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https://www.thestudentroom.co.uk/showthread.php?t=5223336 | 1,540,307,923,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516194.98/warc/CC-MAIN-20181023132213-20181023153713-00474.warc.gz | 1,097,338,638 | 43,309 | You are Here: Home >< GCSEs
# Maths watch
1. The length of a rectangle is three times its width.
The perimeter of the rectangle is 96cm.
Label the diagram with the correct length and width for the rectangle ?
3. (Original post by lara36)
The length of a rectangle is three times its width.
The perimeter of the rectangle is 96cm.
Label the diagram with the correct length and width for the rectangle ?
There are 2 x length and 2 x width in the perimeter
Let the width be x, so the length = 3x
Form an equation in terms of x and solve it to find the width. You can then use this to find the length.
Spoiler:
Show
2*x + 2*3x = 2x + 6x = 8x
8x = 96
x = 96/8 = 12cm
Width = 12cm, so length = 3 x 12 = 36cm
4. (Original post by lara36)
I'll get you started:
We can deduce that the ratio width:length is 1:3.
So let the width be x. The length would therefore be 3x, due to the ratio.
What you can do now is calculate the perimeter in terms of x. Then equate this to the actual perimeter (96cm), and solve.
Finally, substitute your found value of x into the expressions we made for the length and width of the rectangle. You can do it!
5. Thank you
6. Could you also help with another ?
8. (Original post by lara36)
Could you also help with another ?
Ok
9. (Original post by lara36)
Could you also help with another ?
Sure, as many as u want! I'm online for a while right now.
10. (Original post by lara36)
Imagine pushing the inverted long edge up, and the inverted short edge to the right. You will get a complete rectangle, then it'll be easy!
11. (Original post by lara36)
You can see that the lines parallel to the 16cm side add up to 16 cm and the lines parallel to the 11cm side add up to 11cm. So the perimeter is 2x(11+16)=54
12. (Original post by hanguyen174)
You can see that the lines parallel to the 16cm side add up to 16 cm and the lines parallel to the 11cm side add up to 11cm. So the perimeter is 2x(11+16)=54
You're not allowed to give the answer straight away mate, give hints first
13. (Original post by lara36)
Here's a handy diagram...
Attached Images
14. (Original post by Mehru1214)
You're not allowed to give the answer straight away mate, give hints first
Oh, sorry! Quite new around here, haven’t helped anyone out yet
15. (Original post by hanguyen174)
Oh, sorry! Quite new around here, haven’t helped anyone out yet
S'ok!
16. Thank you very much i now understand.
17. (Original post by lara36)
Thank you very much i now understand.
No problem
18. can you help im useless at this stuff
19. All your useful tips helped and ive figured it for myself but thank you anyway
20. (Original post by lara36)
All your useful tips helped and ive figured it for myself but thank you anyway
Well done!
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http://math.boisestate.edu/ultman175fall14/ | 1,516,537,385,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890582.77/warc/CC-MAIN-20180121120038-20180121140038-00334.warc.gz | 215,508,512 | 11,665 | # Home
## Final Exam — Monday 15 December, 9:30-11:30am, in our usual classroom.
You are allowed to use both sides of three 8.5” x 11” sheets of paper for handwritten notes, this table of basic derivatives & integrals, this table of trigonometric integrals, and a calculator. No other notes or technology will be allowed for this exam. If you wish to use the derivative/integral table and/or trigonometric integral table, you must print out and bring your own copy.
The final exam is comprehensive; it is over all material covered this semester.
Review problems for Chapters 6-10 are posted on WebAssign. For Chapter 11, review the daily problem sets and homework 5.
A few notes:
• You will lose points for not including the differential ($dx$, $dt$, $d\theta$, etc) in your integrals.
• One of the important concepts in Calc I/Calc II is that the definite integral is arrived at through a process of “chopping and adding”. I will check whether you understand this, especially in the context of material covered in Chapter 6.
• Even though the integration method of $u$-substitution was introduced in Calc I, I consider it “testable material” for this class.
• When reviewing the material for sequences and series, pay particular attention to the following:
• Be able to derive and recognize the general formula for terms of a sequence or series, and use sum notion for series.
• Be able to recognize $p$-series and geometric series; know when they converge or diverge.
• Be able to compute a Maclaurin or Taylor series with center $c$ for a given function using the formula for the $n^{\text{th}}$ term: $s_n = \frac{f^{(n)}(c)}{n!}(x – c)^n$. When going through the review problems for computing Maclaurin and Taylor series, make sure you know how to use the formula, and be aware that this is not how WebAssign is presenting the solutions (WebAssign’s solutions depend on manipulation of basic series, which is good to know, but not what I will be testing you on).
#### Wednesday 3 December
slides_12:03_sec11.4_175_14fall
Area and arc length in polar coordinates.
Review problems for Chapter 6 have been posted on WebAssign. Review problems for Chapters 7-10 will be posted tomorrow. Review for Chapter 11 on your own, by looking at the problems sets and homework.
#### A few announcements as we head into the end of the semester:
• Skills Test re-takes will be held in class on Friday 5 December. This is your last chance to re-take the Skills Test; re-takes will not be given during dead week or finals week.
• The final exam will be held in our usual classroom on Monday 15 December from 9:30-11:30am.
• There will be new material taught on the Monday of dead week, but the last two days of class will be used for review for the final exam.
• Student evaluations are open online. I appreciate constructive feedback, and take it into account when planning future classes. Please take a moment to complete the evaluation.
#### Monday 1 December
Welcome back.
slides_12:01_sec11.3_175_14fall
Polar coordinates.
#### Friday 21 November
slides_11:21_sec11.2_175_14fall
Arc length and speed.
#### Wednesday 19 November
slides_11:19_sec11.1_175_14fall
Parametric equations and parameterized curves in the plane.
#### Friday 14 November
There’s a significant possibility that the weather overnight will create hazardous driving conditions Friday morning, so I am postponing the exam until Monday (11/17). There will not be class Friday, but I will be in my office from 9:00-10:15 (if the university is open), and for regular office hours from 11:00-12:00.
Update: My office hours today are also cancelled. See you all on Monday.
#### Monday 10 November
slides_11:10_sec10.7_175_14fall (Updated Thursday 12/13.)
Taylor series.
A note about the problems sets for Taylor series: they have been broken down into two parts. The first part (#22a) involves using various substitution and multiplication techniques to derive Maclaurin series from simpler Maclaurin series. The second part (#22b) involves computing Taylor series directly.
Warning: practice another version and solutions have been disabled for #22b. The solutions presented involve algebraic manipulation of the functions instead of direct computation, and I want you to have practice with the direct computation. Following the solution steps will result in a lot of confusion.
Both problem sets #22a and #22b are due on Friday. You can ask questions about them in class on Wednesday, during office hours, help sessions, etc.
#### Saturday 8 November
Review problems for exam 2 have been posted on WebAssign. There are review problems for sections 7.8, 8.1, 9.1, 9.2, and 10.1–10.5. For sections 10.6 and 10.7, look at the current problem sets and homework.
#### Friday 7 November
slides_11:07_sec10.6_175_14fall
Power series: geometric power series, power series expansions, and integration and differentiation of power series.
#### Wednesday 5 November
slides_11:05_sec10.6_175_14fall
Introduction to power series. Convergence of power series.
A note on daily problems sets: I’ve broken the problem set for power series (sec 10.6 in the text) into two separate assignments: #21a and #21b. Both are due next Monday. However, everything needed to complete problem set #21a was covered in class today, and I highly recommend that you complete it before Friday’s class.
#### Tuesday 4 November
Vote.
Idaho voting information.
#### Monday 3 November
slides_11:03_sec10.5_175_14fall
Ratio and root tests.
Convergence Tests Worksheet
#### Friday 31 October
slides_10:31_sec10.4_175_14fall
Alternating series test, ratio test, and root test.
#### Wednesday 29 October
slides_10:29_sec10.3-10.4_175_14fall
Absolute and conditional convergence.
Note: for problem set #19, you will only have one try for the multiple choice questions (#4-12).
#### Monday 27 October
slides_10:27_sec10.3_175_14fall
Positive series.
#### Friday 24 October
slides_10:24_sec10.2_175_14fall
Introduction to series.
#### Wednesday 22 October
slides_10:22_sec10.1_175_14fall
Sequences: computing convergence and limits.
#### Monday 20 October
slides_10:20_sec10.1_175_14fall
Introduction to sequences.
#### Wednesday 15 October
Slides and the worksheet from today’s lesson on probability — thank you, Rachael!
Unfortunately, there were not any good probability problems on WebAssign…
Also: Rachael’s help session times have been updated in the sidebar.
#### Monday 13 October
Very few people did problem 4b correctly. You will have a chance to improve your score during the first ten minutes of today’s class. This will be your only opportunity to improve your score on this problem.
These are the expectations for the Skills Test on Friday 10/17
#### Friday 10 October
slides_10:10_sec9.1_175_14fall
Introduction to differential equations. Separable differential equations.
#### Wednesday 8 October
slides_10:08_sec8.1_175_14fall
Applications of the integral: arc length and surface area.
#### Monday 6 October
slides_10:06_sec7.8_175_14fall
I have decided to let error bounds go for now. We will see them again when we talk about sequences and series. On today’s problem set, I’ve set part (b) on problems #3 and #4 to zero points. You can try them if you’d like — it will not affect your score. We will talk about arc length and surface area on Wednesday.
#### Monday 29 September
Improper integrals, continued: limit comparison tests, and $p$-integrals.
Here is the worksheet on $p$-integrals.
And, because I tried to hurry the wrap-up, here are notes on part 3 of the worksheet — tying the convergence/divergence behavior of $p$-integrals to the value of $p$.
#### Friday 26 September
slides_09:24_sec7.6_175_14fall
Improper integrals.
We covered everything you need to do the problem set today.
#### Wednesday 24 September
slides_09:24_sec7.5_175_14fall
Techniques of integration: method of partial fraction decomposition – quadratic terms. Completing squares.
#### Monday 22 September
slides_09:22_sec7.5_175_14fall
Techniques of integration: method of partial fraction decomposition – linear terms.
Help Sessions: Rachael’s help sessions for the next week will be:
• Wednesday (9/24) 10:30-11:45 AM in Library Room 205
• Thursday (9/25) 4:30-5:45 PM in Multipurpose Building Room 210
• Tuesday (9/30) 9:00-10:15 AM in Multipurpose Building Room 210
#### Monday 15 Friday 19 September
slides_09:19_sec7.3_175_14fall
Techniques of integration: trigonometric substitution.
A heads-up on today’s problems set (#09 Trigonometric Substitution) — you only get 3 chances on the multiple choice.
#### Monday 15 Wednesday 17 September
slides_09:17_sec7.2_175_14fall
Techniques of integration: trigonometric integrals.
#### Monday 15 September
slides_09:15_sec7.1_175_14fall
Techniques of integration: the method of integration by parts.
Reminder: the first homework assignment is due on Wednesday (5 minutes before class). Rachael is holding a special LA help session today from 12-1:15pm in the Multipurpose Building, Room 207A.
Announcement: If you are interested in providing one-on-one mentorship to a low-income high school student, helping them navigate the college application and financial aid process, contact boise@striveforcollege.org. This is a volunteer organization. My understanding is that the time commitment is one hour per week, and mentors are particularly needed this week and next.
#### Wednesday 10 September
slides_09:10_sec6.5_175_14fall
Application of the definite integral: work.
#### Monday 8 September
slides_09:08_sec6.4_175_14fall
Computing the volume of solids of revolution using the method of cylindrical shells.
Announcement: Help sessions with Rachael have been scheduled. See the sidebar for times and locations.
#### Friday 5 September
slides_09:05_sec6.3_175_14fall
Computing the volume of solids of revolution using the disk/washer method.
#### Thursday 4 September
I am extending the deadline for the third problem set until midnight tomorrow (Friday 9/4). The last half hour of tomorrow’s class will be open for questions about the assignment.
#### Monday Wednesday 3 September
slides_09:03_sec6.2_175_14fall
Applications of the definite integral: computing the volume of solids whose cross-sectional area can be written as a function of a single variable.
Worksheet (corrected): volumes of solids sec6.2 175 14fall
Notes on the example and the worksheet problems:
The notes give a more linear presentation of the examples and worksheet problems from this morning’s class.
#### Friday 29 August
slides_08:29_sec6.1_175_14fall
Applications of the definite integral: area between curves, and mass (integrating linear mass density functions).
Announcements:
• I spoke with a WebAssign representative about the problem of being forcibly logged out by WebAssign when trying to access homework. He suggested clearing the cookies and cache of the web browser. Apparently, this fixes several WebAssign issues, so it’s worth trying first when you run into problems. Ideally, you should be logged out of WebAssign when you do it.
• Have a great weekend.
#### Wednesday 27 August
slides_08-27_integration_review_175_14fall
Review of antiderivatives, definite integrals, and the method of $u$-substitution.
Announcements:
• You can check Blackboard to see if your clicker registered in today’s class.
• Please fill out the survey for finding days and times for help sessions with Rachael.
• The first help session with Rachael will be on Friady 29 August from 7:30-8:45am in Albertosns Library Rm 205.
• We will begin class on Friday with questions about WebAssign
#### Monday 25 August
slides_08-25_welcome_175_14fall
There was a clear preference for daily problem sets to comprise 15% of the course grade, and homework to comprise 5%. The syllabus will be updated to reflect this.
The first help session with Rachael will be on Friady 29 August from 7:30-8:45am in Albertosns Library Rm 205.
See you on Wednesday
#### Welcome to Math 175
Welcome to Math 175, section 001, Fall 2014. Our class meets MWF 9:00–10:15am, Interactive Learning Center (ILC), Room 302.
In this course, we continue to study the integral, learning new techniques of integration, exploring a few applications of the integral, and defining indefinite integrals. This will be followed by an introduction to sequences and series. Finally, we will briefly consider parameterized curves in the plane.
Please take a few minutes to familiarize yourself with the course website, which includes: announcements, course slides, and notes (which will be posted on this home page throughout the course); the syllabus (this is where you can see how grades will be calculated); information about the textbook and WebAssign (the online homework platform); and exam information.
My email address and links to the log-in pages for Blackboard and WebAssign can be found in the sidebar at the top right side (or, on tablets, at the bottom) of each page of this site.
Blackboard will be used to register clickers, host a class discussion board, and post grades (except for homework and daily problem set grades, which will be posted on WebAssign).
Information on registering for WebAssign can be found here. Information on registering clickers can be found here. | 3,190 | 13,318 | {"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} | 3.703125 | 4 | CC-MAIN-2018-05 | latest | en | 0.897798 |
https://testbook.com/question-answer/newtons-second-law-of-motion-states-t--5fc4d8c0735b2e868ab4f5ea | 1,638,514,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00233.warc.gz | 596,633,634 | 32,563 | # Newton’s second law of motion states that
This question was previously asked in
ISRO IPRC Technical Assistant Mechanical held on 10/12/2016
View all ISRO Technical Assistant Papers >
1. Force is equal to the product of mass and acceleration
2. Every action has an equal and opposite reaction
3. Force is proportional to the product of mass and acceleration
4. None of the above
Option 3 : Force is proportional to the product of mass and acceleration
Free
CT 1: Growth and Development - 1
53972
10 Questions 10 Marks 10 Mins
## Detailed Solution
Explanation:
Newton’s Second Law of Motion:
According to newton's second law "The force acting to any object is directly proportional to the rate of change of linear momentum of the object."
Hence we can write:
$$Force \propto \frac{{change\;in\;momentum}}{{time}}$$
$$F\propto\frac{d(p)}{dt}$$
We know Momentum (p) = m × v
where m = Mass of the object, v = Velocity of an object
$$F\propto\frac{d(m\times v)}{dt}$$
$$F\propto (m\frac{d(v)}{dt}+v\frac{d(m)}{dt})$$
From the conservation of mass principle, Mass of the object (m) = constant
so $$\frac{dm}{dt}=0$$
∴ $$F\propto m\frac{d(v)}{dt}$$
We know, Acceleration is the rate of change of velocity so, $$a=\frac{dv}{dt}$$
$$F\propto (ma)$$
Hence it is clear from newton's 2nd law that force is proportional to the product of mass and acceleration.
Newton’s First law:
A body continues to be in its state of rest or of uniform motion along a straight line unless it is acted upon by some external force to change the state.
• If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot accelerate.
• Newton’s first law defines inertia and is rightly called the law of inertia.
Newton’s Third Law:
To every actionthere is always an equal (in magnitude) and opposite (in direction) reaction.
• When a body exerts a force on any other body, the second body also exerts an equal and opposite force on the first.
• Forces in nature always occur in pairs. A single isolated force is not possible. | 571 | 2,052 | {"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} | 4.40625 | 4 | CC-MAIN-2021-49 | latest | en | 0.82464 |
https://nl.mathworks.com/matlabcentral/cody/problems/3-find-the-sum-of-all-the-numbers-of-the-input-vector/solutions/2105776 | 1,582,794,223,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00343.warc.gz | 477,986,440 | 16,095 | Cody
# Problem 3. Find the sum of all the numbers of the input vector
Solution 2105776
Submitted on 26 Jan 2020 by Charles Neal
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(vecsum(x),y_correct))
y = 0 y = 1
2 Pass
x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
y = 0 y = 1 y = 3 y = 6 y = 11
3 Pass
x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
y = 0 y = 1 y = 3 y = 6 y = 11
4 Pass
x = 1:100; y_correct = 5050; assert(isequal(vecsum(x),y_correct))
y = 0 y = 1 y = 3 y = 6 y = 10 y = 15 y = 21 y = 28 y = 36 y = 45 y = 55 y = 66 y = 78 y = 91 y = 105 y = 120 y = 136 y = 153 y = 171 y = 190 y = 210 y = 231 y = 253 y = 276 y = 300 y = 325 y = 351 y = 378 y = 406 y = 435 y = 465 y = 496 y = 528 y = 561 y = 595 y = 630 y = 666 y = 703 y = 741 y = 780 y = 820 y = 861 y = 903 y = 946 y = 990 y = 1035 y = 1081 y = 1128 y = 1176 y = 1225 y = 1275 y = 1326 y = 1378 y = 1431 y = 1485 y = 1540 y = 1596 y = 1653 y = 1711 y = 1770 y = 1830 y = 1891 y = 1953 y = 2016 y = 2080 y = 2145 y = 2211 y = 2278 y = 2346 y = 2415 y = 2485 y = 2556 y = 2628 y = 2701 y = 2775 y = 2850 y = 2926 y = 3003 y = 3081 y = 3160 y = 3240 y = 3321 y = 3403 y = 3486 y = 3570 y = 3655 y = 3741 y = 3828 y = 3916 y = 4005 y = 4095 y = 4186 y = 4278 y = 4371 y = 4465 y = 4560 y = 4656 y = 4753 y = 4851 y = 4950 y = 5050 | 703 | 1,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-10 | latest | en | 0.319605 |
https://answers.yahoo.com/question/index?qid=20190616031413AA2wmyQ&sort=N | 1,575,574,109,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00064.warc.gz | 274,452,646 | 21,251 | Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago
# The sum of the first 2 million primes is even or odd?
Relevance
• 6 months ago
There will be one even number (2) in the sum, then the remaining 1,999,999 primes will be odd (because of they were even they'd have 2 as a factor and not be prime).
The sum of an odd number of odd numbers will be odd. Adding 2 to that will not change the parity, so the final sum will still be odd.
• TomV
Lv 7
6 months ago
All primes greater than 2 are going to be odd numbers. The sum of a sequence of odd numbers can be either even or odd. The sum of the first 2 million primes will be the sum of the 1,999,999 primes greater than 2 plus two.
The sum of any sequence of odd numbers will be odd if the count of the sequence is odd and will be even if the count of the sequence is even. Adding 2 to that sum will not change the "odd/even" nature of the sum.
The sum of the first 1,999,999 primes greater than 2 will be odd. Adding 2 to the sum leaves the sum an odd number.
Ans: Odd
• 6 months ago
Even. Once you get to TWO (2) all the subsequent numbers that are prime are odd. So when you add two odd numbers you get an even. The last of the first two million prime numbers is 32,452,843. If you want to see how to get the EVEN answer, see the source below.
• 6 months ago
The sum is odd. 2 is the only even prime number, all other primes are odd. You have a total of 1,999,999 odd primes; if you have an odd number of odd primes being added, the sum is odd, and adding an even number to this sum means the grand sum is odd; odd + even = odd.
• Brian
Lv 6
6 months ago
Odd is should be. as another odd fact, the one millionth digit of Pi after the decimal is 1.
• alex
Lv 7
6 months ago
2 , 3 , 5 , 7 , 11 , ...
2 + (1 999 999 s odd) = odd
• 6 months ago
Oddly enough the oddly worded question's answer is odd. | 521 | 1,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | latest | en | 0.905115 |
https://questions.examside.com/past-years/jee/question/let-the-definite-integral-be-defined-by-the-formula-intlimit-jee-advanced-2006-marks-5-f2o9xhn61vrrhjwo.htm | 1,713,494,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00239.warc.gz | 441,311,132 | 40,076 | 1
IIT-JEE 2006
+5
-1.25
Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).}$$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } }$$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).}$$
$$\int\limits_0^{\pi /2} {\sin x\,dx = }$$
A
$${\pi \over 8}\left( {1 + \sqrt 2 } \right)$$
B
$${\pi \over 4}\left( {1 + \sqrt 2 } \right)$$
C
$${\pi \over {8\sqrt 2 }}$$
D
$${\pi \over {4\sqrt 2 }}$$
2
IIT-JEE 2006
+5
-1.25
Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).}$$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } }$$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).}$$
If $$\mathop {\lim }\limits_{x \to a} {{\int\limits_a^x {f\left( x \right)dx - \left( {{{x - a} \over 2}} \right)\left( {f\left( x \right) + f\left( a \right)} \right)} } \over {{{\left( {x - a} \right)}^3}}} = 0,\,\,$$ then $$f(x)$$ is
of maximum degree
A
$$4$$
B
$$3$$
C
$$2$$
D
$$1$$
3
IIT-JEE 2006
+5
-1.25
Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).}$$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } }$$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).}$$
If $$f''\left( x \right) < 0\,\forall x \in \left( {a,b} \right)$$ and $$c$$ is a point such that $$a < c < b,$$ and
$$\left( {c,f\left( c \right)} \right)$$ is the point lying on the curve for which $$F(c)$$ is
maximum, then $$f'(c)$$ is equal to
A
$${{f\left( b \right) - f\left( a \right)} \over {b - a}}$$
B
$${{2\left( {f\left( b \right)} \right) - f\left( a \right)} \over {b - a}}$$
C
$${{2f\left( b \right) - f\left( a \right)} \over {2b - a}}$$
D
$$0$$
4
IIT-JEE 2005 Screening
+3
-0.75
The area bounded by the parabola $$y = {\left( {x + 1} \right)^2}$$ and
$$y = {\left( {x - 1} \right)^2}$$ and the line $$y=1/4$$ is
A
$$4$$ sq. units
B
$$1/6$$ sq. units
C
$$4/3$$ sq. units
D
$$1/3$$ sq. units
EXAM MAP
Medical
NEET | 1,351 | 3,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-18 | latest | en | 0.393685 |
http://arnienumbers.blogspot.com/2010/12/on-sorlis-conjecture-on-odd-perfect.html | 1,532,359,711,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596542.97/warc/CC-MAIN-20180723145409-20180723165409-00236.warc.gz | 26,173,945 | 17,808 | ## 6.12.10
### On Sorli's Conjecture on Odd Perfect Numbers
Let $N = {p^k}{m^2}$ be an Odd Perfect Number (OPN) with Euler prime $p$. Note that:
(1) $p \equiv k \equiv 1 \pmod 4$
(2) $\gcd(p, m) = 1$
I also proved (in 2008, p. 112 [Theorem 4.2.5]) the following result:
(3) $\sigma(p^k)/m^2 <= 2/3$
(The proof begins with the fact that prime-powers are solitary numbers, and then by observing that $\sigma(m^2)/p^k$ is always odd (for odd primes $p$), (3) follows from an "old" result by Dandapat, Hunsucker and Pomerance [1975]. See the list of References below (the very last one) for the hyperlink to the document.)
In particular, this means that:
(4) $$1 + p^k <= \sigma(p^k) <= (2/3){m^2}$$
But:
(5) $$\log(\sigma(p^k)) = \log(\sum_{i = 0}^{k}{p^i})$$
But then:
(6) $$\sigma(p^k) = \sum_{i = 0}^{k}{p^i} = \frac{p^{k + 1} - 1}{p - 1}$$
Therefore:
(7) $$\log(\sigma(p^k)) = \log(p^{k + 1} - 1) - \log(p - 1)$$
(Note: We also know that, for the Euler prime $p$, $\sigma(p) = p + 1$ divides $\sigma(p^k)$. We give a quick proof below.)
Since $p \equiv k \equiv 1 \pmod 4$, we have an even number of addends in the sum:
$$\sigma(p^k) = 1 + p + \ldots + p^{k - 1} + p^{k}$$
$$= (1 + p)(1 + \ldots + p^{k - 1})$$
$$= (1 + p){\sigma(p^{2a})}$$
(since we know that $k \equiv 1 \pmod 4$, then $4 \mid k - 1$), where $a = (k - 1)/2$.
Thus:
(8) $$\log(\sigma(p^k)) = \log(1 + p) + \log(\sigma(p^{k - 1}))$$
Note that $k - 1 < k$ ( i.e. (8) is a logarithmic relation recursively defining $\sigma(p^k)$ in terms of $\sigma(p^{k - 1})$).
