
``````['a', 'b', 'c', [1, 2], [1, 2]]
``````
In this code, we have used the extend() method to add the list [1, 2] to the my_list twice. The generator expression [1, 2] for i in range(2) returns the list [1, 2] two times.
When using the list.extend() method, it’s important to note that you need to pass an iterable object to the method. An iterable object is an object that can be looped over, such as a list, tuple, dictionary, or generator object.
If you pass a non-iterable object like a string or an integer to the method, it will raise a TypeError.
``````my_list = ['a', 'b', 'c']
my_list.extend(1 for i in range(3))
print(my_list)
``````
### Output:
``````TypeError: 'int' object is not iterable
``````
In this code, we tried to add the integer ‘1’ to the my_list three times using the generator expression 1 for i in range(3). However, this raises a TypeError because an integer is not iterable.
## Conclusion
In this article, we explored one method to append an item to a list multiple times using a generator expression and the list.extend() method. This method is more memory-efficient than creating a list with the same value repeated multiple times using other methods.
We also learned that the list.extend() method requires an iterable object to be passed to it and raises a TypeError if a non-iterable object is used instead. By knowing this method, you can efficiently add multiple items to a list in Python.
In this informative article, we delved into different ways of creating lists with repeated values and efficiently appending an item to a list multiple times in Python. We explored methods such as using the multiplication operator, list comprehension, itertools.repeat(), for loop, and the generator expression with list.extend().
We learned that the generator expression with list.extend() method is more memory-efficient and can be used to append different types of objects to a list. It is important to note that the list.extend() method requires an iterable object and raises a TypeError when passed a non-iterable object.
Knowing these methods and techniques can be highly beneficial for developers who often work with lists in Python and are looking to optimize their code. | 1,966 | 8,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.778506 |
https://math.stackexchange.com/questions/3023649/definition-of-generator-in-an-abelian-category/3023682 | 1,558,575,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00521.warc.gz | 564,552,536 | 32,714 | Definition of generator in an abelian category.
Let $$\mathcal A$$ be an abelian category. Let an object $$G$$ in $$\mathcal A$$ be such that $$Hom\left(G,\unicode{x2013} \right)$$ is a faithful functor from $$\mathcal A$$ to the category of sets.(I assume that I am working with a category where any family of objects has their coproduct existing inside the category)
Why is the above equivalent to the fact that every object $$X$$ in $$\mathcal A$$ admits an epimorphism $$G^I$$ to $$X$$? (where $$I$$ is the index category and is arbitrary)(where,$$G^I$$ is coproduct of copies of $$G$$ which exists in that category)
I do not see how injectivity of maps getting translated to epimorphism and vice versa? I guess I should start with appropriate short exact sequence and apply appropriate functor to make injectivity translated to epimorphism but I have no clue how to proceed practically, i.e how to make it precise.
Any help from anyone is welcome.
• I'm not sure it is. In general Abelian categories one cannot always take arbitrary powers of objects. – Lord Shark the Unknown Dec 3 '18 at 5:01
• @lord-shark-the-unknown,sorry I should have mentioned that I am assuming in that category every family of objects(even possibly infinite) has its coproduct in that category. – HARRY Dec 3 '18 at 5:11
• $G^I$ is a weird notation for a possibly infinite coproduct – Max Dec 3 '18 at 8:11
Suppose there is an epimorphism $$p:G^I\to X$$ and suppose $$f:X\to Y$$ is a morphism which becomes $$0$$ after applying the functor $$\operatorname{Hom}(G,-)$$. This means that for every morphism $$g:G\to X$$, $$fg=0$$. In particular, by taking $$g$$ to be each of the inclusion maps $$G\to G^I$$ composed with $$p$$, this implies $$fp=0$$. Since $$p$$ is an epimorphism, this implies $$f=0$$. Thus if such an epimorphism $$p$$ exists for every $$X$$, $$\operatorname{Hom}(G,-)$$ is faithful.
Conversely, suppose $$\operatorname{Hom}(G,-)$$ is faithful and let $$X$$ be an object. Let $$I=\operatorname{Hom}(G,X)$$ and let $$p:G^I\to X$$ be the unique morphism such that for each $$i\in I$$, the composition of $$p$$ with the $$i$$th inclusion map $$G\to G^I$$ is $$i:G\to X$$. I claim $$p$$ is an epimorphism. To show this, it suffices to show that if $$f:X\to Y$$ is a morphism such that $$fp=0$$, then $$f=0$$. But given any such $$f$$, by composing with the inclusion maps $$G\to G^I$$ we see that $$fi=0$$ for all $$i:G\to X$$. This means that $$\operatorname{Hom}(G,-)$$ sends $$f$$ to $$0$$ and thus $$f=0$$ since $$\operatorname{Hom}(G,-)$$ is faithful.
• I first thought your answer was wrong because of the notation $G^I$ which seems weird for a (possibly infinite) coproduct. Perhaps you could write a word about it ? – Max Dec 3 '18 at 8:12
• @Max,I tried to put $(I)$ there but somehow it did not work.sorry for the notation,but later I explained what it stands for. – HARRY Dec 3 '18 at 8:40
• @HARRY : yes of course, that's how I realized that the answer was actually correct (and having seen Eric's answers over time it seemed weird that he'd make this sort of mistake) – Max Dec 3 '18 at 8:50 | 901 | 3,106 | {"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": 49, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-22 | latest | en | 0.895039 |
http://www.thegeekstuff.com/2014/03/cpp-prime-number-checker/ | 1,513,095,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517350.12/warc/CC-MAIN-20171212153808-20171212173808-00115.warc.gz | 462,387,488 | 12,964 | # How to Implement Prime Number Check Algorithm Using a C++ Program Example
The set {1, 2, 3, …} is known as the set of natural numbers, they are usually signed as N numbers.
This tutorial is about prime numbers. So what are prime numbers?
Let us take number 15, which could be represented as shown below. This is not a prime number.
15 = 1 * 3 * 5
;
Let us take number 13, which could be represented as shown below. This is a prime number.
13 = 1 * 13
In case of number 13, you will not be able to find any natural numbers beside 1 and 13 that will divide number 13 without some left over.
What do we mean by left over? Let us take number 17, and if you divide 17 with 7, you could represent 17 as shown below. The left over is 3 in this case.
17 = 7 *2 + 3.
The following is the list of few primer numbers and it is sometimes noted as P set.
{2, 3, 5, 7, 11, 13, …}
One more thing, the numbers 5 and 7 are the twin primes, they are represented like: 6n – 1 and 6n + 1. This time n is equal to 1.
All prime numbers are represented in that way, if they are bigger than 3. But all numbers of 6n+1 or 6n-1 type are not prime numbers.
Last number that could be candidate to make tested number not prime, is not bigger than sqrt(n).
Also one very important fact about prime number is that 1 is not prime number.
### Prime Number Checker Program
The following C++ example code will check whether the given number is a prime number or not.
```#include <iostream>
#include <cmath>
using namespace std;
bool IsPrime (int);
int
main(void)
{
cout<<"The program checks if the given number is prime!"<<endl;
for(int i= 0; i<=44; i++, cout<<'_');
cout<<endl;
do
{
cout<<"Do you wish to test next number y/n->";
char cRespond;
cin>>cRespond;
if ((cRespond == 'y')||(cRespond == 'Y'))
{
cout<<"Enter the number:->";
int myInput;
cin>> myInput;
(IsPrime(myInput)==true)? cout<<"It is":cout<<"It is not";
cout<<" prime number!"<<endl;
continue;
}
break;
}
while(true);
return EXIT_SUCCESS;
}
bool
IsPrime (int n)
{
if((n==2)||(n==3)) return true;
int iResidum = n % 6;
if(!((iResidum == 5) || ( iResidum == 1))) return false;
if(n %3 == 0) return false;
for(int i=1;
6*i <= (int)sqrt(double(n))+6;
i++)
{
if( n % 6*i-1==0) return false;
if( n % 6*i+1==0) return false;
}
return true;
}```
### Explanation of the Algorithm
First we will analyse the main function and then we will go in IsPrime() function.
The main program does the following:
1. Write the header where we explain what we do in this program.
2. We create “do wile” circle that will enter the numbers to be examined.
3. We ask the user should we stop testing the numbers or should we continue with the testing.
4. If the respond is y or Y, we will test the next number with function IsPrime, otherwise we stop with the checking.
5. If the logical function returns the true we print message that the number is prime, but if the function returns false we print the message that the number is not a prime number.
The function does the following:
1. Test if the number is: 2 or 3, because they are not of 6n+1 or 6n-1 form.
2. We divide the potentially prime number with 6, and if we get remain that is different than 5 or 1 we don’t have potentially prime number. The function returns false.
3. In the “for” we test all potentially prime numbers. If they could be disassembled into composite numbers then those numbers will be of 6n+1 or 6n-1 form. We will have few tests that are not necessary but if one wishes it could find those numbers in the list of prime numbers that would be constructed. That way, those tests would be meaningless. Think why.
4. The last potentially divisor is not greater than sqrt(n) +6. Aldo, I am not sure is it possible to use just sqrt(n).
The above approach is not bad for smaller numbers, but when the number that we are checking becomes to big it could slow down the program. Also, there is one more trick, to generate accidental numbers and divide the candidate number with those numbers, but this way we will not get number that is prime for sure. To benefit from this approach we could insert this before “ for” in the function. That could sometimes catch the numbers that are not prime numbers.
One more idea is to create the list of prime numbers and search is the number in the list. Also, if you really like to create the faster solution, you could keep the numbers in some data structure that could outperform simple vector in this problem.
1. Try to enlist all prime numbers smaller than the given one.
2. Print all prime numbers in range [a..b], where a is less than b.
3. Use Sieve of Eratosthenes to list all prime numbers smaller than n.
4. Find the prime numbers that will divide n without the left over.
5. Find how many prime numbers divide n without left over and how many divide n with left over.
6. Check is the pair of numbers: 6*i + 1 or 6*i -1 for some i couple of prime numbers.
7. Break the number n into sum of prime numbers, if possible.
• The prime numbers will not include 1.
• The prime numbers will include 1.
• Any prime number will be used only once.
• The prime numbers could be used more times.
• The prime numbers are the biggest prime numbers that could break the number into the sum of prime numbers that will use only once each prime number, or use the prime numbers more times.
• Lev Lafayette March 14, 2014, 1:03 am
Interesting results. 🙂
lev@owl ~/Desktop \$ ./prime
The program checks if the given number is prime!
_____________________________________________
Do you wish to test next number y/n->y
Enter the number:->2
It is prime number!
Do you wish to test next number y/n->y
Enter the number:->3
It is prime number!
Do you wish to test next number y/n->y
Enter the number:->5
It is prime number!
Do you wish to test next number y/n->y
Enter the number:->7
It is not prime number!
Do you wish to test next number y/n->y
Enter the number:->11
It is prime number!
Do you wish to test next number y/n->y
Enter the number:->13
It is not prime number!
Do you wish to test next number y/n->y
Enter the number:->17
It is prime number!
Do you wish to test next number y/n->y
Enter the number:->19
It is not prime number!
Do you wish to test next number y/n->y
Enter the number:->23
It is prime number!
• Gurudutt March 14, 2014, 1:51 am
So, this program does not work is it?
• duskoKoscica March 14, 2014, 2:37 am
Ok, “interesting” , it looks like iResidum and the if that goes with it should be adjusted, so if the residum is in { 0, 2, 3, 4} the number is not prime for sure!
• duskoKoscica March 14, 2014, 3:08 am
To solve this there are at least two possible solutions> you could code somthing like this if( iResidue==0) or ( iResidue==2) or (iResidue ==3) or (iResidue==4)) return false with iso646 or iso464, or you could use if((iResidue==1) or (iResidue==5)) the code that checks is the number prime} else { return false;} . But ther is few more things to improve!
• Ron March 14, 2014, 7:10 am
Interesting article.
using c++ , object oriented language, and coding in as a procedural language like ‘C’ or perl should be considered as a crime.
Many new developers visiting the site will pick up bad habits and then again all code should be written so that it could be easily unit tested.
I have seen many such articles are written where the author writes articles to hon their language skills without keep bigger picture in mind.
Keeping that in mind the subject is quite informative.
Thanks.
Ron.
• duskoKoscica March 14, 2014, 8:20 am
So with bigger picture in mind, try to see how would you handle this, the task for more advanced, this is like you wanna use friend function or how, I would like to hear more about idea you have in mind.
• Ron March 14, 2014, 8:35 am
duskoKoscica,
this is what I’d suggest.
Create a class().
add methods(), keeping each method smaller and simpler for testablity.
reduce cyclomatic complexity.
Write in an object oriented manner.
• duskoKoscica March 14, 2014, 10:53 am
Ok now I know how would I do it but I would like to see how would you do it!
By the way, happy Pi day, if that thing is not a joke, it seams cool!
For further studies I would recomend that you wisit the wikipedia and find more about the prime numbwers and yes there is one very interesting site about prime numbere of different form. There are, also the sites where you should find some algorithms that work better with certain types of prime numbers, tha algorrithmes that count how manny prime numbers there are etc..
• DirkM March 15, 2014, 3:18 am
Boring! Prime numbers in this way? It is a subject of no practical mening at all!
• duskoKoscica March 17, 2014, 8:02 am
I have had some time on this week to look more carefully in those functions and there are few more corrections. Also, I need to say that I am not happiest how it looks like, but it was tested till 10, 100, 1000. Now I need to say that it should be inspected more carefully but that is up to reader.
Those who like to fix the things have had enough time, so now for those who could not fix it on they own!
bool
IsPrime( unsigned long long n)
{
if((n==2)||(n==3)) return true;
int iResidum = n%6;
if((iResidum==0)||(iResidum==2)||(iResidum==3)||(iResidum==4)) return false;
for(unsigned long long i =1;
6*i <= sqrt(i) +1;
++i)
{
if(n%(6*i-1)==0) return false;
if(n%(6*i+1)==0) return false;
}
return true;
}
If some of you have noticed some errors, THX a lot!
• Mike March 21, 2014, 3:59 am
I don’t know why DirkM said that.
My teacher said that prime numbers are important!
I don’t know why! But he said Ok to.
• Ana March 22, 2014, 4:25 am
DirkM you never listen to techer and all you have are those storyies that you don’t like something or somethig like that.
Me, Miranda, Johanna and Betty don’t like you.
You should liten to techer, ana pay more attention to subjects.
Next when you say something like that in class I will tell you off.
• duskoKoscica April 3, 2014, 11:58 pm
To checke if the big number is prime number there are algorithms like: Polard-Rool, Dickson, and few other methods that are known to common people and few that are not. There are few forms of prime numbers, ok a lot of forms of prime numbers, and also one could even get the money revord if has success in finding bigest prime. This prime numbers are never ending story, it is not like Fermat big theorem, when they solve the thing after long time and that’s it. Prime numbers are always in.
• duskoKoscica December 1, 2014, 6:33 am
There are some other forms of prime numbers like 6n+1 and 6n+5, it would be nice to find few conditions that would tell you very fast if the number is prime or not, that way you would not need to divide numbers.
Yes there are some ways to say if certain type of prime number is prime to.,
• Alain February 26, 2015, 1:21 pm
all algorithms are buggy!
I have tested the 500th prime which should be 3557 according to Wikipedia ( http://en.wikipedia.org/wiki/List_of_prime_numbers#The_first_500_prime_numbers ).
There are a lot of multiple of 5 declared as primes!
I have presently no time to look after the errors..
• DuskoKoscica July 4, 2015, 6:10 am
@Alain
Hi!
First, I will notice that this function is created for single testing, rather than for range of tests, this argument would be obvious from performance point of view, however it could be modified for that purpose as well. In OOP it would be nice to have…
Second, if you could figure out that there is a problem with numbers that are multiples of five, I am sure that you would be able to figure out how to exclude them as well.
It would be interesting to find what did cause that problem.
Zero, this is a blog. | 3,100 | 11,668 | {"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} | 3.796875 | 4 | CC-MAIN-2017-51 | latest | en | 0.883536 |
https://brainsanswers.co.uk/mathematics/which-of-the-following-values-is-no-12926142 | 1,601,594,240,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402132335.99/warc/CC-MAIN-20201001210429-20201002000429-00735.warc.gz | 277,639,194 | 32,605 | , 09.07.2019 17:32 emily4984
# Which of the following values is not change by moving the center of an ellipse? a or b b c a b or c
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What are the possible rational zeros of [tex]f(x) = x^4+2x^3-3x^2-4x+18[/tex] | 210 | 616 | {"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-2020-40 | latest | en | 0.858412 |
https://manuals.pqstat.pl/en:przestrzenpl:gestoscpl | 1,722,931,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00195.warc.gz | 309,982,817 | 15,913 | # PQStat - Baza Wiedzy
### Pasek boczny
en:przestrzenpl:gestoscpl
### Density analysis
To conduct density analysis on the basis of a Map data we should have at our disposal a point, multipoint, or polygonal file. In the case of an analysis of a polygonal file, calculations are based on centroids of polygons, and in the case of a multipoint file they are based on centers of objects.
Graphically, this method is a generalization of a histogram, or one-dimensional analysis, to a two-dimensional case. Building a histogram we have one variable, which we divide into intervals of equal length and give the number of cases in each interval. When building a grid of squares, we have two variables on which we build the grid and give the number of cases in each grid square (DPS – Dot Per Square). The ratio of this number to the area of a square determines the intensity of the color in which a given grid square is colored.
Based on the number of casess in the grid squares, we can study their spatial distribution. If there are the same number of points in each square, it means perfectly uniform distribution. When the opposite is true, when the variation in the number of points in the squares is very large, it means that there are squares with a much larger number of points, that is, clusters are formed.
If we denote by the number of points of the study area and by the number of squares into which the study area is divided, then we can determine the mean, variance, and standard deviation of the number of points per square:
where – is the number of squares with the number of points equal to .
Coefficient
The most important information is provided by the variance-mean ratio – the coefficient, which is the quotient of the variance and the mean:
A value of indicates too little variation in the number of points in squares which suggests a uniform dispersion effect, indicates too much variation in the number of points in squares and therefore a clustering effect, and a value close to 1 indicates an average variation in the number of points in squares which implies a random distribution of points.
The Index of Cluster Size (ICS) is often considered in the literature:
The expected value of assuming random points is 0. A positive value indicates a clustering effect and a negative value indicates a regular distribution of points.
Significance of the coefficient
The coefficient significance test is used to verify the hypothesis that the observed point counts n the squares are the same as the expected counts that would occur for a random distribution of points.
Hypotheses:
The test statistic has the form:
This statistic has an asymptotically distribution with degrees of freedom.
The p-value, designated on the basis of the test statistic, is compared with the significance level :
Note
The result of the analysis depends to a large extent on the density of the grid and thus on the number/size of squares into which the analyzed area is divided. In the test options window, you can set the grid that will be used to divide the test area into squares by specifying the number of squares vertically and horizontally.
The window with the settings for the quadrat count method is launched via the menu Spatial AnalysisSpatial StatisticsQuadrat analysis
EXAMPLE (file squares.pqs)
Using the datasheet, generate two point maps and perform a density analysis of these points. Answer the question: are the points randomly distributed in each of these maps?
You create the point maps using the formulas: menu DataFormulas…
This will result in two new sheets containing maps. For each of these sheets, we perform a quadrat analysis.
Hypotheses:
The results for Map 1 indicate a significant variation in the number of points in the squares, that is, a clustering effect (value p=<0.0001). This effect persists for different grid densities. For a grid density of 10:10 the VMR ratio is as high as 12.5, the entire report is included below:
For map 2, the situation is quite different. For the 10:10 density grid, we have a lack of statistical significance (value p=0.9585) and the value of the coefficient VMR=0.77 indicate that the distribution of points is random.
Using the button in the report, we move to the Map Manager to select the analysis grid from the displayed list of layers and obtain a graphic interpretation of the results.
2022/02/09 12:56
#### Kernel density estimator
##### Two-dimensional kernel estimator
The two-dimensional kernel estimator (like the one-dimensional estimator) allows the distribution of the data, expressed by the method of squares, to be approximated by smoothing.
The two-dimensional kernel density estimator approximates the density of a data distribution by creating a smoothed density plane in a non-parametric manner. It produces a better density estimate than is given by the traditional method of squares, whose squares form a step function.
As in the one-dimensional case, this estimator is defined based on appropriately smoothed summed kernel functions (see description in the PQStat User Manual). There are several smoothing methods to choose from and several kernel functions described for the one-dimensional estimator (Gaussian, uniform, triangular, Epanechnikov, quartic/biweight). While the kernel function has little effect on the resulting plane smoothing, the smoothing factor does.
For each point in the range defined by the data, the density or kernel estimator is determined. It is obtained by summing the product of the kernel function values at that point:
If we give the individual cases weights , then we can construct a weighted nuclear density estimator defined by the formula:
The window with settings for the kernel 2D density estimator ptions is launched via the menu Spatial analysisSpatial statisticsKernel density estimator 2D
EXAMPLE (snow.pqs plik)
Currently, the main problem in presenting point data on the location of people is the need to protect them. Data protection prohibits publishing research results in such a way, that it would be possible to recognize a given person on their basis. A good solution in this case is a point density estimator.
We will present point data illustrating the cholera epidemic in London in 1854 using such an estimator. To do so, we will use a map of points (deaths due to cholera) with layers already overlaid to illustrate both streets and water pumps, and the result of an analysis by physician John Snow.
In the analysis window for the point map, we will stay with the Gaussian (normal) distribution kernel and the SNR smoothing factor. The grid density will be set to 80:80 and the boundaries will be increased so that the edges do not have a sharp edge by entering 300 as the minimum value for the X and Y coordinates and 1100 as the maximum value.
Using the button in the report, we go to the Map Manager, where we can add a layer representing this estimator (the last item in the list of layers).
After applying the nuclear density estimator layer, edit it o remove the grid lines and change the yellow color to the natural background color (white in this case). The layer thus obtained is moved up g_kolejnosc_warstw, so that it is drawn at the beginning. We turn off the points layer (Base Map).
EXAMPLE cont. (squares.pqs file)
Using the kernel estimator, we represent the point density for map 1 - obtained in the earlier part of the task.
In the analysis window, we set the grid density to 50:50 and the kernel type as normal distribution and include a graph. We perform the analysis three times while changing the User smoothing factor: h (10:10), then h (10:20) and h (20:20). The obtained results presented on the map (via Map Manager) and on the 3D graph are shown below:
##### Three-dimensional kernel estimator
The three-dimensional kernel estimator (like the one-dimensional estimator and the two-dimensional estimator) allows you to approximate the distribution of the data by smoothing it.
The three-dimensional kernel density estimator approximates the density of the data distribution by creating a smoothed density plane in a non-parametric way. Graphically, we can represent it by plotting the first two dimensions in layers created by the third dimension. As in the one-dimensional case (see description in the PQStat User's Guide) and the two-dimensional estimator, this estimator is defined based on appropriately smoothed summed kernel functions. There are several smoothing methods to choose from and several kernel functions described for the one-dimensional estimator (Gaussian, uniform, triangular, Epanechnikov, quartic/biweight). While the kernel function has little effect on the resulting plane smoothing, the smoothing factor does.
For each point in the range defined by the data, the density that is the kernel estimator is determined. It is formed by summing the product of the kernel function values at that point:
If we give the individual cases weights , then we can construct a weighted kernel density estimator defined by the formula:
The window with settings for the kernel 3D density estimator options is launched via the menu Spatial analysisSpatial statisticsKernel 3D density estimator
Note
Displaying subsequent layers of the estimator, determined by the third dimension, is possible by editing the layer in the map Manager window and selecting the appropriate layer index.
2022/02/09 12:56
en/przestrzenpl/gestoscpl.txt · ostatnio zmienione: 2022/02/16 13:03 przez admin | 1,946 | 9,504 | {"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} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.925808 |
https://gpl4you.com/mbaquestion.php?q=mcq&type=Mathematical%20Skills | 1,718,878,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00758.warc.gz | 267,582,061 | 15,938 | ## Company
• 1.
### How many kg of pure salt must be added to 30 kg of a 2% solution of salt and water to increase it to a 10% solution? (a) 2 23 kg (b) 15 kg(c) 3 kg(d) 14 kg
last reply by CpjJwWHV • 11 years ago • asked in MAT
• 2.
### Two persons are walking in the same direction at rates 3 km/hr and 6 km/hr. A train comes running from behind and passes them in 9 and 10 seconds. The speed of the train is(a) 22 km/hr (b) 40 km/hr3) 33 km/hr (d) 35 km/hr
last reply by CpjJwWHV • 12 years ago • asked in MAT
• 3.
### A person travels 285 km in 6 hrs in two stages. In the first part of the journey, he travels by bus at the speed of 40 km per hr. In the second part of the journey, he travels by train at the speed of 55 km per hr. How much distance did he travel by train?(a) 165 km (b) 145 km(c) 205 km (d) 185 km
last reply by CpjJwWHV • 12 years ago • asked in MAT
• 4.
### The average monthly salary of employees, consisting of officers and workers of an organisation is Rs 3000. The average salary of an officer is Rs 10,000 while that of a worker is Rs 2,000 per month. If there are total 400 employees in the organisation, find the number of officers and workers separately.(a) 50, 350 (b) 350, 450(c) 50, 275 (d) 325, 350
last reply by CpjJwWHV • 11 years ago • asked in MAT
• 5.
### How many kg of sugar costing Rs 5.75 per kg should be mixed with 75 kg of cheaper sugar costing Rs 4.50 per kg so that the mixture is worth Rs 5.50 per kg ?(a) 350 kg (b) 300 kg(c) 250 kg (d) 325 kg
last reply by CpjJwWHV • 11 years ago • asked in MAT | 540 | 1,585 | {"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} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.874844 |
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# Factor the expression $4+4a+a^2$
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### Main Topic: Polynomial Factorization
They are a group of techniques that help us rewrite polynomial expressions as a product of factors. | 413 | 785 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.446533 |
https://stacks.math.columbia.edu/tag/00JU | 1,725,785,610,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00830.warc.gz | 529,350,031 | 6,057 | Lemma 10.57.8. Let $S$ be a graded ring.