We now state Sorli's conjecture for OPNs:
[Sorli, 2003] If $N$ is an OPN given in the Eulerian form $N = {p^k}{m^2}$ (i.e. $p$ is the Euler prime and $\gcd(p, m) = 1$), then $k = 1$.
Here we give a characterization for odd numbers $N = {p^k}{m^2}$ to be perfect.
Proof (to appear - perhaps in IJNT):
Assume there is an OPN $N$. Then $N$ has the Eulerian form $N = {p^k}{m^2}$, with $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.
Prime powers are deficient because, in general, the abundancy index for prime powers $q >= 2$ satisfies the inequality:
(9) $1 < I(q^s) < 3/2$ for all exponents s
(Note that (9) follows from the inequality
(10) $(q + 1)/q <= I(q^s) < (q - 1)/q$
which is true for all primes $q$ and exponents $s$.)
Thus, it is not possible that $m = p$.
Case 1: Assume $m < p$. Since $k \equiv 1 \pmod 4$ [in particular, we know that $k \ge 1$], we then have:
$m < p <= p^k$
By the upper bound in (3): $p^k < (2/3){m^2} < m^2$
But squaring both sides of the inequality $m < p$, we get:
$m^2 < p^2$
We now have the inequality:
$p \le p^k < p^2$
(i.e. the prime-power $p^k$ is "trapped" between two consecutive powers of the prime $p$).
This implies that $1 <= k < 2$. Since $k \equiv 1 \pmod 4$, we conclude that $k = 1$.
We now consider Case 2.
Case 2: Assume $p < m$.
From Case 1, we proved the implication
"$m < p^k$ implies $k = 1$".
The contrapositive is "$k > 1$ implies $p^k < m$". (Note that $m$ is not equal to $p^k$ because of (4).)
Suppose $k > 1$. By the contrapositive result mentioned, $p^k < m$, and trivially thus: $p < p^k < m$.
Observe that Case 1 gives a criterion for having $k = 1$.
This is because the conditions
$p \le p^k < m$
and
$m < p \le p^k$
are mutually exclusive.
(Of course, the [related] conditions
$k > 1$
and
$k = 1$
are also mutually exclusive.)
Thus, to avoid any contradictions in (any series of) our (logical) implications, since Case 1 holds (for the OPN conjecture) under the assumption of Sorli's conjecture, we must have logical equivalences between:
Case 1: $m < p \le p^k$ and $k = 1$
Case 2: $p \le p^k < m$ and $k \ge 1$
(Note that, in Case 2, we can actually take $k \geq 5$.)
Additionally, in a post over MathOverflow, I was able to show results in the direction of proving the following implication:
MathOverflow (posted over the weekend): Sorli's Conjecture implies the OPN Conjecture.
That is, if Sorli's Conjecture is proved, then there are no odd perfect numbers.
(I am now hereby editing this post in response to Pace Nielsen's request for "backing up my claim".)
To recap, assuming $k = 1$ will make the sum $I(p^k) + I(m^2)$, which satisfies the inequality
$2.85 = 57/20 < L(p) < I(p^k) + I(m^2) \leq U(p) < 3$
attain the upper bound $U(p) = (3p^2 + 2p + 1)/(p(p + 1))$.
The lower bound $L(p)$ is given by $L(p) = (3p^2 - 4p + 2)/(p(p - 1))$. Note that:
$I(p) = (p + 1)/p$
and
$I(m^2) = 2/[I(p)] = 2p/(p + 1)$.
Now, let us call a prime $p$ an "admissible" Euler prime if it can be the Euler prime of an OPN.
That is, a prime $p \equiv 1 \pmod 4$ is an "admissible" Euler prime if it comes from the "functional equation" $I(p)I(m^2) = 2$.
On the other hand, given a particular "admissible" Euler prime $p$, by computing $u = U(p)$ and then computing the integers $q$ satisfying $L(q) < u$, you get a $q$ that CANNOT BE EQUAL to $p$. Thus, the "admissible" Euler prime $p$ gives rise to an "inadmissible" Euler prime $q$. But please do take note that the underlying assumption is that $p = q$, whence you get a contradiction.
Indeed, it suffices to check that the implication
"$L(q) < u = U(p)$ implies $q$ is NOT equal to $p$"
holds for all "admissible" Euler primes $p \ge 5$, just by checking the inequality
$L(q) < U(p) = U(5) = (6/5) + (5/3)$
for $p = 5$. Why?
This is because $L(p)$ and $U(p)$ are both increasing functions of $p$ in the interval $[5, \infty)$, with $L(p) < U(p)$ for all $p$ (note the strict inequality), with both $L(p)$ and $U(p)$ approaching $3$ as $p$ becomes arbitrarily large.
In particular, note that:
$$L(p) = \frac{3p^2 - 4p + 2}{p(p - 1)} = 3 - \frac{p - 2}{p(p - 1)}$$
and
$$U(p) = \frac{3p^2 + 2p + 1}{p(p + 1)} = 3 - \frac{p - 1}{p(p + 1)}$$
So if $L(q) < U(p)$, then
$$3 - \frac{q - 2}{q(q - 1)} < 3 - \frac{p - 1}{p(p + 1)}$$
which implies that
$$\frac{p - 1}{p(p + 1)} < \frac{q - 2}{q(q - 1)}$$
$$q(p - 1)(q - 1) < p(p + 1)(q - 2)$$
$$pq^2 - pq - q^2 + q < {p^2}q - 2{p^2} + pq - 2p$$
Furthermore, since $q$ and $p$ are integers (in fact, they are prime numbers), then
$$2q - p \geq 4$$
Suppose $2q - p = 4$. Since $p$ and $q$ are both Euler primes,
$$p \equiv q \equiv 1 \pmod 4$$
Thus, $2q \equiv 2 \pmod 4$ and $p \equiv 1 \pmod 4$ imply that $$2q - p \equiv 1 \pmod 4$$
This contradicts $2q - p = 4$.
Thus, $2q - p \geq 5$.
$2q \geq p + 5$
$2q \geq p + 5 \geq 5 + 5 = 10$
$q \geq 5$
$(p - 1)(q - 1) < (p + 1)(q - 2)$ for all possible pairs $(p, q)$
In particular, WolframAlpha gives the solutions:
(i) $p > -1, q > (p + 3)/2$
(ii) $p > -1, q < 1$
(iii) $p > -1, (p + 3)/2 < q < 1$
We only need to consider Case (i):
(i) $p > -1, q > (p + 3)/2$
Since we know that the Euler primes $p$ satisfy $p \equiv 1 \pmod 4$, automatically, we have $p \ge 5$, and therefore the first condition is satisfied.
Checking the second condition:
First, we show that it is true that $(p + 3)/2 < p$ if $p \ge 5$. Assume to the contrary that $(p + 3)/2 \ge p$. Then we have $p + 3 \ge 2p$. This leads to the contradiction $p \le 3$. (Essentially, what we have shown is that, hypothetically, $q < (p + 3)/2 < p$ follows from $p \geq 5$).
Now, we compare $q$ and $p$. (Recall: We need to show that $q$ is not equal to $p$.)
Since $p \ge 5$, it follows that $q > (p + 3)/2 \ge (5 + 3)/2 = 8/2 = 4$. Thus, $q \ge 5$.
It thus remains to tackle the case $p = q = 5$.
It is (actually) a more fundamental problem to establish that $q = (p + 3)/2$ leads to a contradiction when $p$ and $q$ are both Euler primes. This is because:
If $p$ and $q$ are both Euler primes, then $p \equiv q \equiv 1 \pmod 4$.
Yet $q = (p + 3)/2$ implies that
$$q = (p + 3)/2 \equiv (1 + 3)/2 \equiv 4/2 = 2 \pmod 4$$
Another way of saying this is that if $p$ and $q$ are both Euler primes, then the equality $q = (p + 3)/2$ cannot happen since it will contradict $p \equiv q \equiv 1 \pmod 4$. That is why it was necessary to have the strict inequality $q > (p + 3)/2$.
But then again it begs the question:
Why must $p$ and $q$ necessarily be distinct Euler primes, if OPNs are to exist?
In other words, it does appear (heuristically) that Euler primes are in one-to-one correspondence with OPNs, in the same way that Mersenne primes are in one-to-one correspondence with even perfect numbers (EPNs).
For the case $q < p$
WolframAlpha gives the solutions:
(i) $p > 3, (1/2)(p + 3) < q < p$
(ii) $p > 3, q < 1$
(iii) $-1 < p \le 1, q < p$
(iv) $1 < p \le 3, q < 1$
For the case $p < q$
WolframAlpha gives the solutions:
(i) $p > 3, q > p$
(ii) $-1 < p \le 1, q > (1/2)(p + 3)$
(iii) $-1 < p \le 1, p < q < 1$
(iv) $1 < p \le 3, q > (1/2)(p + 3)$
(v) $p < -1, (1/2)(p + 3) < q < 1$
For the case $p = q$
WolframAlpha gives the solutions:
(i) $p > 3, q = p$
(ii) $-1 < p < 1, q = p$
Integer solution: $p = 0, q = 0$
Assuming WolframAlpha's numerical methods of computation are correct for $p \ne q$, we need to tackle this remaining case: $p = q = 5$.
Lastly, in order to prove that the OPN conjecture is true (in full generality), we need to consider the case:
(Case 2) $N$ is an OPN in the Eulerian form $N = {p^k}{m^2}$ with $k > 1$ AND $p^k < m$.
In other words, we now have the following characterization theorem for OPNs:
Theorem [December 2010]: If $N$ is an OPN with the Eulerian form $N = {p^k}{m^2}$, then $p^k < m$ and $k >= 5$.
The Theorem gives sufficient and necessary conditions for odd numbers of the form
$N = {p^k}{m^2}$ to be perfect.
On some final notes: The even perfect numbers also have a form similar to the OPNs:
$M$ is an even perfect number if and only if
$$M = {2^{r - 1}}(2^r - 1) = (1/2){M_r}(M_r + 1)$$
where $M_r$ is a Mersenne prime with exponent $r$.
(The "exponent" $k'$ on the Mersenne prime [ i.e. if you write it as ${M_r}^{k'} = (2^r - 1)^{k'}$ ] is of course $k' = 1$.)
Note that $(M_r + 1)/2 < M_r$.
Additionally, $M_r \equiv 3 \pmod 4$ for all Mersenne primes with exponent $r$. (Compare that with $p^k \equiv 1 \pmod 4$ for all Euler primes with exponent $k$.)
Lastly, it is known that the set of perfect numbers is a (proper) subset of the set of triangular numbers.
In the coming days/weeks, I will try to tackle the remaining case (Case 2) for OPNs.
References:
(A) Ronald Sorli [2003], "Algorithms in the Study of Multiperfect and Odd Perfect Numbers"
(B) Arnie Dris [2008], "Solving the Odd Perfect Number Problem: Some Old and New Approaches"
(C) MathOverflow post [2010], "On Sorli’s Conjecture Re: OPNs (Circa 2003)"
(F) Melvin Bernard Nathanson [2000], "Elementary Methods in Number Theory"
(G) G. G. Dandapat, J. L. Hunsucker, and Carl Pomerance [1975], "Some New Results on Odd Perfect Numbers" | 4,076 | 10,837 | {"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} | 3.90625 | 4 | CC-MAIN-2018-30 | latest | en | 0.749052 |
https://doodlelearning.com/maths/skills/subtraction/repeated-subtraction | 1,726,794,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00460.warc.gz | 189,681,322 | 84,627 | # What is repeated subtraction?
Repeated subtraction is a useful way to introduce students to division. By using this method to solve division problems, students gain a deeper understanding of how to divide numbers into equal groups.
Author
Katie Wickliff
Published
February 2024
# What is repeated subtraction?
By using repeated subtraction to solve division problems, students gain a deeper understanding of how to divide numbers into equal groups.
Author
Katie Wickliff
Published
February 2024
# What is repeated subtraction?
Repeated subtraction is a useful way to introduce students to division. By using this method to solve division problems, students gain a deeper understanding of how to divide numbers into equal groups.
Author
Katie Wickliff
Published
February 2024
Key takeaways
• The repeated subtraction method helps students understand division as creating equal groups.
• The repeated subtraction method can help students learn division because it asks them to perform a skill (subtraction) that they have already learned.
• Repeated subtraction is one of several effective division strategies.
When your child is learning division, it helps to have several strategies to strengthen their understanding and boost their confidence. One effective introductory strategy is repeated subtraction. The repeated subtraction method provides students with a concrete way to see the relationship between division and subtraction.
This article outlines how to perform repeated subtraction to solve simple division problems. We also provide several examples, practice problems, and additional practice resources. Let’s get started!
## What is repeated subtraction?
Repeated subtraction is a way to solve certain division problems in maths. This method takes away the same number from a larger number until you reach zero— or a number smaller than the one being subtracted. The number of times you “jump back” is the answer to the division problem.
## Division by repeated subtraction
Using the repeated subtraction method is helpful when introducing kids to division because it allows them to use a familiar maths skill to help solve a new type of problem. Children feel like they are building on their knowledge, rather than learning something completely unfamiliar.
Using the repeated subtraction process as a foundation can help students gain the confidence to tackle more complex division problems with varying strategies, such as short or long division!
## Repeated subtraction examples
Let’s look at a few examples of division as repeated subtraction.
### Example 1
16÷4= ?
To divide a number using repeated subtraction, start with the number you are dividing. This number is also called the dividend. In this example, that number is 16.
Then, subtract the number you are dividing by. This number is also called the divisor. In this example, that number is 4.
16-4= 12
There are more than 4 left, so we subtract again.
12-4=8
We still have more than 4 left, so we’ll subtract until we get to zero or a number smaller than four:
8-4=4
One last time:
4-4= 0
Now that we’ve reached zero, we count the number of times we subtracted 4 from 16. We subtracted 4 times, so our answer (or quotient) is 4.
16÷4=4
### Example 2
Piper has 24 stickers. She gives an equal amount of stickers to 6 friends. How many stickers does each friend get?
To solve this division problem using repeated subtraction, we first look at the dividend. In this example, that number is 24.
Then, we repeatedly subtract the divisor until we reach zero or a number smaller than the divisor. In this example, that number is 6.
24-6=18
18-6=12
12-6=6
6-6=0
Now that we reached zero, we count the number of times we subtracted the divisor (6) from the dividend (24).
We subtracted the divisor 4 times.
24 ÷ 6= 4
Piper gave 6 friends 4 stickers apiece.
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## Explore repeated subtraction with DoodleMaths
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## Repeated subtraction practice questions
Solve each of these problems using the repeated subtraction method. Be sure to show your work! When you’ve worked through all of these, find more helpful practice problems with DoodleLearning’s interactive maths app.
John has 8 crayons. If he shares them equally with 4 friends, how many crayons does each friend get?
Solve 20÷ 4 using repeated subtraction.
Stella brought 16 oranges to the football game. If there are 8 players on the team, how many oranges does each player get?
Twenty-one students are participating in the school relay race. If each relay team can have 3 members, how many relay teams will there be altogether?
If 36 is the dividend, and 9 is the divisor, what is the quotient?
8-4=4
4-4=0
8÷ 4=2
John gave each friend 2 crayons.
20÷ 4=?
20-4=16
16-4=12
12-4=8
8-4=4
4-4=0
20 ÷ 4 = 5
16÷ 8=?
16-8=8
8-8=0
16÷ 8=2
Each football player gets 2 oranges.
21÷ 3=?
21-3=18
18-3=15
15-3=12
12-3=9
9-3=6
6-3=3
3-3=0
21÷ 3=7
There will be 7 relay teams altogether.
36÷ 9=?
36-9=27
27-9=18
18-9=9
9-9=0
36÷ 9=4
The quotient is 4.
Repeated subtraction is a method students can use to solve some division problems. This method asks students to repeatedly subtract the divisor from the dividend until they reach zero or a lesser number.
Yes, division is repeated subtraction. Using repeated subtraction for division problems allows students to visualise how to break numbers down into equal groups.
An example of repeated subtraction is solving 12÷ 3 this way:
12-3=9
9-3=6
6-3=3
3-3=0
In this repeated subtraction example, 3 is subtracted 4 times, so the answer to 12÷ 3 is 4.
Lesson credits
Katie Wickliff
Katie holds a master’s degree in Education and a bachelor’s degree in both Journalism and English. She has over 15 years of experience as a teacher and is also a certified tutor. Most importantly, Katie is the mother of an 8 and 11-year-old. She's passionate about education and firmly believes that the right tools and support can help every child reach their full potential.
Katie Wickliff
Katie holds a master’s degree in Education and a bachelor’s degree in both Journalism and English. She has over 15 years of experience as a teacher and is also a certified tutor. Most importantly, Katie is the mother of an 8 and 11-year-old. She's passionate about education and firmly believes that the right tools and support can help every child reach their full potential.
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Are you a parent or teacher? | 1,617 | 6,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.945878 |
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# Light
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### Light
1. 1. LIGHT By:Kurniawati Ahmida 09330063 / 5B
2. 2. OBJECTIVIES OF LEARNING Student can to explain the properties of light Student can to explain the law of light reflection, draw image formation on the flat mirror, concave mirror, and convex mirror Student can to explain the law of light refraction, draw image formation in concave lens, and convex lens and explain the image properties in concave lens and convex lens, calculate the power of lens
3. 3. Light is an elektromagnetic wave, namely thewave the vibration of which is electric field andmagnetic field. Different with sound, light can spreadwithout medium, so light can spread in vacuum.That’s why, sunlight can arrive to the eartheventhough between the sun and the earth is foundvacuum space
4. 4. While several properties of light among other are:• Spreading according to straigt Line• Has energy• Can be seen• Radiated in the from of radiation• Has spreading direction perpendicular to vibration direction• Can experience reflection, refraction, inter ference, diffraction, and polaritation
5. 5. LIGHT REFLECTION Law of Light Reflection: Incident angle a. Incide angle, normal line and reflected ray mirror lie on one flat plane i = sudut datang b. Incident angle is equal to reflected angle (i r = sudut pantul Normal line = r) Reflected ray lieThe Kinds of Light Reflection Specular Reflection Diffuse Reflection
6. 6. Light Reflection on flat Mirror a. Virtual image b. Upright as the body c. Same size with the body d. Facing perverted with the body e. Distance of the body to the mirror is equal to distance of image to the mirror
7. 7. Light Reflection in Concave Mirror 1. Incident ray which is paralles to main axis ef the mirror will be reflected passing through the focal point F. 2. The incident ray passing through the focal point F will be reflected parallel to main axis. 3. The incident ray passing through center point of curvature of mirror P will be reflected back through that center point of curvature.
8. 8. Formation of a shadow on the concave mirrorConcave mirror equation or
9. 9. An object is placed 20 cm infront of a concave mirrorwith radius of curvature of 30 cm. What is thedistance of image to the mirror and the imagemagnification ?Solution:Given: S = 20 cm R = 30 cmAsked: S’ and M ?Answer: So, the image magnification is 3 timeSo, the image distance is 60 cm in front of mirror
10. 10. Light Reflection in Convex Mirror 1. The incident ray parallel to main axis is reflected seemed comes from focal of the mirror (F). 2. Incident ray to focal point (F) is reflected parallel to main axis. 3. Incident ray to center of curvature of mirror P is reflected back seemed comes from the center point of curvature (on the same line).
11. 11. Convex Mirror Equation or
12. 12. Light Refraction Light refraction is phenomenon of bending of light that spreads from one medium to another medium which density are different. Law of refraction: 1. Incident ray, refracted ray, and normal line lie on one flat plane and those three intersect at one point. 2. The ratio of projection of incident ray refracted ray at boundary plane between those two medium is constant number is defined as index of refraction.
13. 13. Convex Lense 1. Incident ray parallel to main axis is refracted through active focus F1. 2. Incident ray through positive focal point F2 is refracted parallel to main axis. 3. Incident ray through lens center point O is continued without experiencing refraction.
14. 14. Concave Lens 1. Incident ray Parallel to main axis is refracted seemed comes from active focal point F1. 2. Incident ray seemed goes to passive focal point F2 is refracted parallel to main axis. 3. Incident that goes to optic center point of lens is continued without refraction.
15. 15. An object is put 20 cm infront of a convex lens. If focus distance of the lensis 15 cm, what is:a. Image distance from the lensb. Image magnificationc. Properties of imageSolution:Given: S = 20 cm f = 15 cmAsked: a. S’..? b. M..? c. Properties of image..?
16. 16. Answer: So, image magnification is 3 times Properties of image is real, inversed enlarged.So, image distance from thelens is 60 cm behind thelens | 1,072 | 4,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-05 | latest | en | 0.816391 |
https://www.physicsforums.com/threads/light-intensity-through-2-slits.466749/ | 1,591,021,320,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347417746.33/warc/CC-MAIN-20200601113849-20200601143849-00380.warc.gz | 831,991,769 | 16,119 | # Light intensity through 2 slits
## Homework Statement
Monochromatic light of wavelength 620nm is going through a very narrow gap through the first curtain,then encounters a second curtain that is parallel with the first in which there are two parallel narrow slits S1 and S2 as shown in Fig. Slit S1 is located at the point of the second curtain that is closest point S, and S2 is d away from the S1. At point P, which is equidistant from the S1 and S2 we measure the intensity of the light and get the same intensity in both cases when only one of the leaked S1 and S2 is open, while in the case when both are open, we get 3 times greater the intensity. Determine the minimum distance d of slits, when the slits S and S1 are at distance L = 12dm.
## Homework Equations
path/vave length=phase difference/2*pi
The intensity is proportional to the electric wave squared
## The Attempt at a Solution
I tried to solve this using phasors, but in the end I am left with the equation containing the unknown term of phase difference (which contains the distance between the slits) and the intensity of a single slit. Am I doing it right?
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
Delphi51
Homework Helper
Welcome to PF! You have a difficult problem here and I am only offering a suggestion because it is about to fall off the first page and the real experts may not see it on page 2. If there was no phase difference for the two slits, you would get double the E field strength so, 4 times the intensity, wouldn't you? I'm sure you remember this better than I do! If so, then the next question is what phase difference will give 3 times the intensity or sqrt(3) times the E. It will be something like
E+E*sin(θ) = E*sqrt(3) I would think. Solve that for θ and then figure out what distance d will give you that phase angle for the S to S2 minus the S to S1 distance.
Last edited: | 459 | 1,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | longest | en | 0.949394 |
null | null | null | null | null | null | Lamb Cooking Introduction
Dry Heat Cooking | Moist Heat Cooking
There are two general methods used for cooking lamb (and most other meats):
Dry Heat Cooking and Moist Heat Cooking.
Dry Heat Cooking
When cooking lamb using dry heat cooking, the meat is in direct contact with a hot surface or close to the heat source. High heat quickly browns the surface of the meat; then the heat is most often turned down to a somewhat lower temperature to finish the cooking process. This method works best for tender cuts, although tougher cuts, which have been tenderized (as with a marinade), can be cooked successfully with dry heat.
Grilling, roasting, rotisserie, broiling, sautéing, and pan-frying are common dry heat cooking methods.
Lamb chops, boneless leg roasts, rib roasts, and tenderloins are popular tender cuts of lamb that are most often cooked with dry heat cooking methods.
Lamb Chop
Boneless Leg Roast
Rib Roast or Rack of Lamb
(Single Rack of Lamb)
Rib Roast or Guard of Honor
(Double Rack of Lamb)
Moist Heat Cooking
With moist heat methods, the meat is cooked in contact with hot liquid, usually at a lower temperature than dry heat cooking methods. The hot liquid tenderizes the meat, and it also acts as a flavoring source. Moist heat cooking methods are most often used when cooking tougher lamb cuts. Tender cuts of lamb can also be cooked with moist heat methods (but it is not preferred), but care must be taken to prevent the meat from being overcooked and becoming mushy. Braising, stewing, and slow cooking are among the most popular moist heat cooking methods.
Lamb cuts that benefit from moist heat cooking methods include the shank, boneless shank half of leg, shoulder, and neck.
Rear Leg Shank
Boneless Shank Half of Leg
(rolled and tied)
Lamb Cooking Introduction Reviews
© Copyright 2015 Tecstra Systems, All Rights Reserved, RecipeTips.com | null | null | null | null | null | null | null | null | null |
http://mathhelpforum.com/advanced-algebra/217966-limit-infina-ratio-test-root-test-print.html | 1,495,588,830,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607726.24/warc/CC-MAIN-20170524001106-20170524021106-00197.warc.gz | 234,577,848 | 3,466 | Limit infina of ratio test and root test
• Apr 21st 2013, 06:16 PM
phys251
Limit infina of ratio test and root test
For a strictly positive sequence, prove that $\liminf (a_j^{\frac{1}{j}}) \geq \liminf (\frac{a_{j+1}}{a_j})$.
I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf), but beyond that, I am drawing a blank. I don't even know where to begin.
• Apr 22nd 2013, 12:16 AM
Gusbob
Re: Limit infina of ratio test and root test
Hint: For $x>0$, $\lim_{n\to \infty} \sqrt[n]{x}=1$. In particular, $\liminf_{n\to \infty} \sqrt[n]{x}=\lim_{n\to \infty} \sqrt[n]{x}=1$.
Now think about why $\liminf \left(\frac{a_{n+1}}{a_n}\right)\leq 1$. (It may be useful to also think about why $\limsup \left(\frac{a_{n+1}}{a_n}\right) \geq 1$). In particular, what happens if the sequence is eventually constant? What if it never converges?
As an aside, this:
Quote:
I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf)
is not the way to think about it. I'm not even sure if makes sense. Anyways, the following is true for $a_n >0$:
$\liminf \left(\frac{a_{n+1}}{a_n}\right)\quad\leq\quad \liminf \sqrt[n]{a_n}\quad\leq \quad \limsup \sqrt[n]{a_n} \quad \leq\quad \limsup\left(\frac{a_{n+1}}{a_n}\right)$
• Apr 22nd 2013, 02:17 PM
phys251
Re: Limit infina of ratio test and root test
I still do not understand. At all.
• Apr 22nd 2013, 09:17 PM
Gusbob
Re: Limit infina of ratio test and root test
Do you understand the definition of liminf? Can you see that if a sequence a_n converges, liminf a_n = lim a_n = limsup a_n?
• Apr 23rd 2013, 07:35 AM
phys251
Re: Limit infina of ratio test and root test
Infinum means greatest lower bound. I.e., if A = (0,1), inf A = 0, because 0 <= all elements in A, and 0 is the greatest possible number that satisfies that inequality.
Lim inf x_n = lim (n -> infinity) {a_n, a_n+1, a_n+2, ...}. Basically, lim inf is the infinum of the sequence infinitely far from the starting element.
All of that I get. What I don't get is how to put it together in a meaningful way on this proof.
• Apr 23rd 2013, 03:08 PM
Gusbob
Re: Limit infina of ratio test and root test
If the sequence $a_n$ is not eventually constant, there are infinitely many points where $a_{n+1} < a_n$. Then $\limsup \frac{a_{n+1}}{a_n}\leq 1 = \liminf \sqrt[n]a_n = \lim \sqrt[n]a_n$. The case where $a_n$ is eventually constant is obvious. | 825 | 2,464 | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-22 | longest | en | 0.785241 |
null | null | null | null | null | null | An Unfinished Term Paper
‘I hate college,” Keith thought miserably. He had a term paper due in Intro to Space Plasma Physics tomorrow, and it was already midnight. ‘Not tomorrow anymore, and that’s what I get for making the idiot decision to major in Astrophysics,’ he grumbled to himself. ‘Three years of college so far, wasted if I don’t get this stupid paper done tonight.’ Sighing, he sat down at the computer and turned it on, waiting for it to boot. The machine was old, however, and took nearly five minutes just to start up, and during that time Keith heard his roommate, Adrian, walk into the apartment. ‘Of course, he took Aviation as his major, so he’s been out partying all night while I sat here trying to finish the homework for that stupid Optics class…’
Keith has long been jealous of his roommate, who, it seemed, never had any serious homework to do and was always out at parties hanging out with people from around school. No lonely nights for him, and no sitting up until 5am writing term papers, either. Keith cheerfully ignored that fact that he could have started writing the term paper last weekend, instead choosing to silently bash his roomie for having a good time. Engrossed in his silent litany of Adrian’s “faults” and busy waiting for the computer to boot, Keith jumped in surprise when Adrian wandered into his room.
“Bit jumpy aren’t you? You know it’s midnight…” Adrian grinned.
Keith rolled his eyes. “Hey. No, I didn’t notice.”
“Term paper again?” Keith nodded silently. Adrian laughed and walked over to him, sitting down on the floor next to Keith’s chair. “You know what your problem is?”
“I wasn’t aware I had a problem,” Keith growled, annoyed at the turn the conversation had just taken.
“Yep, it’s a big one. Know what it is?”
“Obviously not…”
“You’re too tense, that’s your problem. Of course, you’re also an idiot for taking something like Astrophysics, but mainly you’re just too tense. You should go to more parties or something.”
Keith, unhappy to hear his own thoughts repeated back to him, looked away bitterly. “I’m sorry, I’m just the geek who sits around doing homework all the time. I don’t get invited to parties.”
Adrian laughed. “You would if anyone thought you’d go. Being a geek is part of your charm you know.”
“I don’t have charm.”
“Do too.”
“Do – you know, this is just a stupid conversation. And anyway, you’re wrong, no one would ask me to go to a party.”
“Bet you I can prove you wrong.”
“I can!”
“Whatever, go for it.”
Keith was expecting to get hauled out of his chair and dragged off to a party somewhere; the term paper wasn’t going to get done anyway, so why not? He definitely wasn’t expecting for his roommate to stand up next to him, grab him by the back of the head, and pull him up into a full kiss on the mouth.
For a split second, Keith could do nothing but sit there in shock. The shock, however, wore away as Adrian continued kissing him and showed no signs of intending to stop, and Keith melted into the soft sensation of Adrian’s warm lips upon his. Pulling Adrian’s body down closer, Keith shifted to allow their chests to touch, and their bodies molded together in that sensuous embrace. Lost on the passion of that kiss, Keith allowed the misty feeling of Adrian’s lips to envelop him, forgetting about everything else. He forgot that only moments ago he had been panicking at the thought that he wasn’t going to get his term paper finished. He forgot that for the past five years of his life he had tried obliterating any feelings like this, viciously crushing any desire for the same sex, and at the same time, unwittingly crushing any desire for any sort of happiness. Now, however, nothing mattered beyond the blissful sensation of Adrian’s tender lips against his own.
Vaguely, as though from a great distance, Keith heard someone moaning and wondered in a small corner of his mind who it might be. It never occurred to him that it was himself. With a sigh of pleasure, the two of them separated, eyelashes brushing against each other as their eyes slowly opened. With their chests still pressed together, Keith could feel Adrian’s heart beating in time with his own, and a soft, wondering smile curved his lips.
Adrian smiled back, taking a few slow, deep breaths. Then he tangled his fingers in Keith’s hair and pulled Keith’s lips to his again.
Keith had time to suck in a quick breath before Adrian’s lips, now hard and burning with lust and desire, locked themselves to his. Then Adrian pulled back, for a second only, and kissed him again, showering Keith’s lips and face with rough kisses. Gasping in breaths between Adrian’s kisses, Keith began talking in a low voice.
“Take me to bed with you.”
Startled by Keith’s sudden boldness, Adrian drew back for a moment, gazing steadily at him. “Are you sure that’s what you want?”
Keith nodded, determined. He was tired of letting his fears dictate the way he ran his life. He had hurt himself with unfeeling for long enough; this outbreak of passion made him feel more alive than he’d felt in years, and now that he’d had a taste of being alive, he wasn’t about to let it go.
Adrian looked into his eyes for a minute longer, then smiled slowly, pleased by the determination and ardor that he saw there. Leaning forward to rest his chest against Keith’s again, he allowed his fingers to slide out of Keith’s hair and slowly, sensuously, down his back. He brought his lips up close to Keith’s ear and whispered softly.
“I am going to give you a night like you’ve never had before.” Keith shivered with pleasure at the sensation of those long, slender fingers trailing slowly down his back, and Adrian smiled, drawing his hands around to the front to move firmly up Keith’s chest, slowly swirling his fingertips around Keith’s nipples. Keith groaned, fully aroused, and Adrian took the opportunity to kiss him again, slipping his tongue into Keith’s open mouth. Excited by the new sensation, Keith slipped his own tongue into Adrian’s mouth. Chuckling, Adrian began sucking and teasing Keith’s tongue, swirling his own tongue around it and alternately sucking and blowing on it. Keith moaned again, and Adrian shoved their hips together, pressing his pulsing cock into Keith’s groin and moaning himself as he felt Keith’s dick grind into him.
Panting, Adrian pulled his mouth away from Keith’s again, still teasing his nipples with his fingertips. “Come on babe, let’s move this to the bedroom.”
They both stood, shivering at the sudden cold after being pressed together for so long. Unwilling to have their bodies separated for even such a short time, Keith quickly stepped forward and crushed himself back against Adrian, wrapping his arms around him. Adrian allowed his hands to wander back to the other side of Keith’s body, grasping his tight ass and Adrian pulled him into another passionate kiss, suckling on his tongue. Completely engrossed in each other, their bodies, slick with sweat, sliding up against each other’s and erections grinding together, the short trip down the hall to Adrian’s bedroom took them several minutes.
Once in the bedroom, Adrian broke off the kiss and shoved Keith away from him and onto the bed. Startled by the abrupt movement, but pleased to find himself sitting with his legs spread on the bed, he grinned mischievously up at Adrian, leaning back on one hand and using the other to stroke slowly down the length of his erection. Laughing at Keith’s brashness, Adrian stalked over, his own erect cock bouncing slightly with each step. Transfixed by the sight of that huge erection, Keith stopped paying attention long enough for Adrian to reach the bed and shove him flat onto his back.
“Oh, you’re a gorgeous little tease, aren’t you babe?” Grinning again in agreement, Keith languidly moved one hand up to tangle in Adrian’s hair, wrapping hi
s legs around him as Adrian lowered himself onto Keith’s body, kissing him again and running firm hands up and down Keith’s ches
t, flicking his nipples lightly on each pass. With each flick Adrian ground his hips into Keith’s, and the combined assault soon had Keith moaning wildly and bucking up and down beneath Adrian’s body. Thrilled with Keith’s enthusiastic response, Adrian continued this until he sensed Keith coming to the brink of an orgasm, then pulled away until Keith moved slowly back from the brink.
Keith sobbed at the sudden absence of Adrian’s body and opened his eyes, wondering what was wrong. Smiling gently, Adrian locked eyes with him and slowly lowered his mouth to Keith’s lips, lightly teasing them with his tongue as he watched Keith’s sea-green eyes flicker with pleasure. Smiling again, Adrian slowly moved down from Keith’s mouth, lapping lightly at the sweat that trickled down the groove of Keith’s chest. Pausing only to give each of Keith’s nipples a single, vicious lash with his tongue, actions that made Keith cry out and writhe in ecstasy, Adrian continued all the way down Keith’s body, stopping only when the length of Keith’s erection forced him to turn his head sideways to continue to its base.
Stopping there, Adrian slowly lipped at the base of Keith’s cock, massaging his balls slowly and gently and Keith’s erection pulsed hotly beneath his mouth. Absorbing Keith’s moans of pleasure, Adrian slowly, torturously, moved his lips up Keith’s shaft until he reached the tip of his penis, already leaking pre-cum as Keith’s entire body arched in rapture. Focused entirely on Keith’s erection now, Adrian cupped his balls in his hands and devoted his attention to Keith’s throbbing cock, taking just the tip of it into his mouth and swirling his tongue slowly around it, enjoying the salty taste of Keith’s pre-cum. Adrian bobbed his head shallowly up and down for several minutes, delighting in the taste and fell of the cock pulsating in his mouth, and smiling around it as he heard Keith whimper for more.
Happy to oblige his lover, Adrian allowed his mouth to slide down the length of Keith’s shaft, swallowing every inch of it. Keith’s whimpers quickly turned back into moans as Adrian pulled back and then slid his mouth down again, and again, picking up speed each time he repeated the action. Before long, Keith was pumping his hips in time with the pumping of Adrian’s mouth, and he felt himself quickly rising to the brink again. Adrian, feeling the tightening of Keith’s muscles and scrotum, didn’t pull away this time, instead rubbing at Keith’s balls and the base of his erection as he pumped his head up and down even faster.
Keith came hard, shooting a thick load into Adrian’s mouth. Swallowing it all and licking up what little escaped onto his lips, Adrian continued pumping at Keith’s cock until all of his spurts had subsided. Saving the cum from those last spurts in his mouth, Adrian slid back up Keith’s body, parting his lips with his tongue and letting him get a small mouthful of himself, which they passed back and forth with their tongues before finally swallowing it all.
Sighing with pleasure, Keith’s body went lax, until Adrian lowered his body back onto his and Keith realized that Adrian still had a massive erection. Laughing softly at the look of surprise on Keith’s face, Adrian murmured into his ear again. “I promised you a night like you’d never forget… That blow job certainly wasn’t all you’re getting from me tonight.” Keith gasped as Adrian brought his fingers down to rub slowly at Keith’s opening, and Adrian smiled at him again. “Time for some fun of a rougher sort, babe.”
Sliding off of Keith, Adrian reached over to the end table next to the bed and pulled a bottle of lotion out of the top drawer. He lathered the lotion onto his stiff cock as Keith watched, open-mouthed, and then he indicated that Keith should roll over. “On all fours, babe.” Keith complied quickly, and Adrian, with lotion still slathered all over his hands and fingers, began rubbing around Keith’s opening with his fingers. After Keith relaxed a little bit, Adrian slid one well-lubricated finger up into his tight opening, smiling at Keith’s grunt and at the tightness of the hole. Adrian quickly slid a second finger into the hole to join the first, pressing them in and out and spreading them slightly to loosen up the opening. At first, Keith was merely uncomfortable and a bit sore at the intrusion; then Adrian’s fingers bumped up against his prostate. Gasping and jerking at the sudden shock of pleasure, Keith’s muscles tightened up with excitement and he pressed himself farther onto Adrian’s fingers. Adrian grinned at Keith’s eagerness and slipped his fingers out.
Keith looked over his shoulder to glare at Adrian, but his frustration was short-lived as Adrian replaced his fingers with the tip of his cock, pressing it slowly inside. Keith shuddered with satisfaction as Adrian pumped his tip in and out of Keith’s opening, allowing some pre-cum to build up there. After doing this for a few moments, Adrian suddenly thrust his hips forward, letting the pre-cum lubricate the way as he slammed his entire cock deep into Keith’s ass. Keith cried out sharply in pain, and Adrian paused for a moment before continuing so that Keith could adjust. A moment was all he gave, however, and he quickly pulled out all the way to the tip again and then, like before, slammed his shaft into Keith, gripping Keith’s hips and jerking them back on his cock. Keith’s cry this time was mingled with pleasure as he noted the sensation of that burning, quivering shaft sliding through him and pounding in. Grinning at the pleasure in Keith’s voice, Adrian began thrusting rhythmically, unable to contain his own shouts of ecstasy as he pounded in. By this point, Keith was beyond feeling any pain, and he shouted in time with Adrian, gripping the headboard of the bed tightly and trying to shove himself back onto Adrian’s dick each time Adrian thrusted forward and jerked on his hips.
Keith’s own cock was fully erect again, bouncing each time Adrian’s massive tool slammed into him. Adrian took one hand off of Keith’s hips, where bruises in the form of fingerprints were starting to form, and closed his fist around Keith’s cock, jacking it off in time with his own plunging dick. Blood rushing in their ears, the two could barely hear their own ecstatic shouts as Adrian fucked Keith as hard as he could. Keith was the first to cum, shooting his load all over Adrian’s pillows. The tightening of the muscles in his ass when he came set Adrian off, and he shot his load deep into Keith’s as they both collapsed onto the bed, their heads resting on the cum-covered pillows.
Wrapping one leg around Keith, Adrian laid there with his cock still deep inside Keith’s ass, with cum leaking out around it and dribbling slowly onto the bedspread. Keith sighed, wriggling backwards as close as he could to Adrian, his hair sticking to his face as his own cum smeared across it. He hadn’t even gotten his breath back yet, but he already wished Adrian were hard again so that he could start ramming him again. Adrian, guessing where Keith’s thoughts lay, smiled and nibbled on his earlobe, reaching around him and teasing Keith’s nipples with his fingers. “Don’t worry babe,” he whispered, as his lips curved up into a soft smile. “We’ll do this again real soon.”
Sighing contentedly, Keith relaxed against him, letting the fogginess of sleep overtake him. ‘I never did do that term paper…”
To be continued…
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https://www.cleariitmedical.com/2021/07/straight-lines-quiz-4.html | 1,718,848,129,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00731.warc.gz | 602,721,052 | 122,560 | ## Straight Lines Quiz-4
The unit Straight Line holds sheer significance in the JEE Advanced and other engineering exams. It has a weightage of 6% pairing with Circles. With focused practice good marks can be fetched from this section. These questions are important in achieving your success in JEE and Other Engineering Exams..
Q1. A beam of light is sent along the linex-y=1, which after refraction from the x-axis enters the opposite side by running through 30° towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is
• (x-1)^2+y^2=4
• x+y=3
• 2x-y=0
• None of these
Solution
Q2.A line of fixed length 2 units moves so that its ends are on the positive x-axis and that part of the line x+y=0 which lies in the second quadrant. Then the locus of the midpoint of the line has the equation
• x2+5y2+4xy-1=0
• x2+5y2+4xy+1=0
• x2+5y2-4xy-1=0
• 4x2+5y2+4xy+1=0
Solution
Q3. The number of integral values of m, for which the x-coordinate of the point of intersection of the lines 3x+4y=9 and y=mx+1 is also n integer is
• 2
• 0
• 4
• 1
Solution
Q4. If u=a1 x+b1 y+c1=0,v=a2 x+b2 y+c2=0 and a1/a2=b1/b2=c1/c2, then the curve u+kv=0 is
• The same straight line u
• Different straight line
• Not a straight line
• None of these
Solution
Q5.If x-2y+4=0 and 2x+y-5=0 are the sides of a isosceles triangle having are 10 sq. units the equation of third side is
• 3x-y=-9
• 3x-y+11=0
• x-3y=19
• 3x-y+15=0
Solution
Q6. The locus of the point which moves such that its distance from the point (4, 5) is equal to its distance from the line 7x-3y-13=0 is
• A straight line
• A circle
• A parabola
• An ellipse
Solution
Q7.If the vertices P and Q of a triangle PQR are given by (2, 5) and (4,-11), respectively, and the point R moves along the line N given by 9x+7y+4=0, then the locus of the centriod of the triangle PQR is a straight line parallel to
• PQ
• QR
• RP
• N
Solution
Q8.If x1,x2,x3 as well as y1,y2,y3 are in G.P. with same common ratio, then the points P(x1,y1 ),Q(x2,y2) and R(x3,y3)
• Lie on a straight line
• Lie on an ellipse
• Lie on a circle
• Are vertices of a triangle
Solution
Q9.The coordinates of two consecutive vertices A and B of a regular hexagon ABCDEF are (1, 0) and (2, 0), respectively. The equation of the diagonal CE is
• √3 x+y=4
• x+√3 y+4=0
• x+√3 y=4
• None of these
Solution
Q10. If sum of the distance of a point from two perpendicular lines in a plane is 1, then its locus is
• A square
• A circle
• A straight line
• Two intersecting lines
Solution
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Straight Lines-Quiz-4 | 1,024 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.84416 |
null | null | null | null | null | null | Wiimote problems- just smack it
by: Chuck -
If your Wii remote is acting up it seems like the best way to fix it is to turn the controller button side down and smack it into the palm of your hand. That's not my advice, that's actually the advice from Nintendo tech support (my advice would be to just throw the thing at the wall but that's just me). I'm not sure what's funnier, the fact that this is a Nintendo approved solution or that it actually works.
Hat tip to Kotaku for the link.
comments powered by Disqus | null | null | null | null | null | null | null | null | null |
https://www.exactlywhatistime.com/time-ago/310-years-ago-from-today | 1,717,032,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00314.warc.gz | 653,297,577 | 7,684 | # What was the date 310 years ago?
## Wednesday May 30, 1714
0
310 years in the past was 30 May 1714, a Wednesday. Subtracting 310 years in the past is usually simple. Anything under a decade can usually be counted on one hand. The biggest challenge will be skipping decades behind or even centuries. Additionally, we’re 30 days from the end of May, so being in the end of of the month, you'll need to consider monthly changes as well. Weekly and daily changes most likely won't impact 310 years ago.
## How we calculated 310 years before today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how 310 years ago gets calculated on each visit:
1. We started with date inputs: used current day of 30 May, then set the calculation - 310 years, and factored in the year 2024
2. Noted your current time of year: 310 years in May will bring us back to April or further.
3. Counted backwards years from current day: date - 30 May, factoring in the 30 days left in May to calculate Wednesday May 30, 1714
4. Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the fiscal calendar
### Tips when solving for May 30
• Current date: 30 May
• Day of the week: Wednesday
• New Date: Wednesday May 30, 1714
• New Date Day of the week: Wednesday
• Counting backward from May could put you back in Q1 or even the previous year.
• The solution crosses into a different year.
## Ways to calculate 310 years ago
1. Calculate it: Start with a time ago calculator. 310 years is easiest solved on a calculator. For ours, we've already factored in the 30 days in May + all number of days in each month and the number of days in . Simply add your years and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use May's calendar: Begin by identifying 30 May on a calendar, note that it’s Wednesday, and the total days in April (trust me, you’ll need this for smaller calculations) and days until last year (double trust me, you'll need this for larger calculations). From there, count backwards 310 times years by years, subtracting years from until your remainder of years is 0.
3. Use excel: For more complex years calculations or if you h8 our site (kidding), I use Excel functions like =TODAY()-310 to get or =WORKDAY(TODAY()), -310, cell:cell) for working years.
## Working years in 310 calendar years
310 years is Wednesday May 30, 1714 or could be if you only want workdays. This calculation takes 310 years and only subtracts by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Wednesday May 30, 1714 date.
Work years Solution
Monday
Tuesday
310 years back
Wednesday
May 30
Thursday
Friday
Saturday
Sunday
## The past 310 years is equivalent to:
Counting back from today is Wednesday May 30, 1714 using a full calendar, and is also 2715600 hours ago and 41.1% of the year.
Did you know?
Wednesday Wednesday May 30, 1714 was the 150 day of the year. At that time, it was 41.1% through 1714.
## In 310 years, the average person Spent...
• 24304620.0 hours Sleeping
• 3231564.0 hours Eating and drinking
• 5295420.0 hours Household activities
• 1575048.0 hours Housework
• 1737984.0 hours Food preparation and cleanup
• 543120.0 hours Lawn and garden care
• 9504600.0 hours Working and work-related activities
• 8744232.0 hours Working
• 14311212.0 hours Leisure and sports
• 7766616.0 hours Watching television
## What happened on May 30 (310 years ago) over the years?
### On May 30:
• 1987 North American Philips Company unveils compact disc video
• 1911 1st Indianapolis 500: Ray Harroun driving a Marmon Wasp for Nordyke & Marmon Company comes out of retirement, wins inaugural event; average speed: 74.602 mph (120.060 km/h) | 1,000 | 3,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.940298 |
https://web2.0calc.com/questions/solve_55 | 1,611,599,155,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00441.warc.gz | 614,296,811 | 5,963 | +0
# solve
0
97
2
Find real numbers x and y such that x + y = 6 and x^3 + y^3 = 60.
Sep 26, 2020
#1
+10834
+1
Find real numbers x and y such that x + y = 6 and x^3 + y^3 = 60.
Hello Guest!
$$x+y=6\\ y=6-x\\ x^3+y^3=60\\ x^3+(6-x)^3=60\\ x^3+(36-12x+x^2)(6-x)=60$$
$$x^3+216-36x-72x+12x^2+6x^2-x^3=60\\ 18x^2-108x+216=60$$
$$18x^2-108x+156=0$$
$$x = {-108 \pm \sqrt{11664-4\cdot 18\cdot 156} \over 2\cdot 18}\\ x=\frac{-108\pm 20.785}{36}$$
$$x_1=-2.423\\ x_2=-3.577$$ $$y_1=8.423\\ y_2=9.577$$
!
Sep 26, 2020
edited by asinus Sep 27, 2020
#2
0
x + y = 6.......................(1)
x^3 + y^3 =60.............(2)
y =6 - x
Solve for x:
(6 - x)^3 + x^3 = 60
Expand out terms of the left hand side:
18 x^2 - 108 x + 216 = 60
Divide both sides by 18:
x^2 - 6 x + 12 = 10/3
Subtract 12 from both sides:
x^2 - 6 x = -26/3
x^2 - 6 x + 9 = 1/3
Write the left hand side as a square:
(x - 3)^2 = 1/3
Take the square root of both sides:
x - 3 = 1/sqrt(3) or x - 3 = -1/sqrt(3) | 493 | 992 | {"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} | 4.28125 | 4 | CC-MAIN-2021-04 | latest | en | 0.619272 |
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/16%3A_Electric_Charges_and_Fields/16.05%3A_Summary | 1,642,831,772,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00444.warc.gz | 508,488,808 | 24,681 | $$\require{cancel}$$
# 16.5: Summary
## Key Takeaways
Objects can acquire a net charge if they acquire a net excess or deficit of electrons. Charges are never created, they are only transferred from one object to another. One can charge an object by friction, conduction, or induction. Materials can be classified broadly as conductors, where electrons can move freely in a material, or conductors, in which electrons remain tightly bound to the atoms in the material. If a conducting object acquires a net charge, those charges will migrate to the surface of the conductor.
Coulomb was the first to quantitatively describe the electric force exerted on a point charge, $$Q_1$$, by a second point charge, $$Q_2$$, located a distance, $$r$$, away: \begin{aligned} \vec F_{12}=k\frac{Q_1Q_2}{r^2}\hat r_{21}=\frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{r^2}\hat r_{21}\end{aligned} where $$\hat r_{21}$$ is the unit vector from $$Q_2$$ to $$Q_1$$. One can write the force using either Coulomb’s constant, $$k$$, or the permitivity of free space, $$\epsilon_0$$. Coulomb’s force is attractive if the product $$Q_1Q_2$$ is negative, and repulsive if the product is positive. Thus, charges of the same sign exert a repulsive force on each other, whereas opposite charges exert an attractive for on each other.