1. Any minimal prime of $S$ is a homogeneous ideal of $S$.
2. Given a homogeneous ideal $I \subset S$ any minimal prime over $I$ is homogeneous.
Proof. The first assertion holds because the prime $\mathfrak q$ constructed in Lemma 10.57.7 satisfies $\mathfrak q \subset \mathfrak p$. The second because we may consider $S/I$ and apply the first part. $\square$
There are also:
• 6 comment(s) on Section 10.57: Proj of a graded ring
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 182 | 674 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-38 | latest | en | 0.76651 |
https://math.stackexchange.com/questions/3584649/the-category-of-c-algebras-has-cokernels | 1,712,956,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00113.warc.gz | 361,539,128 | 35,740 | # The category of C* algebras has cokernels
Let $$\mathcal{C}^*$$ be the category defined as follows:
$$1.\text{Ob}(\mathcal{C}^*)=\{C^*\text{-algebras}\}$$
$$2.$$ For $$A,B\in\mathcal{C}^*$$ set $$\text{Hom}_{\mathcal{C}^*}(A,B)=\{\varphi:A\to B\;|\; \varphi \text{ is a *-homomorphism} \}$$
This is a category (it can be easily verified) and we know that $$\text{Hom}_{\mathcal{C}^*}(A,B)$$ is exactly the set of norm-decreasing homomorphisms (also easily proved). With these homomorphisms, $$\mathcal{C^*}$$ actually is NOT additive, as we cannot simply add $$*$$-homomorphisms (the result is not a $$*$$-homomorphism). However, we do have kernels. My question is: does $$\mathcal{C}^*$$ have cokernels? I believe so, and here is my work:
Let $$A\xrightarrow{\varphi}B$$ be a morphism in $$\mathcal{C}^*$$. Let $$I$$ denote the closed (double) ideal generated by $$\varphi(A)$$. Then $$B/I$$ (with quotient norm) is a $$C^*$$-algebra and we have the quotient $$*$$-homomorphism $$\pi:B\to B/I$$. I claim that $$\text{cokernel}(A\xrightarrow{\varphi}B)=B\xrightarrow{\pi}B/I$$.
The description of $$I$$ is the closed linear span of the set $$S:=\{\varphi(a),\varphi(a)b, b\varphi(a), b\varphi(a)b': a\in A, b,b'\in B\}$$ (maybe this can be simplified if we consider the unitization of $$B$$, but I'd like to avoid it right now). If $$B\xrightarrow{\psi}D$$ is a $$*$$-homomorphism such that $$\psi\circ\varphi=0$$, then we define $$\bar{\psi}:B/I\to D$$ by $$\bar{\psi}(b+I)=\psi(b)$$.
First we need to show that $$\bar{\psi}$$ is a well defined $$*$$-homomorphism. The only non-trivial thing is that $$\bar{\psi}$$ is well defined. But if $$b-b'\in I$$, we need to show that $$\psi(b-b')=0$$. But $$b-b'=\displaystyle{\lim_{n\to\infty}u_n}$$ where $$u_n\in\overline{\text{span}(S)}$$. By $$\psi$$'s continuity and preservation of sum and scale, it suffices to show that $$\psi\equiv0$$ on $$S$$. but this is true, since $$\psi(\varphi(a))=0, \psi(\varphi(a)b)=\psi(\varphi(a))\psi(b)=0,\dots$$ and so on.
Obviously $$\bar{\psi}\circ\pi=\psi$$ and if $$\bar{\psi}':B/I\to D$$ was another such morphism, then $$\bar{\psi}'(b+I)=\bar{\psi}'\circ\pi(b)=\psi(b)=\bar{\psi}(b+I)$$, so we have uniqueness.
This is the universal property of cokernels, so we are done. Does my proof contain mistakes? If someone can confirm that there are no mistakes, I suggest editing the title and taking the question mark off.
A comment: Note that $$\varphi(A)$$ is not equal to the image in the categorical setting: an obvious reason: the image is defined as the kernel of the cokernel. but kernels are always ideals, $$\varphi(A)$$ is not necessarily an ideal. Also, we know that $$\varphi(A)\cong A/\ker(\varphi)$$, the co-image of $$\varphi$$.
• It looks fine for me. Mar 17, 2020 at 21:27
• @Berci Thank you for your reply Mar 18, 2020 at 0:45
• I don't think that $\text{Hom}_{\mathcal{C}^*}(A,B)$ is an abelian group under any operations. Additionally, you can show that the ideal generated by $\varphi(A)$ is the closed span of $\{b\varphi(a)b':a\in A,b\in B\}$ using an approximate identity argument. Mar 22, 2020 at 1:51
• @Aweygan I see, I rushed on this and didnt notice that the sum of algebra homomorphisms is not an algebra homomorphism. Thank you for noticing, I will edit my post soon. Mar 22, 2020 at 2:02 | 1,118 | 3,315 | {"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": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.830939 |
https://www.studypool.com/discuss/5212/multiple-regression | 1,521,917,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650764.71/warc/CC-MAIN-20180324171404-20180324191404-00506.warc.gz | 928,747,083 | 23,582 | Multiple regression
Anonymous
account_balance_wallet \$20
Question description
Multiple regression analysis is widely used in business
research in order to forecast and predict purposes. It is also used to
determine what independent variables have an influence on dependent
variables, such as sales.
Sales can be attributed to quality, customer service,
and location. In multiple regression analysis, we can determine which
independent variable contributes the most to sales; it could be quality
or customer service or location.
Now, consider the following scenario. You have been
assigned the task of creating a multiple regression equation of at least
three variables that explains Microsoft’s annual sales.
Use a time series of data of at least 10 years. You can search for this data using the Internet.
Before running the regression analysis , predict
what sign each variable will be and explain why you made that
prediction.
Run three simple linear regressions by considering one independent variable at a time
After running each of the three linear regressions, interpret the regression.
Does the regression fit the data well?
Run a multiple regression using all three independent variables.
Interpret the multiple regression. Does the regression fit the data well?
Does each predictor play a significant role in explaining the significance of the regression?
Are some predictors not useful?
If so, did you consider removing those and rerunning the regression?
Are the predictors related too significantly to one another? What is the coefficient of correlation “r”? Do you think this “r” value suggests a strong correlation among the predictors ( the independent variables?
On a separate page, cite all sources using the APA guidelines.
skywalkerus
School: University of Maryland
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Wow this is really good.... didn't expect it. Sweet!!!!
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https://zyxyvy.wordpress.com/2011/09/ | 1,547,996,382,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00627.warc.gz | 986,052,839 | 14,997 | The Calculus of Dreams
Ever since 2005, I have been pondering the subject of dreaming greatly, and have finally came to some useful philosophical conclusions (but I still haven’t found sure proof that this world is real), but one particular milestone is when I watched Inception, the second greatest movie I’ve seen. Out of many interesting things I noted while watching the movie, by far the greatest idea was a model for dreams that I came up with while watching it. It is the representation of dream levels as functions, and the representation of dream transactions as calculus.
Let’s, for purposes of introduction, call real life 1, and by “1” I mean the function f(x)=1, not the number 1. This is a sensible assignment, initially because the simplicity of “1” represents how reality makes sense straightforwardly, but also because of reasons to be explained later. What happens after you die? Well, nonexistence quickly leads to thinking of “0” (or d(x)=0) to represent what occurs when life ends.
What would be a good expression to represent your dream? (I’ll just call it ‘dream’ for now and ‘Dream Level 1’ later.) A dream is mysterious and doesn’t always make sense—much of it is a representation of unknown, so a sensible expression for a dream would be “x” (as a function, g(x)=x).
Of course, some things about “x” don’t make sense, like what the slope represents in terms of properties of the perceived mental world; for example, what does the slope represent? Insanity?
In any case, let’s go back on track. What if we had a dream within a dream? Would that just be x²? That would make an intuitive guess, but these new dreams are very weird…they are more unstable and have greater time dilation. Perhaps h(x)=x²/2 is a better representation for Dream Level 2.
At this point, something just has to be noticed. h'(x)=g(x), g'(x)=f(x), and f'(x)=d(x). That’s right, the relation between worlds of existence is analogous to the relation between dream levels. Philosophically, it makes sense, in terms of the derivations of spontaneous actions from the unconscious. The factor that the power rule tells us to bring around represents closely the shift in apparent time speed, although in this case it’s by factorials rather than powers of 2, but both of these functions show the correct exponential growth of time dilation in deeper dream levels. Knowing this, it can be seen that entering a dream is not just inception, but also integration—integration of our secret emotional reserves into our interaction with the world around. The world of “1” is thus distinctly different from dream worlds, because attempting to derive the source of entering the f(x)=1 world leads to death (or perhaps limbo, but see later), where more attempts of deriving are futile for change and integration from 0 (reincarnation?) leads to too much uncertainty for returning to where one was. The integral of d(x)=0 is just “C,” and who knows which world will be picked for you: 2? 9001? π?
Once a full connection is made though, the next thing to do is extrapolation. What exactly is limbo? Think about the delicate nature of how the process of transferring between levels uses the discrete package of a difference of “1” between dream levels. When the framework of meta-dream mechanics is sabotaged or vandalized, perhaps the error of inputting a fractional difference could occur. What does this imply? One could end in a world with a fractional exponent of x, and unless or until another such mistake occurs of coincidental fractional fix, one must wander this stack of functions related by integrals and derivatives. This is like limbo. Why? There is only mindless wandering here. Unlike the case with dream worlds of positive integral degree, looking down the dream stack reveals a well with no bottom. Differentiating again and again leads to worlds that are indeed one exponential degree apart, but there is no stopping at zero, no getting rid of the x in representation, and an infinite descent down the negative degrees. There isn’t really any dream level more dreamy than another now, as they are all infinitely far from any world that can be distinguished as reality. All the worlds are equally dreamy: it is dreams all the way up and dreams all the way down.
But what if this, instead of representing limbo, represented an alternative structure for a universe over the parameter of dreams? It shows that perhaps there isn’t really any reality (which I could talk about more but would be tangential here), thus in some way nulling the meaningfulness of the distinction of a world as “dreaming.” That would be a pretty weird alternative world, but one worth thinking about.
Yet we could extrapolate further. Consider each function as possible of representing a world in reference to dreaminess. What happens in an f(x)=sin x world? In that interesting hypothetical world, entering a dream consecutively four times leads you back to where one was—a cyclical dream stack. That is a possibility that deserves some thinking as well. What about an f(x)=e^x world? A dreamless world? What does integrating in f’(x)=1/x to f(x)=ln|x| represent? What would doing these things in polar coordinates represent?
And most importantly, what do all of these imply about the juices that mold life and its meanings and properties—the elusive top that has been spinning in the minds of humans for millennia?
Nightmareboard
So as we all know, QWERTY is pretty inefficient. Incidentally, it’s actually by far not the most painful layout possible, so I made a personal attempt to produce a truly egregious keyboard layout. In this keyboard, I tried my best to put the keys in the most unbearable positions possible. Try typing (okay, maybe hovering your fingers over the keys to represent typing the keys) something using this keyboard layout, like perhaps your name, or a quote you like. Enjoy the torture.
On the Boredom Problem
When I was 5, my parents introduced me to the concept of death. Yes, I was very frightened, because dying sounded very scary (you literally just end and then you can do nothing afterwards)—more scary to me apparently than to other children. In fact, it scared me so much that I started trying to find all ways possible to prolong death as much as can be done, yet grieving over the fact that it must eventually happen. When I was 10, however, on a day in February in 2005, I was in the bathroom thinking and reached some dramatic conclusions about life, the first of which was the realization that death is a great thing. Without death, we will suffer the alternative, a never-ending consciousness of the world which eventually intuitively will get painfully boring and miserable. But then I realized that death is the equivalent as well (although, thankfully, we won’t be conscious of what’s happening, although I was worried where our consciousness goes after death), and I reached a horrible dilemma that haunted me for two years: both death and living forever seemed like abysmally bleak results—are we completely doomed in this world?
As I said, the problem was painfully in my mental back burner for two years before I finally reached a comforting conclusion. I realized that a more concrete explanation of the problem with death is that it is actually foreverness in a sense—one is in the same boring state with no end in sight (or, heck, in reality). The source of the problem of both cases is the eventual settling to an unwanted existence: boredom is the actual greatest enemy of the conscious. But this boredom seemed inevitable until I discovered that there is a possibility of an alternative.
Here’s how the alternative comes about: imagine the set of the decimal numbers. Death can be represented as a terminating decimal—for example, 4.62—the 2 marks the point of death after which the decimal halts. Non-terminating life can be represented as a repeating decimal—for example, 3.33333333333… (I mean 10/3 in decimal form)—there is no end to the series of 3’s. First, we can notice that this confirms the notion of how death and non-terminating life are related in boredom: the 4.62 is actually 4.6200000000000…, but 0 represents nonexistence. In both cases, the problem is having to deal with something with no end in sight, a horrid prospect.
But more importantly, we can finally be consoled that there is a third path, as there exists irrational numbers. A π-life would not have these horrendous issues. Of course, it was hard for me to imagine what such a life would be like, but the analogy was strong enough to convince me and finally have some rest in the part of my brain that until then was continuously stomped on by this problem. I was relieved even though this dogma is highly unrealistic (although later, when I was 15, I found an alternate theory that is as convincing of relief and already is in place in the “real” world). But of course, as you probably know, when picking a random number, chances are it’s irrational.
In fact, if you put the mathematics aside, think for a moment: the world contains irrationality after irrationality. Can you imagine how horrible life would be without them? | 1,976 | 9,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-04 | latest | en | 0.940492 |
https://blog.heinemann.com/choosing-rich-math-tasks | 1,718,634,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00494.warc.gz | 122,998,163 | 16,371 | <img height="1" width="1" alt="" style="display:none" src="https://www.facebook.com/tr?id=940171109376247&ev=PageView&noscript=1">
# Dedicated to Teachers
Just as conferring is one part of the readers’ and writers’ workshop and could not be implemented in isolation, conferring in mathematics must take place on a broader instructional stage. As teachers, we hope to talk with students about their thinking as they struggle with big ideas and ways of thinking mathematically—in the midst of this kind of grappling there is quite a lot to talk about. But if tasks or expectations in the classroom don’t demand deep thinking, we’re left with thin conversations about answers. So, what does it take to create a mathematical environment ripe for conferring?
Rich math tasks create space for students to struggle with concepts, develop strategies, make mistakes and connections, and engage in math practices. In the messiness of developing understanding through problems, we can have interesting and fruitful conversations with students about their thinking while that thinking is still forming. These are prime moments for nudging students down a productive pathway. The right task creates these opportunities and in this chapter, we’ll look at what makes such a task.
Although students have always been asked to solve problems in math classrooms, in the past these problems have shared some features that make them less productive for learning and discussion. These problems typically have a single prescribed procedure for solving them and a single correct answer, which all students were expected to mirror. Success was defined by the uniformity of students’ work, and any talk was often limited to the reporting and evaluation of answers. This is a desiccated version of mathematics; with just the withered husk of procedures and answers there is not much to discuss.
## What Is a Rich Task?
What makes a task rich depends in part on your students' daily experiences, what your students already understand, and what they are ready to learn. However, rick tasks share several features that can guide you when selecting and writing problems for your students:
1. Rich tasks are open-ended, encouraging multiple solutions.
2. They allow students to struggle and make sense of important mathematical ideas.
3. They require justification and the use of other mathematical practices.
In life, mathematics is messy and full of authentic questions and choices. This is what rich math looks like, the ordinary and yet complex problems we face every day as we navigate the real world. Our challenge as teachers is to bring this lively, interconnected, dynamic experience of mathematics to students through rich tasks.
••• | 528 | 2,730 | {"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-2024-26 | latest | en | 0.961788 |
https://www.myguruedge.com/our-thinking/gmat-blog/gmat-tutor-tips-gmat-time-management-part-2 | 1,709,410,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00632.warc.gz | 893,779,328 | 16,379 | # GMAT & MBA Admissions Blog
In our previous article on effective GMAT time management, we covered how the GMAT is scored. We noted that due to its adaptive nature, effective time management is critical. We also discussed how the test is not primarily about academic skills, but rather the ability to appy those skills and use critical reasoning and logic to allocate your time and arrive at the best answer. Hopefully, that created a foundational of effecive GMAT time management. In this article, we’ll more specifically discuss how to manage your time when answering questions on the GMAT test itself.
To score well on the GMAT, you should also intentionally build not only your reading speed but your sense of time. You can use various timing apps, including the timer and lap function on your cell phone, to build an internal clock. By setting a timer to go off at one-minute intervals while doing the independent reading we mention above, you can condition your inner clock to ding at one-minute intervals, which will be extremely useful as you take your official GMAT.
Why is the 1-minute mark important?
For the multiple-choice questions on the VR, QR, and IR sections, 1-minute represents the point at which you should have an informed idea of how difficult a question is, whether you think you’ll be able to arrive at an accurate solution, and how much time it will take.
Generally speaking, if a question takes more than 3 minutes to answer, make an educated guess and move forward.
Time is at too much of a premium to spend more than 3 minutes on any given question, especially since the 2-minute average means any question you spend more than 2 minutes on takes time away from other questions, which will compel you to rush and guess even more in order to finish all the questions on time.
It would be great if GMAT prep were as simple as mastering every single concept. But as we’ve said, the GMAT actually isn’t designed to be 100% mastered. Rather, the test is designed to gauge your decision-making skills while working at the edge of your comprehension, which means you need to study where that edge is for you.
While there are many areas of life where dogged determination is valuable, on the GMAT, your persistence can work against you. By the time you’ve done all your GMAT prep, you should have a solid sense of what kinds of problems give you the most trouble and require the most time for you to answer, if you can even answer them correctly at all. You might be surprised to realize that to achieve a score above 700 on the GMAT it might be more important to learn how to manage your weaknesses by quickly guessing on problems you know are difficult than it is to build your skills so that these types of problems aren’t as difficult. Of course, if you don’t understand probability or ratio problems at all, you’ll get a great return on learning these principles so that you can at least find your way around those types of problems. However, if you can learn to identify that you are facing a very challenging ratio problem and those always seem to trip you up, you can then guess intelligently and move on to problems you have a much better chance of answering correctly.
If and when you encounter a kind of problem you know you tend to struggle with—and you can usually tell if a given problem falls into this category within the first 30 seconds of reading through it—make an educated guess and move on.
## LEARN TO GUESS EFFECTIVELY
Making strategic guesses with limited resources and incomplete information is an integral component of GMAT success and success in other managerial contexts.
There are several strategies you can implement to make stronger guesses on the GMAT.
The first is to eliminate wrong answers before guessing. This may not always be possible or efficient, but you can often quickly get rid of one or more obviously wrong answers even if you don’t know how to solve the problem in its entirety, or if you just don’t have enough time to do so.
After quickly eliminating whatever wrong answers you can, the other strategy you can use is to always choose the same letter to guess whenever you don’t have time to work toward a single solution. This will help distribute your guesses and increase the likelihood of some of them being right. Before you even begin, just pick your favorite letter from A to E and stick with that one for the duration of the test.
This is also an area where knowing yourself comes in handy. By the time you’re ready to take your official test, you should have a sense of how much you tend to struggle with the time limit from your timed practice tests. This should give you an idea of how much guessing you’ll have to do in order to answer all the questions effectively. Let’s say you find you typically only have enough time to fully work through 24 of the 32 Quant problems. This means 1 out of every 4 questions will require an educated guess. Rather than waiting to guess on the final 8 questions, keep your guessing strategies close at hand and be ready to implement a guess on roughly every fourth question as you go, selecting the most difficult ones you encounter.
You should also keep in mind that the GMAT penalizes you more heavily for answering easy questions incorrectly and for answering several consecutive questions incorrectly. Therefore, if you have had to guess on multiple questions in a row, be on the lookout for an easier question, which might be worth investing slightly more time in to try to get a correct solution and to avoid steeper penalties. This means breaking up your guesses effectively.
## FINALLY, WORK WITH AN EXPERT GMAT TUTOR
While there are virtually limitless resources for conducting GMAT prep—from a variety of test-prep texts, many of which specialize in certain sections and problem types, to test prep software and group classes, there is no better way to prep for the test than working one-on-one with an expert tutor. A skilled and experienced GMAT tutor will not only be familiar with all the concepts the GMAT tests—they will also have an intricate understanding of the necessary time management skills you’ll need. And most importantly, expert tutors like MyGuru’s will be able to work closely with you to help you design a personally tailored GMAT prep plan so you can learn to implement those time management skills yourself. | 1,289 | 6,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.965723 |
https://forum.kirupa.com/t/3d-working-out-x-y-z-coordinates/298996 | 1,716,019,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00788.warc.gz | 230,049,638 | 9,430 | # 3D - Working out x,y,z coordinates
I am working on a function that will convert a theoretical 3d point into x and y coordinates on the screen. But I need to know where the 3d point is in relation to the camera.
My question is: How would I work out the position of a theoretical 3d point in relation to the camera’s x, y and z coordinates and the cameras pitch, roll and yaw?
It would be in the form of a function like this:
``````convert3Dto2D = function(pointIn3D:Object) {
//pointIn3D has x, y and z variables
3d.x = ????? //this is the x in relation to the camera
3d.y = ????? //this is the y in relation to the camera
3d.z = ????? //this is the z in relation to the camera
2d.x = //This section I already know but it needs 3d's x, y and z
2d.y = //This section I already know but it needs 3d's x, y and z
return 2d
}
``````
I am using flash 8 by the way. | 250 | 865 | {"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-2024-22 | latest | en | 0.923849 |
https://www.edge.org/response-detail/27098 | 1,719,085,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00623.warc.gz | 663,046,775 | 12,906 | # 2017 : WHAT SCIENTIFIC TERM OR CONCEPT OUGHT TO BE MORE WIDELY KNOWN?
Theoretical Physicist, Caltech; Author, Something Deeply Hidden
Bayes's Theorem
You’re worried that your friend is mad at you. You threw a dinner party and didn’t invite them; it’s just the kind of thing they’d be annoyed about. But you’re not really sure. So you send them a text: “Want to hang out tonight?” Twenty minutes later you receive a reply: “Can’t, busy.” How are we to interpret this new information?
Part of the answer comes down to human psychology, of course. But part of it is a bedrock principle of statistical reasoning, known as Bayes’s Theorem.
We turn to Bayes’s Theorem whenever we’re uncertain about the truth of some proposition, and new information comes to light that affects the probability of that proposition being true. The proposition could be our friend’s feelings, or the outcome of the World Cup, or a presidential election, or a particular theory about what happened in the early universe. In other words: we use Bayes’s Theorem literally all the time. We may or may not use it correctly, but it’s everywhere.
The theorem itself isn’t so hard: the probability that a proposition is true, given some new data, is proportional to the probability it was true before that data came in, times the likelihood of the new data if the proposition were true.
So there are two ingredients. First, the prior probability (or simply “the prior”) is the probability we assign to an idea before we gather any new information. Then, the likelihood of some particular piece of data being collected if the idea is correct (or simply “the likelihood”). Bayes’s theorem says that the relative probabilities for different propositions after we collect some new data is just the prior probabilities times the likelihoods.
Scientists use Bayes’s Theorem in a precise, quantitative way all the time. But the theorem—or really, the idea of “Bayesian reasoning” that underlies it—is ubiquitous. Before you sent your friend a text, you had some idea of how likely it was they were mad at you or not. You had, in other words, a prior for the proposition “mad” and another one for “not mad.” When you received their response, you implicitly did a Bayesian updating on those probabilities. What was the likelihood they would send that response if they were mad, and if they weren’t? Multiply by the appropriate priors, and you can now figure out how likely it is that they’re annoyed with you, given your new information.
Behind this bit of dry statistical logic lurk two enormous, profound, worldview-shaping ideas.
One is the very notion of a prior probability. Whether you admit it or not, no matter what data you have, you implicitly have a prior probability for just about every proposition you can think of. If you say, “I have no idea whether that’s true or not,” you’re really just saying, “My prior is 50%.” And there is no objective, cut-and-dried procedure for setting your priors. Different people can dramatically disagree. To one, a photograph that looks like a ghost is incontrovertible evidence for life after death; to another, it’s much more likely to be fake. Given an unlimited amount of evidence and perfect rationality, we should all converge to similar beliefs no matter what priors we start with—but neither evidence nor rationality are perfect or unlimited.
The other big idea is that your degree of belief in an idea should never go all the way to either zero or one. It’s never absolutely impossible to gather a certain bit of data, no matter what the truth is—even the most rigorous scientific experiment is prone to errors, and most of our daily data-collecting is far from rigorous. That’s why science never “proves” anything; we just increase our credences in certain ideas until they are almost (but never exactly) 100%. Bayes’s Theorem reminds us that we should always be open to changing our minds in the face of new information, and tells us exactly what kind of new information we would need. | 865 | 4,014 | {"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} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.919185 |
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Explorer
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Registered: 11-21-2013
## Synchronization stage of FIFO IP with independent clock
Hi, Dear All,
I generated an FIFO with independent clock from the IP catalog, and it has a option of selecting synchronization stage from 2 up to N (N>2).
I am using this FIFO with the same data width, but with different clocks. From what i understand, this value i should choose for this synchronization stage is related to the differences of clocks.
So what does it actually mean?
for synchronization stage 2, does it mean that after writing data, you can read the same data after 2 slower clock cycles?
Historian
3,333 Views
Registered: 01-23-2009
## Re: Synchronization stage of FIFO IP with independent clock
When a signal is sampled by an asynchronous clock, the flip-flop doing the sampling may go metastable; taking a value that is neither a one nor a zero. When that occurs, it takes a probabilistic amount of time for the metastability to resolve.
Using a metastable signal in your logic can cause system failure. So you have to protect your system against the metastable event. This is done by providing "time" in the form of a number of back to back synchronization flip-flops. Even if the first flip-flop goes metastable, by the time the second flip-flop samples the output of that flip-flop, there is a probability that the metastability will have resolved, and hence the output of the second flip-flop will be stable.
Depending on clock frequencies, the number of signal changes per second, the structure of the flip-flop and a few other things, the time provided by two back to back flip-flops may not be sufficient to reduce the probability of metastability of the final output to a level that is acceptable for your system. If it isn't sufficient, then you can increase the number of back to back flip-flops to 3 or 4 or even more - all to get the mean time between failures (MTBF) high enough to be acceptable for your system.
The consequence of adding metastability flip-flops is that it does increase the latency of the clock crossing. In a FIFO, this means a longer time between the push (wr_en) and the deassertion of empty, as well as between the pop (rd_en) and the deassertion of full. The time of these conditions will be longer than the length of the metastability chain, since there is some additional logic outside the metastability chain as well, but each additional FF will increase the time.