Mathematically, Coulomb’s Law is identical to the gravitational force in Newton’s Universal Theory of Gravity, which implies that it is conservative. The electric field vector at some position in space is defined to be the electric force per unit charge at that position in space. That is, at some position in space where the electric field vector is $$\vec E$$, a charge, $$q$$, will experience an electric force: \begin{aligned} \vec F=q\vec E\end{aligned} much like a mass, $$m$$, will experience a gravitational force, $$m\vec g$$, in a position in space where the gravitational field is $$\vec g$$. A positive charge will experience a force in the same direction as the electric field, whereas a negative charge will experience a force in the direction opposite of the electric field. The electric field at position, $$\vec r$$, from a point charge, $$Q$$, located at the origin, is given by: \begin{aligned} \vec E = k\frac{Q}{r^2}\hat r\end{aligned} One can visualize an electric field by using “field lines”. The field vector at any point in space has a magnitude that is proportional to the number of field lines at that point, and a direction that is tangent to the field lines at that point.
We can model the electric field from a continuous charged object (i.e. not a point charge) by modeling the object as being made up of many point charges. Often, it is easiest to model an $$N$$-dimensional object as being the sum of objects of dimension $$N-1$$ and an infinitesimal length in the remaining dimension. For example, we modeled a line of charge as the sum of point charges that have an infinitesimal length, and we modeled a disk of charge as the the sum of rings that have an infinitesimal thickness. In general, the strategy to model the electric field from a continuous distribution of charge is the same:
1. Make a good diagram.
2. Choose a charge element $$dq$$.
3. Draw the electric field element, $$d\vec E$$, at the point of interest.
4. Write out the electric field element vector, $$d\vec E$$, in terms of $$dq$$ and any other relevant variables.
5. Think of symmetry: will any of the component of $$d\vec E$$ sum to zero over all of the $$dq$$?
6. Write the total electric field as the sum (integral) of the electric field elements.
7. Identify which variables change as one varies the $$dq$$ and choose an integration variable to express $$dq$$ and everything else in terms of that variable and other constants.
8. Do the sum (integral).
Finally, we introduced the electric dipole, which is an object comprised of two equal and opposite charges, $$+Q$$ and $$-Q$$, held at fixed distance, $$l$$, from each other. One can model an electric dipole using its dipole vector, $$\vec p$$, defined to point in the direction from $$-Q$$ to $$+Q$$, with magnitude: \begin{aligned} p=Ql\end{aligned} When a dipole is immersed in a uniform electric field, $$\vec E$$, it will experience a torque given by: \begin{aligned} \vec\tau=\vec p\times \vec E\end{aligned} The torque will act such as to align the vector $$\vec p$$ with the electric field vector. We can define a potential energy, $$U$$, to model the energy that is stored in a dipole when it is not aligned with the electric field: \begin{aligned} U=-\vec p \cdot \vec E\end{aligned} The point of lowest potential energy corresponds to the case when $$\vec p$$ and $$\vec E$$ are parallel, whereas the point of highest potential energy is when the two vectors are anti-parallel.
## Important Equations
### Electric field:
\begin{aligned} \vec E = k \frac{Q}{r^2}\vec r \\ \vec E = \int d \vec E \\\end{aligned}
### Electric force:
\begin{aligned} \vec F = q \vec E \\\end{aligned}
### Electric dipole moment:
\begin{aligned} p = Ql \\\end{aligned}
### Torque on a dipole:
\begin{aligned} \vec \tau = \vec p \times \vec E \\\end{aligned}
### Potential energy stored in a dipole:
\begin{aligned} U = -\vec p \cdot \vec E \\\end{aligned}
## Important Definitions
Definition
Charge: An object will have a charge if it has an excess or deficit of electrons. SI units: $$[\text{C}]$$. Common variable(s): $$Q$$, $$q$$.
Definition
Electric field: The electric field is defined to be the electric force per unit charge. SI units: $$[\text{N/C},\text{V/m}]$$. Common variable(s): $$\vec E$$.
Definition
Coulomb’s constant: A fundamental physical constant which relates charge and distance to electric field. SI units: $$[\text{Nm}^{2}C^{-2}]$$. Common variable(s): $$k$$.
Definition
Electric dipole moment: A vector used to represent an electric dipole. SI units: $$[\text{Cm}]$$. Common variable(s): $$\vec p$$.
Definition
Linear charge density: The charge per unit length of an object. SI units: $$[\text{C/m}]$$. Common variable(s): $$\lambda$$.
Definition
Surface charge density: The charge per unit area of an object. SI units: $$[\text{Cm}^{-2}]$$. Common variable(s): $$\sigma$$.
Definition
Volume charge density: The charge per unit volume of an object. SI units: $$[\text{Cm}^{-3}]$$. Common variable(s): $$\rho$$. | 1,632 | 6,322 | {"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} | 3.671875 | 4 | CC-MAIN-2022-05 | latest | en | 0.867729 |
https://socratic.org/questions/how-do-you-write-the-equation-of-the-circle-with-center-point-2-4-and-passing-th | 1,713,325,157,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817128.7/warc/CC-MAIN-20240417013540-20240417043540-00630.warc.gz | 495,803,855 | 5,921 | # How do you write the equation of the circle with center point (2,4) and passing through point (4,-1)?
Jun 22, 2016
${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 29$
#### Explanation:
The radius of the circle is
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(2 - 4\right)}^{2} + {\left(- 1 - 4\right)}^{2}} = \sqrt{29}$
The general formula for a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
In this case
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 29$ | 248 | 628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-18 | latest | en | 0.600654 |
null | null | null | null | null | null | not loaded not loaded not loaded
Sandusky, OH
about +
I am a college student majoring in computer science, but I have spent a large amount of my time working on video and audio projects with my friends. "A Series of One Takes" came about from the desire of Chris S. to record songs in one take (or as few as possible). Many of the songs relate to a video project we were also working on at the time - "Wolfie's Keyboard of Doom 2."
I enjoy acoustic guitar, and learned to play about 6 years ago. I am not a seriously committed artist, as I usually use the excuse that I have too much else to do, but I enjoy making instrumental songs or riffs when I can.
music +
upcoming shows + | null | null | null | null | null | null | null | null | null |
https://elsenaju.eu/Functions/Complex-Functions.htm | 1,709,470,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00589.warc.gz | 215,066,042 | 8,908 | # Domain colouring plots of complex functions
## Basic settings for the plotter
### Hue
The hue is selected according to the angle.
### Lightness
The brightness is determined according to the following diagram.
In the interval [0,0.5) applies val = a1 * k + b1
In the interval [0.5,1) applies val = a2 * k + b2
It is: min ≤ val ≤ max
### Saturation
The saturation is determined according to the following diagram.
In the interval [0,0.5) applies sat = a1 * k + b1
In the interval [0.5,1) applies sat = a2 * k + b2
It is: min ≤ sat ≤ max
Selecting a color scheme:
Angle offset: φ0=
Set width and height of the plot:
Width in px=
Height in px=
a1=
b1=
a2=
b2=
min=
max=
a1=
b1=
a2=
b2=
min=
max=
## Representation of complex functions
### Linear complex function
$f\left(z\right)=z=x+iy$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)=x$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)=y$
Amount of f(z):
$|f\left(z\right)|=\sqrt{{x}^{2}+{y}^{2}}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=\mathrm{atan}\frac{x}{y}$
### Square complex function
$f\left(z\right)={z}^{2}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)={x}^{2}-{y}^{2}$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)=2xy$
Amount of f(z):
$|f\left(z\right)|={x}^{2}+{y}^{2}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=\mathrm{atan}\frac{2xy}{{x}^{2}-{y}^{2}}$
### Fractional rational complex function
$f\left(z\right)=\frac{1}{z-a}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a=\alpha +i\beta$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Parameter
α=
β=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)$$=\frac{x-\alpha }{{\left(x-\alpha \right)}^{2}-{\left(y-\beta \right)}^{2}}$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)$$=\frac{-\left(y-\beta \right)}{{\left(x-\alpha \right)}^{2}-{\left(y-\beta \right)}^{2}}$
Amount of f(z):
$|f\left(z\right)|$$=\frac{1}{\sqrt{{\left(x-\alpha \right)}^{2}-{\left(y-\beta \right)}^{2}}}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=\mathrm{atan}\frac{y-\beta }{\alpha -x}$
### Fractional linear function
$f\left(z\right)=\frac{z+a}{z+b}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a=\alpha +i\beta \text{;}\phantom{\rule{1em}{0ex}}b=\gamma +i\delta$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Parameter
α=
β=
γ=
δ=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)$$=\frac{\left(x+\alpha \right)\left(x+\gamma \right)+\left(y+\beta \right)\left(y+\delta \right)}{{\left(x+\gamma \right)}^{2}+{\left(y+\delta \right)}^{2}}$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)$$=\frac{\left(y+\beta \right)\left(x+\gamma \right)-\left(y+\delta \right)\left(x+\alpha \right)}{{\left(x+\gamma \right)}^{2}+{\left(y+\delta \right)}^{2}}$
$f\left(z\right)=\frac{1}{z-a}+\frac{1}{z-b}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a=\alpha +i\beta \text{;}\phantom{\rule{1em}{0ex}}b=\gamma +i\delta$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Parameter
α=
β=
γ=
δ=
The function f is a composition of the previous f=f1+f2 and so the real and imaginary parts result from the addition of the individual functions f1 and f2.
Real part of f(z):
$\mathfrak{Re}\left({f}_{1}+{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)+\mathfrak{Re}\left({f}_{2}\right)$
Imaginary part of f(z):
$\mathfrak{Im}\left({f}_{1}+{f}_{2}\right)=\mathfrak{Im}\left({f}_{1}\right)+\mathfrak{Im}\left({f}_{2}\right)$
### Complex e-function
$f\left(z\right)={e}^{z}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)={e}^{x}\mathrm{cos}y$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)={e}^{x}\mathrm{sin}y$
Amount of f(z):
$|f\left(z\right)|={e}^{x}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=y+2n\pi$
### exp(z) / (z-a)
$f\left(z\right)=\frac{{e}^{z}}{z-a}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a=\alpha +i\beta$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Parameter
α=
β=
The function f is a product of two functions f=f1 * f2 and then the real and imaginary result follows:
Real part of f(z):
$\mathfrak{Re}\left({f}_{1}*{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)\mathfrak{Re}\left({f}_{2}\right)-\mathfrak{Im}\left({f}_{1}\right)\mathfrak{Im}\left({f}_{2}\right)$
Imaginary part of f(z):
$\mathfrak{Im}\left({f}_{1}*{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)\mathfrak{Im}\left({f}_{2}\right)+\mathfrak{Re}\left({f}_{2}\right)\mathfrak{Im}\left({f}_{1}\right)$
### exp(z)/(z-a) + (z+b)/(z+c)
$f\left(z\right)=\frac{{e}^{z}}{z-a}+\frac{z+b}{z+c}$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a=\alpha +i\beta \phantom{\rule{1em}{0ex}}\text{;}\phantom{\rule{1em}{0ex}}b=\gamma +i\delta \phantom{\rule{1em}{0ex}}\text{;}\phantom{\rule{1em}{0ex}}c=\epsilon +i\zeta$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Parameter
α=
β=
γ=
δ=
ε=
ζ=
The function f is a product of two functions f=f1 * f2 and then the real and imaginary result follows:
Real part of f(z):
$\mathfrak{Re}\left({f}_{1}*{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)\mathfrak{Re}\left({f}_{2}\right)-\mathfrak{Im}\left({f}_{1}\right)\mathfrak{Im}\left({f}_{2}\right)$
Imaginary part of f(z):
$\mathfrak{Im}\left({f}_{1}*{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)\mathfrak{Im}\left({f}_{2}\right)+\mathfrak{Re}\left({f}_{2}\right)\mathfrak{Im}\left({f}_{1}\right)$
The function f is also the sum of two functions f=f1+f2 and so the real- and imaginary part results from addition of f1 and f2.
Real part of f(z):
$\mathfrak{Re}\left({f}_{1}+{f}_{2}\right)=\mathfrak{Re}\left({f}_{1}\right)+\mathfrak{Re}\left({f}_{2}\right)$
Imaginary part of f(z):
$\mathfrak{Im}\left({f}_{1}+{f}_{2}\right)=\mathfrak{Im}\left({f}_{1}\right)+\mathfrak{Im}\left({f}_{2}\right)$
### Complex sine function
$f\left(z\right)=\mathrm{sin}z$
$\text{with}\phantom{\rule{1em}{0ex}}z\phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)=\mathrm{sin}x\mathrm{cosh}y$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)=\mathrm{cos}x\mathrm{sinh}y$
Amount of f(z):
$|f\left(z\right)|=\sqrt{{\mathrm{sin}}^{2}x+{\mathrm{sinh}}^{2}y}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=\mathrm{atan}\left(\mathrm{cot}x\mathrm{tanh}y\right)$
### x2 + i y2
$f\left(z\right)={x}^{2}+i{y}^{2}$
Axes ranges
x-min=
x-max=
y-min=
y-max=
Real part of f(z):
$\mathfrak{Re}\left(f\left(z\right)\right)={x}^{2}$
Imaginary part of f(z):
$\mathfrak{Im}\left(f\left(z\right)\right)={y}^{2}$
Amount of f(z):
$|f\left(z\right)|=\sqrt{{x}^{2}+{y}^{2}}$
Argument φ of f(z):
$\phi \left(f\left(z\right)\right)=\mathrm{atan}\frac{{y}^{2}}{{x}^{2}}$
## General
The function theory investigates functions of complex variable functions so complex numbers whose values range are also complex numbers. The complex numbers are an extension of the real numbers in the two-dimensional space. Many computational rules of real numbers can be applied to complex numbers. Was justified, the theory of complex functions mainly by Augustin-Louis Cauchy, Bernhard Riemann, and Karl Weierstrass.
## Domain Colouring
The color wheel method is a method to graphically represent complex functions. Complex functions represent the two-dimensional complex plane in turn, the real and imaginary values. The color circle method used amount r = |f(z)| and angle φ the complex function value f(z) aroand the display color of the function value set. According to r and φ the function value is selected the color from the color wheel. The amount defines the saturation and modulo is mapped to intervals . The first interval is 0 .. 1 then follow the intervals ( 1 .. e] , (e. .. e 2 ] , (e 2 ... e 3 ], etc. the color is defined by the angle and in 6 color zones starting with split blue from 0° to 60° and ending with green from 300° to 360°. the method is designed to that the function values are close together are also displayed similar color. mapping the sums on intervals of the power of e corresponds to a logarithmic representation.
## Colour Wheel
A compilation of a color wheel can be put together from different points depending on which state of affairs is to be visualized. The basis for the color circle the perception of similar colors. Leaving subjects with normal color pattern according to the sensation on similarity sort, which hues are usually brought in the same order. Beginning and end of the series are aroand so similar that the series can be closed to form a circle.
## Gauss Plane
The complex numbers are two-dimensional and can be used as vectors in the Gaussian plane of numbers represent. On the horizontal axis (Re) of the real part and on the vertical axis is applied (Im) of the imaginary part of the complex number. Vectors may also be similar to either the complex number in Cartesian coordinates (x, y) or polar coordinates (r, φ) can be expressed. In the color circle method polar coordinates are used and the color wheel is placed on the manner interval Gaussian-number plane. | 3,686 | 10,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 58, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.513924 |
http://thebeautifullmind.com/2012/05/24/lead-compensator-design-with-bode-plot/ | 1,386,925,944,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164921422/warc/CC-MAIN-20131204134841-00000-ip-10-33-133-15.ec2.internal.warc.gz | 184,056,120 | 21,169 | # Lead Compensator design with Bode plot
The steps to design the Lead Compensator are:
1. Determine K from the error constants given
2. Sketch the bode plot
3. Determine phase margin
4. The amount of phase angle to be contributed by lead network is
, where is the required phase margin and is 5 initially. if the angle is greater than 60then we have to design the compensator as 2 cascade compensator with lead angle as
5. calculate
from bode plot find such that it is the frequency corresponding to the gain
6. calculate
7. a lead compensator has the form
8. form the complete transfer function with the lead compensator added in series to th original system
9. plot the new Bode plot and determine phase margin and observe that it is the required phase margin
10. if not satisfactory repeat steps from step 4 by changing value of by 5
Now to do this In Matlab let us take a question. I will be solving the question number 6.2 in Control Systems By Nagoor Kani. In the text book the question has been solved without using matlab you can go through it to understand the steps better.
The question is
the phase margin should be atleast 33 and velocity error constant is 30.
Solution:
The velocity error constant is
On solving we get K =9600
The code below shows the matlab commands to obtain the design.
clear all % clear all variables
close all % close all previous graphs
K=9600 % value of K calculated from error constants
G=tf(K,conv([1 4 0],[1 80])) % make the transfer function
[Gm,Pm]=margin(G) %get the gain margin and phase margin
margin(G) %plot the bode plot
Pd=input('Enter Desired Phase margin')
ep=input('Enter value e') % enter the value of epsilon initially start with 5
ph=Pd-Pm+ep
if ph>60 % check if angle needed is greater than 60
phm=ph/2
else
phm=ph
end
alpha=(1-sind(phm))/(1+sind(phm)) % calculate alpha
db=-20*log10(1/sqrt(alpha)) %calculate the gain in db
wm=input(‘Enter frequency’) % from plot obtain frequency corresponding to above gain
T=1/(wm*sqrt(alpha))
Gc=tf([1 1/T],[1 1/(alpha*T)]) % form transfer function of compensator
‘Open Loop Transfer Function is’
if ph==phm
‘Single’
Go=Gc*G/alpha
else
Go=Gc*Gc*G
%if angle was greater than 60 we designed compensator for half value so we cascade them and their attenuation is not nullified and should be retained
end
margin(Go)
pause
comp=feedback(Go,1,-1);
uncomp=feedback(G,1,-1);
step(uncomp) % observe step response
hold on
step(comp)
legend(‘Uncompensated’,'Compensated’)
Step response of compensated and Uncompensated system | 670 | 2,523 | {"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} | 3.6875 | 4 | CC-MAIN-2013-48 | longest | en | 0.838183 |
https://math.stackexchange.com/questions/2967680/basic-combinatorial-probability-with-dice | 1,716,963,401,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00472.warc.gz | 322,167,610 | 35,759 | # Basic Combinatorial Probability with Dice
How many times do you have to roll a pair of dice such that you've got a 50% chance of at least one roll being double sixes?
My own intuition (and mentioned in the video cited below as a common wrong answer) is 18: $$\frac{1}{36}$$ chance of rolling double sixes, therefore after 18 rolls you've got $$\frac{18}{36}=0.5$$.
I understand mathematically how the correct answer is $$\frac{\log{\left(0.5\right)}}{\log{\left(\frac{35}{36}\right)}}\approx24.6$$, but I'm missing how this number comes about intuitively. How could I explain this result to say, my family around the dinner table, without exponents or logarithms?
Relatedly, I made a spreadsheet calculating the number of rounds required to solve this problem for any $$n$$ (in this question $$n=36$$), and found that if you divide the correct answer (24.6) by the "intuitive" answer (18), the result appears to converge at 1.39. Does this value mean anything?
$$\lim_{n\to\infty}\frac{2\log\left(0.5\right)}{n\log\left(\frac{n-1}{n}\right)}\approx1.3874032818...$$
(The source of this question is a Vsauce2 video https://www.youtube.com/watch?v=Uyw7d579nxY&feature=youtu.be&t=141)
• Not sure what you mean by saying $24.6$ is the correct answer. Obviously, you can't roll the dice $24.6$ times. The correct answer must be the least integer exceeding this, hence $25$. I've posted more detail below.
– lulu
Oct 23, 2018 at 14:24
The probability that a given roll yields double sixes is $$\frac 1{36}$$. It follows that the probability that a given roll yields something else is $$\frac {35}{36}$$. Thus, the probability that $$n$$ given rolls fails to yield any double sixes is $$\left(\frac {35}{36}\right)^n$$. So you want the least $$n$$ such that $$\left(\frac {35}{36}\right)^n≤.5$$
You can now proceed by trial and error, using a calculator of course. Not hard to find $$n=25$$ as the solution. To do it analytically you can solve $$\left(\frac {35}{36}\right)^x=.5$$ using logs. We get $$x\times (\log {35}-\log {36})=\log {.5}\implies x\approx 24.61$$
which again leads us to $$n=25$$.
To your second question, since $$\lim_{n\to \infty} \left( \frac {n-1}n\right)^n=e^{-1}$$ you are just computing $$-2\log {.5}=\boxed {1.386294361}$$. | 673 | 2,254 | {"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": 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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.865861 |
https://www.coursehero.com/file/6293268/solution2A/ | 1,527,401,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00579.warc.gz | 719,002,036 | 32,862 | {[ promptMessage ]}
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solution2A
# solution2A - [email protected] Be sure to bubble the answers to...
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Unformatted text preview: w @NOTE: Be sure to bubble the answers to questions 1 throuthO on ' your scantron. ' _ Part I: 2 points each In problems 1 through 5 indicate Whether each of the following statements 7 is true or false. 1. The equation :32 + 7 = 5:2} hasno real solution. X2“§?‘~ “a? “a {:3 rue dis cwméflgffll ;. (fly); .- (f- ‘ 7 ‘5: “" 3 6: - b False _ 2-. The inequality 3 W5 is equivalent. to the combined. inequality —'5 3 CB 3 5. ' 7 _ 7 _ . f _ l“) E “"3” Ha; mo «agflxcaiormlwm a. True ' _ fihfiFalse I . - (£15 ' Q/Wmffisa _ 3" Q) 3. (f3)2=(\/—+3)2 ffim __ Wrifi/ ‘3‘: True ZXE-fwox'f'7fig b. False _‘ I diam“:V‘akrtur‘eryrifi‘f” !. m" {ffla ‘71 r: m 5. The graph of the equation y = is symmetric with respect to the :c—axis. _ . , i j if? P a. True {‘fimaema an fro rams? a! ... ' wfifimgk l b. False r a. égb, 6. What is the solution set to the equation \$2 — :0 —|— 1 2' 0 in the complex numbers? . _ l _ - hr 1 fi- _i .7 5 j: 77,} _ _ . . 8. Which of the following equations describes a circle-With center (1, ——2) and radius 3? ' ' a. (a:+1)2—:—(y-2)2=3 .b(\$*U”[email protected]+2F%3 9. Which of the points A =' (—4,6) or B m (—1,7) is closest to the I origin? - ' ' ' ' _ _ . - . ' * . . ' dew) W .5 a. A is closest and has distance 2x/1'3form the origin 3; mg; g b. Ais closest and has distance-10 from the origin 0! ii rm} ext-0% 7’2 @B is closest and has distance 5\/§ from the origin { ligciw ( five“? .-= - - e \, so d. B is closest and has distance 8 from the origin if gr 4? f _ = 57 - J? 3 10. The domain of MIL-«23; is a -— 3.x 2 a } w -aewwa v9xa°t who (—00, 2] a a”; E c. [—22] d. [2,00) Part III: 3 points each - 11. Which Of the following statements'is / are true? A.If-0<a<bthen%>%. 1-way ' ' _ B. The complex conjugate of m2 is 2. #fm- ‘ "'2" =3 '" 2“ _. C. If a) graph of an equation is symmetric with reapect to the origin and _ it contains the point (—3, 5) then it also contains the point (3, —5). 4min a a. A, s and c ' b. B only A and C 011in d. C only e. B and C only I 12; If M = (1, 3) is the midpoint of the line segment joining A and B and A=(.—1,w1),then g! X 7 _ (In?) 3(W#"+ '3‘ i ' l*-"~"‘t.2§- " .z'miqs»; dB=ma afiwwx tw~we " e.' B: 3 fl I s In? in. QR I 13. T he surface area S of a cylinder is given by S = 27773 + 277%, Where h ‘ is the height and r is the radius Of the cylinder. If the" surface area is 48% square inches-and the height is 2 inches, then What is the "radius r of the cylinder in inches? 64-8 ‘1? v: a?“ r2“ «#4 QWF’A / z; a???” as a rim-an as -' ' a. r 2: —6 ' b. r = 6 - - A. _ - {r wag-(“mtg C. T —4 ' we ‘3; if! I my. ex? swig! _- ' d' T : 4 - Mai» ; mafia: as t, 6% Va M5 F” “3'3 ' -516; e. This cannot be uniquely determined from the given information. 14. The inequality 2 +(551: + 3)‘1 2 has What solution set? ' a, (—00, mg] -2.- «r» file‘s—t3.- szfi 2; _§) ('33:. 43,)“; 4; o 16. What is true about the equation (m —i— 302/3 = 4? «a? W3 2‘: 5‘ r: i 2; an? . 3 ‘ '_ a. It has exactly one real solution and this solution is negative. _ b. It has exacty one real solution and this solution is positive. 9‘ a“? “3 V " It has exactly two real solutions and both are contained in the 7 interval [512,6]. ' ' d. It has exactly two real solutions and both are contained in the ~ . . ' . a . w' interval [-—6,12]. _ 65%;; {Mag} at w {may a (fig; a}; . V/ ' . I L 3;, a a: 6‘. s e. It has no real solution. “ (ft-3) 3 e 8" 4 £1 ‘ 5 17. The cost forl'producing 3: units of "a product is C' = 35 + 235'. The revenue from selling :0 units is given by the formula R = 1433 — \$2. ' Then the company breaks even - 5 g3 ,1 g g My; «ex 2“ a. . .only if it sells exactly 5 units ‘of the product. K 2* w ’22“ "5'3"? fl " ’ ' ' {Kwéfl}{x~?) \$53 ' - b. . . . only if it: sells exactly 7 units of the product. - M _ - - ' - ' x 2:: 5‘” are «it s/ c. . .l . only if it sells exactly 5 or exactly 7 units of the produCt. d. . . ;only if it sells exactly 6 units of the product.- e. . . . only if it sells more than 7 units of the product. 18. The graph of the equation 3; = 23:2 —_ 533 — 3 has gash; I . 1 l. '53 “‘2’ ":2- a- .y‘lnterceptSI —§, 3 m-1ntercept: ~3 _ _ ’ l I r K a» lelemmg its : b. —' t t : —l — ; . . V 3; 1n ercep s 2, 3 a: intercept _‘ 3 1 g gig; “blami- W3 c; -' te t: . :1:-'nte ce ts: ——~ ~3 ytin'rcepfl 3 V1 1‘. p 2: afiéxfyfxgfi) .- __ y—intercept: '—3 - x—intercepts: —%, 3 5 - ' . Kg: a; as?» 5% “:7 sin 'e. y~intercept2 —3 :c—intercepts: 2i @ in“ _ 19. In an isosceles triangle the vertices have the coordinates A 2 (1,1), B = (O, 6) and C: (—4, 2). What is the length of the height through _ the point A in that triangle? - ' . (Hint: The height is the vlength'of the segment AM where M is the midpoint of the segment BC.) - i: 18 - M flag»! 9. c; (k V” (fwr‘wa2fi" f5: {fa-96); C. r I “WWW I d. V I_ r 3' . ‘f‘ m ffi+fijl e. 26 _ I figs“? 20. (Bonus!) Given the equation 330‘2 w 73: +4 2 0. The following are the first two steps in solving the equation for :c by. completing the square: Step 1: \$2~gw+§=fl . . 2 7 _ 4 Step 2. a: —— 5:13 * —§ Which of the following equations is a future step inthe process of completing the sQuare? ; 4 . _ . (“‘2 "’ , Mm a ,v i f - viii Isl ‘ -- g all}? 2 _ r Ann _:L e' a“ “7 337+ 6 — 34+ 6 . t a g i. e 2_Z I _E Z l b :13 3513+?) 3+3 2_Z Z _é Z c a: 3\$+6 3+5 MAC 1140, Test 2 A, Part II Summer 2007 Section number: I Name: UF ID: ' _ ' Signature: _-W" 1. Find all real solutions to the equation I _\/5-—2I:1:—-\/2—3:'=1. (:5) if?” ’3' 2.v2_.*;? + 2 as: 1%) w} “:3 £3?» ‘2’}; ém) X 2. a; if £57} I K *3: '2.“ a?“ X 3%“ m“ .. Wflmw WWW fig?” w M"? Maggy" E flay-ma )7 w xii, w {we}? w V’s we viva t gym W frag“? “3 M E; W” E 9/ “a '“ ““~w-=..-. I. "ma; ‘wmflm M hm? hi fi '1? v Vg‘g 2’ Q \g’ (I ‘5 {1 w {a i“ V??? wt? ; f The solutlon set 18: .1. . . . . . n; , , , 2. Find all real solutions to the equation" .(\$+UU?—w+¢PM—6=o. (Hint: Use substitution!) r I. - :'\ I I I! if" . 2" I {Ks/f} 6'; ' f *f'”‘9 = C3 ma firm: at stmé. V‘s..." _ I ‘ 7...; rm; sfiaflsfiéufi (gm: Mgfi‘ffi‘) ' Egg, 7’: ‘ . - ‘ . I 5‘ fl 2 mum-m mm s l ‘1, x I VET}? m 5: H 33 ' ‘3? w» by “V The solution set is:' ._ 9;}. . . . . . . . . 3 3. Solve” the inequality and express your answer in interval notation: 2—%|5:};+31g—1 ' L2. w a: [5x -_ elf-(“£2 __; - {5"}: “:3: 5g; Sax "é a g: m g my 6 {— §x+3 Cm? affix gage}. me- E54333: @> K I; m- fix ~ - " é? I ,3;- '. Thesolution set 13%": 525?; 7.33.6? gag”) _ 10 4. Peter rides his bike from his apartment to school at an average speed __ of 12 miles per hour. His roommate J ack leaves 20 minutes after Peter and he drives his carto school at an average rate of 36 miles per hour. J ack and Peter'arrive at school at the same time. How long does it take Peter to drive to school? ' HOW far is the commute from the apartment to school? m r . . v 4 {a e - ‘7? m 2:: 9%,, ' massages P gsyfigmre p drier,er WE _ ; § . f w . ggwfi 2:: gss ~ g g I - s. ““ ll g; ,x I - “2;: “*" 5 2m {"2 “"— X i .7 Cd; {- '2 8a " "“ 3”" [H v:- EQ WW “an? a A f” lime/If”- gha >2 a? . J. u m , I Peter’s tlme IS: . .37.. . 2"“. WW. . .. The distance 1s: . . t‘ffifizg . . . . . . . . . . 11 5. A fruit punch contains 15 percent apple juice. Apple juice concentrate contains 95 percent apple juice. How much of the puHCh and how much of the concentrate needs. to be mixed to get 40 gallons of a new drink'that Contains 20" percent apple juice? ' total amount amount of apple in gallons juice in gallons % of juice punch concentrate new drink. Concentrate needed: . @éfQEitél . 12 ...
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https://www.maplesoft.com/support/help/maple/view.aspx?path=Physics/Coefficients&L=E | 1,716,419,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00272.warc.gz | 762,360,757 | 24,270 | Coefficients - Maple Help
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Home : Support : Online Help : Physics : Coefficients
Physics[Coefficients] - extract all coefficients of a multivariate polynomial involving commutative and anticommutative variables
Calling Sequence Coefficients(p, x, N, onlynonzero)
Parameters
p - algebraic expression, or relation between them, or a set or list of them x - variable, can be a name, function, product, power, or a list of them N - optional, default to all, can be an integer, or a range of integers (n..m), or any of the keywords leading, trailing, all; indicates whether to compute all the coefficients or some or only one onlynonzero - optional, can be true or false (default), to receive only the coefficients not equal to zero in the returned sequence
Description
• The Coefficients command extracts the coefficients of x in the polynomial p, where x and p can involve anticommutative variables. Coefficients is a one-command generalization of coeff, coeffs, lcoeff and tcoeff, that works with commutative noncommutative and anticommutative variables in equal footing.
• The first argument, p, can also be a relation between polynomials, or a set or list of them, in which case Coefficients maps itself over the elements of the relation, set or list. For example, if p is an equation, then Coefficients(p, x) returns the equation obtained by computing Coefficients(lhs(p), x) = Coefficients(rhs(p), x), where lhs(p) and rhs(p) respectively represent the left and right hand sides of p.
• The second argument, x, can be a name, function, product, power, or a list of them. When x is a power, say as in a^n, Coefficients(p, x) returns the same as Coefficients(p, a, n), that is the coefficient of the nth power.
• The third argument, N, is optional, and indicates whether to extract all the coefficients (default behavior when N is not given, this is as coeffs) or the one of the Nth power when N is an integer (this is how coeff works), or the leading or trailing coefficient (pass N as the corresponding word, this produces the equivalent of lcoeff and tcoeff results), or a sequence of coefficients when N is a range of integers. The case N = all is then equivalent to the range case N = lower_degree .. higher_degree.
• Unlike coeff and coeffs, when x is a single variable and more than one coefficient is requested (for example, you call Coefficients with just two arguments, or with a third argument as a range), Coefficients returns the sequence of coefficients in ascending order, including those that are equal to 0. To receive only the non-zero coefficients use the optional argument onlynonzero. When N is omitted it is assumed equal to all.
• When x is a product, say a * b, Coefficients(p, a * b, N) returns the equivalent of taking a * b as a single variable - say c. Note: when a and b are anticommutative, Coefficients(p, a*b) returns the same as - Coefficients(p, b*a); likely, Coefficients(a*b, a) = - Coefficients(b*a, a).
• When x is a list, say [a, b], Coefficients(p, [a, b], N) returns the equivalent of recursively computing the coefficients with respect to each of the elements of the list, that is the same as op(map(Coefficients, [Coefficients(p, a, N)], b, N)). Note that when a and b are anticommutative, Coefficients(p, [a, b]) returns the same as - Coefficients(p, [b, a]); likely, Coefficients(a*b, [a, b]) = - Coefficients(b*a, [a, b]).
• Related to extracting coefficients, to compute the Degree of an expression with respect to anticommutative variables use the Physics:-Library:-Degree command.
Examples
> $\mathrm{with}\left(\mathrm{Physics}\right):$
> $\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$
$\left[{\mathrm{mathematicalnotation}}{=}{\mathrm{true}}\right]$ (1)
First set theta as an identifier to work with type/anticommutative variables (see Setup)
> $\mathrm{Setup}\left(\mathrm{anticommutativepre}=\mathrm{θ}\right)$
$\mathrm{* Partial match of \text{'}}\mathrm{anticommutativepre}\mathrm{\text{'} against keyword \text{'}}\mathrm{anticommutativeprefix}\text{'}$
$\mathrm{_______________________________________________________}$
$\left[{\mathrm{anticommutativeprefix}}{=}\left\{{\mathrm{\theta }}\right\}\right]$ (2)
> $a{\mathrm{θ}}_{1}{\mathrm{θ}}_{2}+b$
${a}{}{{\mathrm{\theta }}}_{{1}}{}{{\mathrm{\theta }}}_{{2}}{+}{b}$ (3)
The following three input lines have the same meaning, returning a sequence with all the coefficients
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1}\right)$
${b}{,}{a}{}{{\mathrm{\theta }}}_{{2}}$ (4)
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1},\mathrm{all}\right)$
${b}{,}{a}{}{{\mathrm{\theta }}}_{{2}}$ (5)
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1},0..1\right)$
${b}{,}{a}{}{{\mathrm{\theta }}}_{{2}}$ (6)
When the third argument is an integer, Coefficients returns the coefficient of the corresponding power
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1},0\right)$
${b}$ (7)
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1},1\right)$
${a}{}{{\mathrm{\theta }}}_{{2}}$ (8)
Computing the coefficient or the 1st power of an anticommutative variable is the same as differentiating with respect to it
> $\frac{\partial }{\partial {\mathrm{θ}}_{1}}$
${a}{}{{\mathrm{\theta }}}_{{2}}$ (9)
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{2},1\right)$
${-}{a}{}{{\mathrm{\theta }}}_{{1}}$ (10)
> $\frac{\partial }{\partial {\mathrm{θ}}_{2}}$
${-}{a}{}{{\mathrm{\theta }}}_{{1}}$ (11)
The coefficient of a product: note the change in sign when you reverse the order of the anticommutative variables in the coefficient product-variable
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{1}{\mathrm{θ}}_{2}\right)$
${b}{,}{a}$ (12)
> $\mathrm{Coefficients}\left(,{\mathrm{θ}}_{2}{\mathrm{θ}}_{1}\right)$
${b}{,}{-}{a}$ (13)
The coefficients of a list of variables - note the zeros in the output
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right]\right)$
${b}{,}{0}{,}{0}{,}{a}$ (14)
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],\mathrm{all}\right)$
${b}{,}{0}{,}{0}{,}{a}$ (15)
To receive only the nonzero coefficients use the onlynonzero optional argument
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],\mathrm{onlynonzero}\right)$
${b}{,}{a}$ (16)
The leading and trailing coefficients
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],\mathrm{leading}\right)$
${a}$ (17)
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{2},{\mathrm{θ}}_{1}\right],\mathrm{trailing}\right)$
${b}$ (18)
When the third argument, N, is an integer, and the second argument is a list, the coefficients are computed recursively;
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],0\right)$
${b}$ (19)
When N is equal to 1, this is also equivalent to differentiation
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],1\right)$
${a}$ (20)
> $\frac{{\partial }^{2}}{\partial {\mathrm{θ}}_{2}\partial {\mathrm{θ}}_{1}}$
${a}$ (21)
Reversing the order of the anticommutative variables in the list,
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{2},{\mathrm{θ}}_{1}\right],1\right)$
${-}{a}$ (22)
> $\frac{{\partial }^{2}}{\partial {\mathrm{θ}}_{1}\partial {\mathrm{θ}}_{2}}$
${-}{a}$ (23)
When the variables are anticommutative, their square is zero,
> ${\mathrm{θ}}_{1}^{2}$
${0}$ (24)
Hence,
> $\mathrm{Coefficients}\left(,\left[{\mathrm{θ}}_{1},{\mathrm{θ}}_{2}\right],2\right)$
${0}$ (25) | 2,291 | 7,637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 53, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-22 | latest | en | 0.736447 |
https://www.bankersadda.com/quantitative-aptitude-quiz-for-canara-2/ | 1,596,627,014,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00072.warc.gz | 603,622,568 | 49,627 | # Quantitative Aptitude Quiz For Canara Bank PO: 25th November 2018
Dear Aspirants,
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
Directions (1-5): Given below is the table which shows the various items purchased by shopkeeper from wholesaler, % markup price per kg by shopkeeper, List price per kg marked by shopkeeper and % discount offered by shopkeeper on list price.
Q1. What is the ratio of total profit obtained on selling whole quantity of rice to the total profit obtained on selling whole quantity of Maize.
(a) 3 : 4
(b) 7 : 2
(c) 9 : 2
(d) 5 : 2
(e) None of these
Q4. If 12 kg of Jowar is spoiled, then by what percent he should increase his List price per kg so that there is no profit and no loss.
(a) 12.5%
(b) 22.5%
(c) 8.5%
(d) 16.5%
(e) 13.75%
Q5. If shopkeeper uses faulty weight of 800 gm instead of 1 kg while selling, then what will be the total profit in selling Jowar
(a) 850
(b) 625
(c) 1250
(d) 1050
(e) 900
Directions (6-10): In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer (a) if x > y
Give answer (b) if x ≥ y
Give answer (c) if x < y
Give answer (d) if x ≤ y
Give answer (e) if x = y or the relationship cannot be established.
Directions (11-15): What should come in place of question mark (?) in the following questions?
Q11. 18.6 × 3 + 7.2 – 16.5 = ? + 21.7
(a) 35.7
(b) 21.6
(c) 24.8
(d) 27.6
(e) 31.5
Q12. [(140)2÷70×16]÷8=14× ?
(a) 38
(b) 22
(c) 55
(d) 40
(e) 31
Q13. 56% of 225 + 20% of 150 =? – 109
(a) 49
(b) 103
(c) 53
(d) 47
(e) 265
Q14. 0.5% of 674 of 0.8% of 225 =?
(a) 7.066
(b) 9.12
(c) 6.066
(d) 5.17
(e) 3.17
Q15. 68% of 625 + ? % of 185 = 499
(a) 42
(b) 40
(c) 28
(d) 25
(e) 15
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null | null | null | null | null | null | by Walt Becker
Nothing missing here: This debut novel smoothly mixes Indiana Jones-style adventure with X-Files spookiness to create an irresistible sci-fi thriller.
In West Africa, paleoanthropologists Jack Austin and Samantha Colby unearth an artifact that explains why ancient cultures were able to build pyramids and chart the stars and why so many of them share similar legends. It's a history-making find—if the pair survive attacks by superstitious tribesmen, a greedy arms dealer and some centuries-old booby traps left by aliens to make their discovery public. Becker bogs down his dialogue with wordy scientific soliloquies. But Link's awe-inspiring premise—and the obvious research that went into it—make for a close encounter of a very cool kind. (Morrow, $25)
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The Inside Story of the Making and Unmaking of the Franken Presidency
by Al Franken
If it is truly better to laugh than cry in the face of crisis, then America truly needs this witty send-up of Presidential politics by Al Franken, the former Saturday Night Live writer who scored a hit with 1996's Rush Limbaugh Is a Big, Fat Idiot and Other Observations. This time, Franken chronicles his own fanciful bid for the White House in 2000, with hilarious spoofs of campaign ads, memos, news clips and the candidate's diary. Running a single-issue campaign based on opposition to cash-machine fees ("ATM=America's Terrible Menace" goes one slogan), the candidate hits a snag when he learns that New Hampshire has no more than five ATMs, but miraculously, he emerges triumphant—only to self-destruct in office. Merging Tip O'Neill's adage ("All politics is local") with the James Carville slogan ("It's the economy, stupid"), Franken concludes, "It's very simple. 'All politics are the local economy, stupid!' " Though the premise is stretched a bit thin over 289 pages, Franken proves again that he's one of our savviest satirists. (Delacorte, $23.95)
Bottom Line: Deserves your vote
by Sandra Bernhard
Could it be that Sandra Bernhard, queen of the perpetual sneer and sarcastic jab, has reformed? To judge by this collection of musings and poems, something (new motherhood, perhaps?) has caused the comedian to trade in her signature spitefulness for a calm and centered spirituality. The surprise here is how skillfully and sweetly she manages to weave her message of love, truth, patience and, yes, kindness, into a series of anecdotes—whose subjects range from a trip to Morocco to keeping watch over a sleeping lover—without sounding sappy. Even the comic bits about celebrities who ignore her (such as Streisand and Oprah) have been toned down. Where the old Bernhard would have torn into someone like Courtney Love for her perfect looks, here the new Sandra is a pussycat: "You know, Courtney, you may live to be a hundred, and never have a wrinkle.... But what plastic surgeon is going to sew up the wounds in your heart?" (Rob Weisbach/Morrow, $24)
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Move over, Cecil B. De Mille and The Prince of Egypt: When it comes to enlivening the Bible, Ken Davis is the real miracle man. In the latest of his Don't Know Much About books he again shows a knack for making esoteric ideas (like empty piety or original sin) understandable, with powerful results. In addition to clearing up several little mysteries—such as why the Hebrew and Christian bibles are arranged differently, or whether Jesus was born on Dec. 25 (he wasn't) and the real difference between a disciple and an apostle—the author throws in plenty of information about the discoveries of biblical scholars and about historical events taking place during the millennium it took to assemble the book we call the Bible. Davis makes clear that he has no religious agenda, and he never condescends. Reading him is like returning to the classroom of the best teacher you ever had. (Eagle Brook/Morrow, $25)
Bottom Line: Great look into the Good Book
by Bret Easton Ellis
Glamorama is a pretty funny 250-page novel that loosens its belt, puts up its feet and lingers for 482 pages, growing so violently nutty that you feel like tiptoeing away and hiding. Initially, the satire is clever, the observations wicked fun. Actor-model Victor Ward wafts through fashion-crazed downtown Manhattan as aimlessly as Ellis's plot, his brain a gossip-column dump of celebrity names and Top 40 song lyrics, his friends "debating the color yellow" and proving they have as much substance as their designer waters. At times the intentionally affected style veers from smart to smarmy, especially when Ellis (author of 1985's Less Than Zero) pays homage to himself by dropping in the names of his own real-life pals and of characters from his other novels, but it's hard not to chuckle at the trés Holly Golightly Victor—who, after pleading that "life is tacky," is told, "But you don't need to win first prize." Then people's limbs start getting blown off. Gears clanking, Ellis lurches into a ridiculous, gory and tiresome Euro-anarchist terror plot (Bret, hel-lo? You aren't John le Carré) as Victor is forced to plumb the depths of his own shallowness. (Knopf, $25)
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by Michael Connelly
Page-turner of the week
Days before Howard Elias's latest suit representing an alleged victim of police brutality goes to trial, the LAPD is dreading the battle against the charismatic African-American attorney. Then they're hit with an even more pressing problem: Elias's bullet-riddled body is discovered on a historic downtown railway.
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As aficionados of Connelly's previous mysteries featuring Detective Bosch already know, he's a dogged, let-the-chips-fall-where-they-may kind of guy. His struggle here to get a grip on this slippery, potentially career-breaking case—while wrestling with personal problems, including a disintegrating marriage—makes for compelling reading, though not with quite the haunting power of the author's previous Blood Work or The Poet (both, as it happens, departures from the Bosch series). Although Angels rarely takes wing, it's still a flight well worth taking. (Little, Brown, $25)
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BREASTS Meema Spadola Inspired by the 1996 Cinemax documentary of the same name, a collection of interviews with women on the physical and emotional significance of their most public private parts. (Wildcat Canyon, $13.95)
• Contributors:
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• Pam Lambert. | null | null | null | null | null | null | null | null | null |
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Category: Entertainment
## Presentation Transcript
### slide 1:
QNT 275 Week 2 Homework Problem Set Excel File For more course tutorials visit www.tutorialrank.com 1. In a hearing test subjects estimate the loudness in decibels of a sound and the results are: 69 67 71 72 65 75 68 68 83 73 68. Calculate the measures of central tendency Mean median mode and the measures of dispersion range standard deviation variance. 2.
### slide 2:
The local amusement park was interested in the average wait time at their most popular roller coaster at the peak park time 2 p.m.. They selected 13 patrons and had them get in line between 2 and 3 p.m. Each was given a stopwatch to record the time they spent in line. The times recorded were as follows in minutes. 117 123 110 117 99 120 148 118 119 120 45 130 118 What is the 72d percentile 3. The average life of Canadian women is 73.90 years and the standard deviation of the life expectancy of Canadian women is 9 years. Based on Chebyshevs Theorem determine the upper and lower bounds on the
### slide 3:
average life expectancy of Canadian women such that at least 95 percent of the population is included if the sample size is 100 women. 4. The local amusement park was interested in the average wait time at their most popular roller coaster at the peak park time 2 p.m.. They selected 13 patrons and had them get in line between 2 and 3 p.m. Each was given a stopwatch to record the time they spent in line. The times recorded were as follows in minutes: 118 121 114 116 110 120 145 118 119 121 45 135 118. Calculate the measures of central tendency Mean median mode and the measures of dispersion range standard deviation variance. 5.
### slide 4:
The average lateness for one of the top airline companies is 10 minutes. The variance of the lateness measure is calculated as 8. An airplane arrived 12 minutes after the stated arrival time. Calculate the z-score for the lateness of this particular airplane. 6. According to a survey of the top 15 employers in a major city in the Midwest a worker spends an average of 400 minutes a day on the job. Suppose the standard deviation is 20 minutes and the time spent is approximately a normal distribution. What are the times within which approximately 99.73 percent of all workers will fall 7.
### slide 5:
Recently an advertising company called 200 people and asked them to identify the company that was in an ad running nationwide. The following results were obtained. What percentage of those surveyed could not correctly recall the company 8. A local electronics retailer recently conducted a study on purchasers of large screen televisions. The study recorded the type of television and the credit account balance of the customer at the time of purchase. They obtained the following results. What percentage of purchases were plasma televisions by customers with the smallest credit balances 9.
### slide 6:
The following is a partial relative frequency distribution of grades in an introductory statistics course. Find the relative frequency for the B grade. 10 A CFO is looking at what percentage of a companys resources are spent on computing. He samples companies in the pharmaceutical industry and develops the following stem-and-leaf graph. What would be the class length used in creating a frequency histogram ..................................................................................................................... ......................................... QNT 275 Week 3 Homework Problem Set Excel File For more course tutorials visit www.tutorialrank.com 1.
### slide 7:
A survey is made in a neighborhood of 90 voters. 75 are Democrats and 15 are Republicans. Of the Democrats 30 are women while 7 of the Republicans are women. If one subject from the group is randomly selected find the probability the individual is a male Republican. 2. Container 1 has 8 items 3 of which are defective. Container 2 has 5 items 3 of which are defective. If one item is drawn from each container what is the probability that only one of the items is defective 3. A letter is drawn from the alphabet of 26 letters. What is the probability that the letter drawn is a vowel
### slide 8:
4. A family has two children. What is the probability that both are girls given that at least one is a girl 5. If n 29 and p .6 then the standard deviation of the binomial distribution is
### slide 9:
6. Consider a Poisson distribution with an average of 4 customers per minute at the local grocery store. If X the number of arrivals per minute find the probability of more than 6 customers arriving within a minute. 7. An important part of the customer service responsibilities of a cable company is the speed with which trouble in service can be repaired. Historically the data show that the likelihood is 0.70 that troubles in a residential service can be repaired on the same day. For the first four troubles reported on a given day what is the probability that all four will be repaired on the same day
### slide 10:
8. Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 29 minutes and a standard deviation of 6 minutes. Suppose that in an effort to provide better service to the public the director of the local office is permitted to provide discounts to those individuals whose waiting time exceeds a predetermined time. The director decides that 10 percent of the customers should receive this discount. What number of minutes do they need to wait to receive the discount 9. An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern state. The amount of juice obtained from each of these apples is approximately normally distributed with a mean of 2.38 ounces and a standard deviation of 0.1 ounce. What is the probability that a randomly selected apple will contain more than 2.40 ounces 10
### slide 11:
While conducting experiments a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 8.00 m. What is the probability that a randomly selected depth is between 2.25 m and 5.00 m ..................................................................................................................... ......................................... QNT 275 Week 4 Homework Problem Set Excel File For more course tutorials visit www.tutorialrank.com 1. A sample of 100 items has a population standard deviation of 5.9 and a mean of 32. Construct a 95 percent confidence interval for μ. 2. At the end of 1990 1991 and 1992 the average prices of a share of stock in a portfolio were 34.75 34.65 and 31.25 respectively. To
### slide 12:
investigate the average share price at the end of 1993 a random sample of 75 stocks was drawn and their closing prices on the last trading day of 1993 were observed with a mean of 33.78 and a standard deviation of 14.25. Estimate the average price of a share of stock in the portfolio at the end of 1993 with a 90 percent confidence interval. 3. A research study investigated differences between male and female students. Based on the study results we can assume the population mean and standard deviation for the GPA of male students are µ 3.3 and σ 0.3. Suppose a random sample of 1200 male students is selected and the GPA for each student is calculated. Find the interval that contains 90 percent of the sample means for male students. 4. A manufacturing company measures the weight of boxes before shipping them to the customers. If the box weights have a population mean of 80 lb and standard deviation of 6 lb respectively then based on a sample size of 100 boxes what is the probability that the average weight of the boxes will exceed 83 lb
### slide 13:
5. A random sample of size 100 is taken from a population with mean 64 and standard deviation 5.2. Find Px bar 60. | 1,717 | 7,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.940693 |
http://www.newhappyholidays.com/faq/faq-72-hrs-is-how-many-days.html | 1,656,498,916,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00647.warc.gz | 95,187,071 | 10,398 | ## Is 72 hours the same as 3 days?