Vivado has the ability to analyze the MTBF of synchronization circuits (in UltraScale devices and beyond) using the report_synchronizer_mtbf command. If you annotate the synchronizer activity properly, you can get the MTBF number from the tool and detetmine if 2 or 3 or 4 FFs is sufficient.
Since the FIFO is being implemented from an IP core, the tool will manage the application of the ASYNC_REG property to these flip-flops as well as the clock crossing constraints that are required.
Avrum | 716 | 3,253 | {"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-2019-18 | latest | en | 0.885854 |
https://www.saralstudy.com/study-eschool-ncertsolution/mathematics/relations-and-functions/994-show-that-the-relation-r-in-the-set-a-1-2-3 | 1,632,640,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00146.warc.gz | 962,856,055 | 11,881 | CBSE Sample Papers NCERT Books
Question 8
# Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
A = {1, 2, 3, 4, 5}
R = { (a,b) ; |a – b| is even}
It is clear that for any element a ∈A, we have |a -a| = 0(which is even).
∴R is reflexive.
Let (ab) ∈ R.
=> |a –b| is even.
=> |- (a –b)| = |b - a| is also even.
=> (b, a) ∈ R is even.
A = {1, 2, 3, 4, 5}
R = { (a, b) : | a – b| is even}
It is clear that for any element a ∈A, we have |a - a | = 0 (which is even).
∴R is reflexive.
Let (ab) ∈ R.
⇒ |a –b| is even.
⇒ |- (a –b)| = |b - a| is also even.
⇒ (b, a) ∈ R is even.
∴R is symmetric.
Now, let (ab) ∈ R and (bc) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (ac) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
∴R is symmetric.
Now, let (ab) ∈ R and (bc) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (ac) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
## Recently Viewed Questions of Class 12 Mathematics
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• NCERT Chapter | 942 | 2,766 | {"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} | 4.59375 | 5 | CC-MAIN-2021-39 | longest | en | 0.878075 |
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# Bucket argument
Isaac Newton's rotating bucket argument (also known as "Newton's Bucket") attempts to demonstrate that true rotational motion cannot be defined as the relative rotation of the body with respect to the immediately surrounding bodies. It is one of five arguments from the "properties, causes, and effects" of true motion and rest that support his contention that, in general, true motion and rest cannot be defined as special instances of motion or rest relative to other bodies, but instead can be defined only by reference to absolute space. Alternatively, these experiments provide an operational definition of what is meant by "absolute rotation", and do not pretend to address the question of "rotation relative to what?".
## Background
These arguments, and a discussion of the distinctions between absolute and relative time, space, place and motion, appear in a Scholium at the very beginning of his great work, The Mathematical Principles of Natural Philosophy (1687), which established the foundations of classical mechanics and introduced his law of universal gravitation, which yielded the first quantitatively adequate dynamical explanation of planetary motion. See the Principia on line at Andrew Motte Translation pp. 77-82.
Despite their embrace of the principle of rectilinear inertia and the recognition of the kinematical relativity of apparent motion (which underlies whether the Ptolemaic or the Copernican system is correct), natural philosophers of the seventeenth century continued to consider true motion and rest as physically separate descriptors of an individual body. The dominant view Newton opposed was devised by René Descartes, and was supported (in part) by Gottfried Leibniz. It held that empty space is a metaphysical impossibility because space is nothing other than the extension of matter, or, in other words, that when one speaks of the space between things one is actually making reference to the relationship that exists between those things and not to some entity that stands between them. Concordant with the above understanding, any assertion about the motion of a body boils down to a description over time in which the body under consideration is at t1 found in the vicinity of one group of "landmark" bodies and at some t2 is found in the vicinity of some other "landmark" body or bodies.
Descartes recognized that there would be a real difference, however, between a situation in which a body with movable parts and originally at rest with respect to a surrounding ring was itself accelerated to a certain angular velocity with respect to the ring, and another situation in which the surrounding ring was given a contrary acceleration with respect to the central object. With sole regard to the central object and the surrounding ring, the motions would be indistinguishable from each other assuming that both the central object and the surrounding ring were absolutely rigid objects. However, if neither the central object nor the surrounding ring were absolutely rigid then the parts of one or both of them would tend to fly out from the axis of rotation. People who have noticed a train originally at rest beside them in the railway station pulling away from them, and have soon thereafter noticed with surprise that it is not their train that remains parked at the station, have experienced the basic nature of the Descartes experiment. Frequently these observers first question their initial impressions when they sense g forces from the acceleration of their own train.
For contingent reasons having to do with the Inquisition, Decartes spoke of motion as both absolute and relative. However, his real position was that motion is absolute.
A contrasting position was taken by Mach, who contended that all motion was relative.
## The argument
Newton discusses a bucket filled with water hung by a cord. If the cord is twisted up tightly on itself and then the bucket is released, it begins to spin rapidly, not only with respect to the experimenter, but also in relation to the water it contains. (This situation would correspond to diagram B above.)
Although the relative motion at this stage is the greatest, the surface of the water remains flat, indicating that the parts of the water have no tendency to recede from the axis of relative motion, despite proximity to the pail. Eventually, as the cord continues to unwind, the surface of the water assumes a concave shape as it acquires the motion of the bucket spinning relative to the experimenter. This concave shape shows that the water is rotating, that despite the fact that the water is at rest relative to the pail. In other words, it is not the relative motion of the pail and water that causes concavity of the water, contrary to the idea that motions can only be relative, and that there is no absolute motion. (This situation would correspond to diagram D.) Possibly the concavity of the water shows rotation relative to something else: say absolute space? The argument is incomplete, as it limits the participants relevant to the experiment to only the pail and the water, which has not been established. In fact, the concavity of the water clearly involves gravitational attraction, and by implication the Earth also is a participant. Here is a critique due to Mach:
In the 1846 Andrew Motte translation of Newton's words:
## Rotating spheres
Newton remained concerned to address the problem of how it is that we can experimentally determine the true motions of bodies in light of the fact that absolute space is not something that can be perceived. Such determination, he says, can be accomplished by observing the causes of motion (that is, forces) and not simply the apparent motions of bodies relative to one another (as in the bucket experiment). As an example where causes can be observed, if two globes, floating in space, are connected by a cord, measuring the amount of tension in the cord, with no other clues to assess the situation, alone suffices to indicate how fast the two objects are revolving around the common center of mass. (This experiment involves observation of a force, the tension). Also, the sense of the rotation —whether it is in the clockwise or the counter-clockwise direction— can be discovered by applying forces to opposite faces of the globes and ascertaining whether this leads to an increase or a decrease in the tension of the cord (again involving a force). Alternatively, the sense of the rotation can be determined by measuring the apparent motion of the globes with respect to a system background bodies that, according to the preceding methods, have been established already not to be in a state of rotation, as an example from Newton's time, the fixed stars.
In the 1846 Andrew Motte translation of Newton's words:
To summarize this proposal, here is a quote from Born:
Mach took some issue with the argument, pointing out that the rotating sphere experiment could never be done in an empty universe, where possibly Newton's laws do not apply, so the experiment really only shows what happens when the spheres rotate in our universe, and therefore, for example, may indicate only rotation relative to the entire mass of the universe. An interpretation that avoids this conflict is to say that the rotating spheres experiment does not really define rotation relative to anything in particular (for example, absolute space or fixed stars); rather the experiment is an operational definition of what is meant by the motion called absolute rotation.
## Bibliography
Brian Greene (2004). The Fabric of the Cosmos: space, time, and the texture of reality. A A Knopf. | 1,499 | 7,671 | {"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} | 2.671875 | 3 | CC-MAIN-2015-22 | latest | en | 0.937581 |
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File trac_7575.patch, 65.8 KB (added by Robert Miller, 13 years ago)
depends on #8184, #8155 (and possibly #8124)
• ## new file sage/schemes/elliptic_curves/BSD.py
# HG changeset patch
# User Robert L. Miller <rlm@rlmiller.org>
# Date 1263943454 28800
# Node ID 8a90b87fca0ce33dded2bd986d413441ac42ff0f
# Parent 1ca8aa94810aa7b13cd1671e651b3ed6a365db76
#7575: introduce BSD.py module, catch errors, fix height bounds in heegner_index_bound, mwrank interface & doc fixes
diff -r 1ca8aa94810a -r 8a90b87fca0c sage/schemes/elliptic_curves/BSD.py
- # -*- coding: utf-8 -*- #import ell_point #import formal_group #import ell_torsion #from ell_generic import EllipticCurve_generic, is_EllipticCurve #from ell_number_field import EllipticCurve_number_field #import sage.groups.all import sage.rings.arith as arith import sage.rings.all as rings from sage.rings.all import ZZ def prove_BSD(self, verbosity=0, simon=False, proof=None, secs_hi=30): r""" Attempts to prove the Birch and Swinnerton-Dyer conjectural formula for E, returning a list of primes p for which this function fails to prove BSD(E,p). Here, BSD(E,p) is the statement: "the Birch and Swinnerton-Dyer formula holds up to a rational number coprime to p." INPUT: - verbosity - int, how much information about the proof to print. - 0 - print nothing - 1 - print sketch of proof - 2 - print information about remaining primes - simon - bool (default False), whether to use two_descent or simon_two_descent at p=2. - proof - bool or None (default: None, see proof.elliptic_curve or sage.structure.proof). If False, this function just immediately returns the empty list. - secs_hi - maximum number of seconds to try to compute the Heegner index before switching over to trying to compute the Heegner index bound. (Rank 0 only!) NOTE: When printing verbose output, phrases such as "by Mazur" are referring to the following list of papers: REFERENCES: .. [Cha] B. Cha. Vanishing of some cohomology goups and bounds for the Shafarevich-Tate groups of elliptic curves. J. Number Theory, 111:154- 178, 2005. .. [Jetchev] D. Jetchev. Global divisibility of Heegner points and Tamagawa numbers. Compos. Math. 144 (2008), no. 4, 811--826. .. [Kato] K. Kato. p-adic Hodge theory and values of zeta functions of modular forms. Astérisque, (295):ix, 117-290, 2004. .. [Kolyvagin] V. A. Kolyvagin. On the structure of Shafarevich-Tate groups. Algebraic geometry, 94--121, Lecture Notes in Math., 1479, Springer, Berlin, 1991. .. [LumStein] A. Lum, W. Stein. Verification of the Birch and Swinnerton-Dyer Conjecture for Elliptic Curves with Complex Multiplication (unpublished) .. [Mazur] B. Mazur. Modular curves and the Eisenstein ideal. Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 33--186 (1978). .. [Rubin] K. Rubin. The "main conjectures" of Iwasawa theory for imaginary quadratic fields. Invent. Math. 103 (1991), no. 1, 25--68. .. [SteinWuthrich] W. Stein and C. Wuthrich. Computations about Tate-Shafarevich groups using Iwasawa theory. http://wstein.org/papers/shark, February 2008. .. [SteinEtAl] G. Grigorov, A. Jorza, S. Patrikis, W. Stein, C. Tarniţǎ. Computational verification of the Birch and Swinnerton-Dyer conjecture for individual elliptic curves. Math. Comp. 78 (2009), no. 268, 2397--2425. EXAMPLES:: sage: EllipticCurve('11a').prove_BSD(verbosity=2) p = 2: True by 2-descent True for p not in {2, 5} by Kolyvagin. True for p=5 by Mazur [] sage: EllipticCurve('14a').prove_BSD(verbosity=2) p = 2: True by 2-descent True for p not in {2, 3} by Kolyvagin. Remaining primes: p = 3: reducible, not surjective, good ordinary, divides a Tamagawa number [3] sage: EllipticCurve('14a').prove_BSD(simon=True) [3] A rank two curve:: sage: E = EllipticCurve('389a') We know nothing with proof=True:: sage: E.prove_BSD() Set of all prime numbers: 2, 3, 5, 7, ... We (think we) know everything with proof=False:: sage: E.prove_BSD(proof=False) [] A curve of rank 0 and prime conductor:: sage: E = EllipticCurve('19a') sage: E.prove_BSD(verbosity=2) p = 2: True by 2-descent True for p not in {2, 3} by Kolyvagin. True for p=3 by Mazur [] sage: E = EllipticCurve('37a') sage: E.rank() 1 sage: E._EllipticCurve_rational_field__rank {True: 1} sage: E.analytic_rank = lambda : 0 sage: E.prove_BSD() Traceback (most recent call last): ... RuntimeError: It seems that the rank conjecture does not hold for this curve (Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field)! This may be a counterexample to BSD, but is more likely a bug. We test the consistency check for the 2-part of Sha:: sage: E = EllipticCurve('37a') sage: S = E.sha(); S Shafarevich-Tate group for the Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field sage: S.an = lambda : 4 sage: E.prove_BSD() Traceback (most recent call last): ... RuntimeError: Two descent => ord_2(#Sha[2]) = 0, but ord_2(#sha_an) = 2. An example with a Tamagawa number at 5:: sage: E = EllipticCurve('123a1') sage: E.prove_BSD(verbosity=2) p = 2: True by 2-descent True for p not in {2, 5} by Kolyvagin. Remaining primes: p = 5: reducible, not surjective, good ordinary, divides a Tamagawa number [5] A curve for which 3 divides the order of the Shafarevich-Tate group:: sage: E = EllipticCurve('681b') sage: E.prove_BSD(verbosity=2) # long time p = 2: True by 2-descent True for p not in {2, 3} by Kolyvagin. ALERT: p = 3 left in Kolyvagin bound 0 <= ord_p(#Sha) <= 2 ord_p(#Sha_an) = 2 Remaining primes: p = 3: irreducible, surjective, non-split multiplicative [3] A curve for which we need to use heegner_index_bound:: sage: E = EllipticCurve('198b') sage: E.prove_BSD(verbosity=1, secs_hi=1) p = 2: True by 2-descent Timeout stopped Heegner index computation... Proceeding to use heegner_index_bound instead. True for p not in {2, 3} by Kolyvagin. [3] TESTS: This was fixed by trac #8184 and #7575:: sage: EllipticCurve('438e1').prove_BSD(verbosity=1) p = 2: True by 2-descent True for p not in {2} by Kolyvagin. [] :: sage: E = EllipticCurve('960d1') sage: E.prove_BSD(verbosity=1) p = 2: True by 2-descent Timeout stopped Heegner index computation... Proceeding to use heegner_index_bound instead. True for p not in {2} by Kolyvagin. [] """ if proof is None: from sage.structure.proof.proof import get_flag proof = get_flag(proof, "elliptic_curve") else: proof = bool(proof) if not proof: return [] two_tor_rk = self.two_torsion_rank() Sha = self.sha() sha_an = Sha.an() if simon: rank_lower_bd, two_sel_rk_bd = self.simon_two_descent()[:2] two_sel_rk_bd -= two_tor_rk sha2_lower_bd = 0 else: MWRC = self.mwrank_curve() two_sel_rk_bd = MWRC.selmer_rank() two_sel_rk_bd -= two_tor_rk rank_lower_bd = MWRC.rank() sha2_lower_bd = two_sel_rk_bd - MWRC.rank_bound() sha2_upper_bd = two_sel_rk_bd - rank_lower_bd # note: two_sel_rk_bd will include sha rank = None if rank_lower_bd > 1: # We do not know BSD(E,p) for even a single p, since it's # an open problem to show that L^r(E,1)/(Reg*Omega) is # rational for any curve with r >= 2. from sage.sets.all import Primes return Primes() if sha2_upper_bd == sha2_lower_bd: rank = rank_lower_bd if sha_an.ord(2) != sha2_lower_bd: raise RuntimeError("Two descent => ord_2(#Sha[2]) = %d, but ord_2(#sha_an) = %d."%(sha2_lower_bd,sha_an.ord(2))) if verbosity > 0: print 'p = 2: True by 2-descent' two_proven = True else: if sha2_upper_bd < sha2_lower_bd: raise RuntimeError("Apparent contradiction: ord_2(#Sha[2]_an) == %d, rank(2-Sel)-rank(2-tor) = %d, rank >= %d, ord_2(#Sha[2]) >= %d"%(sha_an.ord(2),two_sel_rk_bd,rank_lower_bd,sha2_lower_bd)) if two_sel_rk_bd - rank_lower_bd == sha_an.ord(2): if verbosity > 0: print 'p = 2: ord_2(#Sha[2]_an) == %d >= ord_2(#Sha[2]).'%sha_an.ord(2) elif two_sel_rk_bd - rank_lower_bd > sha_an.ord(2): if verbosity > 0: print 'p = 2: ord_2(#Sha[2]_an) == %d, and ord_2(#Sha[2]) <= %d.'%(sha_an.ord(2),two_sel_rk_bd - rank_lower_bd) else: raise RuntimeError("Apparent contradiction: ord_2(#Sha[2]_an) == %d, rank(2-Sel)-rank(2-tor) = %d, rank >= %d"%(sha_an.ord(2),two_sel_rk_bd,rank_lower_bd)) two_proven = False if verbosity > 0: print 'Looking rank up in database...' rank = self.rank(use_database=True, only_use_mwrank=False) if rank != self.analytic_rank(): raise RuntimeError("It seems that the rank conjecture does not hold for this curve (%s)! This may be a counterexample to BSD, but is more likely a bug."%(self)) # We replace self by the optimal curve, which we can do since # truth of BSD(E,p) is invariant under isogeny. self = self.optimal_curve() N = self.conductor() # reduce set of remaining primes to a finite set import signal remaining_primes = [] kolyvagin_primes = [] heegner_index = None if self.rank() == 0: try: old_alarm = signal.alarm(secs_hi) old_alarm_set = (old_alarm != 0) for D in self.heegner_discriminants_list(10): I = None while I is None: dsl=15 try: I = self.heegner_index(D, descent_second_limit=dsl) except RuntimeError as err: if err.args[0][-33:] == 'Generators not provably computed.': dsl += 1 else: raise RuntimeError(err) J = I.is_int() if J[0] and J[1]>0: I = J[1] else: J = (2*I).is_int() if J[0] and J[1]>0: I = J[1] else: I = None if I is not None: if heegner_index is None: heegner_index = I break # no big long loops just yet... old_alarm_sub = signal.alarm(0) if old_alarm_set: old_alarm -= old_alarm_sub except KeyboardInterrupt: if signal.alarm(0)==0: print 'Timeout stopped Heegner index computation...' print 'Proceeding to use heegner_index_bound instead.' else: raise KeyboardInterrupt except RuntimeError as err: if err.args[0][:37] == 'End Of File (EOF) in read_nonblocking': print 'Computing Heegner index failed due to mwrank interface: see #7575.' else: raise RuntimeError(err) old_alarm_sub = signal.alarm(0) if old_alarm_set: old_alarm -= old_alarm_sub if old_alarm <= 0: raise KeyboardInterrupt signal.alarm(old_alarm) if old_alarm_set: # in case alarm was already set... if old_alarm <= 0: raise KeyboardInterrupt signal.alarm(old_alarm) if heegner_index is None: for D in self.heegner_discriminants_list(100): max_height = 12 heegner_primes = -1 while heegner_primes == -1: max_height += 1 heegner_primes, _ = self.heegner_index_bound(D, max_height=max_height) if isinstance(heegner_primes, list): break if not isinstance(heegner_primes, list): raise RuntimeError("Tried 100 Heegner discriminants, and heegner_index_bound failed each time.") if 2 in heegner_primes: heegner_primes.remove(2) else: heegner_primes = [p for p in arith.prime_divisors(heegner_index) if p!=2] else: # rank 1 for D in self.heegner_discriminants_list(10): I = self.heegner_index(D) J = I.is_int() if J[0] and J[1]>0: I = J[1] else: J = (2*I).is_int() if J[0] and J[1]>0: I = J[1] else: continue heegner_index = I break heegner_primes = [p for p in arith.prime_divisors(heegner_index) if p!=2] if self.has_cm(): # ensure that CM is by a maximal order non_max_j_invs = [-12288000, 54000, 287496, 16581375] if self.j_invariant() in non_max_j_invs: for E in self.isogeny_class()[0]: if E.j_invariant() not in non_max_j_invs: Sha = E.sha() sha_an = Sha.an() if verbosity > 0: print 'CM by non maximal order: switching curves' break else: E = self if E.analytic_rank() == 0: if verbosity > 0: print 'p >= 5: true by Rubin' remaining_primes.append(3) else: K = rings.QuadraticField(E.cm_discriminant(), 'a') D_K = K.disc() D_E = E.discriminant() if len(K.factor(3)) == 1: # 3 does not split in K remaining_primes.append(3) for p in arith.prime_divisors(D_K): if p >= 5: remaining_primes.append(p) for p in arith.prime_divisors(D_E): if p >= 5 and D_K%p and len(K.factor(p)) == 1: # p is inert in K remaining_primes.append(p) for p in heegner_primes: if p >= 5 and D_E%p != 0 and D_K%p != 0 and len(K.factor(p)) == 1: # p is good for E and inert in K kolyvagin_primes.append(p) assert sha_an in ZZ and sha_an > 0 for p in arith.prime_divisors(sha_an): if p >= 5 and D_K%p != 0 and len(K.factor(p)) == 1: if E.is_good(p): if verbosity > 2 and p in heegner_primes and heegner_index is None: print 'ALERT: Prime p (%d) >= 5 dividing sha_an, good for E, inert in K, in heegner_primes, should not divide the actual Heegner index' # Note that the following check is not entirely # exhaustive, in case there is a p not dividing # the Heegner index in heegner_primes, # for which only an outer bound was computed if p not in heegner_primes: raise RuntimeError("p = %d divides sha_an, is of good reduction for E, inert in K, and does not divide the Heegner index. This may be a counterexample to BSD, but is more likely a bug. %s"%(p,self)) if verbosity > 0: print 'True for p not in {%s} by Kolyvagin (via Stein & Lum -- unpublished) and Rubin.'%str(list(set(remaining_primes).union(set(kolyvagin_primes))))[1:-1] else: # no CM E = self # do some tricks to get to a finite set without calling bound_kolyvagin remaining_primes = [p for p, reason in E.non_surjective()] for p in heegner_primes: if p not in remaining_primes: remaining_primes.append(p) assert sha_an in ZZ and sha_an > 0 for p in arith.prime_divisors(sha_an): if p not in remaining_primes: remaining_primes.append(p) if 2 in remaining_primes: remaining_primes.remove(2) if verbosity > 0: print 'True for p not in {' + str([2]+list(remaining_primes))[1:-1] + '} by Kolyvagin.' primes_to_remove = [] for p in remaining_primes: if E.is_surjective(p)[0] and not E.has_additive_reduction(p): if E.has_nonsplit_multiplicative_reduction(p): if E.rank() > 0: continue if p==3: if (not (E.is_ordinary(p) and E.is_good(p))) and (not E.has_split_multiplicative_reduction(p)): continue if E.rank() > 0: continue if verbosity > 1: print 'p = %d: Trying p_primary_bound'%p p_bound = Sha.p_primary_bound(p) if sha_an.ord(p) == 0 and p_bound == 0: if verbosity > 0: print 'True for p=%d by Stein-Wuthrich.'%p primes_to_remove.append(p) else: print 'Analytic %d-rank is '%p + str(sha_an.ord(p)) + ', actual %d-rank is at most %d.'%(p, p_bound) print ' by Stein-Wuthrich.\n' for p in primes_to_remove: remaining_primes.remove(p) kolyvagin_primes = [] for p in remaining_primes: if E.is_surjective(p)[0]: kolyvagin_primes.append(p) for p in kolyvagin_primes: remaining_primes.remove(p) # apply other hypotheses which imply Kolyvagin's bound holds bounded_primes = [] D_K = rings.QuadraticField(D, 'a').disc() assert 2 not in remaining_primes # Cha's hypothesis for p in remaining_primes: if D_K%p != 0 and N%(p**2) != 0 and E.is_irreducible(p): if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Cha.'%p kolyvagin_primes.append(p) # Stein et al. if not E.has_cm(): L = arith.lcm([F.torsion_order() for F in E.isogeny_class()[0]]) for p in remaining_primes: if p in kolyvagin_primes: continue if L%p != 0: if len(arith.prime_divisors(D_K)) == 1: if D_K%p != 0: if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Stein et al.'%p kolyvagin_primes.append(p) else: if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Stein et al.'%p kolyvagin_primes.append(p) for p in kolyvagin_primes: if p in remaining_primes: remaining_primes.remove(p) prime_bounds = [] # apply Kolyvagin's bound primes_to_remove = [] for p in kolyvagin_primes: if sha_an.ord(p) == 0 and p not in heegner_primes: if verbosity > 0: print 'True for p = %d by Kolyvagin bound.'%p primes_to_remove.append(p) continue if heegner_index is not None: # p must divide heegner_index ord_p_bound = 2*heegner_index.ord(p) # Here Jetchev's results apply. m_max = max([E.tamagawa_number(q).ord(p) for q in N.prime_divisors()]) if m_max > 0 and verbosity > 0: print 'Jetchev\'s results apply (at p = %d) with m_max ='%p, m_max ord_p_bound -= 2*m_max if ord_p_bound == 0: if sha_an.ord(p) != 0: raise RuntimeError("p = %d: ord_p_bound == 0, but sha_an.ord(p) == %d. This appears to be a counterexample to BSD, but is more likely a bug."%(p,sha_an.ord(p))) if verbosity > 0: print 'True for p = %d by Kolyvagin bound.'%p primes_to_remove.append(p) continue elif p not in heegner_primes: ord_p_bound = 0 else: from sage.rings.infinity import Infinity ord_p_bound = Infinity if verbosity > 0: print 'p = %d may divide the Heegner index, for which only a bound was computed.'%p if verbosity > 0: print 'ALERT: p = %d left in Kolyvagin bound'%p print ' 0 <= ord_p(#Sha) <=', ord_p_bound print ' ord_p(#Sha_an) =', sha_an.ord(p) for p in primes_to_remove: kolyvagin_primes.remove(p) remaining_primes = list( set(remaining_primes).union(set(kolyvagin_primes)) ) # Kato's bound if rank == 0 and not E.has_cm(): L_over_Omega = E.lseries().L_ratio() kato_primes = Sha.bound_kato() primes_to_remove = [] for p in remaining_primes: if p not in kato_primes: if verbosity > 0: print 'Kato further implies that #Sha[%d] is trivial.'%p primes_to_remove.append(p) if p not in [2,3] and N%p != 0: if E.is_surjective(p)[0]: if verbosity > 1: print 'Kato might apply nontrivially for %d'%p # ordp(sha) <= ordp(L_over_omega) for p in primes_to_remove: remaining_primes.remove(p) # Mazur if N.is_prime(): for p in remaining_primes: if E.is_reducible(p): remaining_primes.remove(p) if verbosity > 0: print 'True for p=%s by Mazur'%p if two_proven is False: remaining_primes.append(2) # print some extra information remaining_primes.sort() if verbosity > 1: if len(remaining_primes) > 0: print 'Remaining primes:' for p in remaining_primes: s = 'p = ' + str(p) + ': ' if not E.is_reducible(p): s += 'ir' s += 'reducible, ' if not E.is_surjective(p)[0]: s += 'not ' s += 'surjective, ' a_p = E.an(p) if E.is_good(p): if a_p%p != 0: s += 'good ordinary' else: s += 'good, non-ordinary' else: assert E.is_minimal() if a_p == 0: s += 'additive' elif a_p == 1: s += 'split multiplicative' elif a_p == -1: s += 'non-split multiplicative' if E.tamagawa_product()%p==0: s += ', divides a Tamagawa number' print s return remaining_primes
• ## sage/schemes/elliptic_curves/ell_rational_field.py
diff -r 1ca8aa94810a -r 8a90b87fca0c sage/schemes/elliptic_curves/ell_rational_field.py