What is considered 3 days? The 3-day period is the 3 days before the flight’s departure. The Order uses a 3-day timeframe instead of 72 hours to provide more flexibility to the traveler. By using a 3-day window, test validity does not depend on the time of the flight or the time of day that the test was administered.
## How do you convert 72 hours to days?
72 Hours is 3 Days.
## How many hours is 48 hours?
48 Hours is 2 Days.
## How many hours make 2 days?
So a full day is 24 hours. So that means two full days 48 hours.
## Does within 3 days include today?
But people often take today to be the first day of the count, so if on Monday someone says “within 3 days” they are thinking day 1=today, Monday; day 2=Tuesday, day 3= Wednesday.
## How many hours are in a week?
How Many Hours are in a Week? There are 168 hours in a week, which is why we use this value in the formula above.
You might be interested: Readers ask: How Many Days Can You Take Benzonatate?
## How many years are there in 24 months?
24 months in years: 24 months = 2 years.
## What is the 48 hour rule?
The “48-hour rule” is one key to success. The 48-hour rule, simply stated, stipulates that to more effectively seize a new opportunity you should follow up or perform an action within 48 hours after interest has been established.
## How many minutes is two hours?
There are 60 minutes in 1 hour. To convert from minutes to hours, divide the number of minutes by 60. For example, 120 minutes equals 2 hours because 120/60=2.
## How long is 12 hour?
The 12-hour clock runs from 1am to 12 noon and then from 1pm to 12 midnight. The 24-hour clock uses the numbers 00:00 to 23:59 (midnight is 00:00).
## How minutes are in an hour?
Since there are 60 minutes in one hour, that’s the conversion ratio used in the formula.
## How much is 500 hours in work days?
1 Expert Answer 500 person hours / 40 person hours per day = 12.5 days to complete the job.
## How do you calculate 12 hours?
Add 12 to the first hour of the day and include “AM.” In 24-hour time, midnight is signified as 00:00. So, for midnight hour, add 12 and the signifier “AM” to convert to 12-hour time. This means that, for example, 00:13 in 24-hour time would be 12:13 AM in 12-hour time. | 615 | 2,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.946926 |
http://de.metamath.org/mpeuni/cygznlem3.html | 1,718,228,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00030.warc.gz | 10,690,823 | 15,342 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cygznlem3 Structured version Visualization version GIF version
Theorem cygznlem3 19737
Description: A cyclic group with 𝑛 elements is isomorphic to ℤ / 𝑛ℤ. (Contributed by Mario Carneiro, 21-Apr-2016.)
Hypotheses
Ref Expression
cygzn.b 𝐵 = (Base‘𝐺)
cygzn.n 𝑁 = if(𝐵 ∈ Fin, (#‘𝐵), 0)
cygzn.y 𝑌 = (ℤ/nℤ‘𝑁)
cygzn.m · = (.g𝐺)
cygzn.l 𝐿 = (ℤRHom‘𝑌)
cygzn.e 𝐸 = {𝑥𝐵 ∣ ran (𝑛 ∈ ℤ ↦ (𝑛 · 𝑥)) = 𝐵}
cygzn.g (𝜑𝐺 ∈ CycGrp)
cygzn.x (𝜑𝑋𝐸)
cygzn.f 𝐹 = ran (𝑚 ∈ ℤ ↦ ⟨(𝐿𝑚), (𝑚 · 𝑋)⟩)
Assertion
Ref Expression
cygznlem3 (𝜑𝐺𝑔 𝑌)
Distinct variable groups: 𝑚,𝑛,𝑥,𝐵 𝑚,𝐺,𝑛,𝑥 · ,𝑚,𝑛,𝑥 𝑚,𝑌,𝑛,𝑥 𝑚,𝐿,𝑛,𝑥 𝑥,𝑁 𝜑,𝑚 𝑛,𝐹,𝑥 𝑚,𝑋,𝑛,𝑥
Allowed substitution hints: 𝜑(𝑥,𝑛) 𝐸(𝑥,𝑚,𝑛) 𝐹(𝑚) 𝑁(𝑚,𝑛)
Proof of Theorem cygznlem3
Dummy variables 𝑎 𝑏 𝑖 𝑗 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2610 . . . 4 (Base‘𝑌) = (Base‘𝑌)
2 cygzn.b . . . 4 𝐵 = (Base‘𝐺)
3 eqid 2610 . . . 4 (+g𝑌) = (+g𝑌)
4 eqid 2610 . . . 4 (+g𝐺) = (+g𝐺)
5 cygzn.n . . . . . 6 𝑁 = if(𝐵 ∈ Fin, (#‘𝐵), 0)
6 hashcl 13009 . . . . . . . 8 (𝐵 ∈ Fin → (#‘𝐵) ∈ ℕ0)
76adantl 481 . . . . . . 7 ((𝜑𝐵 ∈ Fin) → (#‘𝐵) ∈ ℕ0)
8 0nn0 11184 . . . . . . . 8 0 ∈ ℕ0
98a1i 11 . . . . . . 7 ((𝜑 ∧ ¬ 𝐵 ∈ Fin) → 0 ∈ ℕ0)
107, 9ifclda 4070 . . . . . 6 (𝜑 → if(𝐵 ∈ Fin, (#‘𝐵), 0) ∈ ℕ0)
115, 10syl5eqel 2692 . . . . 5 (𝜑𝑁 ∈ ℕ0)
12 cygzn.y . . . . . 6 𝑌 = (ℤ/nℤ‘𝑁)
1312zncrng 19712 . . . . 5 (𝑁 ∈ ℕ0𝑌 ∈ CRing)
14 crngring 18381 . . . . 5 (𝑌 ∈ CRing → 𝑌 ∈ Ring)
15 ringgrp 18375 . . . . 5 (𝑌 ∈ Ring → 𝑌 ∈ Grp)
1611, 13, 14, 154syl 19 . . . 4 (𝜑𝑌 ∈ Grp)
17 cygzn.g . . . . 5 (𝜑𝐺 ∈ CycGrp)
18 cyggrp 18114 . . . . 5 (𝐺 ∈ CycGrp → 𝐺 ∈ Grp)
1917, 18syl 17 . . . 4 (𝜑𝐺 ∈ Grp)
20 cygzn.m . . . . 5 · = (.g𝐺)
21 cygzn.l . . . . 5 𝐿 = (ℤRHom‘𝑌)
22 cygzn.e . . . . 5 𝐸 = {𝑥𝐵 ∣ ran (𝑛 ∈ ℤ ↦ (𝑛 · 𝑥)) = 𝐵}
23 cygzn.x . . . . 5 (𝜑𝑋𝐸)
24 cygzn.f . . . . 5 𝐹 = ran (𝑚 ∈ ℤ ↦ ⟨(𝐿𝑚), (𝑚 · 𝑋)⟩)
252, 5, 12, 20, 21, 22, 17, 23, 24cygznlem2a 19735 . . . 4 (𝜑𝐹:(Base‘𝑌)⟶𝐵)
2612, 1, 21znzrhfo 19715 . . . . . . . 8 (𝑁 ∈ ℕ0𝐿:ℤ–onto→(Base‘𝑌))
2711, 26syl 17 . . . . . . 7 (𝜑𝐿:ℤ–onto→(Base‘𝑌))
28 foelrn 6286 . . . . . . 7 ((𝐿:ℤ–onto→(Base‘𝑌) ∧ 𝑎 ∈ (Base‘𝑌)) → ∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖))
2927, 28sylan 487 . . . . . 6 ((𝜑𝑎 ∈ (Base‘𝑌)) → ∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖))
30 foelrn 6286 . . . . . . 7 ((𝐿:ℤ–onto→(Base‘𝑌) ∧ 𝑏 ∈ (Base‘𝑌)) → ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗))
3127, 30sylan 487 . . . . . 6 ((𝜑𝑏 ∈ (Base‘𝑌)) → ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗))
3229, 31anim12dan 878 . . . . 5 ((𝜑 ∧ (𝑎 ∈ (Base‘𝑌) ∧ 𝑏 ∈ (Base‘𝑌))) → (∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗)))
33 reeanv 3086 . . . . . . 7 (∃𝑖 ∈ ℤ ∃𝑗 ∈ ℤ (𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) ↔ (∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗)))
3419adantr 480 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → 𝐺 ∈ Grp)
35 simprl 790 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → 𝑖 ∈ ℤ)
36 simprr 792 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → 𝑗 ∈ ℤ)
372, 20, 22iscyggen 18105 . . . . . . . . . . . . . 14 (𝑋𝐸 ↔ (𝑋𝐵 ∧ ran (𝑛 ∈ ℤ ↦ (𝑛 · 𝑋)) = 𝐵))
3837simplbi 475 . . . . . . . . . . . . 13 (𝑋𝐸𝑋𝐵)
3923, 38syl 17 . . . . . . . . . . . 12 (𝜑𝑋𝐵)
4039adantr 480 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → 𝑋𝐵)
412, 20, 4mulgdir 17396 . . . . . . . . . . 11 ((𝐺 ∈ Grp ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ ∧ 𝑋𝐵)) → ((𝑖 + 𝑗) · 𝑋) = ((𝑖 · 𝑋)(+g𝐺)(𝑗 · 𝑋)))
4234, 35, 36, 40, 41syl13anc 1320 . . . . . . . . . 10 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝑖 + 𝑗) · 𝑋) = ((𝑖 · 𝑋)(+g𝐺)(𝑗 · 𝑋)))
4311, 13syl 17 . . . . . . . . . . . . . . 15 (𝜑𝑌 ∈ CRing)
4421zrhrhm 19679 . . . . . . . . . . . . . . 15 (𝑌 ∈ Ring → 𝐿 ∈ (ℤring RingHom 𝑌))
45 rhmghm 18548 . . . . . . . . . . . . . . 15 (𝐿 ∈ (ℤring RingHom 𝑌) → 𝐿 ∈ (ℤring GrpHom 𝑌))
4643, 14, 44, 454syl 19 . . . . . . . . . . . . . 14 (𝜑𝐿 ∈ (ℤring GrpHom 𝑌))
4746adantr 480 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → 𝐿 ∈ (ℤring GrpHom 𝑌))
48 zringbas 19643 . . . . . . . . . . . . . 14 ℤ = (Base‘ℤring)
49 zringplusg 19644 . . . . . . . . . . . . . 14 + = (+g‘ℤring)
5048, 49, 3ghmlin 17488 . . . . . . . . . . . . 13 ((𝐿 ∈ (ℤring GrpHom 𝑌) ∧ 𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ) → (𝐿‘(𝑖 + 𝑗)) = ((𝐿𝑖)(+g𝑌)(𝐿𝑗)))
5147, 35, 36, 50syl3anc 1318 . . . . . . . . . . . 12 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐿‘(𝑖 + 𝑗)) = ((𝐿𝑖)(+g𝑌)(𝐿𝑗)))
5251fveq2d 6107 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘(𝐿‘(𝑖 + 𝑗))) = (𝐹‘((𝐿𝑖)(+g𝑌)(𝐿𝑗))))
53 zaddcl 11294 . . . . . . . . . . . 12 ((𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ) → (𝑖 + 𝑗) ∈ ℤ)
542, 5, 12, 20, 21, 22, 17, 23, 24cygznlem2 19736 . . . . . . . . . . . 12 ((𝜑 ∧ (𝑖 + 𝑗) ∈ ℤ) → (𝐹‘(𝐿‘(𝑖 + 𝑗))) = ((𝑖 + 𝑗) · 𝑋))
5553, 54sylan2 490 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘(𝐿‘(𝑖 + 𝑗))) = ((𝑖 + 𝑗) · 𝑋))
5652, 55eqtr3d 2646 . . . . . . . . . 10 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘((𝐿𝑖)(+g𝑌)(𝐿𝑗))) = ((𝑖 + 𝑗) · 𝑋))
572, 5, 12, 20, 21, 22, 17, 23, 24cygznlem2 19736 . . . . . . . . . . . 12 ((𝜑𝑖 ∈ ℤ) → (𝐹‘(𝐿𝑖)) = (𝑖 · 𝑋))
5857adantrr 749 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘(𝐿𝑖)) = (𝑖 · 𝑋))
592, 5, 12, 20, 21, 22, 17, 23, 24cygznlem2 19736 . . . . . . . . . . . 12 ((𝜑𝑗 ∈ ℤ) → (𝐹‘(𝐿𝑗)) = (𝑗 · 𝑋))
6059adantrl 748 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘(𝐿𝑗)) = (𝑗 · 𝑋))
6158, 60oveq12d 6567 . . . . . . . . . 10 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝐹‘(𝐿𝑖))(+g𝐺)(𝐹‘(𝐿𝑗))) = ((𝑖 · 𝑋)(+g𝐺)(𝑗 · 𝑋)))
6242, 56, 613eqtr4d 2654 . . . . . . . . 9 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → (𝐹‘((𝐿𝑖)(+g𝑌)(𝐿𝑗))) = ((𝐹‘(𝐿𝑖))(+g𝐺)(𝐹‘(𝐿𝑗))))
63 oveq12 6558 . . . . . . . . . . 11 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (𝑎(+g𝑌)𝑏) = ((𝐿𝑖)(+g𝑌)(𝐿𝑗)))
6463fveq2d 6107 . . . . . . . . . 10 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (𝐹‘(𝑎(+g𝑌)𝑏)) = (𝐹‘((𝐿𝑖)(+g𝑌)(𝐿𝑗))))
65 fveq2 6103 . . . . . . . . . . 11 (𝑎 = (𝐿𝑖) → (𝐹𝑎) = (𝐹‘(𝐿𝑖)))
66 fveq2 6103 . . . . . . . . . . 11 (𝑏 = (𝐿𝑗) → (𝐹𝑏) = (𝐹‘(𝐿𝑗)))
6765, 66oveqan12d 6568 . . . . . . . . . 10 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → ((𝐹𝑎)(+g𝐺)(𝐹𝑏)) = ((𝐹‘(𝐿𝑖))(+g𝐺)(𝐹‘(𝐿𝑗))))
6864, 67eqeq12d 2625 . . . . . . . . 9 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → ((𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏)) ↔ (𝐹‘((𝐿𝑖)(+g𝑌)(𝐿𝑗))) = ((𝐹‘(𝐿𝑖))(+g𝐺)(𝐹‘(𝐿𝑗)))))
6962, 68syl5ibrcom 236 . . . . . . . 8 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏))))
7069rexlimdvva 3020 . . . . . . 7 (𝜑 → (∃𝑖 ∈ ℤ ∃𝑗 ∈ ℤ (𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏))))
7133, 70syl5bir 232 . . . . . 6 (𝜑 → ((∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗)) → (𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏))))
7271imp 444 . . . . 5 ((𝜑 ∧ (∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗))) → (𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏)))
7332, 72syldan 486 . . . 4 ((𝜑 ∧ (𝑎 ∈ (Base‘𝑌) ∧ 𝑏 ∈ (Base‘𝑌))) → (𝐹‘(𝑎(+g𝑌)𝑏)) = ((𝐹𝑎)(+g𝐺)(𝐹𝑏)))
741, 2, 3, 4, 16, 19, 25, 73isghmd 17492 . . 3 (𝜑𝐹 ∈ (𝑌 GrpHom 𝐺))
7558, 60eqeq12d 2625 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝐹‘(𝐿𝑖)) = (𝐹‘(𝐿𝑗)) ↔ (𝑖 · 𝑋) = (𝑗 · 𝑋)))
762, 5, 12, 20, 21, 22, 17, 23cygznlem1 19734 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝐿𝑖) = (𝐿𝑗) ↔ (𝑖 · 𝑋) = (𝑗 · 𝑋)))
7775, 76bitr4d 270 . . . . . . . . . . . 12 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝐹‘(𝐿𝑖)) = (𝐹‘(𝐿𝑗)) ↔ (𝐿𝑖) = (𝐿𝑗)))
7877biimpd 218 . . . . . . . . . . 11 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝐹‘(𝐿𝑖)) = (𝐹‘(𝐿𝑗)) → (𝐿𝑖) = (𝐿𝑗)))
7965, 66eqeqan12d 2626 . . . . . . . . . . . 12 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → ((𝐹𝑎) = (𝐹𝑏) ↔ (𝐹‘(𝐿𝑖)) = (𝐹‘(𝐿𝑗))))
80 eqeq12 2623 . . . . . . . . . . . 12 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (𝑎 = 𝑏 ↔ (𝐿𝑖) = (𝐿𝑗)))
8179, 80imbi12d 333 . . . . . . . . . . 11 ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → (((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏) ↔ ((𝐹‘(𝐿𝑖)) = (𝐹‘(𝐿𝑗)) → (𝐿𝑖) = (𝐿𝑗))))
8278, 81syl5ibrcom 236 . . . . . . . . . 10 ((𝜑 ∧ (𝑖 ∈ ℤ ∧ 𝑗 ∈ ℤ)) → ((𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → ((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏)))
8382rexlimdvva 3020 . . . . . . . . 9 (𝜑 → (∃𝑖 ∈ ℤ ∃𝑗 ∈ ℤ (𝑎 = (𝐿𝑖) ∧ 𝑏 = (𝐿𝑗)) → ((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏)))
8433, 83syl5bir 232 . . . . . . . 8 (𝜑 → ((∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗)) → ((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏)))
8584imp 444 . . . . . . 7 ((𝜑 ∧ (∃𝑖 ∈ ℤ 𝑎 = (𝐿𝑖) ∧ ∃𝑗 ∈ ℤ 𝑏 = (𝐿𝑗))) → ((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏))
8632, 85syldan 486 . . . . . 6 ((𝜑 ∧ (𝑎 ∈ (Base‘𝑌) ∧ 𝑏 ∈ (Base‘𝑌))) → ((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏))
8786ralrimivva 2954 . . . . 5 (𝜑 → ∀𝑎 ∈ (Base‘𝑌)∀𝑏 ∈ (Base‘𝑌)((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏))
88 dff13 6416 . . . . 5 (𝐹:(Base‘𝑌)–1-1𝐵 ↔ (𝐹:(Base‘𝑌)⟶𝐵 ∧ ∀𝑎 ∈ (Base‘𝑌)∀𝑏 ∈ (Base‘𝑌)((𝐹𝑎) = (𝐹𝑏) → 𝑎 = 𝑏)))
8925, 87, 88sylanbrc 695 . . . 4 (𝜑𝐹:(Base‘𝑌)–1-1𝐵)
902, 20, 22iscyggen2 18106 . . . . . . . . 9 (𝐺 ∈ Grp → (𝑋𝐸 ↔ (𝑋𝐵 ∧ ∀𝑧𝐵𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋))))
9119, 90syl 17 . . . . . . . 8 (𝜑 → (𝑋𝐸 ↔ (𝑋𝐵 ∧ ∀𝑧𝐵𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋))))
9223, 91mpbid 221 . . . . . . 7 (𝜑 → (𝑋𝐵 ∧ ∀𝑧𝐵𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋)))
9392simprd 478 . . . . . 6 (𝜑 → ∀𝑧𝐵𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋))
94 oveq1 6556 . . . . . . . . . 10 (𝑛 = 𝑗 → (𝑛 · 𝑋) = (𝑗 · 𝑋))
9594eqeq2d 2620 . . . . . . . . 9 (𝑛 = 𝑗 → (𝑧 = (𝑛 · 𝑋) ↔ 𝑧 = (𝑗 · 𝑋)))
9695cbvrexv 3148 . . . . . . . 8 (∃𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋) ↔ ∃𝑗 ∈ ℤ 𝑧 = (𝑗 · 𝑋))
9727adantr 480 . . . . . . . . . . . . 13 ((𝜑𝑧𝐵) → 𝐿:ℤ–onto→(Base‘𝑌))
98 fof 6028 . . . . . . . . . . . . 13 (𝐿:ℤ–onto→(Base‘𝑌) → 𝐿:ℤ⟶(Base‘𝑌))
9997, 98syl 17 . . . . . . . . . . . 12 ((𝜑𝑧𝐵) → 𝐿:ℤ⟶(Base‘𝑌))
10099ffvelrnda 6267 . . . . . . . . . . 11 (((𝜑𝑧𝐵) ∧ 𝑗 ∈ ℤ) → (𝐿𝑗) ∈ (Base‘𝑌))
10159adantlr 747 . . . . . . . . . . . 12 (((𝜑𝑧𝐵) ∧ 𝑗 ∈ ℤ) → (𝐹‘(𝐿𝑗)) = (𝑗 · 𝑋))
102101eqcomd 2616 . . . . . . . . . . 11 (((𝜑𝑧𝐵) ∧ 𝑗 ∈ ℤ) → (𝑗 · 𝑋) = (𝐹‘(𝐿𝑗)))
103 fveq2 6103 . . . . . . . . . . . . 13 (𝑎 = (𝐿𝑗) → (𝐹𝑎) = (𝐹‘(𝐿𝑗)))
104103eqeq2d 2620 . . . . . . . . . . . 12 (𝑎 = (𝐿𝑗) → ((𝑗 · 𝑋) = (𝐹𝑎) ↔ (𝑗 · 𝑋) = (𝐹‘(𝐿𝑗))))
105104rspcev 3282 . . . . . . . . . . 11 (((𝐿𝑗) ∈ (Base‘𝑌) ∧ (𝑗 · 𝑋) = (𝐹‘(𝐿𝑗))) → ∃𝑎 ∈ (Base‘𝑌)(𝑗 · 𝑋) = (𝐹𝑎))
106100, 102, 105syl2anc 691 . . . . . . . . . 10 (((𝜑𝑧𝐵) ∧ 𝑗 ∈ ℤ) → ∃𝑎 ∈ (Base‘𝑌)(𝑗 · 𝑋) = (𝐹𝑎))
107 eqeq1 2614 . . . . . . . . . . 11 (𝑧 = (𝑗 · 𝑋) → (𝑧 = (𝐹𝑎) ↔ (𝑗 · 𝑋) = (𝐹𝑎)))
108107rexbidv 3034 . . . . . . . . . 10 (𝑧 = (𝑗 · 𝑋) → (∃𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎) ↔ ∃𝑎 ∈ (Base‘𝑌)(𝑗 · 𝑋) = (𝐹𝑎)))
109106, 108syl5ibrcom 236 . . . . . . . . 9 (((𝜑𝑧𝐵) ∧ 𝑗 ∈ ℤ) → (𝑧 = (𝑗 · 𝑋) → ∃𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎)))
110109rexlimdva 3013 . . . . . . . 8 ((𝜑𝑧𝐵) → (∃𝑗 ∈ ℤ 𝑧 = (𝑗 · 𝑋) → ∃𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎)))
11196, 110syl5bi 231 . . . . . . 7 ((𝜑𝑧𝐵) → (∃𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋) → ∃𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎)))
112111ralimdva 2945 . . . . . 6 (𝜑 → (∀𝑧𝐵𝑛 ∈ ℤ 𝑧 = (𝑛 · 𝑋) → ∀𝑧𝐵𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎)))
11393, 112mpd 15 . . . . 5 (𝜑 → ∀𝑧𝐵𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎))
114 dffo3 6282 . . . . 5 (𝐹:(Base‘𝑌)–onto𝐵 ↔ (𝐹:(Base‘𝑌)⟶𝐵 ∧ ∀𝑧𝐵𝑎 ∈ (Base‘𝑌)𝑧 = (𝐹𝑎)))
11525, 113, 114sylanbrc 695 . . . 4 (𝜑𝐹:(Base‘𝑌)–onto𝐵)
116 df-f1o 5811 . . . 4 (𝐹:(Base‘𝑌)–1-1-onto𝐵 ↔ (𝐹:(Base‘𝑌)–1-1𝐵𝐹:(Base‘𝑌)–onto𝐵))
11789, 115, 116sylanbrc 695 . . 3 (𝜑𝐹:(Base‘𝑌)–1-1-onto𝐵)
1181, 2isgim 17527 . . 3 (𝐹 ∈ (𝑌 GrpIso 𝐺) ↔ (𝐹 ∈ (𝑌 GrpHom 𝐺) ∧ 𝐹:(Base‘𝑌)–1-1-onto𝐵))
11974, 117, 118sylanbrc 695 . 2 (𝜑𝐹 ∈ (𝑌 GrpIso 𝐺))
120 brgici 17535 . 2 (𝐹 ∈ (𝑌 GrpIso 𝐺) → 𝑌𝑔 𝐺)
121 gicsym 17539 . 2 (𝑌𝑔 𝐺𝐺𝑔 𝑌)
122119, 120, 1213syl 18 1 (𝜑𝐺𝑔 𝑌)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∀wral 2896 ∃wrex 2897 {crab 2900 ifcif 4036 ⟨cop 4131 class class class wbr 4583 ↦ cmpt 4643 ran crn 5039 ⟶wf 5800 –1-1→wf1 5801 –onto→wfo 5802 –1-1-onto→wf1o 5803 ‘cfv 5804 (class class class)co 6549 Fincfn 7841 0cc0 9815 + caddc 9818 ℕ0cn0 11169 ℤcz 11254 #chash 12979 Basecbs 15695 +gcplusg 15768 Grpcgrp 17245 .gcmg 17363 GrpHom cghm 17480 GrpIso cgim 17522 ≃𝑔 cgic 17523 CycGrpccyg 18102 Ringcrg 18370 CRingccrg 18371 RingHom crh 18535 ℤringzring 19637 ℤRHomczrh 19667 ℤ/nℤczn 19670 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-inf2 8421 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 ax-addf 9894 ax-mulf 9895 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-se 4998 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-isom 5813 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-tpos 7239 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-omul 7452 df-er 7629 df-ec 7631 df-qs 7635 df-map 7746 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-sup 8231 df-inf 8232 df-oi 8298 df-card 8648 df-acn 8651 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-4 10958 df-5 10959 df-6 10960 df-7 10961 df-8 10962 df-9 10963 df-n0 11170 df-z 11255 df-dec 11370 df-uz 11564 df-rp 11709 df-fz 12198 df-fl 12455 df-mod 12531 df-seq 12664 df-exp 12723 df-hash 12980 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-dvds 14822 df-struct 15697 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-mulr 15782 df-starv 15783 df-sca 15784 df-vsca 15785 df-ip 15786 df-tset 15787 df-ple 15788 df-ds 15791 df-unif 15792 df-0g 15925 df-imas 15991 df-qus 15992 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-mhm 17158 df-grp 17248 df-minusg 17249 df-sbg 17250 df-mulg 17364 df-subg 17414 df-nsg 17415 df-eqg 17416 df-ghm 17481 df-gim 17524 df-gic 17525 df-od 17771 df-cmn 18018 df-abl 18019 df-cyg 18103 df-mgp 18313 df-ur 18325 df-ring 18372 df-cring 18373 df-oppr 18446 df-dvdsr 18464 df-rnghom 18538 df-subrg 18601 df-lmod 18688 df-lss 18754 df-lsp 18793 df-sra 18993 df-rgmod 18994 df-lidl 18995 df-rsp 18996 df-2idl 19053 df-cnfld 19568 df-zring 19638 df-zrh 19671 df-zn 19674 This theorem is referenced by: cygzn 19738
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null | null | null | null | null | null | Monday, February 02, 2009
The Sweat Descends
On Saturday night I had a beer in a church. The Epiphany Episcopal Church to be precise, on the outskirts of Chicago's west loop. A fantastic venue for a concert and my favorite band in the world, Les Savy Fav, did not disappoint. Oddly enough, the beer tasted just like regular beer despite being sanctified, at least, locationally.
Walking inside I was greeted by the ticket folk, then allowed to enter the nave where tables were set up in the back corners hawking beer and water and hard liquor. The pews were removed and jammed against the walls where people were tossing their coats, hats and winter gear. About 400 people turned up, which is of course an estimate, and I was surprised by the number of women in the audience. probably close to 40/60, which is simply unheard of at these kind of shows. So either Chicago women are turned on by a front-man with a great big beard and belly (I can't say I blame them, Tim is quite charming), or just love great music.
The bands (LSF was accompanied by Jai Alai Savant) performed upon the raised chancel which was backed by a large floral mural, vaguely edenic, with a bunch of saints and our Lord and savior. There was barely any dressing of the stage although I'm not sure if it was prohibited or even necessary in the first place with such are great background. A bunch of lights were all that was needed to illuminate the murals, band, and audience in turn with your typical supersaturated concert reds and greens. Jai Alai Savant proved to be skilled and cheerful openers, and did the trick of warming everyone up. By the time they finished up I was in a knot of folks located just right of center, four or five people deep from the band.
Tim Harrington came out dressed as a priest, an outfit that didn't last long, which he used to deliver a somewhat boring if not blasphemous prayer akin to 'rub-a-dub-dub, thanks for the grub.' And when they began to play the crowd went apeshit, at least where I was standing. You know everything is well and good when you leap up and gravity doesn't restore you back to earth because everyone else is jumping and pressed up tight together. Tim did his audience participation thing, jumping down into the crowd and wandering about, sometimes mucking around quite deep, which can be disorienting for folks, since it seems like many people focus soley on the singer, and you have much of the audience in the front with their backs to the band while other concert-goers, by far the minority, continue to cheer on the rest of the musicians.
I've always thought of Tim as kind of an overgrown toddler, who is fascinated by everything and gets easily distracted from the music by alternately being a showman and audience member himself. At times it seems to exasperate the rest of the band, I noticed the bassist, Syd giving knowing glances to the tech guys, as if to say "It's just not worth reigning him back, just let him go." At one point Syd had to remind Tim what song was next, but at least he received a delighted, "I love that song!' in response. But in general Syd and the rest of the guys seem content to be, well, not forgotten, they're too skilled for that.
As always there was a lot of physicality in Tim's showing off. the tight clothes, the wigs, the near nudity, at one point he wore a green spandex number with leopard fringe that Harrington claimed was taken from a rapper named 'middle weezy' or 'medium-sized jeezy' or something like that. And his crowning moment came in the audience where he knelt down in front of the guy next to me, unzipped his fly, and tucked the mic in the hole and began to sing his lyrics as if giving head. As many as a dozen or so cell-phone cameras were snapping pictures and recording the moment for posterity.
My only lament was that I am old and very quickly became pretty beaten up and broken down, just halfway threw the set or so. Before that I was nearly hysterical, as were those around me. Tim even gave us a shout saying, "You guys ought to spread out through the rest of the crowd." Sadly I couldn't keep up. I was dehydrated before things even began, and when I started seeing stars I thought it best to get some air. I watched the remainder of the show from the sidelines, with people gently bobbing their heads and singing along.
Yet i must have been doing something right. It's days later and I'm still sore.
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At 9:05 PM , Blogger The Cranky TA said...
So, I'm up on the latest journal articles, but I just got around to downloading Inches (to pair with Go Forth). I'm really liking it. Especially, Knowing How the World Works. If "all of them" is not a valid answer, what's your recommendation for a next purchase?
At 8:47 AM , Blogger Les Savy Ferd said...
Yeah, Inches is a great collection. Probably the strongest set they have. I especially Like 'Yawn Yawn Yawn'.
But as far as 'all of them' is concerned, for a band that has been around as long as they have, their catalog is pretty slim.
Their newest album, "Let's Stay Friends" is the most accessible. There are even female guest vocalists. You can get a good idea if you like the album by listening to the song "the Equestrian." Do a youtube search as the band had a contest for fans to do their own video for the song and there is an amazing one with a pink haired little girl playing with My Little Ponies (equestrian... clever?).
The tiny EP "ROME," sometimes written "ROME (upside down)" as, well, it's written upside down, has only 5 songs but they are all amazing. "In These Woods" is probably my favorite Fav song.
The other 2 albums, '3/5' and 'Cat and the Cobra' are both terrific but more for completists, especially the former. They are rougher, purposely dissonant; the music of art punks who have just got out of college.
Finally, there is a rocky Live record that is download only called "After the Balls Drop." A concert performed in NYC on New Year's. There are some performances on there that are nigh unlistenable, but some redemption towards the end when you get some neat covers of 'Sliver' 'Astro-Zombies' and 'Debaser' But by no means a must get, lot a chaff on it.
At 2:17 AM , Blogger The Cranky TA said...
Awesome annotated discography. thanks.
At 12:06 PM , Blogger LNE said...
I love that first paragraph, F-man! What a great post/review.
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### 1.1. A bullet going 481 m/s embeds in a stationary block of wood. The bullet and block combo are going 5.27 m/s after the collision, and the combo has a mass of 12.1 kg (Bullet and block). What was the mass of the bullet? (0.133 kg)
Solution goes here
### 2. 2. A 65 kg person dives 3.68 m/s to the right off of a 23 kg cart. What is the velocity of the cart if the cart and person were initially at rest? (10. m/s to the left)
Solution goes here
### 3. 3. 68 kg-Francois running 7.8 m/s jumps on a 45.3 kg cart already rolling at 2.3 m/s in the same direction. What speed are they going after he jumps on? (5.6 m/s)
Solution goes here
### 4. 4. A 1240 kg Toyota Camry going 12.0 m/s to the east, strikes a 2530 kg SUV going west at 16.3 m/s. What is the velocity of the wreckage after the collision? (6.99 m/s to the west)
Solution goes here
### 5. 5. A 65 kg person is riding a 23 kg cart to the right at 3.15 m/s. What speed must he dive off the cart, and in what direction, to give the cart a velocity of 22.3 m/s to the right? (3.6 m/s to the left)
Solution goes here | 384 | 1,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-35 | latest | en | 0.911013 |
http://www.slidesearchengine.com/slide/11th-physics-notes-magnetism | 1,542,076,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741192.34/warc/CC-MAIN-20181113020909-20181113042242-00021.warc.gz | 478,121,624 | 7,111 | # 11th Physics Notes - Magnetism
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Published on February 4, 2014
Author: ednexa
Source: slideshare.net
Magnetism Important Terms and Definitions Two types of magnets: - natural and artificial Natural magnets The stones which possess the property to attract other stones are called load-stones. Where load stones are suspended freely, it remained in equilibrium in north-south direction. Magnetic dipole An arrangement of two magnetic poles of equal and opposite strengths separated by a finite distance is called a magnetic dipole or a bar magnet. The two magnetic poles of dipole are called N-pole & S-pole An isolated pole does not exist.
Poles are Fictitious :A bar magnet is assuming to have 2 opposite poles (N & S) situated near its ends. If we try to separate the two poles by cutting the magnet into two pieces, opposite poles are found to appear at the broken ends and each piece is again a new magnet with opposite poles. Even if the magnet is broken to the atomic level, it exists with two poles. This shows that the two poles of a magnet can not be separated from each other. Thus, a single magnetic pole i.e. monopole cannot exist. So, magnetic poles always appear in pairs and hence any magnet is called a dipole and it is said that, a magnetic monopole doesn’t exist or poles of a magnet cannot be separated.
Magnetic length The distance between the two poles of a bar magnet is called the magnetic length of the magnet. It is a vector. Direction S-pole to N-pole Magnetic length = 5/6 (Geom. length)
Note 1. Pole strength (m) The ability of a magnetic pole to produce magnetic field around it is called magnetic pole strength. 2. Axis of a magnetic dipole A straight line passing through magnetic poles of a magnetic dipole is called axis of a magnetic dipole. 3. Equator of a magnetic dipole A straight line passing through the centre of a magnetic dipole and perpendicular to its axis is called equator of a magnetic dipole.
Magnetic dipole moment (M) The product of the strength of either pole and the magnetic length of the magnet is called magnetic dipole moment. Magnitude : M = m × 2ℓ, where m = pole strength of N-pole Direction : S-pole of the magnet to its N- : Ampere meter2 (Am2) pole Unit
Magnetic field The space around a magnet (or a conductor carrying current) in which its magnetic effect can be experienced, is called the magnetic field. The SI unit of strength of magnetic field (also known as magnetic induction or magnetic flux density) is tesla (T) 1 tesla (T) = 1 weba / metre2 = 1 wb/m2 1 tesla (T) = 1 N/Am........ {B = F/m, which is the B due to a role strength} 10-4 tesla (T) = 1 gauss (G)
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This section includes the course lecture notes, prepared in LaTeX by James Silva, an MIT student, based upon handwritten notes. | 1,067 | 4,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-47 | latest | en | 0.917181 |
https://www.collimator.ai/reference-guides/what-is-gram-schmidt-orthonormalization-process | 1,696,256,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00100.warc.gz | 761,431,397 | 16,622 | July 6, 2023
# What is gram schmidt orthonormalization process?
The gram schmidt orthonormalization process is a mathematical technique used to transform a set of vectors into a new set of orthonormal vectors. In this article, we will explore the basics of orthonormalization, its history, detailed explanation of the gram schmidt process, applications, and advantages/disadvantages of this process.
## Understanding the Basics of Orthonormalization
In mathematics, orthonormalization is the process of constructing or finding a set of orthonormal vectors from a given set of linearly independent vectors. The main goal of orthonormalization is to create a new basis that simplifies calculations and provides better representation of vectors.
### Definition of Orthonormalization
Orthonormalization is the procedure to transform a set of linearly independent vectors into a set of vectors that are orthogonal (perpendicular) to each other and have unit length. The resulting set of vectors is called orthonormal.
Orthonormalization is a fundamental concept in linear algebra. It plays a crucial role in various mathematical applications, enabling the analysis and manipulation of vectors in a more efficient and meaningful way.
When a set of vectors is orthonormal, it means that each vector is perpendicular to every other vector in the set. This property allows for easier computations and simplifies the representation of vectors in terms of their components.
Moreover, the unit length of the orthonormal vectors ensures that they have a magnitude of 1. This normalization factor is essential for preserving the relative magnitudes and directions of vectors when performing operations like dot products or projections.
### Importance of Orthonormalization in Mathematics
Orthonormalization has numerous applications in different branches of mathematics, such as linear algebra, functional analysis, and signal processing. It simplifies calculations involving vectors, enhances numerical stability, and enables better understanding of vector spaces.
In linear algebra, orthonormal bases are particularly useful for representing vectors as linear combinations of basis vectors. This representation allows for straightforward computations and provides a clear geometric interpretation of vector operations.
Functional analysis, a field that studies vector spaces of functions, heavily relies on orthonormalization techniques. By orthonormalizing a set of functions, one can construct an orthonormal basis for the function space, facilitating various analyses and computations.
In signal processing, orthonormal bases are employed for efficient data compression, noise reduction, and signal reconstruction. By representing signals in terms of orthonormal basis functions, one can extract relevant information while minimizing the impact of noise and interference.
Overall, orthonormalization serves as a powerful tool in mathematics, enabling the simplification of calculations, enhancing the stability of numerical algorithms, and providing deeper insights into the properties of vector spaces.
## The History of Gram Schmidt Orthonormalization Process
The gram schmidt orthonormalization process is named after its developers - Jørgen Pedersen Gram and Erhard Schmidt, who independently introduced it in the early 20th century. Let's delve into the history of these mathematicians and the evolution of this process.
### Who is Gram Schmidt?
Jørgen Pedersen Gram was a Danish mathematician born in 1850. He made significant contributions to the fields of linear algebra and geometry. Gram developed the gram schmidt process as a fundamental tool in linear algebra, forging a path for its application in various disciplines.
Gram's interest in mathematics began at a young age. He displayed exceptional talent and a deep understanding of mathematical concepts. As he pursued his studies, Gram became captivated by the intricacies of linear algebra and its applications in solving complex problems.
During his academic career, Gram worked closely with renowned mathematicians, exchanging ideas and collaborating on research projects. His dedication and relentless pursuit of mathematical excellence led him to develop the gram schmidt orthonormalization process.
The gram schmidt process revolutionized the field of linear algebra by providing a method to transform a set of linearly independent vectors into a set of orthonormal vectors. This process has since become a cornerstone of numerous mathematical and scientific disciplines.
### Evolution of the Orthonormalization Process
Initially, the concept of orthonormalization was introduced by Erhard Schmidt, a German mathematician. Schmidt's method aimed to transform a set of linearly independent vectors into a set of orthogonal vectors. However, his approach had limitations in terms of computational complexity.
Gram recognized the potential of Schmidt's work but sought to refine and improve the process. He devised a modified version of the orthonormalization process, which not only produced orthogonal vectors but also ensured their normalization, making them orthonormal.
Gram's modification on Schmidt's work made the process more practical and efficient. His method involved a step-by-step algorithm that systematically orthogonalized and normalized the vectors, resulting in a set of orthonormal vectors that were easier to work with in various mathematical applications.
The gram schmidt orthonormalization process gained widespread recognition and adoption in the mathematical community due to its simplicity and effectiveness. It became an essential tool in solving linear algebraic problems, particularly in fields such as signal processing, quantum mechanics, and computer graphics.
Over the years, researchers and mathematicians have built upon Gram and Schmidt's work, further refining and expanding the orthonormalization process. Their contributions have led to the development of advanced techniques and algorithms that have enhanced the efficiency and applicability of the gram schmidt orthonormalization process in various scientific and engineering domains.
## Detailed Explanation of Gram Schmidt Process
Now, let's take a closer look at the step-by-step guide to the gram schmidt orthonormalization process and its mathematical representation.
The gram schmidt process starts with a set of linearly independent vectors. It involves three main steps:
1. Orthogonalization: The first vector in the set remains unchanged. For each subsequent vector, it is projected onto the subspace spanned by the previously processed orthogonal vectors and subtracted to ensure orthogonality.
2. Normalization: After achieving orthogonality, each vector is divided by its length to attain unit length. This step ensures that the resulting vectors are not only orthogonal but also have a magnitude of 1.
3. Iterative Process: The process is repeated for all vectors until the entire set has been orthonormalized.
Let's dive deeper into each step of the gram schmidt process.
### Step 1: Orthogonalization
The first step of the gram schmidt process is orthogonalization. It ensures that each vector in the set becomes orthogonal to all the previously processed vectors.
Starting with the first vector, which remains unchanged, we move on to the subsequent vectors. Each vector is projected onto the subspace spanned by the previously processed orthogonal vectors and subtracted. This subtraction ensures that the new vector is orthogonal to the previous vectors.
Let's consider an example to illustrate this step. Suppose we have a set of vectors {v1, v2, v3}. We start by keeping v1 as it is. Then, we proceed to v2.
To make v2 orthogonal to v1, we need to subtract the projection of v2 onto v1 from v2. This ensures that v2 is orthogonal to v1.
Next, we move on to v3. To make v3 orthogonal to both v1 and v2, we subtract the projections of v3 onto v1 and v2 from v3. This step guarantees that v3 is orthogonal to both v1 and v2.
This process continues for all the vectors in the set until all vectors are orthogonal to each other.
### Step 2: Normalization
Once the orthogonalization step is complete, we move on to the normalization step. This step ensures that each vector in the set has a magnitude of 1.
To normalize a vector, we divide it by its length. This division scales the vector to have a unit length.
Continuing with our example, after orthogonalizing the vectors {v1, v2, v3}, we divide each vector by its length to normalize them.
Normalization is essential because it allows us to compare the magnitudes of different vectors accurately. It also simplifies calculations involving dot products and other vector operations.
### Step 3: Iterative Process
The gram schmidt process is an iterative process that repeats the orthogonalization and normalization steps for each vector in the set.
Following our example, we would repeat the orthogonalization and normalization steps for each vector in the set {v1, v2, v3}. This iterative process ensures that all vectors in the set become orthonormal.
By the end of the process, we obtain a set of orthonormal vectors {u1, u2, u3} that span the same subspace as the original set {v1, v2, v3}.
### Mathematical Representation of the Process
In mathematical terms, the gram schmidt orthonormalization process can be represented as follows:
Let {v1, v2, ..., vn} be the set of linearly independent vectors. The resulting orthonormal set can be obtained as:
u1 = v1 / ||v1||
u2 = (v2 - proju1(v2)) / ||(v2 - proju1(v2))||
u3 = (v3 - proju1(v3) - proju2(v3)) / ||(v3 - proju1(v3) - proju2(v3))||
...
un = (vn - proju1(vn) - proju2(vn) - ... - projun-1(vn)) / ||(vn - proju1(vn) - proju2(vn) - ... - projun-1(vn))||
## Applications of Gram Schmidt Orthonormalization Process
The gram schmidt orthonormalization process finds extensive applications in various fields of study. Let's explore two prominent areas where this process plays a crucial role.
### Use in Linear Algebra
In linear algebra, the gram schmidt process is utilized for orthogonalizing vectors, finding orthonormal bases, and solving systems of linear equations. It simplifies calculations involving inner products, matrix transformations, and eigenvalue problems.
One application of the gram schmidt orthonormalization process in linear algebra is in computer graphics. In 3D computer graphics, orthonormal bases are crucial for representing objects in three-dimensional space. By applying the gram schmidt process to a set of vectors, we can obtain an orthonormal basis that can be used to perform efficient transformations, such as rotations and translations, on objects in a computer-generated scene.
Another application of the gram schmidt process is in signal processing. In this field, it is common to work with sets of vectors that represent signals. By orthogonalizing these vectors using the gram schmidt process, we can eliminate any correlation between the signals, which can be useful for noise reduction or extracting specific components of a signal.
### Role in Quantum Mechanics
The gram schmidt orthonormalization process is fundamental to quantum mechanics, a branch of physics. It facilitates the description and analysis of quantum states, wavefunctions, and observable quantities. The orthonormality of quantum states is a prerequisite for accurate predictions and calculations.
In quantum mechanics, the gram schmidt process is used to construct orthonormal bases for representing quantum states. These bases are essential for understanding the behavior of particles and systems in the quantum realm. By applying the gram schmidt process to a set of basis vectors, we can ensure that they are mutually orthogonal and normalized, allowing us to make precise measurements and predictions about the quantum system under study.
Furthermore, the gram schmidt orthonormalization process is utilized in quantum chemistry. In this field, the process is used to construct orthonormal molecular orbitals, which are crucial for describing the electronic structure of molecules. By orthogonalizing the atomic orbitals using the gram schmidt process, we can obtain a set of orthonormal molecular orbitals that accurately represent the electron density and energy levels of a molecule, enabling us to study its chemical properties and reactions.
## Advantages and Disadvantages of Gram Schmidt Process
Like any mathematical technique, the gram schmidt orthonormalization process has its own set of advantages and limitations. Let's discuss them briefly.
### Benefits of Using Gram Schmidt Process
• The resulting orthonormal vectors simplify calculations and reduce computational complexity.
• Orthonormal bases obtained through this process provide a better representation of vectors and enable easier manipulation.
• The gram schmidt process enhances numerical stability by eliminating any linear dependence or correlation between vectors.
### Limitations of the Process
• Under certain conditions, the gram schmidt process may suffer from numerical instability, leading to inaccuracies in computations.
• Applying the process to a large set of vectors can be computationally intensive and time-consuming, especially for high-dimensional spaces.
• The resulting orthonormal vectors may lose important structural or contextual information, depending on the specific application.
In conclusion, the gram schmidt orthonormalization process is a valuable mathematical technique used across various disciplines. Its ability to transform a set of vectors into an orthonormal basis has profound implications in linear algebra, quantum mechanics, and other fields. By understanding the basics, history, detailed explanation, applications, and advantages/disadvantages of this process, we can appreciate its significance in modern mathematics and science. | 2,726 | 13,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-40 | latest | en | 0.897812 |
null | null | null | null | null | null | Friday, 12 December 2008
Janine di Giovanni embelishes her story about France and the French with arsenic and lies
SuperFrenchie reports a bad reporting, a case of writing in bad faith by an American French-bashing journalist, who it seems, if the Wiki blurb were to be believed, is a 'decorated' war correspondent.
I read the purported beastly article in question and true enough, Janine di Giovanni's arsenic-laced piece of journalism is an exercise in frustration and it does "take the cake for bad faith reporting."
Giovanni's piece nearly had me flummoxed. I mean, some of the garbage she said about France are just completely false. An example: zero credit card. She writes, "No one lives on credit in France because banks don’t allow overdrafts and zero percent credit cards do not exist."
Where did she get that piece of rubbish? So blatantly misleading is she that klancymiller, one of the commenters in her site, was prompted to ask quite rightly: "Where do you do your research?" and went on to point Giovanni's spins.
No one lives on credit in France and don't allow overdrafts, etc., etc.? Utter falsehood. My brother, who is a student in an Ecole de commerce in Paris, survives virtually on his credit card overdrafts. I can tell you right here and now that my parents have just (very recently matter of fact) paid off his 1500-euro overdraft after his bank threatened to do something drastic about his burgeoning debt.
Giovanni pushed her reporting too far when she wrote, “Last week Lanvin, the French couture house favored by Hollywood celebrities, gave a private sale and I stood in the midst of hysterical women clamoring for fur coats. Fur coats? In the midst of the global meltdown?”
I Googled the Lanvin sale and found what she might have been talking about in L'Express which is a pretty reputable French weekly (by the way, I am familiar with these Faubourg-Saint Honoré soldes):
Lanvin- 40% sur le prêt-à-porter automne-hiver, les chaussures, les sacs et les ceintures sauf manteaux en fourrure et certains bijoux cristaux, colliers plastron et perles. La bonne affaire: la sublime robe longue du soir en jersey gris chiné retenue par de fines bretelles, avec sur le devant un nœud de velours noir à 925 € au lieu de 1 540 €. Ou bien, petit «must-have» de la collection, les ballerines plates à 330 € soldées 198 €. 15, rue du Faubourg-Saint-Honoré, VIIIe. 01-44-71-33-33.
Translation of those in bold: 40% on ready to wear Fall-Winter collection, shoes, bags, belts except fur coats and some crystal jewelry, plastron and pearl necklaces. °
In other words, Lanvin fur coats were not part of the private sale, hence the women shoppers couldn't have been clamouring for fur coats in that private sale precisely because they weren't included in the private sale!
So, is Giovanni (shown right in picture) LYING?
Well, what's clear to me is that she has embelished her story to shock and awe her readers but why did she have to make up, i.e., lie about, some of the things she was reporting on? Of course, there are more anneries in her artitcle but this is just one spin too many so I will leave it at that. Go, read it and try not to puke!