a from sage.rings.padics.precision_error import PrecisionError import heegner import BSD from lseries_ell import Lseries_ell OUTPUT: Nothing - nothing is returned (though much is printed unless verbose=False) EXAMPLES:: sage: E=EllipticCurve('37a1') sage: E.two_descent(verbose=False) # no output """ self.mwrank_curve().two_descent(verbose, selmer_only, first_limit, second_limit, n_aux, second_descent) Returns True if the descent succeeded, i.e. if the lower bound and the upper bound for the rank are the same. In this case, generators and the rank are cached. A return value of False indicates that either rational points were not found, or that Sha[2] is nontrivial and mwrank was unable to determine this for sure. EXAMPLES:: sage: E=EllipticCurve('37a1') sage: E.two_descent(verbose=False) True """ misc.verbose("Calling mwrank C++ library.") C = self.mwrank_curve() C.two_descent(verbose, selmer_only, first_limit, second_limit, n_aux, second_descent) if C.certain(): self.__gens[True] = [self.point(x, check=True) for x in C.gens()] self.__gens[True].sort() self.__rank[True] = len(self.__gens[True]) return C.certain() #################################################################### # Etc. def simon_two_descent(self, verbose=0, lim1=5, lim3=50, limtriv=10, maxprob=20, limbigprime=30): r""" Given a curve with no 2-torsion, computes (probably) the rank of the Mordell-Weil group, with certainty the rank of the 2-Selmer group, and a list of independent points on the curve. Computes a lower bound for the rank of the Mordell-Weil group E(Q), the rank of the 2-Selmer group, and a list of independent points on E(Q)/2E(Q). INPUT: OUTPUT: - integer - "probably" the rank of self - integer - lower bound on the rank of self - integer - the 2-rank of the Selmer group - list - list of independent points on the curve. quotient E(Q)/2E(Q). IMPLEMENTATION: Uses Denis Simon's GP/PARI scripts from """ t = simon_two_descent(self, verbose=verbose, lim1=lim1, lim3=lim3, limtriv=limtriv, maxprob=maxprob, limbigprime=limbigprime) prob_rank = rings.Integer(t[0]) rank_low_bd = rings.Integer(t[0]) two_selmer_rank = rings.Integer(t[1]) prob_gens = [self(P) for P in t[2]] return prob_rank, two_selmer_rank, prob_gens gens_mod_two = [self(P) for P in t[2]] if rank_low_bd == two_selmer_rank - self.two_torsion_rank(): gens = [P for P in gens_mod_two if P.additive_order() != 2] self.__gens[True] = gens self.__gens[True].sort() self.__rank[True] = len(self.__gens[True]) return rank_low_bd, two_selmer_rank, gens_mod_two two_descent_simon = simon_two_descent def rank(self, use_database=False, verbose=False, only_use_mwrank=True, algorithm='mwrank_shell', algorithm='mwrank_lib', proof=None): """ Return the rank of this elliptic curve, assuming no conjectures. if algorithm == 'mwrank_lib': misc.verbose("using mwrank lib") C = self.mwrank_curve() if self.is_integral(): E = self else: E = self.integral_model() C = E.mwrank_curve() C.set_verbose(verbose) r = C.rank() if not C.certain(): del self.__mwrank_curve raise RuntimeError, "Unable to compute the rank with certainty (lower bound=%s). This could be because Sha(E/Q)[2] is nontrivial."%C.rank() + "\nTrying calling something like two_descent(second_limit=13) on the curve then trying this command again. You could also try rank with only_use_mwrank=False." if C.certain(): proof = True else: if proof: print "Unable to compute the rank with certainty (lower bound=%s)."%C.rank() print "This could be because Sha(E/Q)[2] is nontrivial." print "Try calling something like two_descent(second_limit=13) on the" print "curve then trying this command again. You could also try rank" print "with only_use_mwrank=False." del E.__mwrank_curve raise RuntimeError, 'Rank not provably correct.' else: misc.verbose("Warning -- rank not proven correct", level=1) self.__rank[proof] = r elif algorithm == 'mwrank_shell': misc.verbose("using mwrank shell") return self.__rank[proof] def gens(self, verbose=False, rank1_search=10, algorithm='mwrank_shell', algorithm='mwrank_lib', only_use_mwrank=True, proof = None, use_database = True): use_database = True, descent_second_limit=12): """ Compute and return generators for the Mordell-Weil group E(Q) *modulo* torsion. HINT: If you would like to control the height bounds used in the 2-descent, first call the two_descent function with those height bounds. However that function, while it displays a lot of output, returns no values. TODO: Allow passing of command-line parameters to mwrank. .. warning:: If the program fails to give a provably correct result, it - use_database - bool (default True) if True, attempts to find curve and gens in the (optional) database - descent_second_limit - (default: 16)- used in 2-descent OUTPUT: # end if (not_use_mwrank) if algorithm == "mwrank_lib": misc.verbose("Calling mwrank C++ library.") C = self.mwrank_curve(verbose) if not self.is_integral(): xterm = 1; yterm = 1 ai = self.a_invariants() for a in ai: if not a.is_integral(): for p, _ in a.denom().factor(): e = min([(ai[i].valuation(p)/[1,2,3,4,6][i]) for i in range(5)]).floor() ai = [ai[i]/p**(e*[1,2,3,4,6][i]) for i in range(5)] xterm *= p**(2*e) yterm *= p**(3*e) E = constructor.EllipticCurve(list(ai)) else: E = self; xterm = 1; yterm = 1 C = E.mwrank_curve(verbose) if not (verbose is None): C.set_verbose(verbose) C.two_descent(verbose=verbose, second_limit=descent_second_limit) G = C.gens() if proof is True and C.certain() is False: del self.__mwrank_curve raise RuntimeError, "Unable to compute the rank, hence generators, with certainty (lower bound=%s, generators found=%s). This could be because Sha(E/Q)[2] is nontrivial."%(C.rank(),G) + \ "\nTrying calling something like two_descent(second_limit=13) on the curve then trying this command again." "\nTry increasing descent_second_limit then trying this command again." else: proof = C.certain() G = [[x*xterm,y*yterm,z] for x,y,z in G] else: # when gens() calls mwrank it passes the command-line # parameter "-p 100" which helps curves with large """ return len(self.gens(proof = proof)) def regulator(self, use_database=True, proof=None, precision=None): def regulator(self, use_database=True, proof=None, precision=None, descent_second_limit=12, verbose=False): """ Returns the regulator of this curve, which must be defined over Q. - precision - int or None (default: None): the precision in bits of the result (default real precision if None) - descent_second_limit - (default: 16)- used in 2-descent - verbose - whether to print mwrank's verbose output EXAMPLES:: # Next we find the gens, taking them from the database if they # are there and use_database is True, else computing them: G = self.gens(proof=proof, use_database=use_database) G = self.gens(proof=proof, use_database=use_database, descent_second_limit=descent_second_limit, verbose=verbose) # Finally compute the regulator of the generators found: - height_limit (float) - bound on naive height (at most 21, - or mwrank overflows - see below) - verbose (bool) - (default: False) OUTPUT: points (list) - list of independent points which generate the subgroup of the Mordell-Weil group generated by the points found and then p-saturated for p20. found and then p-saturated for p \leq 20. .. warning:: height_limit is logarithmic, so increasing by 1 will cause the running time to increase by a factor of approximately 4.5 (=exp(1.5)). The limit of 21 is to prevent overflow, but in any case using height_limit=20 takes rather a long time! IMPLEMENTATION: Uses Cremona's mwrank package. At the heart of this function is Cremona's port of Stoll's ratpoints program (version 1.4). 4.5 (=exp(1.5)). IMPLEMENTATION: Uses Michael Stoll's ratpoints library. EXAMPLES:: sage: E=EllipticCurve('389a1') sage: E.point_search(5, verbose=False) [(0 : -1 : 1), (-1 : 1 : 1)] [(-1 : 1 : 1), (-3/4 : 7/8 : 1)] Increasing the height_limit takes longer, but finds no more points:: sage: E.point_search(10, verbose=False) [(0 : -1 : 1), (-1 : 1 : 1)] [(-1 : 1 : 1), (-3/4 : 7/8 : 1)] In fact this curve has rank 2 so no more than 2 points will ever be output, but we are not using this fact. :: sage: E.saturation(_) ([(0 : -1 : 1), (-1 : 1 : 1)], '1', 0.152460172772408) ([(-1 : 1 : 1), (-3/4 : 7/8 : 1)], '1', 0.152460172772408) What this shows is that if the rank is 2 then the points listed do generate the Mordell-Weil group (mod torsion). Finally, sage: E.rank() 2 """ height_limit = float(height_limit) if height_limit > _MAX_HEIGHT: raise OverflowError, "height_limit (=%s) must be at most %s."%(height_limit,_MAX_HEIGHT) c = self.mwrank_curve() mw = mwrank.mwrank_MordellWeil(c, verbose) mw.search(height_limit, verbose=verbose) v = mw.points() return [self(P) for P in v] from sage.libs.ratpoints import ratpoints from sage.functions.all import exp from sage.rings.arith import GCD H = exp(float(height_limit)) # max(|p|,|q|) <= H, if x = p/q coprime coeffs = [16*self.b6(), 8*self.b4(), self.b2(), 1] points = [] a1 = self.a1() a3 = self.a3() new_H = H*2 # since we change the x-coord by 2 below for X,Y,Z in ratpoints(coeffs, new_H, verbose): if Z == 0: continue z = 2*Z x = X/2 y = (Y/z - a1*x - a3*z)/2 d = GCD((x,y,z)) x = x/d if max(x.numerator().abs(), x.denominator().abs()) <= H: y = y/d z = z/d points.append(self((x,y,z))) return self.saturation(points, verbose=verbose, max_prime=20)[0] def two_torsion_rank(self): r""" x+=1 return ans def prove_BSD(self, verbosity=0, simon=False, proof=None, secs_hi=30): r""" Attempts to prove the Birch and Swinnerton-Dyer conjectural formula for E, returning a list of primes p for which this function fails to prove BSD(E,p). Here, BSD(E,p) is the statement: "the Birch and Swinnerton-Dyer formula holds up to a rational number coprime to p." INPUT: - verbosity - int, how much information about the proof to print. - 0 - print nothing - 1 - print sketch of proof - 2 - print information about remaining primes - simon - bool (default False), whether to use two_descent or simon_two_descent at p=2. - proof - bool or None (default: None, see proof.elliptic_curve or sage.structure.proof). If False, this function just immediately returns the empty list. - secs_hi - maximum number of seconds to try to compute the Heegner index before switching over to trying to compute the Heegner index bound. (Rank 0 only!) NOTE: When printing verbose output, phrases such as "by Mazur" are referring to the following list of papers: REFERENCES: .. [Cha] B. Cha. Vanishing of some cohomology goups and bounds for the Shafarevich-Tate groups of elliptic curves. J. Number Theory, 111:154- 178, 2005. .. [Jetchev] D. Jetchev. Global divisibility of Heegner points and Tamagawa numbers. Compos. Math. 144 (2008), no. 4, 811--826. .. [Kato] K. Kato. p-adic Hodge theory and values of zeta functions of modular forms. Astérisque, (295):ix, 117-290, 2004. .. [Kolyvagin] V. A. Kolyvagin. On the structure of Shafarevich-Tate groups. Algebraic geometry, 94--121, Lecture Notes in Math., 1479, Springer, Berlin, 1991. .. [LumStein] A. Lum, W. Stein. Verification of the Birch and Swinnerton-Dyer Conjecture for Elliptic Curves with Complex Multiplication (unpublished) .. [Mazur] B. Mazur. Modular curves and the Eisenstein ideal. Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 33--186 (1978). .. [Rubin] K. Rubin. The "main conjectures" of Iwasawa theory for imaginary quadratic fields. Invent. Math. 103 (1991), no. 1, 25--68. .. [SteinWuthrich] W. Stein and C. Wuthrich. Computations about Tate-Shafarevich groups using Iwasawa theory. http://wstein.org/papers/shark, February 2008. .. [SteinEtAl] G. Grigorov, A. Jorza, S. Patrikis, W. Stein, C. Tarniţǎ. Computational verification of the Birch and Swinnerton-Dyer conjecture for individual elliptic curves. Math. Comp. 78 (2009), no. 268, 2397--2425. EXAMPLES:: sage: EllipticCurve('11a').prove_BSD(verbosity=2) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2, 5} by Kolyvagin. True for p=5 by Mazur Remaining primes: p = 2: irreducible, surjective, good, non-ordinary [2] sage: EllipticCurve('14a').prove_BSD(verbosity=2) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2, 3} by Kolyvagin. Remaining primes: p = 2: reducible, surjective, non-split multiplicative, divides a Tamagawa number p = 3: reducible, surjective, good ordinary, divides a Tamagawa number [2, 3] A rank two curve:: sage: E = EllipticCurve('389a') We know nothing with proof=True:: sage: E.prove_BSD() Set of all prime numbers: 2, 3, 5, 7, ... We (think we) know everything with proof=False:: sage: E.prove_BSD(proof=False) [] A curve of rank 0 and prime conductor:: sage: E = EllipticCurve('19a') sage: E.prove_BSD(verbosity=2) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2, 3} by Kolyvagin. True for p=3 by Mazur Remaining primes: p = 2: irreducible, surjective, good, non-ordinary [2] sage: E = EllipticCurve('37a') sage: E.rank() 1 sage: E._EllipticCurve_rational_field__rank {True: 1} sage: E._EllipticCurve_rational_field__rank = {True:0} sage: E.prove_BSD() Traceback (most recent call last): ... RuntimeError: It seems that the rank conjecture does not hold for this curve (Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field)! This may be a counterexample to BSD, but is more likely a bug. We check the error message indicating that this code doesn't yet use Simon 2-descent in case of rational 2-torsion (though it could with a little more work):: sage: E = EllipticCurve('14a') sage: E.prove_BSD(simon=True) Traceback (most recent call last): ... RuntimeError: Simon two-descent only valid for curves without two torsion. We test the consistency check for the 2-part of Sha:: sage: E = EllipticCurve('37a') sage: S = E.sha(); S Shafarevich-Tate group for the Elliptic Curve defined by y^2 + y = x^3 - x over Rational Field sage: S.an = lambda : 4 sage: E.prove_BSD() [2] An example with a Tamagawa number at 5:: sage: E = EllipticCurve('123a1') sage: E.prove_BSD(verbosity=2) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2, 5} by Kolyvagin. Remaining primes: p = 2: irreducible, surjective, good, non-ordinary p = 5: reducible, surjective, good ordinary, divides a Tamagawa number [2, 5] A curve for which 3 divides the order of the Shafarevich-Tate group:: sage: E = EllipticCurve('681b') sage: E.prove_BSD(verbosity=2) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2, 3} by Kolyvagin. ALERT: p = 3 left in Kolyvagin bound 0 <= ord_p(#Sha) <= 2 ord_p(#Sha_an) = 2 Remaining primes: p = 2: reducible, surjective, good ordinary, divides a Tamagawa number p = 3: irreducible, surjective, non-split multiplicative [2, 3] A curve for which we need to use heegner_index_bound:: sage: E = EllipticCurve('198b') sage: E.prove_BSD(verbosity=1, secs_hi=1) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank Timeout stopped Heegner index computation... Proceeding to use heegner_index_bound instead. True for p not in {2, 3} by Kolyvagin. [2, 3] TESTS:: sage: E = EllipticCurve('438e1') sage: E.prove_BSD(verbosity=1) p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank True for p not in {2} by Kolyvagin. [2] :: sage: E = EllipticCurve('960d1') sage: E.prove_BSD(verbosity=1) # long time p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank Timeout stopped Heegner index computation... Proceeding to use heegner_index_bound instead. True for p not in {2} by Kolyvagin. [2] """ if proof is None: from sage.structure.proof.proof import get_flag proof = get_flag(proof, "elliptic_curve") else: proof = bool(proof) if not proof: return [] two_tor_rk = self.two_torsion_rank() if simon: if two_tor_rk > 0: raise RuntimeError("Simon two-descent only valid for curves without two torsion.") rank_lower_bd, two_sel_rk = self.simon_two_descent()[:2] if rank_lower_bd == two_sel_rk: rank = rank_lower_bd else: raise RuntimeError("Rank can't be computed precisely using Simon's program.") else: two_sel_rk_bd = self.rank_bound() rank = self.rank() if rank > 1: # We do not know BSD(E,p) for even a single p, since it's # an open problem to show that L^r(E,1)/(Reg*Omega) is # rational for any curve with r >= 2. from sage.sets.all import Primes return Primes() if rank != self.analytic_rank(): raise RuntimeError("It seems that the rank conjecture does not hold for this curve (%s)! This may be a counterexample to BSD, but is more likely a bug."%(self)) # We replace self by the optimal curve, which we can do since # truth of BSD(E,p) is invariant under isogeny. self = self.optimal_curve() Sha = self.sha() sha_an = Sha.an() N = self.conductor() # p = 2 if two_sel_rk_bd > rank: if verbosity > 0: print 'p = 2: mwrank did not achieve a tight bound on the Selmer rank.' two_proven = False elif two_sel_rk_bd < rank: raise RuntimeError("MWrank seems to have computed an incorrect upper bound of %d on the rank."%two_sel_rk_bd) else: # until we can easily access the computed rank of Sha[2]: two_proven = False if verbosity > 0: print 'p = 2: Unverified since it is difficult to access the rank bound for Sha[2] computed by MWrank' # reduce set of remaining primes to a finite set import signal remaining_primes = [] kolyvagin_primes = [] heegner_index = None if self.rank() == 0: try: old_alarm = signal.alarm(secs_hi) old_alarm_set = (old_alarm != 0) for D in self.heegner_discriminants_list(10): I = self.heegner_index(D) J = I.is_int() if J[0] and J[1]>0: I = J[1] else: J = (2*I).is_int() if J[0] and J[1]>0: I = J[1] else: I = None if I is not None: if heegner_index is None: heegner_index = I break # no big long loops just yet... old_alarm_sub = signal.alarm(0) if old_alarm_set: old_alarm -= old_alarm_sub except KeyboardInterrupt: if signal.alarm(0)==0: print 'Timeout stopped Heegner index computation...' print 'Proceeding to use heegner_index_bound instead.' else: raise KeyboardInterrupt if old_alarm_set: # in case alarm was already set... if old_alarm <= 0: raise KeyboardInterrupt signal.alarm(old_alarm) if heegner_index is None: for D in self.heegner_discriminants_list(100): heegner_primes, _ = self.heegner_index_bound(D) if isinstance(heegner_primes, list): break if not isinstance(heegner_primes, list): raise RuntimeError("Tried 100 Heegner discriminants, and heegner_index_bound failed each time.") if 2 in heegner_primes: heegner_primes.remove(2) else: heegner_primes = [p for p in arith.prime_divisors(heegner_index) if p!=2] else: # rank 1 for D in self.heegner_discriminants_list(10): I = self.heegner_index(D) J = I.is_int() if J[0] and J[1]>0: I = J[1] else: J = (2*I).is_int() if J[0] and J[1]>0: I = J[1] else: continue heegner_index = I break heegner_primes = [p for p in arith.prime_divisors(heegner_index) if p!=2] if self.has_cm(): # ensure that CM is by a maximal order non_max_j_invs = [-12288000, 54000, 287496, 16581375] if self.j_invariant() in non_max_j_invs: for E in self.isogeny_class()[0]: if E.j_invariant() not in non_max_j_invs: Sha = E.sha() sha_an = Sha.an() if verbosity > 0: print 'CM by non maximal order: switching curves' break else: E = self if E.analytic_rank() == 0: if verbosity > 0: print 'p >= 5: true by Rubin' remaining_primes.append(3) else: K = rings.QuadraticField(E.cm_discriminant(), 'a') D_K = K.disc() D_E = E.discriminant() if len(K.factor(3)) == 1: # 3 does not split in K remaining_primes.append(3) for p in arith.prime_divisors(D_K): if p >= 5: remaining_primes.append(p) for p in arith.prime_divisors(D_E): if p >= 5 and D_K%p and len(K.factor(p)) == 1: # p is inert in K remaining_primes.append(p) for p in heegner_primes: if p >= 5 and D_E%p != 0 and D_K%p != 0 and len(K.factor(p)) == 1: # p is good for E and inert in K kolyvagin_primes.append(p) assert sha_an in ZZ and sha_an > 0 for p in arith.prime_divisors(sha_an): if p >= 5 and D_K%p != 0 and len(K.factor(p)) == 1: if E.is_good(p): if verbosity > 2 and p in heegner_primes and heegner_index is None: print 'ALERT: Prime p (%d) >= 5 dividing sha_an, good for E, inert in K, in heegner_primes, should not divide the actual Heegner index' # Note that the following check is not entirely # exhaustive, in case there is a p not dividing # the Heegner index in heegner_primes, # for which only an outer bound was computed if p not in heegner_primes: raise RuntimeError("p = %d divides sha_an, is of good reduction for E, inert in K, and does not divide the Heegner index. This may be a counterexample to BSD, but is more likely a bug. %s"%(p,self)) if verbosity > 0: print 'True for p not in {%s} by Kolyvagin (via Stein & Lum -- unpublished) and Rubin.'%str(list(set(remaining_primes).union(set(kolyvagin_primes))))[1:-1] else: # no CM E = self # do some tricks to get to a finite set without calling bound_kolyvagin remaining_primes = [p for p, reason in E.non_surjective()] for p in heegner_primes: if p not in remaining_primes: remaining_primes.append(p) assert sha_an in ZZ and sha_an > 0 for p in arith.prime_divisors(sha_an): if p not in remaining_primes: remaining_primes.append(p) if 2 in remaining_primes: remaining_primes.remove(2) if verbosity > 0: print 'True for p not in {' + str([2]+list(remaining_primes))[1:-1] + '} by Kolyvagin.' primes_to_remove = [] for p in remaining_primes: if E.is_surjective(p)[0] and not E.has_additive_reduction(p): if E.has_nonsplit_multiplicative_reduction(p): if E.rank() > 0: continue if p==3: if (not (E.is_ordinary(p) and E.is_good(p))) and (not E.has_split_multiplicative_reduction(p)): continue if E.rank() > 0: continue p_bound = Sha.p_primary_bound(p) if sha_an.ord(p) == 0 and p_bound == 0: if verbosity > 0: print 'True for p=%d by Stein-Wuthrich.'%p primes_to_remove.append(p) else: print 'Analytic %d-rank is '%p + str(sha_an.ord(p)) + ', actual %d-rank is at most %d.'%(p, p_bound) print ' by Stein-Wuthrich.\n' for p in primes_to_remove: remaining_primes.remove(p) kolyvagin_primes = [] for p in remaining_primes: if E.is_surjective(p)[0]: kolyvagin_primes.append(p) for p in kolyvagin_primes: remaining_primes.remove(p) # apply other hypotheses which imply Kolyvagin's bound holds bounded_primes = [] D_K = rings.QuadraticField(D, 'a').disc() assert 2 not in remaining_primes # Cha's hypothesis for p in remaining_primes: if D_K%p != 0 and N%(p**2) != 0 and E.is_irreducible(p): if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Cha.'%p kolyvagin_primes.append(p) # Stein et al. if not E.has_cm(): L = arith.lcm([F.torsion_order() for F in E.isogeny_class()[0]]) for p in remaining_primes: if p in kolyvagin_primes: continue if L%p != 0: if len(arith.prime_divisors(D_K)) == 1: if D_K%p != 0: if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Stein et al.'%p kolyvagin_primes.append(p) else: if verbosity > 0: print 'Kolyvagin\'s bound for p = %d applies by Stein et al.'%p kolyvagin_primes.append(p) for p in kolyvagin_primes: if p in remaining_primes: remaining_primes.remove(p) prime_bounds = [] # apply Kolyvagin's bound primes_to_remove = [] for p in kolyvagin_primes: if sha_an.ord(p) == 0 and p not in heegner_primes: if verbosity > 0: print 'True for p = %d by Kolyvagin bound.'%p primes_to_remove.append(p) continue if heegner_index is not None: # p must divide heegner_index ord_p_bound = 2*heegner_index.ord(p) # Here Jetchev's results apply. m_max = max([E.tamagawa_number(q).ord(p) for q in N.prime_divisors()]) if m_max > 0 and verbosity > 0: print 'Jetchev\'s results apply with m_max =', m_max ord_p_bound -= 2*m_max if ord_p_bound == 0: if sha_an.ord(p) != 0: raise RuntimeError("p = %d: ord_p_bound == 0, but sha_an.ord(p) == %d. This appears to be a counterexample to BSD, but is more likely a bug."%(p,sha_an.ord(p))) if verbosity > 0: print 'True for p = %d by Kolyvagin bound.'%p primes_to_remove.append(p) continue elif p not in heegner_primes: ord_p_bound = 0 else: from sage.rings.infinity import Infinity ord_p_bound = Infinity if verbosity > 0: print 'p = %d may divide the Heegner index, for which only a bound was computed.'%p if verbosity > 0: print 'ALERT: p = %d left in Kolyvagin bound'%p print ' 0 <= ord_p(#Sha) <=', ord_p_bound print ' ord_p(#Sha_an) =', sha_an.ord(p) for p in primes_to_remove: kolyvagin_primes.remove(p) remaining_primes = list( set(remaining_primes).union(set(kolyvagin_primes)) ) # Kato's bound if rank == 0 and not E.has_cm(): L_over_Omega = E.lseries().L_ratio() kato_primes = Sha.bound_kato() primes_to_remove = [] for p in remaining_primes: if p not in kato_primes: if verbosity > 0: print 'Kato further implies that #Sha[%d] is trivial.'%p primes_to_remove.append(p) if p not in [2,3] and N%p != 0: if E.is_surjective(p)[0]: if verbosity > 1: print 'Kato might apply nontrivially for %d'%p # ordp(sha) <= ordp(L_over_omega) for p in primes_to_remove: remaining_primes.remove(p) # Mazur if N.is_prime(): for p in remaining_primes: if E.is_reducible(p): remaining_primes.remove(p) if verbosity > 0: print 'True for p=%s by Mazur'%p if two_proven is False: remaining_primes.append(2) # print some extra information remaining_primes.sort() if verbosity > 1: if len(remaining_primes) > 0: print 'Remaining primes:' for p in remaining_primes: s = 'p = ' + str(p) + ': ' if not E.is_reducible(p): s += 'ir' s += 'reducible, ' if not E.is_surjective(p): s += 'not ' s += 'surjective, ' a_p = E.an(p) if E.is_good(p): if a_p%p != 0: s += 'good ordinary' else: s += 'good, non-ordinary' else: assert E.is_minimal() if a_p == 0: s += 'additive' elif a_p == 1: s += 'split multiplicative' elif a_p == -1: s += 'non-split multiplicative' if E.tamagawa_product()%p==0: s += ', divides a Tamagawa number' print s return remaining_primes prove_BSD = BSD.prove_BSD def integral_points(self, mw_base='auto', both_signs=False, verbose=False): """
• ## sage/schemes/elliptic_curves/heegner.py
diff -r 1ca8aa94810a -r 8a90b87fca0c sage/schemes/elliptic_curves/heegner.py
a return IR(alpha-MIN_ERR,alpha+MIN_ERR) * IR(LE1-err_E,LE1+err_E) * IR(LF1-err_F,LF1+err_F) def heegner_index(self, D, min_p=2, prec=5): def heegner_index(self, D, min_p=2, prec=5, descent_second_limit=16, verbose_mwrank=False): r""" Return an interval that contains the index of the Heegner point y_K in the group of K-rational points modulo torsion - min_p (int) - (default: 2) only rule out primes = min_p dividing the index. - verbose (bool) - (default: False); print lots of - verbose_mwrank (bool) - (default: False); print lots of mwrank search status information when computing regulator - prec (int) - (default: 5), use prec\*sqrt(N) + 20 terms of L-series in computations, where N is the conductor. - descent_second_limit - (default: 16)- used in 2-descent when computing regulator of the twist OUTPUT: an interval that contains the index index by 2. Unfortunately, this is not an if and only if condition, i.e., sometimes the index must be multiplied by 2 even though the denominator is not 2. This example demonstrates the descent_second_limit option, which can be used to fine tune the 2-descent used to compute the regulator of the twist. If we set the parameter lower than its usual value, then the point search is not high enough to find what it is looking for:: sage: E = EllipticCurve([0, 0, 1, -34874, -2506691]) sage: E.heegner_index(-8, descent_second_limit=10) Traceback (most recent call last): ... RuntimeError: ... However when we use the default values, we find the points we need:: sage: E.heegner_index(-8) 1.00000? """ # First compute upper bound on height of Heegner point. tm = verbose("computing heegner point height...") if c > _MAX_HEIGHT or F is self: verbose("Doing direct computation of MW group.") reg = F.regulator() reg = F.regulator(descent_second_limit=descent_second_limit, verbose=verbose_mwrank) return self._adjust_heegner_index(ht/IR(reg)) # Do naive search to eliminate possibility that Heegner point
• ## sage/schemes/elliptic_curves/sha_tate.py
diff -r 1ca8aa94810a -r 8a90b87fca0c sage/schemes/elliptic_curves/sha_tate.py
a self.__an_numerical = Sha return Sha def an(self, use_database=False): def an(self, use_database=False, descent_second_limit=12): r""" Returns the Birch and Swinnerton-Dyer conjectural order of Sha as a provably correct integer, unless the analytic rank is > 1, in which case this function returns a numerical value. INPUT: use_database -- bool (default: False); if True, try to use any databases installed to lookup the analytic order of Sha, if possible. The order of Sha is computed if it can't be looked up. INPUT: - use_database -- bool (default: False); if True, try to use any databases installed to lookup the analytic order of Sha, if possible. The order of Sha is computed if it can't be looked up. - descent_second_limit -- int (default: 12); limit to use on point searching for the quartic twist in the hard case This result is proved correct if the order of vanishing is 0 and the Manin constant is <= 2. listed in Cremona's tables, if this curve appears in Cremona's tables. NOTE: If you come across the following error:: sage: E = EllipticCurve([0, 0, 1, -34874, -2506691]) sage: E.sha().an() Traceback (most recent call last): ... RuntimeError: Unable to compute the rank, hence generators, with certainty (lower bound=0, generators found=[]). This could be because Sha(E/Q)[2] is nontrivial. Try increasing descent_second_limit then trying this command again. You can increase the descent_second_limit (in the above example, set to the default, 12) option to try again:: sage: E.sha().an(descent_second_limit=16) 1 EXAMPLES:: sage: E = EllipticCurve([0, -1, 1, -10, -20]) # 11A = X_0(11) self.__an = s return s regulator = E.regulator(use_database=use_database) # this could take a *long* time; and could fail...? regulator = E.regulator(use_database=use_database, descent_second_limit=descent_second_limit) T = E.torsion_subgroup().order() omega = E.period_lattice().omega() Sha = int(round ( (L1 * T * T) / (E.tamagawa_product() * regulator * omega) )) | 14,770 | 48,076 | {"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} | 2.890625 | 3 | CC-MAIN-2022-49 | latest | en | 0.705571 |
https://chris-said.io/2020/12/26/two-things-that-confused-me-about-cross-entropy/ | 1,618,334,456,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00409.warc.gz | 281,248,706 | 4,766 | # Things that confused me about cross-entropy
Every once in a while, I try to better understand cross-entropy by skimming over some Medium posts and StackExchange answers. I always come away only half-understanding it.
A major cause of confusion is that different sources use different notations and conventions. Here are some that tripped me up.
### p and q vs y and p
In information theory, the cross-entropy for an event with $$M$$ discrete outcome classes is
$H(p, q) = -\sum_{j=1}^M{p_j log (q_j)}$
… where $$p$$ is the true distribution of outcomes and $$q$$ is the approximating distribution.
But in machine learning, the cross-entropy is defined as:
$H(y, p) = -\sum_{j=1}^M{y_j log (p(y_j))}$
… where $$y$$ are the true labels and $$p(y)$$ is the approximating distribution, i.e. the classifier’s probability predictions.
Notice that the meaning of $$p$$ has been reversed! In information theory, it is the true distribution. In machine learning, it is the approximating distribution.
This points to a key difference between generic information theory and machine learning applications. In information theory, the target distribution $$p$$ is a probability distribution where the probability mass can be distributed broadly over the classes. In machine learning, the target distribution $$y$$ has all of its mass allocated to a known label.
### Multiple classes vs multiple trials, and indicators vs labels
As shown above, the cross-entropy for a single event with $$M$$ classes is:
$H(y, p) = -\sum_{j=1}^M{y_j log (p(y_j))}$
If there are multiple trials, you just need to sum up the cross-entropies from all the trials so that the total cross-entropy is:
$H(y, p) = -\sum_{i=1}^N \sum_{j=1}^M{y_j^{(i)} log (p(y_j^{(i)}))}$
… where $$N$$ is the number of trials.
This all makes notational sense. Unfortunately, confusion arises when people use non-compact notation for summing over binary classes. When dealing with binary classification, it’s common to see cross-entropy defined as:
$H(y, p) = -\sum_{i=1}^N{y_i log (p(y_i)) + (1-y_i) log (1 -p(y_i))}$
Superficially, this looks a lot like the first formula, but it’s actually just a clever and somewhat confusing way of writing the second formula for binary classes. Its notation differs from previous formulas in a few ways:
• First, $$p(y_i)$$ always refers to the probability of a positive result. In previous formulas, $$p(y_j)$$ referred to the probability of the particular class being considered in the summation loop.
• Second, $$y_i$$ (when not wrapped by a $$p$$) is a label, not an indicator! When the event has an outcome of 1, $$y_i$$ is 1. When it has an outcome of 0, $$y_i$$ is 0. In previous formulas, as you loop through classes you would set $$y_j$$ to be 1 whenever the outcome was the class being considered; otherwise you set it to 0.
• While not a notational difference, this formula sums over events (and then enumerates a sum over classes), whereas the first formula in this section only sums over classes.
The image below summarizes the many confusing differences between these formulas.
### Where to learn about cross-entropy
I don’t know if there is a single best place to learn about cross-entropy, but below are a few places that were helpful. If you read them make sure you think about (a) how they are notating the true distribution and the approximating distribution (b) whether they are writing about single trials or multiple trials (c) whether they are writing about binary classes or multi classes and (d) whether the $$y$$’s are indicators or labels.
• Statistical Rethinking. Information theory notation, single trial case, multiple classes.
• Andrew Webb. Both notations, both the single and multiple trial case, multiple classes.
• Harshith. ML notation, multiple trials, binary classes. | 932 | 3,816 | {"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} | 3.265625 | 3 | CC-MAIN-2021-17 | latest | en | 0.881813 |
http://intermath.coe.uga.edu/tweb/rockdale-geom/jtaylor/write_ups/InterMath%20-%20Write-up%20Biggie%20Size%20This_files/InterMath%20-%20Write-up%20Biggie%20Size%20This.htm | 1,516,108,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00421.warc.gz | 190,966,498 | 5,499 | Intermath | Workshop Support
Write-up
Title
Biggie Size This
Problem Statement
If you double the lengths of each of the sides of a triangle, what happens to the perimeter and the area? Explain why.
How would the results change if you triple the lengths of each of the sides? What if you make the sides ten times their original size? Explain your reasoning.
Problem setup
This problems asks us to double the sides of a triangle and compare the related perimeters and areas. As a further investigation, we are asked to triple the sides, quadruple the sides, etc….
Plans to Solve/Investigate the Problem
To begin I drew similar triangles in GeoSketchPad. I drew one with a k value of 2/1 and a third with a k value of 3/1. I then created an Excel spreadsheet with k values from 2 to 10..
Investigation/Exploration of the Problem
Upon finishing my GeoSketchPad drawings, I calculated the perimeters and areas. As you will see below, as the perimeters increased by a rate of k+1, the areas increased by (k+1)². When I entered the data into an Excel spreadsheet, this held true throughout each k. See Below:
Side A Side B Side C S Perimeter Area P ratio A Ratio 2 3 4 4.5 9 2.904738 4 6 8 9 18 11.61895 1/2 1/4 6 9 12 13.5 27 26.14264 1/3 1/9 8 12 16 18 36 46.4758 1/4 0.0625 10 15 20 22.5 45 72.61844 1/5 0.04 12 18 24 27 54 104.5706 1/6 0.027778 14 21 28 31.5 63 142.3321 1/7 0.020408 16 24 32 36 72 185.9032 1/8 0.015625 18 27 36 40.5 81 235.2837 1/9 0.012346 20 30 40 45 90 290.4738 0.1 0.01
Author & Contact
Jim Taylor
jtaylor1@rockdale.k12.ga.us
Link(s) to resources, references, lesson plans, and/or other materials | 542 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-05 | latest | en | 0.809455 |
http://new.math.uiuc.edu/public402/models/exercisesCh8.html | 1,505,886,341,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686465.34/warc/CC-MAIN-20170920052220-20170920072220-00156.warc.gz | 232,342,241 | 4,083 | Exercises on Moebius Transformations
revised 21apr11, 13jul11\\
\textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department,
University of Illinois}
\begin{document}
\maketitle
\section{Introduction}
This lesson is a commentary on Hvidsten's \textit{Exercises 8.2.1-8.2.15}
on page 325. There is a great deal of geometry hidden in these problems, which
is obscured by the brevity of their statement, or even their solution.
The elaborations here is intended to help you use the exercises to understand
the facts and methods. A "solution" to a problem
should be a mini essay on what the problem means, and not consist of
just cranking out what looks like an answer to a direct question.
In this spirit we proceed.
\section{Discussion of the Exercises}
\subsection{Transformations which form Groups}
\begin{enumerate}
\item \textbf{Ex. 8.2.1} "Show that the $w=e^{i\phi}+b$ form a group, the group of
rigid Euclidean motions (a.k.a. Euclid's congruences) of the plane." This is
special case of the group of \textit{similitudes} treated extensively in the
lessons.
\item Remember, to show that a
subset of the group of all bijective functions of the extended plane is itself
a group, all you need to do is to show that
\begin{itemize}
\item \textit{Closure: }The composition of two such
function can be put into the same canonical form that defines the subset.
\item \textit{Inverse: } The inverse of such a function can be put
into the same canonical form.
\end{itemize}
Be sure you have this example in your Journal and are able to reproduce
the definitions and proof of this theorem on a test.
\item As Hvidsten says: \textbf{Ex 8.2.4 - 8.2.7} are steps in the proof that the set of Moebius
transformations also form a group. Much of this is in the notes. These
4 problems help you to organize this material in a coherent manner in your
Journal. As long as the actual calculations are short, a test question
may ask you to check that a given set of MTs form a group.
\begin{itemize}
\item Ex 8.2.4 this is the closure property and is pure high school algebra.
Substitute the result of the first transformation into the formula for the
second and rewrite in the standard form of Moebius transformation. Use our
definition, where $ad-bc=1$ is the condition on the coefficients. You'll be
happier for it.
\item Ex 8.2.5
As in our notes, the presence of the identity follows from
the presence of the inverse (8.2.6) and closure (8.2.4). However, an
immediate guess will solve this problem directly. But here there is an
easier solution by inspection.
\item Ex 8.2.6 you'll need to solve $w=\frac{az+b}{cz+d}$
for $z =\frac{ez+f}{gz+h}$, by giving $e,f,g,h$ as functions of
$a,b,c,d$. But Hvidsten suggests a simpler approach. For this to hold,
you'll need to use his criterion $ad-bc\ne 0$. Recall why we can
assume either condition. These calculations belong into your Journal.
\item Ex 8.2.7 is done below. The answer is that transformations
always are associative: $(k \circ h) \circ g = k \circ (h \circ g)$.
All you need to remember is how to test that two functions are the same.
\end{itemize}
\item Solution to 8.2.7: An equation of functions holds provided the
equation holds for every value in the domain of the functions. So
applying LHS to an arbitrary complex number $z$ we see that
$(k\circ h)\circ g (z) = (k \circ h)(g(z)) = k(h(g(z)))$ which is
what you get if you plug $z$ into the RHS.
\item \textbf{Ex 8.2.8: } This exercise asks you prove that a particular subset
of transformations which have a particular geometric property,forms
a group. As we have seen, all you need to check is closure and inverses.
It is unfortunate that Hvidsten uses the word "fix" instead of "preserve".
He does not mean that each point on the unit circle remains fixed under
the transformation. (That would mean it is the identity, right? since
there are at least 3 points on the circle.))
By "$f$ fixes or preserves the unit circle" we mean it maps every point
on the circle back to some (possibly different) point on the same circle.
Algebraically, if $|z|=1$ then $|f(z)|=1$ as well.
\end{enumerate}
\subsection{CircLine Geometry}
\begin{enumerate}
\item \textbf{ Ex 8.2.2 }
"Show that the equation of a line can be given as
$\mathfrak{Im}(\alpha z + \beta) =0$." While you can prove this with
just a little arithmetic applied to the definitions, it has a deeper meaning.
Note that for any complex number $w$, $\mathfrak{Im}(w)=0$ is the same
thing as saying $w \in \mathbb{R}.$
\item \textbf{Extension 8.2.2bis} Since a straight line is a special case of a circline, this problem
is a special case of the general theorem that the equation of a
circline through three points $z_0,z_1,z_\infty$is given by
$\mathfrak{Im}CR(z,z_0,z_1,z_\infty)=0$. Equivalently that
$CR(z,z_0,z_1,z_\infty) \in \mathbb{R}$.
\item \textbf{ Ex 8.2.3}
In my edition of the text has a misprint here. The correct
equation of the unit circle is $|z|=1$.
"Find a Moebius transformation that takes the unit circle to the x-axis."
There are, of course, very many solutions. But there is a "canonical"
solution in terms of cross-ratios, if you think of the unit circle as
$\odot(1,i,-1)$. Use the theorem that an MT defined by
$f(z)=CR(z,z_0,z_1,z_\infty)$ takes the circline $\odot(z_0,z_1,z_\infty)$
to the real-line.
\item \textbf{Extension 8.2.3bis}
Incidentally, if the three points are on any circle,
and in positive (counter-clockwise) order, then this MT takes the interior of
the circle to the upper halfplane, and the exterior to the lower halfplane.
\end{enumerate}
\subsection{The Hyperbolic Group}
The set of MTs which "fix" the unit circle form the group of isometries
in the hyperbolic plane, called the \textit{ Hyperbolic Group} for short.
The way this is established is exploit .
\begin{itemize}
\item Calculate the canonical form of every such MT.
\item Use one property of such an MT to define a hyperbolic line.
\item Exploit a property of hyperbolic lines so defined in terms of a cross ratio.
\item Define a ruler, in the sense of Birkhoff, for every line.
\item Verify that Birkhoff's Ruler Axiom is indeed verified.
\end{itemize}
The Exercises in the textbook serve this program.
\begin{enumerate}
\item \textbf{ Ex 8.2.9:} This is the \textit{ transitivity } of the Hyperbolic
Group. By \textif{isometry} here Hvidsten means a
Moebius transformation which preserves the unit circle. In Prop.8.10
he proves that such a Moebius transformation must have the form
\begin{eqnarray*}
w = e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z -1} & \mbox{where} & |\alpha|<1\\
\end{eqnarray*}
So now the problem reduces to finding angle $\theta$ and point $\alpha$
which is moved to the origin under the isometry for given values for $z$ and
$w$. Hvidsten he calls $z=P$ and $w=Q$.
But Hvidsten gives you a hint, by asking the question for the case
that $z$ remains arbitrary, but $w=0$. (Why is it enough to prove it just
for this case? Don't forget to answer this in your solution.)
Finally, for this special case the answer is now very easy,
namely, to solve
\begin{eqnarray*}
0 = e^{i\theta}\frac{P-\alpha}{\bar{\alpha}P -1} & \mbox{where} & |\alpha|<1\\
\end{eqnarray*}
we can take $\theta = 0$, i.e. no rotation, and $\alpha = P$, which is
legitimate, because we are given that $|P| < 1$. The isometry is not unique,
but it is unique up to a rotation.
\end{enumerate}
[The remaining exercises will be discussed here in the future.]
` | 2,144 | 7,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-39 | latest | en | 0.883058 |
http://math.tutornext.com/graphing-linear-functions-one-variable-and-two-variable.html | 1,524,504,289,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00005.warc.gz | 201,173,670 | 5,845 | # Graphing Linear Functions
Sub Topics
In Algebra 2 help ,A linear function with two variables is a variable function, that is, the value of one variable (called as dependent variable) varies with the values assigned to the other variable(called as independent variable). Usually the independent variable is denoted by the letter ‘x’ and the dependent variable is denoted by the letter ‘y’
The graph of a linear function is always a straight line. It is either a vertical line or a horizontal line for a linear function with single variable and an inclined line for a linear function with two variables.
Graph linear functions for the below questions and get practiced to similar kind of problems.
## Graph Linear Function with One Variable
This graph is from algebra answers where a Graph linear function with one variable is expressed either as x = a or, y = a, where ‘a’ is a constant.
The form, y = a, represents a horizontal line in a coordinate grid, parallel to x-axis and at a distance of ‘a’ units from x-axis. To draw a graph of this function, mark ‘a’ units on y-axis and draw a horizontal line passing through that.
Interestingly, the x-axis in a coordinate grid is represented by the linear function, y = 0 and the y-axis in a coordinate grid is represented by the linear function, x = 0
The form, x = a, represents a vertical line in a coordinate grid, parallel to y-axis and at a distance of ‘a’ units from y-axis. To draw a graph of this function, mark ‘a’ units on x-axis and draw a vertical line passing through that.
## Graphing Linear Function with Two Variables
Since the graph of a linear function is a straight line just two points are needed in a grid to draw the graph. First rewrite the function in slope intercept form, if it is in some other form.
Graphing linear function with two variables is all about ploting two points in the garph.
There are two methods to locate two points in the grid.
1) Mark the y-intercept of the given function. This is one point. Use the slope to construct the second point adjacent to the y-intercept. Now draw a line passing through these two points.
This method will be comfortable if the slope and y-intercept are compatible numbers.
2) By guess and check find two compatible ordered pairs. You can try a convenient number as input to get another convenient number as output. Plot these ordered pairs on the grid. Draw a line passing through these points.
This method can be followed if the slope and y-intercept are not compatible numbers.
Although two points are sufficient to draw the graph of a linear equation, you may be more comfortable by plotting additional points from an input-output table of the function.
We use the above methods in solving math problems of linear eqautions in 2 variables.
## Examples of Graphing Linear Function with Two Variables
Draw the graph for the function 2x + 4y = 6
The given function is rewritten as, 4y = -2x + 6 or, y = -(1/2) + 3
The y-intercept is 3.
The slope is -(1/2). That means for every two units of run, the fall is 1.
The graph is drawn as explained below.
## Problem on Graphing a Linear Function with Two Variables
Draw the graph for the function 3 x + 5y = 4
By guess and check, the value of y is (-1) for x = 3 and is 2 for x = -2
Plot two points with ordered pairs (3, -1) and (-2, 2).
Draw a line passing through these points. | 770 | 3,378 | {"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} | 4.75 | 5 | CC-MAIN-2018-17 | latest | en | 0.921894 |
https://quantumcomputing.stackexchange.com/questions/16575/how-to-reason-about-absorbing-pauli-product-rotations-into-measurements/16576 | 1,718,692,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00145.warc.gz | 425,348,009 | 41,165 | # How to reason about absorbing Pauli Product rotations into measurements?
In this image, the orange box refers to the operator $$\exp(-i P \frac{\pi}{4})$$ and the blue box refers to the Pauli Measurement $$P'$$. The idea behind these rules is to show how all the Pauli Product $$\frac{\pi}{4}$$ rotations at the end of the circuit can be absorbed into the measurements.
How would you prove these equations? In particular, I am looking for a proof that will help me reason about what happens if I measure say just the first qubit after performing a multi qubit orange box. For example, how would I absorb the orange box into the measurements in the following?
First of all, let us consider an arbitrary unitary $$U$$. If we first apply $$U$$ and then measure a POVM $$\{E_a\}$$, then this is equivalent to doing nothing and measuring the POVM $$\{U^\dagger E_a U\}$$ instead since $$\mathrm{Pr}[a|\rho] = \mathrm{tr}\left( E_aU \rho U^\dagger \right) = \mathrm{tr}\left( U^\dagger E_a U \rho \right)$$. In particular, a Pauli measurement is given by the projectors $$\frac 1 2 \left( \mathbb{I} \pm P \right)$$ and the altered measurement is described by the Hermitian operator $$O = U^\dagger P U$$.
Now note that $$U=\exp(- i\frac \pi 4 P)$$ is a Clifford unitary if $$P$$ is a Pauli operator. Thus, absorbing the action of $$U$$ into the measurement of a Pauli operator $$Q$$ will result in another Pauli measurement of $$\pm Q'=U^\dagger Q U$$. Note that the possible sign simply relabels the outcomes. Next, it is clear that $$U^\dagger Q U=Q$$ if $$P$$ and $$Q$$ commute, thus we can remove $$U$$ from the circuit. If not, a straightforward calculation shows that $$e^{i\frac \pi 4 P}Qe^{- i\frac \pi 4 P} = i PQ \equiv \pm Q'.$$
If you measure the first qubit after a $$n$$-qubit Clifford gate, than this effectively the same as doing nothing and performing a Pauli measurement on many ($$\leq n$$) qubits (since $$U$$ will generally map a Pauli operator on the first qubit to a Pauli operator supported on $$n$$ qubits).
The measurement in your final example consists of measuring four commuting Pauli operators $$P_1,\dots,P_4$$ (a syndrome measurement). The projectors of the joint measurement are given as products $$P_x = \prod_{i=1}^4 \frac 1 2 (\mathbb I + (-1)^{x_i} P_i)$$ where $$x\in \{0,1\}^4$$. To compute $$UP_x U^\dagger$$, we can thus individually transform the involved projectors.
Let us use the above reasoning sequentially:
1. $$X\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z$$ anti-commutes with $$Y\otimes\mathbb{I}\otimes\mathbb{I}\otimes\mathbb{I}$$, thus we can replace the measurement of $$-Y$$ on the first qubit by the measurement of $$-iXY\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z=Z\otimes\mathbb{I}\otimes\mathbb{I}\otimes Z$$.
2. The Pauli operators on the second and third qubit obviously commute with the unitary, so the measurements are unchanged.
3. The last Pauli operator again anti-commutes, so we can replace it by the measurement of $$X\otimes\mathbb{I}\otimes\mathbb{I}\otimes i ZY=X\otimes\mathbb{I}\otimes\mathbb{I}\otimes X$$.
Note that we obtained an entangling measurement between the first and the fourth qubit. This is in fact a Bell measurement as the joint eigenvectors of $$X\otimes X$$ and $$Z\otimes Z$$ are the four Bell states $$\{|00\rangle \pm |11\rangle, |01\rangle \pm |10\rangle\}$$.
• Hi, I'm still trying to work through understanding your answer but just from my partial understanding so far, might you have assumed that the measurement in the example circuit was a Pauli YYYY measurement? It is in fact 4 distinct single qubit measurements, sorry if this wasn't clear. The first is -Y , the other three are Y. Commented Mar 24, 2021 at 18:00
• @user3717194 I see, sorry for that. Anyway, you can still argue similarly, I'll update the answer. Commented Mar 25, 2021 at 15:03
Here's a sort of simple way to do it, if you have a simulator you trust. Note that a Z-basis measurement is equivalent to a Z-controlled CNOT onto an ancilla qubit and then measuring that ancilla in the Z basis. And that a Y-basis measurement is equivalent to a Y-controlled CNOT onto an ancilla qubit and then measuring that ancilla in the Z basis.
What you can do is assert that moving the rotation before the controlled operation changes the control basis of the controlled operation:
import stim
sim1 = stim.TableauSimulator()
sim1.do(stim.Circuit("""
SQRT_X 0
CNOT 0 99 # 99 is the ancilla where we're putting the measurement result
"""))
sim2 = stim.TableauSimulator()
sim2.do(stim.Circuit("""
YCX 0 99 # Y-controlled X gate
SQRT_X 0
"""))
assert sim1.current_inverse_tableau() == sim2.current_inverse_tableau()
To be safe, you likely want to use the state channel duality to ensure the simulation works for all states instead of just one state.
import stim
sim1 = stim.TableauSimulator()
sim1.do(stim.Circuit("""
H 0
CNOT 0 98 # 98 is the state channel ancilla
SQRT_X 0
CNOT 0 99 # 99 is the ancilla where we're putting the measurement result
"""))
sim2 = stim.TableauSimulator()
sim2.do(stim.Circuit("""
H 0
CNOT 0 98 # 98 is the state channel ancilla
YCX 0 99 # Y-controlled X gate
SQRT_X 0
"""))
assert sim1.current_inverse_tableau() == sim2.current_inverse_tableau() | 1,491 | 5,254 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.856915 |
https://neo4j.com/docs/cypher-manual/current/functions/mathematical/logarithmic/index.html | 1,553,277,852,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00039.warc.gz | 576,719,632 | 4,652 | ## 4.6. Mathematical functions - logarithmic
These functions all operate on numeric expressions only, and will return an error if used on any other values. See also Section 2.7.4, “Mathematical operators”.
Functions:
### 4.6.1. e()
`e()` returns the base of the natural logarithm, `e`.
Syntax: `e()`
Returns:
A Float.
Query.
``RETURN e()``
The base of the natural logarithm, `e`, is returned.
Table 4.60. Result
e()
1 row
`2.718281828459045`
Try this query live. none RETURN e()
### 4.6.2. exp()
`exp()` returns `e^n`, where `e` is the base of the natural logarithm, and `n` is the value of the argument expression.
Syntax: `e(expression)`
Returns:
A Float.
Arguments:
Name Description
`expression`
A numeric expression.
Considerations:
`exp(null)` returns `null`.
Query.
``RETURN exp(2)``
`e` to the power of `2` is returned.
Table 4.61. Result
exp(2)
1 row
`7.38905609893065`
Try this query live. none RETURN exp(2)
### 4.6.3. log()
`log()` returns the natural logarithm of a number.
Syntax: `log(expression)`
Returns:
A Float.
Arguments:
Name Description
`expression`
A numeric expression.
Considerations:
`log(null)` returns `null`. `log(0)` returns `null`.
Query.
``RETURN log(27)``
The natural logarithm of `27` is returned.
Table 4.62. Result
log(27)
1 row
`3.295836866004329`
Try this query live. none RETURN log(27)
### 4.6.4. log10()
`log10()` returns the common logarithm (base 10) of a number.
Syntax: `log10(expression)`
Returns:
A Float.
Arguments:
Name Description
`expression`
A numeric expression.
Considerations:
`log10(null)` returns `null`. `log10(0)` returns `null`.
Query.
``RETURN log10(27)``
The common logarithm of `27` is returned.
Table 4.63. Result
log10(27)
1 row
`1.4313637641589874`
Try this query live. none RETURN log10(27)
### 4.6.5. sqrt()
`sqrt()` returns the square root of a number.
Syntax: `sqrt(expression)`
Returns:
A Float.
Arguments:
Name Description
`expression`
A numeric expression.
Considerations:
`sqrt(null)` returns `null`. `sqrt()` returns `null`
Query.
``RETURN sqrt(256)``
The square root of `256` is returned.
Table 4.64. Result
sqrt(256)
1 row
`16.0`
Try this query live. none RETURN sqrt(256) | 667 | 2,248 | {"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} | 3.671875 | 4 | CC-MAIN-2019-13 | latest | en | 0.450388 |
https://blog.toadworld.com/2017/02/06/sql-different-ways-to-generate-sequence | 1,722,899,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00562.warc.gz | 107,797,398 | 37,857 | There was a question from the op regarding adding a new column to a query output by generating the cyclic sequence numbers from 1 to 3.
`Select A =identity(int,1,1),B,C from table_abs1,A,412,B,133,C,904,D,915,E,986,F,127,G,548,H,16`
For this output, the 4th column generates the Sequence of numbers from 1 to 3 which is shown below
`1,A,41,12,B,13,23,C,90,34,D,91,15,E,98,26,F,12,37,G,54,18,H,16 ,2`
If you are using SQL 2012 then Sequence would be natural choice for any such operations.
### Solution 1
Using Sequence
`CREATE`
`SEQUENCE Seq AS INTEGER START WITH 1 INCREMENT BY 1 MINVALUE 1 MAXVALUE 3 CYCLE;`
`SELECT table_name,NEXT VALUE FOR Seq New_column FROM information_schema.tables`
### Solution 2
Using CTE and Modulus operator
`;with q as( select row_number() over (order by (select null)) A, * from sys.objects)select A, 1+A%3 B, *from q`
### Solution 3
Loops and Temp table
`create table dummyTest(id int,col1 char(1),col2 int)`
```insert into dummyTest values(1,'A',410),(2,'B',411),(3,'c',4111),(4,'d',421),(5,'e',441),(6,'f',451),(7,'g',481),(8,'h',401)create table #dummy
( id int, col1 char(1), col2 int, NewColumn int )
declare @n int,@i int,@limit int set @limit=1
set @i=1
select @n=count(*) from dummyTest while @i<=@n begin
set @limit=1
while @limit<=3
begin
print @limit insert into #dummy
select *,NewColumn=@limit from dummyTest where id=@i set @i=@i+1
set @limit=@limit+1
end
end select * from #dummy```
### Conclusion
The same solution can be derived using the cursor and there may be other solutions as well. At many instances, we opt for any solutions without thinking of data volume that may degrade the performance. This is one of the prime examples of why we need to upgrade to newer version.
### prashanthjayaram
I’m a Database technologist having 8+ years of rich, hands-on experience on Database technologies. I am Microsoft Certified Professional and backed with a Degree in Master of Computer Application. My expertise lies in T-SQL programming, Replication and PowerShell. | 588 | 2,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.734489 |
http://www.infocobuild.com/education/learn-through-videos/physics/TheMechanicalUniverse/lecture-06.html | 1,708,911,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00135.warc.gz | 52,036,725 | 3,779 | # InfoCoBuild
## The Mechanical Universe
Newton's Laws. For all the phenomena in The Mechanical Universe, Isaac Newton laid down the laws. Newton's first law states that every body remains at rest or continues in uniform motion unless a force acts on it. His second law, the most profound statement in classical mechanics, relates the causes to the changes of motion in every object. Newton's third law explains the phenomenon of interactions: for every action, there's an equal and opposite reaction. (from themechanicaluniverse.org)
Lecture 06 - Newton's Laws
Go to The Mechanical Universe Home or watch other lectures:
Lecture 01 - Introduction Lecture 02 - The Law of Falling Bodies Lecture 03 - Derivatives Lecture 04 - Inertia Lecture 05 - Vectors Lecture 06 - Newton's Laws Lecture 07 - Integration Lecture 08 - The Apple and the Moon Lecture 09 - Moving in Circles Lecture 10 - Fundamental Forces Lecture 11 - Gravity, Electricity, Magnetism Lecture 12 - The Millikan Experiment Lecture 13 - Conservation of Energy Lecture 14 - Potential Energy Lecture 15 - Conservation of Momentum Lecture 16 - Harmonic Motion Lecture 17 - Resonance Lecture 18 - Waves Lecture 19 - Angular Momentum Lecture 20 - Torques and Gyroscopes Lecture 21 - Kepler's Three Laws Lecture 22 - The Kepler Problem Lecture 23 - Energy and Eccentricity Lecture 24 - Navigating in Space Lecture 25 - From Kepler to Einstein Lecture 26 - Harmony of the Spheres Lecture 27 - Beyond the Mechanical Universe Lecture 28 - Static Electricity Lecture 29 - The Electric Field Lecture 30 - Potential and Capacitance Lecture 31 - Voltage, Energy, and Force Lecture 32 - The Electric Battery Lecture 33 - Electric Circuits Lecture 34 - Magnetism Lecture 35 - The Magnetic Field Lecture 36 - Vector Fields and Hydrodynamics Lecture 37 - Electromagnetic Induction Lecture 38 - Alternating Currents Lecture 39 - Maxwell's Equations Lecture 40 - Optics Lecture 41 - The Michelson Morley Experiment Lecture 42 - The Lorentz Transformation Lecture 43 - Velocity and Time Lecture 44 - Energy, Momentum, Mass Lecture 45 - Temperature and the Gas Laws Lecture 46 - The Engine of Nature Lecture 47 - Entropy Lecture 48 - Low Temperatures Lecture 49 - The Atom Lecture 50 - Particles and Waves Lecture 51 - From Atoms to Quarks Lecture 52 - The Quantum Mechanical Universe | 524 | 2,333 | {"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-2024-10 | latest | en | 0.726556 |
https://anyquestions.info/philosophy/basic_law_of_logic_what_are_the_foundations_of_the_universe_that_ensure_its_correctness_is_it_possible_to_imagine_a_reality_in_which_this_law_would_not_work/ | 1,685,768,681,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00455.warc.gz | 130,656,092 | 21,669 | 1. alexander_shestakov says:
CONSIDER a few physically completely indistinguishable twins who are, however, mentally different people. Here “A” is represented in more than one instance, and the identity A=A is violated.
2. peter_karlov says:
The law of logic A is A is a consequence of our understanding of the stationarity of the world. It is impossible to imagine a reality where this law does not work. Our world is not real in the classical sense of the word.
3. peter_karlov says:
The dialectical law of anisotropic pair avoidance is discovered. That is, nothing in the world is identical with itself, A + A = A growing old. You are not who you once were, you are getting old.
4. alexander_rodinov says:
None at all. All our ideas about the world that we can spread from person to person are just simplified models based on basic entities (for example, a point, a straight line, an atom, etc.) and set rules for interaction between basic entities (the same logic, for example). Some may be better at describing objective reality, and some may be worse, but they are still just models.
5. george_petrov says:
In the universe (in the World), everything can be counted and measured. Therefore, the answer to this question can be given by natural integers. 1 is 1 whatever we take: a leaf from a tree; a planet; a photon; a window, etc. – all that will be ONE. It's the same with the other number-two or seven.
Numbers are the basis of the universe. (A. F. Losev devoted several of his works to the number.)
6. sergey_sharkov says:
This is called in formal logic the” law ” (or concept) of identity.
At the heart of the universe is (if you believe philosophers and “common sense”) the “law of similarity”: “like … like” (instead of a triple colon, insert any “verb”: known, generated, attracted, treated, understood, etc.). The extremely abstract and ideal formalized expression of which seems to be the “law of identity”.
Such a “wonderful” reality (where there is no identity) does not even need to bother to imagine (it is always around you live!). Any “knowledge” without a knowing subject “does not work” (it does not exist at all without it), and therefore the “law of identity” does not work outside the knowing subject. As the ancients said, “you can't step into the same water twice” (complete ideal identity is possible only in completely abstract and formal “concepts” such as mathematics or geometry, and in real life there is only “similarity”, because everything flows and everything changes). But you, as a cognizing subject, have every right to think otherwise (there is no one “who is right” to judge “objectively”).
7. viktor_gorbatov says:
In short: yes, such a reality can be imagined. Moreover, we live in it, according to physicists.
Currently, there are detailed logical systems that reject the “A is A” law – the so-called “Schrodinger logics”. They are based on the idea of Erwin Schrodinger – one of the pioneers of quantum mechanics and a Nobel laureate – that in particle physics, talking about the identity of objects is often meaningless.
We know that quantum mechanics is often called the “killer” of the fundamental laws of classical logic – for example, the law of non-contradiction (a particle can be in two states that are incompatible from the classical point of view, so that a statement about it can be true and false at the same time), or the law of the excluded third (in addition to truth and falsity, statements about elementary particles often have to be assigned a third meaning – “uncertainty”). But the law of identity, it seems, should work even in the microcosm: how can a particle not be equal to itself?
In fact, we are somewhat deceived by the phrase “itself” – it already implies the identity of the particle in advance. It is as if we are able to decide separately whether there is one particle in front of us, and then check whether it is equal to itself. To get rid of the illusion of evidence, it would be useful to ask: on what grounds do we judge the identity or difference of something?
Traditionally, in matters of identification/distinction, we rely on the famous Leibnizian principle of “identity of the indistinguishable”: if objects have all their properties coincide (that is, they cannot be distinguished by any signs), then these objects are identical. All the usual logic and mathematics treat the concepts of identity and indistinguishability as synonymous. But as soon as we notice that indistinguishability is an epistemological concept (it describes the ability of the knowing subject to distinguish something), and identity is an ontological concept (it describes the characteristic of the object itself), we realize that they do not have to coincide. You never know what objects we can't distinguish – what does it matter to them? Should they “stick together” just because we can't tell them apart?
Let's return to elementary particles. What do we know about them? That their description is possible only in terms of quantum states. They have no properties other than quantum states. So, it turns out that in the subatomic world there are particles for which indistinguishability does not imply identity – these are bosons. An unlimited number of identical bosons can simultaneously exist in a single quantum state. (For fermions, by the way, the opposite is true – there can be no more than one fermion in the same quantum state. In other words, fermions behave like law-abiding citizens of the Leibniz universe, and bosons behave like hooligans who systematically violate the principle of identity of the indistinguishable.)
The bad news is that we have to pay a price for this boson hooliganism – first of all, by abandoning classical set theory. But almost all our logic and mathematics are built on it! By the way, for the first time the idea of the need to abandon classical set theory in describing the quantum world was expressed by our Russian mathematician Yuri Manin in 1976.
The good news is that we have a rough idea of how to do this. In 1980, the Brazilian logician Newton Da Costa developed quasi-set theory, where “quasi-set” refers to a collection of objects that are indistinguishable but not identical. Actually, semantics for Schrodinger logics is usually built on the basis of this theory. And if these logics have clear semantics, then we can really imagine how to reason about a reality in which the law “A is A”does not work.
8. mikhail_kharitonov says:
A = A is not a “law of the universe”. This is the law of thinking. And not thinking at all, but only thinking about statements.
To understand its meaning, one must first understand the meaning of equivalence. A = B if and only if, when substituting in any statement containing the name A, we substitute B, the truth of the statement will not change. Example: “first positive even number”= “2”. If the statement “if you add one to the first positive even number, you get three” is replaced with the statement “if you add one to 2, you get three”, then the truth of the statement does not change (both statements are true). If the statement “the first positive even number is greater than a thousand” is replaced by the statement “2 is greater than a thousand”, the truth value also does not change (both statements are false). And the same applies to all statements where “the first positive even number is greater than a thousand” and “2”appear.
So, if the name A is replaced by itself in any statement, then the statement remains just as true or just as false.
That, in fact, is the whole point of this “basic law”. | 1,682 | 7,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.944305 |
http://www.checkmyknowledge.com/a-level/math/training/as/integration | 1,498,609,046,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321961.50/warc/CC-MAIN-20170627235941-20170628015941-00070.warc.gz | 474,298,594 | 8,693 | ###### 7.1
Given the integral $$\int_5^{11}\frac8x\,\mathrm{d}x$$.
Use the trapezium rule, with $$3$$ strips each of width $$2$$, to estimate the value of the integral. Give your answer correct to $$3$$ significant figures.
$$\int_5^{11}\frac8x\,\mathrm{d}x \approx \frac{1}{2\times2} (\frac85+2\frac87+2\frac89+\frac8{11})\approx 1.598$$
###### 7.2
The gradient of the curve given by $$\frac{dy}{dx}=x^2-2x+5$$. The curve passes through the point $$(3,11)$$.
Find the equation of the curve.
$$y=\int(x^2-2x+5)\,\mathrm dx = \frac13x^3-x^2+5x+c$$. Since the curve passes through the point $$(3,11)$$, we obtain:
$$\frac{3^3}3-3^2+5\times3+c=11\\ 15+c=11\\ c=-4$$
###### 7.3
The diagram shows parts of the curves $$y=x^2+1$$ and $$y=11-\frac{9}{x^2}$$ , which intersect at $$(1,2)$$ and $$(3, 10)$$.
Use integration to find the exact area of the shaded region enclosed between the two curves.
The area under the curve $$y=11-\frac{9}{x^2}$$ for $$1\leqslant x \leqslant 3$$
is $$\int_1^3(11-\frac{9}{x^2})\,\mathrm{d}x = (11x + \frac{9}{x})\Big|_1^3 = 33+3-11-9=16$$ and under the curve $$y=x^2+1$$ for $$1\leqslant x \leqslant 3$$ is $$\int_1^3(x^2+1)\,\mathrm{d}x = ({x^3 \over 3} + x)\Big|_1^3 = 9+3-\frac13-1=\frac{32}3$$. Therefore, the area of the shaded region is $$16-\frac{32}3 = \frac{16}3$$.
###### 7.4
Find the integral $$\int x(x^2-4)\,\mathrm dx$$.
Evaluate $$\int_0^2 x(x^2-4)\,\mathrm dx$$.
$$\int_0^2 x(x^2-4)\,\mathrm dx = \int_0^2 (x^3-4x)\,\mathrm dx = \left.(\frac{x^4}{4}-\frac{4x^2}2)\right|_0^2 = \frac{16}{4}-\frac{4\times4}2 = -4$$
###### 7.5
The diagram shows the curve $$y = \sqrt{4x + 1}$$.
Use the trapezium rule, with strips of width $$0.5$$, to find an approximate value for the area of the region bounded by the curve $$y = \sqrt{4x + 1}$$, the $$x\text{-axis}$$, and the lines $$x = 1$$ and $$x = 3$$. Give your answer correct to $$3$$ significant figures.
The area is approximately equal to $$\int_1^{3}\sqrt{4x+1}\,\mathrm{d}x \approx \frac{2}{2\times4}(\sqrt5 + 2\sqrt{7}+2\sqrt{9}+2\sqrt{11}+\sqrt{13}) \approx 11.883$$
###### 7.6
Given that $$\int_a^\infty \frac{24}{x^3}\, \mathrm dx = 3$$.
Find the value of the positive constant $$a$$.
$$\int_a^\infty \frac{24}{x^3}\, \mathrm dx = \int_a^\infty 24x^{-3}\, \mathrm dx = \left. \frac{24x^{-2}}{-2}\right|_a^\infty = 0 - (-12a^{-2}) = \frac{12}{a^2} = 3$$, so
$$3a^2=12\\ a^2=4$$
Since $$a$$ is positive, $$a=2$$.
###### 7.7
Find the integral $$\int (x^{3 \over 2} -1) \, \mathrm dx$$.
Evaluate $$\int_1^4 (x^{3 \over 2} -1) \, \mathrm dx$$.
$$\int_1^4 (x^{3 \over 2} -1) \, \mathrm dx = \left. \left( \frac{x^{5 \over 2}}{5\over 2} -x \right) \right|_1^4 = \frac{2\times 4^{5 \over 2}}{5} - 4 - \left(\frac{2\times 1^{5 \over 2}}{5} -1\right) = \frac{64}{5} - 3 -\frac25 = 12.4-3 = 9.4$$
###### 7.8
The gradient of a curve is given by $$\frac{dy}{dx} = 6x-4$$. The curve passes through the distinct points $$(2, 5)$$ and $$(p, 5)$$
Find the value of $$p$$.
The equation of the curve is $$y = \int (6x-4)\, \mathrm dx = {6x^2 \over 2} - 4x + c = 3x^2-4x + c$$. Since it passes through the point $$(2, 5)$$, one can conclude that
$$3\times2^2-4\times2 + c=5\\ 12-8+c=5\\ c=1$$
So, $$y = 3x^2-4x + 1$$ and
$$3p^2-4p + 1 = 5\\ 3p^2-4p-4=0\\ D=4^2-4\times3\times(-4) = 16+48 = 64\\ p = \frac{4\pm\sqrt{64}}{6} = \frac{4\pm8}{6}$$
Therefore, $$p = 2$$ or $$p = -\frac23$$, but the given points are distinct, so $$p = -\frac23$$.
###### 7.9
Given the region, enclosed by the curve $$y = x^2+4x$$, the $$x\text{-axis}$$ and the lines $$x = 3$$ and $$x = 6$$.
Use integration to find the exact area of the region.
The area is equal to $$\int_3^6(x^2+4x)\, \mathrm dx = \left. (\frac{x^3}{3} + \frac{4x^2}{2}) \right|_3^6 = \frac{216}{3} + 2\times 36 - \frac{27}{3} - 2\times 9 = 72 + 72 - 9 - 18 = 117$$
###### 7.10
Find the integral $$\int 4\sqrt x \, \mathrm dx$$.
Evaluate $$\int_1^9 4\sqrt x \, \mathrm dx$$.
$$\int_1^9 4\sqrt x \, \mathrm dx = \int_1^9 4x^{\frac12} \, \mathrm dx = \left. \frac{4x^{\frac32}}{3\over2} \right|_1^9 = \frac{8\times9^{\frac32}}{3} - \frac{8\times1^{\frac32}}{3} = \frac{8\times 27}{3} - \frac83 = 72-\frac83 = 69\frac13$$ | 1,890 | 4,184 | {"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} | 4.59375 | 5 | CC-MAIN-2017-26 | longest | en | 0.473043 |
https://unapologetic.wordpress.com/2008/07/18/ | 1,498,739,238,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323970.81/warc/CC-MAIN-20170629121355-20170629141355-00014.warc.gz | 816,260,628 | 33,075 | # The Unapologetic Mathematician
## Unsolvable Inhomogenous Systems
We know that when an inhomogenous system has a solution, it has a whole family of them. Given a particular solution, it defines a coset of the subspace of the solutions to the associated homogenous system. And that subspace is the kernel of a certain linear map.
But must there always be a particular solution to begin with? Clearly not. When we first talked about linear systems we mentioned the example
$3x^1+4x^2=12$
$x^1-x^2=11$
$x^1+x^2=10$
In our matrix notation, this reads
$\displaystyle\begin{pmatrix}3&4\\1&-1\\1&1\end{pmatrix}\begin{pmatrix}x^1\\x^2\end{pmatrix}=\begin{pmatrix}12\\11\\10\end{pmatrix}$
or $Ax=b$ in purely algebraic notation.
We saw then that this system has no solutions at all. What’s the problem? Well, we’ve got a linear map $A:\mathbb{F}^2\rightarrow\mathbb{F}^3$. The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) must equal the dimension of the source. But here this dimension is $2$, and so the rank can be at most $2$, which means there must be some vectors $b\in\mathbb{F}^3$ which can’t be written as $b=Ax$ no matter what vector $x$ we pick. And the vector in the example is just such a vector outside the image of $A$.
The upshot is that we can only solve the system $Ax=b$ if the vector $b$ lies in the image of the linear map $A$, and it might be less than obvious what vectors satisfy this requirement. Notice that this is more complicated than the old situation for single equations of single variables. In that case, the target only has one dimension, and the linear transformation “multiply by the number $A$” only misses this dimension if $A=0$, which is easy to recognize.
July 18, 2008 Posted by | Algebra, Linear Algebra | 3 Comments | 498 | 1,836 | {"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": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2017-26 | longest | en | 0.891919 |
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Inverting Triangles (Posted on 2005-08-15)
A piece of paper had the following diagram:
``` o o o o o
From: o o To: o o o
o o o o o
o o o o o
```
Below it, it read "Given the initial formation of ten coins, move exactly # coins to produce the end formation." It was pretty obvious that # stood for a digit, but it was smudged and couldn't be read. What possible numbers could it have been so the problem was solvable?
To allow explaining the solution, number the coins like this:
``` 0
1 2
3 4 5
6 7 8 9
```
Note: This problem was inspired by a forum question by Nicole.
No Solution Yet Submitted by Erik O. Rating: 2.1667 (6 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): 7 ways - your rules | Comment 8 of 14 |
I have no quarrel with Lisa's solutions, just wondering what her thoughts were, and what rules she gave herself to work with.
Erik O. certainly opened up a forum for conjecture which has been valuable to explore.
Posted by brianjn on 2005-08-16 01:42:35
Search: Search body:
Forums (0) | 329 | 1,230 | {"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-2019-09 | latest | en | 0.94212 |
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Jul 21, 2017
See a solution process below:
#### Explanation:
Step 1) Divide each side of the inequality by $\textcolor{red}{7}$ to eliminate the parenthesis while keeping the inequality balanced:
$\frac{7 \left(x + 4\right)}{\textcolor{red}{7}} > \frac{0}{\textcolor{red}{7}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} \left(x + 4\right)}{\cancel{\textcolor{red}{7}}} > 0$
$x + 4 > 0$
Step 2) Subtract $\textcolor{red}{4}$ from each side of the inequality to solve for $x$ while keeping the inequality balanced:
$x + 4 - \textcolor{red}{4} > 0 - \textcolor{red}{4}$
$x + 0 > - 4$
$x > - 4$ | 221 | 645 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 4.71875 | 5 | CC-MAIN-2021-49 | latest | en | 0.64431 |
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# St. JohnтАЩs, Newfoundland, lies on the
Author Message
Intern
Joined: 07 May 2003
Posts: 9
Location: Dallas
Followers: 0
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St. JohnтАЩs, Newfoundland, lies on the [#permalink]
### Show Tags
21 May 2003, 13:38
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
St. JohnтАЩs, Newfoundland, lies on the same latitude
as Paris, France, but in spring St. JohnтАЩs residents are
less likely to be sitting at outdoor cafes than to be
bracing themselves against arctic chills, shoveling
snow, or seeking
shelter from a raging northeast
storm.
(A) residents are less likely to be sitting at outdoor
cafes than to be bracing themselves against
arctic chills, shoveling snow, or seeking
(B) residents are less likely to sit at outdoor cafes,
and more to brace themselves against arctic
chills, shovel snow, or be seeking
(C) residents are less likely to be sitting at outdoor
cafes, and more likely to be bracing themselves
against arctic chills, shoveling snow, or to be
seeking
(D) residents, instead of their sitting at outdoor
cafes, they are more likely to brace themselves
against arctic chills, shovel snow, or seek
(E) residents, instead of sitting at outdoor cafes, are
more likely to brace themselves against arctic
chills, shovel snow, or to be seeking
Founder
Joined: 04 Dec 2002
Posts: 14946
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3957
Kudos [?]: 25160 [0], given: 4759
### Show Tags
21 May 2003, 13:46
I hope the example above is not copyrighted - it is not a copy of what is copyrighted. We try to keep this place clean as to prolong its life and prevent it from being shut down, thus I strictly enforce copyright rules.
Most of the questions here are made up by people who post them. Stolyar is the leader of course, but that's the way we work. Others modify the originals as not to make them recognizable and remove danger of copyright infringement.
If this question is not clean, I will have to kill it.
-BB
Intern
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Posts: 9
Location: Dallas
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### Show Tags
21 May 2003, 14:49
I dont know whether the problem is copyright or not. Since this problem was forwarded to me by my friend for an answer i posted it. But if u think the problem is a copyright u can always kill it. I will be careful next time.
Manager
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22 May 2003, 00:19
I also prefer to choose A.
Is it A?
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22 May 2003, 00:39
No doubt, it is from the gmatplus. As far as I remember, A should be correct.
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22 May 2003, 01:50
A is fine by me also.
As for BB's concern, if this is from GMATplus, then it is not copyrighted !Since GMATplus is itself a Copy of old GMAT questions !
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22 May 2003, 02:54
I need to be more attentive. I feel terrible, this one was not difficult
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13 Dec 2003, 09:37
Can someone please explain why A ? I was thinking C or E
Senior Manager
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### Show Tags
15 Dec 2003, 15:48
ftoor wrote:
Can someone please explain why A ? I was thinking C or E
C is wrong because the correct usage of "less" is "less...than"
E is not parallel
Re: Explain [#permalink] 15 Dec 2003, 15:48
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Topics:
Qn:979 The failing of the book lies not in a lack of 0 07 Nov 2009, 12:32
1 St. Johns, Newfoundland, lies on the same latitude as Paris, 8 08 Apr 2011, 15:49
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The failing of the book lies not in a lack of attention to 4 02 Nov 2008, 03:52
1 St. John's, Newfoundland, lies on teh same latitude as 3 08 Feb 2008, 02:05
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## What is 53 64 on a tape measure?
1 millimeter = 0.03937 inches….Measurement Conversion Table.
____ Inches = ____ Millimetres
= mm
13/16″ 0.8125″ 20.6375 mm
. 0.8268″ 21.0000 mm
53/64″ 0.8281″ 21.0344 mm
## How thick is 0.093 in inches?
Clear (0000) Acrylic Sheet, 0.093 (3/32 inch), 18 inches x 24 inches.
73 – 76 C
70 – 72 C-
67 – 69 D+
63 – 66 D
## Is a D+ a good grade?
A D+ letter grade is equivalent to a 1.3 GPA, or Grade Point Average, on a 4.0 GPA scale, and a percentage grade of 67–69….List of Common GPA Conversions.
C 73–76 2.0
C- 70–72 1.7
D+ 67–69 1.3
D 65–66 1.0
## How bad is a D+?
A D+ is between a 67% and a 70%. This is below average. It isn’t failing yet, but you are below average in your understanding, as reflected by your work on whatever assignment you got the D+ on. Ideally you want to be in the A or B range.
D+ 71-73 1.33
D 68-70 1.00
D- 65-67 0.67
F Below 65 0.00
## Is D passing in college?
Is a D Considered Passing? A letter grade of a D is technically considered passing because it not a failure. A D is any percentage between 60-69%, whereas a failure occurs below 60%. Even though a D is a passing grade, it’s barely passing.
## What is a D+ GPA?
The top grade is an A, which equals 4.0. You calculate your overall GPA by averaging the scores of all your classes….How to Convert Your GPA to a 4.0 Scale.
C- 70-72 1.7
D+ 67-69 1.3
D 65-66 1.0
E/F Below 65 0.0
## Is a GPA of 2.7 good?
Is a 2.7 GPA Good? A 2.7 GPA is noticeably lower than a 3.0 GPA, which is the national average. It’s almost decent, but still a little low. Luckily, a 2.7 is certainly a GPA that can be pulled up to a solid number.
## What is a 95 GPA?
GPA Percentile. Letter. Grade. 4.0 95-100 A. 3.9 94.
## Is a 95 unweighted GPA good?
Second, notice that an unweighted GPA does not take into account the level of the class. Under this system, an A- in honors or advanced placement course is the same 3.7 as an A- in a lower level class….Unweighted GPA.
A 93-96 4.0
A- 90-92 3.7
B+ 87-89 3.3
B 83-86 3.0
## What is a 95 GPA on 5.0 scale?
The Weighted GPA Conversion Chart | 763 | 2,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.902377 |
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1 / 17
Objectives. By the end of this section you should: know how atom positions are denoted by fractional coordinates be able to calculate bond lengths for octahedral and tetrahedral sites in a cube be able to calculate the size of interstitial sites in a cube
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Objectives
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Presentation Transcript
Objectives
By the end of this section you should:
• know how atom positions are denoted by fractional coordinates
• be able to calculate bond lengths for octahedral and tetrahedral sites in a cube
• be able to calculate the size of interstitial sites in a cube
• know what the packing fraction represents
• be able to define and derive packing fractions for 2 different packing regimes
1.
2.
3.
4.
0, 0, 0
½, ½, 0
½, 0, ½
0, ½, ½
Fractional coordinates
Used to locate atoms within unit cell
Note 1: atoms are in contact along face diagonals (close packed)
Note 2: all other positions described by positions above (next unit cell along)
Octahedral Sites
Coordinate ½, ½, ½
Distance = a/2
Coordinate 0, ½, 0 [=1, ½, 0]
Distance = a/2
In a face centred cubic anion array, cation octahedral sites at:
½ ½ ½, ½ 0 0, 0 ½ 0, 0 0 ½
Tetrahedral sites
Relation of a tetrahedron to a cube:
i.e. a cube with alternate corners missing and the tetrahedral site at the body centre
Can divide the f.c.c. unit cell into 8 ‘minicubes’ by bisecting each edge; in the centre of each minicube is a tetrahedral site
So 8 tetrahedral sites in a fcc
Bond lengths
important dimensions in a cube
Face diagonal, fd
(fd) = (a2 + a2) = a 2
Body diagonal, bd
(bd) = (2a2 + a2) = a 3
Bond lengths:
Octahedral:
half cell edge, a/2
Tetrahedral:
quarter of body diagonal, 1/4 of a3
Anion-anion:
half face diagonal,
1/2 of a2
Sizes of interstitials
fcc / ccp
Spheres are in contact along face diagonals
octahedral site, bond distance = a/2
radius of octahedral site = (a/2) - r
tetrahedral site, bond distance = a3/4
radius of tetrahedral site = (a3/4) - r
Summaryf.c.c./c.c.p anions
4 anions per unit cell at:000½½00½½½0½
4 octahedral sites at:½½½00½½000½0
4 tetrahedral T+ sites at:¼¼¼¾¾¼¾¼¾¼¾¾
4 tetrahedral T- sites at:¾¼¼¼¼¾¼¾¼¾¾¾
A variety of different structures form by occupying T+ T- and O sites to differing amounts: they can be empty, part full or full.
We will look at some of these later.
Can also vary the anion stacking sequence - ccp or hcp
Packing Fraction
• We (briefly) mentioned energy considerations in relation to close packing (low energy configuration)
• Rough estimate - C, N, O occupy 20Å3
• Can use this value to estimate unit cell contents
• Useful to examine the efficiency of packing - take c.c.p. (f.c.c.) as example
So the face of the unit cell looks like:
Calculate unit cell side in terms of r:
2a2 = (4r)2
a = 2r 2
Volume = (162) r3
Face centred cubic - so number of atoms per unit cell =corners + face centres = (8 1/8) + (6 1/2) = 4
Packing fraction
The fraction of space which is occupied by atoms is called the “packing fraction”, , for the structure
For cubic close packing:
The spheres have been packed together as closely as possible, resulting in a packing fraction of 0.74
Group exercise:
Calculate the packing fraction for a primitive unit cell
Primitive
Close packing
• Cubic close packing = f.c.c. has =0.74
• Calculation (not done here) shows h.c.p. also has =0.74 - equally efficient close packing
• Primitive is much lower: Lots of space left over!
• A calculation (try for next time) shows that body centred cubic is in between the two values.
• THINK ABOUT THIS! Look at the pictures - the above values should make some physical sense!
Summary
• By understanding the basic geometry of a cube and use of Pythagoras’ theorem, we can calculate the bond lengths in a fcc structure
• As a consequence, we can calculate the radius of the interstitial sites
• we can calculate the packing efficiency for different packed structures
• h.c.p and c.c.p are equally efficient packing schemes | 1,313 | 4,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-17 | longest | en | 0.860753 |
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Exercise 1: Matlab part
1) Plot the function yx=2∙x-x2+sin(2∙x)∙cos(x)
function problem1 (x)
y=2.*x-x.^2+sin(2.*x).*cos(x)
plot(x,y,'r')
end
2) Print n! from n=2 to 20
function problem2
for k=2:20
n=factorial(k)
end
3) Make a function that calculates RSS for a give vector.
end
4) Make a function that check to see if a number is a prime.
function problem4(n)
flag=0;
for k=[2:1:(n/2)]
a=rem(n,k);
if a==0
flag=1;
end
end
if flag==1
disp ('Number is not prime');
else
disp('Number is prime');
end
5) Calculate a 3-degree nominal to fit the following column. M=[-0.447,1.978,3.11,5.25,5.02,4.66,4.01,4.58,3.45,5.35,9.22] Plot the curve of the nomial with the discrete points in the same graph.
m=[-0.447,1.978,3.11,5.25,5.02,4.66,4.01,4.58,3.45,5.35,9.22]; x1=1:numel(m);
p=polyfit(x1,m,3);
x2=0:0.01:(numel(m)+1);
f=polyval(p,x2);
plot(x1,m,'ko',x2,f,'r-')
legend('n column','3-degNom','location','Best')
grid on
6) Plot the following function as follow.
Rasx=20+x12+x22-10(cos2πx1+cos2πx2)
x1 and x2 are from 0 to 4.
x=0:0.05:4;
y=0:0.05:4;
[x,y]=meshgrid(x,y);
z=20+x.*x+y.*y-10*(cos(2*pi*x)+cos(2*pi*y))
surf(x,y,z)
grid on
7) Given the matrix N and P as follow,
N=124736278123179324191347 P=23193829125849228923414913362248
Please calculate the N+P, N-P, N*P and P*N.
Please calculate the inverse matrix and orthogonal matrix of N.
N=[1,24,7,36;
2,7,8,12;
3,17,9,32
4,19,13,47]
P=[23,19,38,29;
12,58,49,22;
89,23,41,49;
13,36,22,48]
sum=N+P
dif=N-P
mulN=N*P
mulP=P*N
inverse=inv(N)
orthogonal=orth(N)
RESULTS:
N =
1 24 7 36
2 7 8 12
3 17 9 32
4 19 13 47
P =
23 19 38 29
12 58 49 22
89 23 41 49
13 36 22 48
sum =
24 43 45 65
14 65 57 34
92 40 50 81
17 55 35 95
dif =
-22 5 -31 7
-10 -51 -41 -10
-86 -6 -32 -17
-9 -17 -9 -1
mulN =
1402 2868 2293 2628
998 1060 1011 1180
1490 2402 2020 2438
2088 3169 2650 3427
mulP =
291 1882 1032 3635
363 1945 1275 3730
454 3925 1813 7095
343 1850 1201 3860
inverse =
-0.4741 -0.3403 1.2821 -0.4229
0.0291 0.0129 0.1715 -0.1424
0.0858 0.2789 -0.3468 0.0992
0.0049 -0.0534 -0.0825 0.0874
orthogonal =
-0.5489 0.7770 0.0379 0.3058
-0.1952 -0.2831 0.9031 0.2571
-0.4733 0.0005 0.1452 -0.8688
-0.6607 -0.5622 -0.4023 0.2924
8) Design a simple calculator with a GUI.
function varargout = mycalc(varargin)
% MYCALC M-file for mycalc.fig
% MYCALC, by itself, creates a new MYCALC or raises the existing % singleton*.
%
% H = MYCALC returns the handle to a new MYCALC or the handle to % the existing singleton*.
%
% MYCALC('CALLBACK',hObject,eventData,handles,...) calls the local % function named CALLBACK in MYCALC.M with the given input arguments. %
% MYCALC('Property','Value',...) creates a new MYCALC or raises the % existing singleton*. Starting from the left, property value pairs are % applied to the GUI before mycalc_OpeningFunction gets called. An % unrecognized property name or invalid value makes property application % stop. All inputs are passed to mycalc_OpeningFcn via varargin. %
% *See GUI Options on GUIDE's Tools menu. Choose "GUI allows only one % instance to run (singleton)".
%
% Edit the... | 1,304 | 3,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-13 | latest | en | 0.483385 |
http://blog.vrplumber.com/b/2005/02/13/how-quickly-one-forgets/ | 1,571,641,526,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00231.warc.gz | 35,298,741 | 7,222 | ## How quickly one forgets (Basic algebra and geometry...) Written by Mike C. Fletcher on Feb. 13, 2005 in Snaking.
Spent the rest of the evening working on tessellation code, but never did get around to fixing the problem with the Jelle-project tessellation. Instead I realised that I had no test cases for a "combining" tessellation. To understand what that is, consider this shape...
When we render this shape, we want it to look something like this:
To make that happen, we have to create a new vertex at the point where the lines AE and BC intersect. This vertex needs to be a blending of the coordinates, texture-coordinates, colours, and normals of the four contributing vertices. Until this evening, there was some grotty, and incorrect, code in the polygon tessellator which was pretending to do a combine of the vertices, but given that it was broken, I think it's safe to assume we never actually had it run :) . This evening I constructed a test case (which is what's shown above) in order to make sure it works correctly. The shallow error in the combiner for the first three attributes of the vertices was easy to fix, but it took me forever and a day to remember the (trivial) mechanism for weighted combination of normals. I spent a very long time trying to figure out a way to work with the normals using quaternions before I remembered the simple key idea: normals are vectors, you combine them by simply adding them (and then normalising the result). Weighted combinations merely add weighted multiples of the normals. Only complication is that if the normals add to [0,0,0] you have to choose an arbitrary vector perpendicular to something, but I just sort of punted on that and made it perpendicular to one contributing vector in the x,y or x,z plane. I figure you'd really want a calculus solution for perfect results, but given how rarely this code is needed it's probably not worth it to solve the problem this evening. You can see the combination of normals in the top shape. The original points all have normals that project forward and away from the center, but the combined point projects pretty much directly forward, that's why the center point is brighter than the outside (the light is shining in the same direction the camera is pointing). Also, realised that there is an over-optimisation in the polygon tessellation code. It takes any four-vertex contour and converts it to two triangles regardless of the position of the vertices. It really should be checking first to see if the vertices are coplanar, and if they are, whether they are producing two polygons with the same winding order (normal) or not, if not, they should likely be going through the tessellator to produce a shape similar to that above. Oh well, something to work on some other time. | 592 | 2,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-43 | longest | en | 0.948224 |
https://citizenmaths.com/angle/1140.2-gradians-to-quadrants | 1,632,158,487,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00047.warc.gz | 209,746,963 | 11,701 | # 1140.2 Gradians to Quadrants
Gradian
• Angular Mil
• Binary degree
• Centesimal minute of arc
• Centesimal second of arc
• Centiturn
• Degree
• Diameter Part
• Gradian
• Hexacontade
• Hour Angle
• Milliradian
• Milliturn
• Minute of arc
• Minute of time
• Octant
• Point
• Quadrant
• Quarter Point
• Radian
• Second of arc
• Second of time
• Sextant
• Sign
• Turn
=
Quadrant
• Angular Mil
• Binary degree
• Centesimal minute of arc
• Centesimal second of arc
• Centiturn
• Degree
• Diameter Part
• Gradian
• Hexacontade
• Hour Angle
• Milliradian
• Milliturn
• Minute of arc
• Minute of time
• Octant
• Point
• Quadrant
• Quarter Point
• Radian
• Second of arc
• Second of time
• Sextant
• Sign
• Turn
Formula 1,140.2 grad = 1140.2 / 100 quad = 11.402 quad
1,140.2 grad = 11.402 quad
Explanation:
• 1 grad is equal to 0.01 quad, therefore 1140.2 grad is equivalent to 11.402 quad.
• 1 Gradian = 1 / 100 = 0.01 Quadrants
• 1,140.2 Gradians = 1140.2 / 100 = 11.402 Quadrants
## 1140.2 Gradians to Quadrants Conversion Table
Gradian (grad) Quadrant (quad)
1,140.3 grad 11.403 quad
1,140.4 grad 11.404 quad
1,140.5 grad 11.405 quad
1,140.6 grad 11.406 quad
1,140.7 grad 11.407 quad
1,140.8 grad 11.408 quad
1,140.9 grad 11.409 quad
1,141 grad 11.41 quad
1,141.1 grad 11.411 quad
## Convert 1140.2 grad to other units
Unit Unit of Angle
Second of time 246,283.2 s ot
Quarter Point 364.86 qtr point
Point 91.216 point
Minute of time 4,104.72 min ot
Milliturn 2,850.5 mltr
Hour Angle 68.412 HA
Hexacontade 171.03 hc
Diameter Part 1,074.61 Ø dia- part
Centiturn 285.05 centiturn
Centesimal second of arc 11,402,000.0 c SOA
Centesimal minute of arc 114,020.0 c MOA
Angular Mil 18,243.2 µ
Sign 34.206 sign
Octant 22.804 octa
Milliradian 17,910.23 mrad
Turn 2.8505 tr
Sextant 17.103 sxt
Quadrant 11.402 quad
Binary degree 729.73 brad
Second of arc 3,694,248.0 arcsecond
Minute of arc 61,570.8 arcmin
Radian 17.91 rad
Degree 1,026.18 °
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# What is 30% off 210 Pounds
An item that costs £210, when discounted 30 percent, will cost £147
The easiest way of calculating discount is, in this case, to multiply the normal price £210 by 30 then divide it by one hundred. So, the discount is equal to £63. To calculate the sales price, simply deduct the discount of \$63 from the original price £210 then get £147 as the sales price.
### Inputs:
Original Price of the Item: £
Discount Percent (% off): %
### Results:
Amount Saved (Discount): £
Sale / Discounted Price: £
## How to Calculate Discounts - Step-by-Step Solution
To calculate percent off use the following equations:
(1) Amount Saved = Original Price x Discount % / 100
(2) Sale Price = Original Price - Amount Saved
Here are the solutions to the questions stated above:
### 1) What is 30 percent (%) off £210?
Using the formula one and replacing the given values:
Amount Saved = Original Price x Discount % / 100. So,
Amount Saved = 210 x 30 / 100
Amount Saved = 6300 / 100
In other words, a 30% discount for an item with original price of £210 is equal to £63 (Amount Saved).
Note that to find the amount saved, just multiply it by the percentage and divide by 100.
Supose Have you received a ROBLOX promotional code of 30 percent of discount. If the price is £210 what is the sales price:
### 2) How much to pay for an item of £210 when discounted 30 percent (%)? What is item's sale price?
Using the formula two and replacing the given values:
Sale Price = Original Price - Amount Saved. So,
Sale Price = 210 - 63
This means, the cost of the item to you is £147.
You will pay £147 for an item with original price of £210 when discounted 30%. In other words, if you buy an item at £210 with 30% discounts, you pay £210 - 63 = £147
Supose Have you received a amazon promo code of 63. If the price is £210 what was the amount saved in percent:
### 3) 63 is what percent off £210?
Using the formula two and replacing the given values:
Amount Saved = Original Price x Discount % /100. So,
63 = 210 x Discount % / 100
63 / 210 = Discount % /100
100 x 63 / 210 = Discount %
6300 / 210 = Discount %, or
To find more examples, just choose one at the bottom of this page.
## Unit Converters
### Disclaimer
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https://www.cfd-online.com/Forums/ansys-meshing/104032-cylinder-structured-mesh.html | 1,709,134,165,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00106.warc.gz | 709,559,737 | 18,624 | # [GAMBIT] cylinder structured mesh
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July 2, 2012, 08:00
cylinder structured mesh
#1
New Member
masood
Join Date: May 2012
Posts: 18
Rep Power: 13
hi .
how can i generate a mesh like this in gambit ??
i dont know how to add nodes to have such a mesh !!
any help or tutorial appreciated .
Attached Images
dim_2.jpg (9.2 KB, 636 views) 1.jpg (80.0 KB, 163 views)
July 2, 2012, 08:21 #2 Super Moderator Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 3,297 Rep Power: 40 on your first picture you can see a square at center from disk. Let assume you only have the disk (2d) the decomposition will be a square and then join corners from square to the circle (you need to split the circle with some points) Sans titre.jpg __________________ In memory of my friend Hervé: CFD engineer & freerider
July 2, 2012, 10:05
#3
New Member
masood
Join Date: May 2012
Posts: 18
Rep Power: 13
Quote:
Originally Posted by -mAx- on your first picture you can see a square at center from disk. Let assume you only have the disk (2d) the decomposition will be a square and then join corners from square to the circle (you need to split the circle with some points) Attachment 14160
hi max . thanks for yoyr help . i did it .
Attached Images
1.jpg (68.5 KB, 195 views)
July 30, 2012, 17:11 #4 New Member zainab Join Date: Mar 2012 Posts: 17 Rep Power: 13 hi I have the same problem. please, if you get any solution tell me.
July 31, 2012, 05:13
#5
New Member
masood
Join Date: May 2012
Posts: 18
Rep Power: 13
Quote:
Originally Posted by zainab hi I have the same problem. please, if you get any solution tell me.
hi zeinab . while in gambit draw a cylinder then split it into 4 equal parts , make a cube at center of cylinder ,
subtract cube from the cylindrical parts
now use volume mesh >>hex/map
remember to connect the faces .
u can do this in 2d or 3d
August 1, 2012, 11:20 #6 New Member zainab Join Date: Mar 2012 Posts: 17 Rep Power: 13 Hi Masood, thank you for your help I working as you tell me, finally i get a five meshed domains, when I wont to export the mesh to FLUENT , must unite all the volumes to be one cylindrical volume to specify the regions, inlet, outlet, etc. when I do this step the mesh deleted. what can I do
August 1, 2012, 11:55 #7 Super Moderator Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,552 Blog Entries: 6 Rep Power: 53 No need to unite them. Go to boundary condition panel and select all five faces on one side as inlet and corresponding five faces at other side as outlet. Similarly for the outer wall of the cylinder
October 9, 2012, 11:01
please see the picture and help me
#8
New Member
wesam
Join Date: Sep 2012
Posts: 23
Rep Power: 13
Quote:
Originally Posted by masoodina hi zeinab . while in gambit draw a cylinder then split it into 4 equal parts , make a cube at center of cylinder , subtract cube from the cylindrical parts now use volume mesh >>hex/map remember to connect the faces . u can do this in 2d or 3d
....
i need the steps to make a mesh like in pic by gambit . please i need help
Attached Files
c-m.zip (31.5 KB, 101 views)
October 10, 2012, 06:52 #9 Member Tamil Nadu Join Date: Oct 2012 Posts: 44 Rep Power: 13 Hello All, I want to do analysis of heat transfer from water flowing through pipes submerged inside concrete. I am modelling in GAMBIT and wish to analyse it on Ansys FLUENT. Can anybody help me out, how to model and simulate? Does any tutorials exist?
May 27, 2014, 02:26 #10 Member shahrbanoo Join Date: Mar 2013 Posts: 63 Rep Power: 12 Hi I made this mesh on 2d circle, but when I change it to 3d mesh I have negative volume . how can I make it better not to have negative volume? and why I have them?
May 27, 2014, 04:31 #11 Super Moderator Maxime Perelli Join Date: Mar 2009 Location: Switzerland Posts: 3,297 Rep Power: 40 display your geometry __________________ In memory of my friend Hervé: CFD engineer & freerider | 1,147 | 4,021 | {"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-2024-10 | latest | en | 0.861995 |
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# oct chemistry help !! Watch
1. An industrial chemist carries out some research into the NO/O2/NO2 equilibrium used in Stage 2 of the manufacture of nitric acid.
• The chemist mixes together 0.80mol NO(g) and 0.70 mol of O2(g) in a container with a volume of 2.0dm3.
• The chemist heats the mixture and allows it to stand at constant temperature to reach equilibrium.
The container is kept under pressure so that the total volume is maintained at 2.0 dm3.
• At equilibrium, 75% of the NO has reacted.
The question is to work out the KC value.. I'm really stuck
I got the moles at equilibrium of NO2 as 0.60… meaning a 0.20 difference. But when you take 0.20 from 0.70 you get 0.50, but the mark scheme is stating the moles of 02 at equilibrium are 0.40. I'm probably doing this wrong, or being really stupid but any help would really be appreciated
2. (Original post by ellie2996)
An industrial chemist carries out some research into the NO/O2/NO2 equilibrium used in Stage 2 of the manufacture of nitric acid.
• The chemist mixes together 0.80mol NO(g) and 0.70 mol of O2(g) in a container with a volume of 2.0dm3.
• The chemist heats the mixture and allows it to stand at constant temperature to reach equilibrium.
The container is kept under pressure so that the total volume is maintained at 2.0 dm3.
• At equilibrium, 75% of the NO has reacted.
The question is to work out the KC value.. I'm really stuck
I got the moles at equilibrium of NO2 as 0.60… meaning a 0.20 difference. But when you take 0.20 from 0.70 you get 0.50, but the mark scheme is stating the moles of 02 at equilibrium are 0.40. I'm probably doing this wrong, or being really stupid but any help would really be appreciated
Hey, you nearly got it right.
2NO + O2 → 2NO2
I first found the moles of NO that reacted as 0.6. So that means 0.6 moles of NO2 are present at equilibrium.
If 0.6 of of NO reacted, it must mean that 0.2 moles of NO are left at equilibria (0.8-0.6=0.2)
Now all you need to find is the equilibrium moles of oxygen. You know that 0.6moles of NO reacted, and from the equation you can see that only half the moles of oxygen would have reacted. So 0.3 moles of oxygen reacted. Which leaves 0.4 moles of oxygen (0.7-0.3=0.4)
Now you just put it into the Kc equation.
So Kc = [NO2]2 divided by [NO]2 [O2]
make sure when you put the values into the equation, you divide by the volume to find the concentrations.
Hope I helped
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26
If $x$ is binary: Then the "if" condition really means either "$x = 0$" or "$x=1$". To enforce "if $x=0$ then $y=1$": use $$y \ge 1-x.$$ To enforce "if $x=1$ then $y=1$": use $$y \ge x.$$ If you want to require that $y=1$ if and only if the condition holds, then replace the $\ge$s above with $=$s. If $x$ is continuous: In this case, numerical inaccuracy ...
26
Interesting topic (the question was raised several times by my students as well). My short answer is that adding the lower bound through a cut seems a good idea at first glance, but it creates a very large “unnatural” face where your search is trapped for a long while. Essentially you lose the objective function grip, and do not gain anything. Let me ...
20
For books with a focus on industrial applications, see this other question of this forum As textbooks, I would recommend to have a look at: General Intro to OR: W. Winston. Operations Research: Applications and Algorithms (4th Ed.). Brooks/Cole. 2004. Modeling: H.P. Williams. Model building in mathematical programming. John Wiley & Sons, 2013. D. Chen, R....
20
They are not the same thing. Lagrangian decomposition is a special case of Lagrangian relaxation. (Note: I'm talking specifically about integer programming problems in this answer, though some of this answer applies to continuous optimization as well.) Lagrangian relaxation involves removing (relaxing) one or more constraints and penalizing violations of ...
20
Here, in approximate order, are my criteria. Do I need a provably optimal solution (which rules out metaheuristics, other than to generate an initial feasible solution)? Is this something CPLEX can handle (since I have a license for CPLEX and I'm familiar with it)? If CPLEX can handle it, should I consider a heuristic, metaheuristic or constraint solver to ...
18
Here is the advice in the IBM CPLEX documentation. So this pertains to CPLEX. I don't know to what extent it applies to other solvers. First of all, indicator constraints may not be available in all situations: Indicator Constraints in Optimization The constraint must be linear; a quadratic constraint is not allowed to have an indicator constraint. ...
17
I learned very early (this may not be generally true) that I should always prefer binary over integer variables. A reason is that from binary values you can infer logical information, branching on a binary variable fixes its value (=reduces the model) etc. I would go even further. It may be better to have more variables. Why? Of course, this depends on the ...
16
You have asked a broad question, so I will provide a broad answer. Integer programming typically refers to integer linear programming which is a mathematical modeling and solution paradigm. Decisions are modeled as a vector of real numbers, some of which are further constrained to take only integer values. The decision vector is constrained to satisfy a ...
16
Derivation via conjunctive normal form: $$x_1 \implies \underset{i=2}{\overset n{\lor}} x_i \\ \neg x_1 \bigvee \underset{i=2}{\overset n{\lor}} x_i \\ 1 - x_1 + \sum_{i=2}^n x_i \ge 1 \\ x_1 \le \sum_{i=2}^n x_i$$
16
Option 1: Submit as is to a solver which can globally optimize MIQPs having non-convex objective, and which might reformulate to a linearized MILP model under the hood. Such solvers include CPLEX, Gurobi 9.x, and BARON, among others. Option 2: Step 1 Linearize the products of binary variables, per How to linearize the product of two binary variables? . <...
15
For Gurobi there seems to be a dual advantage of using general constraints (http://www.gurobi.com/documentation/8.1/refman/constraints.html#subsubsection:GeneralConstraints): Benefit number one - models are easier to create and can be interpreted easily: If a model contains general constraints, then Gurobi adds the respective MIP formulations for those ...
15
I going to assume that the ratio $L(x)/Q(x)$ is nonnegative. If it can be negative, I think there may be a workaround, but this will complicated enough without dealing with that. I'm also going to assume that $Q(x)$ and $L(x)/Q(x)$ have a priori upper and lower bounds, say $\underline{Q} \le Q(x) \le \overline{Q}$ and $L(x)/Q(x) \in \{1,\dots,N\}$. You can ...
15
As far as I know, it is not possible to fix any variables solely based on a feasible solution without compromising the exactness of your solution method. However, variable fixing is possible when you have both an upper bound and a lower bound on the optimal objective value, using a method called reduced cost fixing (see e.g. Atamtürk, Nemhauser & ...
15
A similar idea as suggested by @ RolfvanLieshout uses Lagrangian duals instead of LP duals, in a Lagrangian-based branch-and-bound scheme. For example, in the uncapacitated fixed-charge location problem (UFLP), the most common Lagrangian approach relaxes the assignment constraints ($\sum_j y_{ij} = 1 \ \forall i$), uses the Lagrangian subproblem to calculate ...
14
Feels like you are asking two things, tractability of convex problems and convexity of integer problems. A first order approximation is that convex programs are tractable, .i.e., most problems you can think of as a layman in the field that are convex, are (probably) tractable to solve. That's why you would be told that in an introductory course on convex ...
14
It is a difference whether one can dualize (or not) or that a duality theory holds (or not). Formally, you can formulate a dual of any integer program, e.g., by considering the linear relaxation, dualizing it, and then enforcing integrality again on the dual variables. It is already trickier which variables to consider as integer in the dual when you dualize ...
14
Recognize that each route can be viewed as being a node on a graph. Edges connect nodes if the routes the nodes represent intersect. This is the canonical graph coloring problem for which there are a number of exact and approximate algorithms. Specifically, you're trying to find a constructive algorithm for determining the chromatic number. For 10 routes ...
13
The slow convergence of the Gomory cuts was well-known and source of frustration for the field up until the 90s. It seemed that Gomory cuts would be a cute idea, but not one that would lead to any real computational success. Then work by Balas, Ceria, Cornuejols, and Natraj rekindled interest in the area, and Gomory cuts became very important in real ...
13
To the best of my knowledge the indicator constraints are just syntactic sugar for the user. Internally these indicator constraints are reformulated using computed big-M formulations or SOS constraints (special ordered set constraints). It might be that you are better at computing the value of the big-M using additional knowledge that the solver does not ...
13
Yes - such a question can be answered by looking at the irreducible inconsistent subsystem (IIS). From the Gurobi documentation: An IIS is a subset of the constraints and variable bounds with the following properties: the subsystem represented by the IIS is infeasible, and if any of the constraints or bounds of the IIS is removed, the subsystem ...
13
In general no, these problems are hard. BUT: You might want to look into totally unimodular matrices and total dual integrality but this requires additional assumptions on the matrix or the problem respectively. If you are lucky, then your problem has these properties and you can solve it efficiently. Totally Unimodular Matrices A totally unimodular ...
13
Maybe I am missing something but it looks like there is no need for a library: \begin{align} \sum_i \sum_j \sum_k x_{ji} y_{kj} cost(i,k)&=\sum_i \sum_j x_{ji} \sum_k y_{kj} cost(i,k) \end{align} Now since $\sum_k y_{kj}=1$, exactly one row is 1, the others zero. We pick the best one: $$=\sum_i \sum_j x_{ji} \max_k cost(i,k)$$ Since $\sum_j x_{ji}=1$ we ...
12
Rather than linearising the logical constraint, I would try the logical constraints built in a solver. Gurobi and SCIP both have indicator constraints. My colleague works with these a lot and he’s finding the indicator constraints in Gurobi perform worse than big-M. He’s in contact with the Gurobi developers so I might be able to get more info if there’s ...
12
These are know as "indicator constraints" or "on/off" constraints. The best formulation is the convex-hull one, it includes the optimal big-M value plus additional non-redundant constraints, here's a note characterizing this formulation. There's also a generalization for convex nonlinear "on/off" constraints here and recent extensions here.
12
Let $M$ be a new parameter (constant) that equals a large number. Greater-than-or-equal-to constraints: The constraint is $a_1x_1 + \cdots + a_nx_n \ge b$. Rewrite it as $$a_1x_1 + \cdots + a_nx_n \ge b - M(1-y).$$ Then, if $y = 1$, the constraint is active, and if $y=0$, it has no effect since the right-hand side is very negative. (If all of the $a_i$ ...
12
This is quite a broad question, and I'm not sure you'll get the answers you need here. To learn how to formulate mathematical optimization problems takes learning and practice, and there's no concrete answer that we can provide on a Q&A site like this. Your best bet is to take a class that teaches or uses optimization modeling, or to learn from a good ...
11
I cannot speak for Gurobi, but CPLEX definitely has this capability, and my guess is that any solver library does. For a standalone solver (as opposed to one with an API), you might have to look for the best bound in the output log and do some text scraping to get it. For the Java API to CPLEX, the function you want is IloCplex.getBestObjValue(), which ...
11
One definition of quadratization (perhaps there is more) is provided in the paper by Boros, 2018. In non-mathematical terms, quadratization is defined as a quadratic reformulation of the nonlinear problem obtained by introducing a set of auxiliary binary variables which can be optimized using quadratic optimization techniques. Rewriting this in ...
11
Branch-and-bound solvers often use node lower bounds to select the next node to process, e.g. in a best-first search. An external lower bound can lead to a different search order, and thus you may have to explore a different number nodes until finding an optimal solution, and proving its optimality. For concreteness imagine a simple depth $4$ binary search ...
11
Since you mentioned videos, I would suggest checking Coursera and edX for free MOOCs (such as the Deterministic Optimization course on edX or the Discrete Optimization course on Coursera). The Coursera site in particular has a nice search engine that makes finding courses easy. I have not done any of the optimization courses there, but I have done a couple ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2,568 | 10,861 | {"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": 1, "equation": 1, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.904898 |
https://www.jiskha.com/display.cgi?id=1262737001 | 1,503,290,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107490.42/warc/CC-MAIN-20170821041654-20170821061654-00426.warc.gz | 918,119,207 | 4,120 | # math
posted by .
a rectangular container with dimensions 10 inches by 15 inches by 20 inches is to be filled with water, using a cylindrical cup whose radius is 2 inches and whose height is 5 inches, what is the maximum number of full cups of water that can be placed in the container without it overflowing?
• Volumes -
Calculate the volume of the container:
Vc = 10*15*20 = 3000 in³
Calculate the volume of the cylindrical cup:
Vp= πr²h = π*2²*5 = 62.832 in³
Maximum number of cups without overflow
= 3000/62.832
= 47.75 cups
Therefore, up to 47 cups, the container will not overflow.
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More Similar Questions | 693 | 2,793 | {"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-2017-34 | latest | en | 0.899291 |
http://avishek.net/blog/?tag=visualisation | 1,563,743,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00180.warc.gz | 15,838,776 | 11,849 | # A Story about Data, Part 1: The shape of the data
Note about the visualisations: All of the plotting was done with Basis-Processing. You’ll find its source here.
The current dataset that I’m working comes from the education domain. Roughly, there are 29000 records, each record lists the following:
• Location of the student’s school
• Language of the student
• Student’s score before intervention
• Student’s score after intervention
# Interacting with Graphs : Mouse-over and lambda-queuer
In the previous post, I described how I’d put together a basic system to drive data selection/exploration through a queue. While generating more graphs, it became evident that the code for mouseover interaction followed a specific pattern. More importantly, using Basis to plot stuff, mandated that I look at the inverse problem; namely, determining the original point from the point under the mouse pointer. In this case, it was pretty simple, since I’m only dealing with 2D points. Here’s a video of how it looks like. The example shows the exploration of a covariance matrix.
# Driving data visualisation over a queue using RabbitMQ and lambda-queuer
One of the things which has bothered me ever since I took the dive into visualisation is the problem of interactivity. The aim of interacting with a visualisation is to drill down or explore areas of the visualisation which are (or seem) interesting. Put another way, we are essentially filtering the data from a visual standpoint. In most cases, mouse interactions may be sufficient. But what if I wanted to be able to filter the data programmatically and have the result reflected in the visualisation?
One way is to simply re-run the code which generates the visualisation each time we use a different filter. This is the simplest, and, in many cases, enough. In this case, the modification to the code is made in an offline fashion. What if we wanted to do the same, but while the program is running? This describes my attempt at one such implementation. Albeit still somewhat primitive, we’ll see where it ends up. For the purposes of demonstration, I used the Parallel Coordinates visualisation, which is available on GitHub. I’ll continue using Processing through Ruby-Processing for this description.
Continue reading Driving data visualisation over a queue using RabbitMQ and lambda-queuer
# A guide to using Basis (updated for v0.6.0+)
This is a quick tour of Basis. Find the source for Basis on GitHub. Installing Basis is pretty simple; just grab it as a gem for your JRuby installation. Brief notes on the installation can be found here.
UPDATE: Starting from version 0.6.0, Basis allows you to specify axis labels. Additionally, you can specify arrays of points instead of plotting points one at a time. When you do this, you can also specify a corresponding legend string, which will show up in a legend guide. See below for more details.
UPDATE: Starting from version 0.5.9, you can turn grid lines on or off. Additionally, the matrix operations implementation has been ported to use the Matrix class in Ruby’s stdlib.
UPDATE: Starting from version 0.5.8, you can customise axis labels, draw arbitrary shapes/text/plot custom graphics at any point in your coordinate system. See below for more details.
UPDATE: With version 0.5.7, experimental support has been added for drawing objects which aren’t points. Interactions with such objects is currently not supported. Additional support for drawing markers/highlighting in custom bases is now in.
UPDATE: Starting from version 0.5.1, Basis has been ported to Ruby 1.9.2, because of the kd-tree library dependency. Currently, there are no plans of maintaining Basis compatibility with Ruby 1.8.x. As an aside, I personally recommend using RVM to manage the mess of Ruby/JRuby installations that you’re likely to have on your machine.
UPDATE: Basis has hit version 0.5.0 with experimental support for mouseover interactivity. More work is incoming, but the demo code below is up-to-date, for now. The code below should be the same as demo.rb on GitHub.
Continue reading A guide to using Basis (updated for v0.6.0+)
# Data interactions in parallel coordinates: 40x-60x speedup
This is an update on the visualisation post on parallel coordinates. Understanding the Processing model made me realise that it probably wasn’t a good idea to draw all the samples each time draw() was called. Of course, every refreshed call of draw() does not clear away the previous frame’s graphics, so that just makes it easier. In the end, I went and explicitly drew only the samples which were under the current mouse position.
The speedup is obvious and massive: whereas the previous version worked well with only 300 samples, the current one processes 18000 samples without breaking a sweat. At 29,000 samples, there is a bit of a slowdown, but only just a bit, you wait 1 second for the highlighting instead of 6-7.
Here’s the new video, using 18k samples. Notice how much denser the mesh is.
# Data interactions in parallel coordinates
Processing is growing on me. Inspired by the different and (very) interesting data visualisation examples I’ve seen, I decided to take a shot at interacting with the parallel coordinates that I generated here. Of course, I had to reduce the number of samples for this demonstration; it’d slow to a unholy crawl otherwise. For this video, I’ve taken 300 samples. The interaction is essentially a mouse-hover highlighting of any sample(s) under it. I fiddled with the colors a bit, but decided that a white-on-greyscale scheme would show up better.
Of course, I still haven’t gotten around to labeling the axes. This I’ll probably pick up next. But as the video demonstrates, there’s a lot to Processing than meets the eye.
PS: By the way, the actual demonstration ends around the halfway mark; I was trying to figure out how to stop the bloody recorder.
# Getting ActiveRecord to behave nicely with Ruby-Processing in JRuby
Really, all I wanted to do was use Processing from Ruby. jashkenas has kindly written a gem which does just that. There was only a slight wrinkle: I wanted to pull my data from MySQL through ActiveRecord. Well, JRuby makes this process slightly more interesting than usual, so I document the process here. To start off with, install the gem with:
`sudo gem install ruby-processing`
Go into the directory where the ruby-processing gem is installed, and from there into {ruby-processing.gem.dir}/lib/core. In my case, this was /usr/lib/ruby/gems/1.8/gems/ruby-processing-1.0.9/lib/core.
Once inside there, you’ll find a file named jruby-complete.jar. Get rid of it, because we’ll be replacing it with a fresh (and different) version of jruby-complete.jar. Download the 1.6.3 version of JRuby-complete jar file. Rename it to jruby-complete.jar and put it in place of the jarfile we just deleted.
One step remains: this jarfile does not contain the activerecord-jdbcmysql-adapter gem. Install that with:
`java -jar jruby-complete.jar -S gem install activerecord-jdbcmysql-adapter --user-install`
You’re good to go now. One more thing, just remember to replace the ActiveRecord adapter string with “jdbcmysql” and allow usage of that gem in your code with:
`require 'arjdbc'`
.
# Parallel Coordinate visualisation of 28k, 5-dimensional data
This is the visualisation of the same dataset that I’ve been working on for a while, for exploring different data mining and visualisation techniques. Currently, the axes aren’t labelled, and the color coding is for different categories. Looks like a really interesting way to explore the data.
# A detour through data visualisation
I should have seen it coming. Text communicates well – up to a point. All the current analyses I’ve been working on, starting from Self-Organising Maps to Decision Trees, are very well served by good, solid visualisation. My current need is a way to visualise data structures effectively, even if it is merely a bunch of nodes which can be expanded/collapsed to show more information. Additionally, it would be nice (not necessary) for this to happen interactively, but I don’t mind a command line-driven approach. In fact, I prefer the command line; makes it easier to drive it through a scripting language like Ruby.
So far, I’ve looked at Processing, d3.js and ProtoVis. I like the idea of d3.js: the data-driven approach makes a lot of sense, but I think I need to refresh quite a bit of CSS-fu to take advantage of its capabilities. Apart from visualisation derived from data mining algorithms, showing the data as-is in an aesthetic manner is also a worthy goal at this point. In particular, the parallel coordinates visualisation caught my eye.
Oh well, at least I know what I’ll be doing for the next few days. | 1,934 | 8,778 | {"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-2019-30 | latest | en | 0.923252 |
http://www.geeksforgeeks.org/find-highest-occurring-digit-prime-numbers-range/ | 1,511,211,301,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00293.warc.gz | 406,513,202 | 17,204 | # Find the highest occurring digit in prime numbers in a range
Given a range L to R, the task is to find the highest occurring digit in prime numbers lie between L and R (both inclusive). If multiple digits have same highest frequency print the largest of them. If no prime number occurs between L and R, output -1.
Examples:
```Input : L = 1 and R = 20.
Output : 1
Prime number between 1 and 20 are 2, 3, 5, 7, 11, 13, 17, 19.
1 occur maximum i.e 5 times among 0 to 9.
```
## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
The idea is to start from L to R, check if the number is prime or not. If prime then increment the frequency of digits (using array) present in the prime number. To check if number is prime or not we can use Sieve of Eratosthenes.
Below is C++ implementation of this approach:
```// C++ program to find the highest occurring digit
// in prime numbers in a range L to R.
#include<bits/stdc++.h>
using namespace std;
// Sieve of Eratosthenes
void sieve(bool prime[], int n)
{
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == false)
for (int i = p*2; i <= n; i+=p)
prime[i] = true;
}
}
// Returns maximum occurring digits in primes
// from l to r.
int maxDigitInPrimes(int L, int R)
{
bool prime[R+1];
memset(prime, 0, sizeof(prime));
// Finding the prime number up to R.
sieve(prime, R);
// Initialse frequency of all digit to 0.
int freq[10] = { 0 };
int val;
// For all number between L to R, check if prime
// or not. If prime, incrementing the frequency
// of digits present in the prime number.
for (int i = L; i <= R; i++)
{
if (!prime[i])
{
int p = i; // If i is prime
while (p)
{
freq[p%10]++;
p /= 10;
}
}
}
// Finding digit with highest frequency.
int max = freq[0], ans = 0;
for (int j = 1; j < 10; j++)
{
if (max <= freq[j])
{
max = freq[j];
ans = j;
}
}
return ans;
}
// Driven Program
int main()
{
int L = 1, R = 20;
cout << maxDigitInPrimes(L, R) << endl;
return 0;
}
```
Output:
```1
```
This article is contributed by >Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
# GATE CS Corner Company Wise Coding Practice
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
2.4 Average Difficulty : 2.4/5.0
Based on 5 vote(s) | 732 | 2,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-47 | latest | en | 0.685912 |
https://byjusexamprep.com/defence/during-an-experiment-the-students-were-asked-to-prepare-a-10-massmass-solution-of-sugar-in-water-ramesh-dissolved-10g-of-sugar-in-100g-of-water-while-sarika-prepared-it-by-dissolving-10g | 1,721,581,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00299.warc.gz | 131,610,790 | 28,232 | # During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
(a) Are the two solutions of the same concentration
(b) Compare the mass % of the two solutions.
Solution:
(a) The two solutions are of the same concentration
(b) The mass % in the case of Sarika is higher than in Ramesh.
examining both solutions:
Let us first consider Ramesh’s solution:
To Ramesh,
• The mass of solute(sugar) = 10g
• The mass of solvent(water) = 100g
We know that:
Mass of solution = mass of solute + mass of solvent
Substituting the values we get:
So, Mass of solution = 10 + 100 = 110g
We know that:
Mass % = mass of solute/ mass of solution x 100
Substituting the values we get:
Mass % = 10/110 x 100 = 9.09%
Now consider Sarika’s solution:
For Sarika,
• The mass of Solute(sugar) = 10g
• The mass of the solution = 100g
We know that:
So, the mass of solvent in the case of Sarika = 100 – 10 = 90g
We know that:
Mass % = mass of solute/ mass of solution x 100
Substituting the values we get:
Mass % = 10/100 x 100 = 10%
Part (a): determining whether the concentrations of the two solutions are equal:
Ramesh’s solution has a concentration of 9.09%.
10% is how concentrated Sarika’s solution is.
As a result, the concentration is different in both circumstances.
Part (b): Compare the mass %
Ramesh’s solution has a mass percentage of 9.09%.
Sarika’s solution has a mass percentage of 10%.
Because of this, Sarika’s mass% is higher than Ramesh’s.
Summary:
## During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.
(a) The two solutions are of the same concentration
(b) The mass % in the case of Sarika is higher than in Ramesh.
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GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com | 634 | 2,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.916176 |
https://www.esaral.com/q/find-the-conjugate-of-each-of-the-following-36492 | 1,726,039,362,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00034.warc.gz | 719,624,838 | 11,529 | # Find the conjugate of each of the following:
Question:
Find the conjugate of each of the following:
$\frac{1}{(4+3 i)}$
Solution:
Given: $\frac{1}{4+3 i}$
First, we calculate $\frac{1}{4+3 i}$ and then find its conjugate
Now, rationalizing
$=\frac{1}{4+3 i} \times \frac{4-3 i}{4-3 i}$
$=\frac{4-3 i}{(4+3 i)(4-3 i)}$
Now, we know that,
$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
So, eq. (i) become
$=\frac{4-3 i}{(4)^{2}-(3 i)^{2}}$
$=\frac{4-3 i}{16-9 i^{2}}$
$=\frac{4-3 i}{16-9(-1)}\left[\because i^{2}=-1\right]$
$=\frac{4-3 i}{16+9}$
$=\frac{4-3 i}{25}$
$=\frac{4}{25}-\frac{3}{25} i$
Hence, $\frac{1}{4+3 i}=\frac{4}{25}-\frac{3}{25} i$
So, a conjugate of $\frac{1}{4+3 i}$ is $\frac{4}{25}+\frac{3}{25} i$ | 334 | 730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.429807 |
https://ars2018.org/and-pdf/1269-list-of-squares-and-cubes-of-numbers-pdf-33-103.php | 1,643,149,138,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00030.warc.gz | 181,448,837 | 7,094 | # List of squares and cubes of numbers pdf
File Name: list of squares and cubes of numbers .zip
Size: 1394Kb
Published: 19.11.2020
## The Square and Cube of First 100 Numbers
A square number, also called a perfect square , is a figurate number of the form , where is an integer. The square numbers for , 1,
### The Square and Cube of First 100 Numbers
Best tricks to calculate cube roots: In the Numerical Ability section of Banking exams, calculation plays an important role to score high. Now, to strengthen it, you have to practice a lot. Practising questions will increase your speed as well as accuracy. Speed is also dependent on short tricks as we use them in the exam to save time for other questions. In the calculation, you must know the tables till 25 at your fingertips. With this, learning squares and cubes till 25, square roots till 20 and cube roots till 10 is must. Remembering them will enhance your calculation skills.
## Cube (algebra)
Anchor the learning of grade 6, grade 7, and grade 8 students with these printable charts on squaring numbers. Serving as a quick reference, these cheat sheets not only support students, but save teachers from having to go over concepts repeatedly. The square of a number is simply a number of the form n 2 or n x n , where n is any integer. Putting it in words: a number raised to the second power or multiplied by itself. Learning the perfect squares by heart is undoubtedly a challenge. The trick, however, is to memorize a few at a time.
Hence, the perfect squares between 1 and are 1, 4, 9, 16, 25, 36, 49, 64, 81, , , , , , , , , , , , and Write 3-digit numbers ending with 0, 1, 4, 5, 6, 9, one for each digit, but none of them is a perfect square. Find the th and th triangular numbers, and find their sum. Verify the Statement 8 for this sum. Therefore, when a perfect number is divided by 5, then the remainder will be 0, 1, 4, 0, 1 or 4. A square yard has area m 2.
### The Square and Cube of First 100 Numbers
At the beginning of the school year each year, I teach square roots and cube roots. This lays the groundwork for approximating irrationals numbers and lets us ease into the school year. I wish that I had time to do more of these square root and cube root activities with students, but the first few days of school are jammed pack with getting to know my students and explaining procedures. So, I only get to do a few of them. I like to have the challenge and it keeps me from getting in a rut.
All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 2 are provided here for you for free. Hence, the perfect squares between 1 and are 1, 4, 9, 16, 25, 36, 49, 64, 81, , , , , , , , , , , , and Write 3-digit numbers ending with 0, 1, 4, 5, 6, 9, one for each digit, but none of them is a perfect square.
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• #### Juventina R. 21.11.2020 at 15:42
This series is based on Simplification.
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• #### Dominic L. 25.11.2020 at 13:52
Squares - Cubes - Square Root Chart number n square n2 cube n3 square root number n square n2 cube n3 square root. 1. 1. 1.
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List squares, cubes, perfect fourths, perfect fifths x x. 2 x. 3 x. 4 x. 5. 1. 1. 1. 1. 1. 2. 4. 8. 3. 9. 4. 1, 5. 3, 6. | 1,280 | 4,316 | {"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} | 4.625 | 5 | CC-MAIN-2022-05 | latest | en | 0.927179 |
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