No matter how you turn her article around, you see it dripping with disdain, contempt, scorn for the French etc. That article contains a historical record of her thoughts about the French and there's no doubt about it, she hates the French.
Why Janine de Giovanni baise français (fucks a Frenchman) boggles the mind but maybe, just maybe, elle souffre de la malbaise! (she's fucked badly!)
Raymond Boucher said...
Mrs di Giovanni should leave France and go back to the states or to Italy!
Why the hell does she want her son to finish his French education if France is such a despicable country?
Next time she wants to write an article about anything, she'd better check her facts since 'The fall of France" is full of errors, not to say lies.
R Boucher (Another uneducated Frenchman who can neither unserstand nor speak a single word of English.)
annetravel said...
I was not impressed with the article. I live in the US where the super rich pay no taxes and the middle class & the poor struggles. Taxes are higher and it is hard for people to get by!. Now regarding the word entrepreneur you repeat what president Bush said when he was in power ( the French don't know the word " entrepreneurs), France has a lot of entrepreneurs , they handle it a different way!. If there is turmoil in this world, then the US has as much as France and is definitely not better off. People are not happy here because it is about the rich. Please don't be arrogant, China owns the US. | null | null | null | null | null | null | null | null | null |
https://www.geeksforgeeks.org/areas-related-to-circles-perimeter-of-circular-figures-areas-of-sector-and-segment-of-a-circle-areas-of-combination-of-plane-figures/?ref=leftbar-rightbar | 1,618,380,750,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00630.warc.gz | 878,990,393 | 19,892 | Related Articles
Areas Related to Circles – Perimeter of circular figures, Areas of sector and segment of a circle & Areas of combination of plane figures
• Last Updated : 21 Dec, 2020
A circle is a shape consisting of a curved line completely surrounded by an area. The circle contains points in the plane which are at the given distance from a given point, the center, equivalently the curve traced out by a point which moves in a plane so the distance from a given point is constant. The distance between any point and the center of the circle is called the radius (r).
### Area of a Circle
The area of the circle is computed by the formula:
πr2
• r is the radius of the circle.
• π is the ratio of the diameter of a circle to its circumference
• The value π is 22/7 (or) 3.14.
### Circumference of a Circle
The perimeter of the circle is the distance along the boundary of the circle. The perimeter is also called the circumference of a circle. Circumference will be the π times of the diameter of the circle.
• Circumference of a Circle is 2πr.
### Sector
A sector of a circle is defined as the region of a circle enclosed by an arc and two radii(r). The smallest area of a circle is called the minor sector and the largest area of a circle is called the major sector.
### Angle of a Sector
• The angle of a sector is that angle which is enclosed within the two radii of the sector. It consists of an arc.
• A sector that has a central angle of 180° is known as a semicircle.
• A minor arc is smaller than a semicircle. A central angle that is subtended by a minor arc has a measure of less than 180°.
• A major arc is larger than a semicircle. A central angle that is subtended by a major arc has a measure larger than 180°.
### Arc
• To find the length of an arc of a circle we use the arc length formula: length = radius * θ
• An arc is a part of a curve.
• It is a portion of the circumference of the circle.
### Sample Problems
Problem 1: Find the area of a sector and arc length of a circle of radius
4cm and the central angle is 2
π/5.
Solution:
Arc length, l = 4*2π/5
= 8π//5cm.
area of a sector = 1/2θr2
= 1/2 * 2π/5 *42
= 16π/5cm2.
### Circumference of a circle when the area is given
The circumference of the circle can be referred as the linear distance around it. If the circle is opened to form a straight line then, the length of that line will be the s circumference of the circle.
To calculate the circumference of a or a given circle, we need to multiply the diameter of the circle with the π.
Problem 2: What is the perimeter of the circle whose surface area is 314.159 sq.cm
Solution:
The formula of the surface area of the circle, we know:
A = π x r2
Now, substituting the value:
314.159 = π x r2
314.159 = 3.14 x r2
r2 = 314.159/3.14
r2 = 100.05
r = √100.05
r = 10 cm
C = 2 * π * r
Substitute value of r
= 2π * 10
= 2 * 3.14 * 10
= 62.8 cm.
### Partial circle area and arc length
The circle area can be also referred to as the circumference. The area of the partial circle which is also called sector can be found by the formula
[(π*r2)/360]θ
The arc length is given by
(θ/360)*(2 * π * r)
Problem 3: A circle of radius 4 units, angle of its sector is 45°. Find the area of the sector.
Solution:
Given,
Angle θ = 45°
Area of the sector
= θ/360o × πr2
= 45/360 × 22/7 × 42 = 6.28sq.units.
### Area of a Shaded Region
The shaded region is the region that can be found when one shape is inscribed within the other shape.
• The area of the shaded region can be calculated by subtracting the area of the inscribed shape area from the area of the shape which is inscribed in it.
• For example, let us assume a circle is inscribed in the triangle then the area of the shaded region is the area of the triangle minus the area of the circle inscribed in the triangle(shaded area = area of a triangle – area of a circle).
Problem 4: Find the area of the shaded region.
Solution:
The area of shaded portion, we have to subtract area of two semicircles from the area of square
Area of shaded portion = Area of square – (Area of semicircle + Area of another semicircle).
= a2 – [(1/2) πr2) + ((1/2) πr2)]
= 82 – πr2
= 64 – (22/7)⋅ (7/2)2
= 64- (22/7)⋅ (7/2)⋅ (7/2)
= 64 – 38.5
= 25.5 cm2
Problem 5: Find the area of the shaded region.
Solution:
Given radius r = 10 cm
area of shaded region = area of square – area of circle inscribed inside square.
area of square= S2
diameter of the circle will be equal to length of the side of the square so
d = S = 2*10
= 20
area of square =(20)2
= 400cm2
area of circle = πr2
= 3.14 * 10 * 10
= 314 cm2
Area of shaded region = area of square – area of circle inscribed inside square
= 400 – 314
= 86cm
Therefore, area of shaded region is 86cm.
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My Personal Notes arrow_drop_up | 1,401 | 4,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-17 | latest | en | 0.935372 |
https://www.qalaxia.com/questions/Find-the-value-of-the-discriminant-of-math-x2-6x | 1,695,812,679,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00314.warc.gz | 1,064,531,698 | 2,899 | Krishna
0
Step 1: Note-down the given quadratic equation and compare it with the standard form ax^2 + bx + c
EXAMPLE: x^2\ +6x\ +11
ax^2 + bx + c
Where a = 1, b = 6 and c = 11
Step 2: Substitute all the values in the discrimination( b^2-4ac ) of the equation.
b^2\ -\ 4ac\ = 6^2-\ 4\ \cdot1\cdot11
= 36 - 44
= -8 | 128 | 320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.582055 |
https://www.hackmath.net/en/math-problem/634?tag_id=21 | 1,576,056,370,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00421.warc.gz | 713,393,180 | 7,819 | # Chocolates
In the market have 3 kinds of chocolates.
How many ways can we buy 14 chocolates?
Result
n = 120
#### Solution:
$n = C'^{ 3}_{ 14} = C_{{ 14}}(16) = \dbinom{ 16}{ 14} = \dfrac{ 16! }{ 14!(16-14)!} = \dfrac{ 16 \cdot 15 } { 2 \cdot 1 } = 120$
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#### Following knowledge from mathematics are needed to solve this word math problem:
See also our combinations with repetition calculator. Would you like to compute count of combinations?
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What is the probability that the two throws of the dice: a) Six falls even once b) Six will fall at least once | 831 | 3,185 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-51 | latest | en | 0.937436 |
https://www.sawaal.com/profit-and-loss-questions-and-answers/an-article-was-sold-at-a-gain-of-18-if-it-had-been-sold-for-rs49-more-then-the-gain-would-have-been-_49010 | 1,627,103,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00533.warc.gz | 1,031,548,631 | 14,431 | 0
Q:
# An article was sold at a gain of 18%. If it had been sold for Rs.49 more, then the gain would have been 25%. The cost price of the article is:
A) ₹ 700 B) ₹ 570 C) ₹ 650 D) ₹ 890
Explanation:
Q:
A sells an article to B at a loss of 20%, B sells it to C at a profit of 12.5% and sells it to D at a loss of 8%. If D buys it for Rs.248.40, then what is the difference between the loss incurred by A and C?
A) Rs. 36.80 B) Rs. 38.40 C) Rs.42.60 D) Rs.39.20
Explanation:
1 193
Q:
A shopkeeper bought 80 kg of rice at a discount of 10%. Besides 1 kg rice was offered free to him on the purchase of every 20 kg rice. If he sells the rice at the marked price, his profit percentage will be:
A) 1623% B) 1513 % C) 1537 % D) 1427%
Explanation:
0 569
Q:
Sudha sold an article to Renu for Rs.576 at a loss of 20%. Renu spent a sum of Rs.224 on its transportation and sold it to Raghu at price which would have given Sudha profit of 24%. The percentage of gain for Renu is:
A) 13.2% B) 10.5% C) 12.9% D) 11.6%
Explanation:
0 1519
Q:
By selling 18 table fans for Rs.11,664 a man incurs a loss of 10%. How many fans should he sell for Rs.17,424 to earn 10% profit?
A) 18 B) 22 C) 20 D) 23
Explanation:
0 1005
Q:
While selling an article of marked price Rs.5,040 at a discount of 40%, if a trader gains 20%, then the profit in Rs is:
A) ₹642 B) ₹504 C) ₹720 D) ₹2,520
Explanation:
0 1644
Q:
The marked price of an item is 25% above its cost price. A shopkeeper sells it, allowing a discount of x % on the marked price. If he incurs a loss of 8%, then the value of x is
A) 25.6% B) 26.8% C) 26.4% D) 25.2%
Explanation:
0 214
Q:
Reema sold 48 articles for Rs. 2,160 and suffered a loss of 10%. How many articles should she sell for Rs. 2,016 to earn a profit of 12%?
A) 36 B) 40 C) 28 D) 32
Explanation:
1 1233
Q:
Sudhir purchased a laptop for Rs.42,000 and a scanner-cum-printer for Rs.8,000. He sold the laptop for a 10% profit and the scanner-cum-printer for a 5% profit. What is his profit percentage?
A) 925°° B) 915°° C) 15°° D) 712°° | 720 | 2,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-31 | latest | en | 0.906357 |
https://justaaa.com/accounting/66370-you-buy-an-8-percent-coupon-10-year-maturity-bond | 1,722,746,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00874.warc.gz | 261,629,517 | 10,193 | Question
# You buy an 8 percent coupon, 10-year maturity bond when its yield to maturity is 9...
You buy an 8 percent coupon, 10-year maturity bond when its yield to maturity is 9 percent. One year later, the yield to maturity is 10 percent. Assume the face value of the bond is \$1,000.
(a) What is the price of the bond today?
(b) What is the price of the bond one year later?
(c) What is your rate of return over the one-year holding period?
Value of the bond is Present value of future cash inflows. So, the value of the bond is calculated by discounting the future cash inflows.
(a) Price of the bond today:
Year Nature Amount PVF@9% Present value 1-10 Coupon \$ 80 PVAF(9%,10)=6.4176 \$513.408 10 Redemption \$1000 PVF(9%,10) =0.4224 \$422.4 Value of the Bond today \$935.808
(b) Price of the bond 1 year later:
Year Nature Amount PVF@10% Present value 1-9 Coupon \$ 80 PVAF(10%,9)=5.7590 \$460.72 9 Redemption \$1000 PVF(10%,9) =0.4241 \$424.1 Value of the Bond after 1 yr \$884.82
(c) Rate of return over 1 yr holding period:
Rate of return = coupon received + (Sale price - purchase price) / Bond purchase price
= (\$80 + (\$884.82 - \$935.808)) /\$935.808
= (\$80 - \$50.98) / \$935.808
= \$29.02 / \$935.808
=0.03101 (or) 3.101%
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 402 | 1,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-33 | latest | en | 0.835662 |
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Broad Topics > Calculations and Numerical Methods > Multiplication & division
### More Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### One O Five
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You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
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What is the least square number which commences with six two's?
### More Mods
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What is the units digit for the number 123^(456) ?
### Vedic Sutra - All from 9 and Last from 10
##### Age 14 to 16 Challenge Level:
Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?
### Why 8?
##### Age 11 to 14 Challenge Level:
Choose any four consecutive even numbers. Multiply the two middle numbers together. Multiply the first and last numbers. Now subtract your second answer from the first. Try it with your own. . . .
### Alphabetti Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku requires you to do some working backwards before working forwards.
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### The Genie in the Jar
##### Age 11 to 14 Challenge Level:
This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Largest Number
##### Age 11 to 14 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
##### Age 11 to 14 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Gabriel's Problem
##### Age 11 to 14 Challenge Level:
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
### A One in Seven Chance
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### As Easy as 1,2,3
##### Age 11 to 14 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Charitable Pennies
##### Age 7 to 14 Challenge Level:
Investigate the different ways that fifteen schools could have given money in a charity fundraiser.
### X Marks the Spot
##### Age 11 to 14 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### The Remainders Game
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Play this game and see if you can figure out the computer's chosen number.
### What Is the Question?
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These pictures and answers leave the viewer with the problem "What is the Question". Can you give the question and how the answer follows?
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Product Sudoku
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The clues for this Sudoku are the product of the numbers in adjacent squares.
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Largest Product
##### Age 11 to 14 Challenge Level:
Which set of numbers that add to 10 have the largest product?
##### Age 7 to 14 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### 3388
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Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Legs Eleven
##### Age 11 to 14 Challenge Level:
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?
### Thirty Six Exactly
##### Age 11 to 14 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
##### Age 11 to 14 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### Slippy Numbers
##### Age 11 to 14 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Funny Factorisation
##### Age 11 to 16 Challenge Level:
Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?
##### Age 11 to 14 Challenge Level:
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Find the number which has 8 divisors, such that the product of the divisors is 331776.
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I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
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What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
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### Going Round in Circles
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Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Factorial
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How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Galley Division
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Can you explain how Galley Division works?
### Multiples Sudoku
##### Age 11 to 14 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Times Right
##### Age 11 to 16 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Integrated Product Sudoku
##### Age 11 to 16 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Ones Only
##### Age 11 to 14 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010.
### So It's Times!
##### Age 7 to 14 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Round and Round and Round
##### Age 11 to 14 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help? | 2,284 | 9,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.855924 |
https://www.sawaal.com/coding-decoding-questions-and-answers/in-a-certain-code-language-concentration-is-written-as-qqjgzigqgdmlx-how-will-nitrification-be-writt_8562 | 1,718,826,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00050.warc.gz | 867,822,349 | 13,961 | 79
Q:
# In a certain code language CONCENTRATION is written as QQJGZIGQGDMLX. How will NITRIFICATION be written in that code language?
A) QQJGZXRIKSGRM B) QQJGYXRIKSGRM C) QQJGZXRIKSGRN D) QQJGZRIKSGSM
Explanation:
The first three letters are coded as same positioned letters from the right end of the alphabet. So are the seventh to tenth letters. The fourth , fifth and sixth letters move one, two and three positions forward respectively in the alphabet. So, do the last three letters. Thus,
CONCENTRATION => XLMDGQGIZGJQQ
Now reverse this and we get QQJGZIGQGDMLX
Q:
If RESEARCH is $#!#%$&@ then SCARE is
A) !&%$# B) !@%$# C) !$%#& D) !@%#$
Explanation:
Filed Under: Coding and Decoding
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13 780
Q:
If 12@18 = -3, 18@6 = 6 and 16@8 = 4, then find the value of 6@10 = ?
A) -10 B) 8 C) 4 D)
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
14 8005
Q:
The following equation is incorrect. Which two signs should be interchanged to correct the equation?
10 + 14 x 20 - 7 ÷ 10 = 40
A) + and x B) + and ÷ C) ÷ and D) - and +
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
13 7422
Q:
In a certain code language, 'x' represents '+', '÷' represents 'x', '-' represents '÷' and '+' represents '-'. Find out the answer to the following question.
18 x 12 ÷ 5 - 10 + 10 = ?
A) 14 B) 6 C) 39 D) 2
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
11 601
Q:
In a certain code language, “NOTES” is written as “141520519” and “MAKAR” is written as “13111118”. How is “LOFTS” written in that code language?
A) 206191316 B) 121562019 C) 12150219 D) 1515121119
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
2 475
Q:
If Q = 10 and FAX = 50, then XEROX = ?
A) 45 B) 46 C) 49 D) 50
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
7 931
Q:
If 60#40 = 10, 40#20 = 6 and 60#90 = 15, then find the value of 40#30 = ?
A) 15 B) 12 C) 18 D) 7
Explanation:
Filed Under: Coding and Decoding
Exam Prep: Bank Exams
6 5709
Q:
The following equation is incorrect. Which two signs should be interchanged to correct the equation?
8 ÷ 2 - 50 x 6 + 16 = 50
A) + and ÷ B) - and + C) + and x D) ÷ and x | 779 | 2,246 | {"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} | 3.9375 | 4 | CC-MAIN-2024-26 | latest | en | 0.791165 |
http://www.physicsforums.com/showthread.php?t=471882 | 1,410,968,396,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657123996.28/warc/CC-MAIN-20140914011203-00069-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 734,529,159 | 8,849 | # Energy Loss vs Energy Delivered and Voltage Drop - Confused!
P: 10 Hi, ok so everytime I think that I have understood the concept of Energy transmission, losses and voltage drop, I get even more confused about things. I have searched several threads on this forum and the physics forum but failed to find anything that directly answers my query. So I kindly request your help on the fundamental concept. I will try to give an example and I would like you guys to correct me where I am wrong -- Thanks in advance!!! ------------------------------------------------------------------------- If a 1 MW load is to be supplied by a 10kV feeder, and the cable resistance is 1 ohms, then the current drawn by the load (or the current travelling on the cable) is: I = P/V = 100 A The line losses are I^2*R = 10 kW and the Voltage Drop = I*R = 100 V Is it right to say that: a) Voltage at supply point is 10kV b) Voltage at consumer end is 9.9kV (10kV - 100V) c) The current through the cable is 100A d) The power required is 1 MW but the actual power being delivered is 0.99 MW [ (P - losses) OR (V*I = 9.9kV * 100A) ] e) hence the efficiency of this system is 0.99MW/1MW * 100 = 99% efficient? ------------------------------------------------------------------------------ Please correct me where I am wrong. Thank you.
P: 247 All looks right to me. Fish
P: 28 nevermind... looks good
Sci Advisor Thanks PF Gold P: 12,256 Energy Loss vs Energy Delivered and Voltage Drop - Confused! @mathological It's near enough OK but, to be accurate, you need to know the actual resistance of the load (assume it's V2/Nominal Power where V is the nominal operating voltage) - that would imply 100Ω load resistance. You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω. The voltage drop across the feeder will be 10kV*1/101. This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). So it's less efficient than you would initially have thought. There's a double whammy in there! Weird huh. Edit - Tidied up the bit about the volts on the load - no change in info- just removed a duplication
P: 28 what if the load was hooked up in parallel?
Sci Advisor Thanks PF Gold P: 12,256 In parallel with what?? You wouldn't connect the cables straight across the mains!!!!
P: 28 true
P: 10
Quote by sophiecentaur @mathological It's near enough OK but, to be accurate, you need to know the actual resistance of the load (assume it's V2/Nominal Power where V is the nominal operating voltage) - that would imply 100Ω load resistance. You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω. The voltage drop across the feeder will be 10kV*1/101. This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). So it's less efficient than you would initially have thought. There's a double whammy in there! Weird huh. Edit - Tidied up the bit about the volts on the load - no change in info- just removed a duplication
You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω.
The voltage drop across the feeder will be 10kV*1/101. Are you using voltage division here?
This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). Sorry I dont get how you calculated this...can you please elaborate? Thanks!
P: 4,512 Look, mathlogical, you have stipulated that your load dissipates 1MW. Now. you have to be clear whether this 10KV pair of wires you call a feeder has 10KV at the source or 10KV at the load. The voltage at the source is not the same as the voltage at the load. Pick one or the other and the answer is determinate.
Sci Advisor Thanks PF Gold P: 12,256 @mathological Yes, I am using voltage division (it's the same old potential divider circuit that we come across everywhere). I was assuming that the original "1MW load" was one which is designed to dissipate 1MW when presented with 10Kv. Few loads adjust themselves to take their specified power; they are mostly 'dumb' resistors, or equivalent. The power dissipated in a resistor is V2/R so the power you get at the end of the cable will be (1/1.01)2 (=0.98) as much as without the cable. Actually, to be fair, the supply (generator) would be delivering less current (also 1/1.01 as much) so, defining efficiency as: power out/power supplied the actual efficiency will not be as low as 98% because less actual power will be put into the system by the generator. But you still get only 98% of the power you wanted.
P: 10 @sophiecentaur Thanks for the clarification! :)
Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 1 Advanced Physics Homework 2 Classical Physics 11 Introductory Physics Homework 3 | 1,221 | 4,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2014-41 | latest | en | 0.929264 |
https://www.whatnumberis.net/cxxxi/ | 1,719,125,772,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00753.warc.gz | 946,349,167 | 3,205 | What number is CXXXI?
Your question is: What numbers are the Roman numerals CXXXI? Learn how to convert the Roman numerals CXXXI into the correct translation of normal numbers.
The Roman numerals CXXXI are identical to the number 131.
CXXXI = 131
How do you convert CXXXI into normal numbers?
In order to convert CXXXI into numbers, the number of position values (ones, tens, hundreds, thousands) is subdivided as follows:
Place valueNumberRoman numbers
Conversion100 + 30 + 1C + XXX + I
Hundreds100C
Dozens30XXX
Ones1I
How do you write CXXXI in numbers?
To correctly write CXXXI as normal numbers, combine the converted Roman numbers. The highest numbers must always be in front of the lowest numbers to get the correct translation, as in the table above.
100+30+1 = (CXXXI) = 131
The next Roman numerals = CXXXII
Convert another Roman numeral to normal numbers. | 222 | 874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.75001 |
https://blog.finxter.com/interview-question-search-insert-position-python/ | 1,719,214,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00417.warc.gz | 112,963,442 | 24,119 | # [Interview Question] How To Search The Insert Position Of Target In A Sorted Array?
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Company tags: Adobe, Uber, Airbnb, Bloomberg
Are you looking to ace your coding interview? If yes! Then this question is one the must-do for you as it has been reportedly asked in numerous interviews by some of the giant organizations like Adobe. Can you solve this problem optimally?
## Problem Statement
Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
Challenge: Can you propose an algorithm with O(log n) runtime complexity?
⚠️Constraints:
1. `1 <= nums.length <= 104`
2. `-104 <= nums[i] <= 104`
3. `nums` contains distinct values sorted in “ascending order”.
4. `-104 <= target <= 104`
## Examples
Let us look at some examples to improve the understanding of the problem:
Now that you have a clear understanding of the problem, let’s dive into various methods to solve the problem:
## Method 1: Linear search
Approach: The most straightforward way to solve the problem would be to iterate through every number in the array. Return the index if the target gets found. Else, check where the target value could be inserted and return that index value.
Algorithm:
1. Check if the array is empty. If yes, return `0`.
2. If the target value is greater than the last element in the array, then the target value would get inserted at the end of the array. Hence, return the length of the array.
3. If the target value is smaller than the first element in the array, the target would be inserted at the beginning of the array. Hence, return `0`.
4. Further, traverse the array. If the current number is greater than or equal to the target value, return the current index.
Solution:
```def search_insert(nums, target):
if not nums:
return 0
if target > nums[-1]:
return len(nums)
if target < nums[0]:
return 0
for i in range(len(nums)):
if nums[i] >= target:
return i```
Test Case Analysis:
Let’s run this solution on our examples:
Yeah! It passed all the test cases.
Complexity Analysis:
• Time Complexity: In the worst-case scenario, you have to visit every number in the array. Hence, the time complexity of this method is O(n).
• Space Complexity: No extra space is used. Hence, the space complexity of this method is O(1).
Discussion: Though this algorithm fetches us the required output, however, it does not ensure that the runtime complexity is log(n) which is also a challenge presented to us. In the next approach, we will find out how to use binary search and reach the optimal solution.
## Method 2: Binary search
Approach: A better approach would be to use binary search as you will be searching for a particular element in the array. You have to initialize two-pointers and calculate the value of `mid`. Compare the mid-value with the target value and return the index if found.
Algorithm:
1. Check if the array is empty. If yes, return `0`.
2. Initialize the variables low and high with `0` and `len(nums)`, respectively.
3. While the “`low`” index is less than “`high`”, calculate the mid-value.
4. Compare the mid-value with the target value.
5. If the target value is greater than the mid-value, then the target value will be to the right. Update `low` to `mid + 1`.
6. Else, if the target value is less than or equal to the mid-value, update `high` to `mid`.
7. When you exit the loop, the position of the `low` pointer is either at the position equal to the target value or at the position where you have to insert the target value. Thus, return the value pointed by `low`.
Consider the following illustration to understand the approach better:
Solution:
```def search_insert(nums, target):
if not nums:
return 0
low, high = 0, len(nums)
while low < high:
mid = (low + high) // 2
if target > nums[mid]:
low = mid + 1
else:
high = mid
return low```
Test Case Analysis:
Let’s run this solution on our examples:
Yeah! It passed all the test cases.
Complexity Analysis:
• Time Complexity: As this method uses binary search, you have to traverse only half the array. Hence, the time complexity of this method is O(log(n)).
• Space Complexity: No extra space is used. Hence, the space complexity of this method is O(1).
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## Bonus Method: Using Bisect Module
Approach: You can use the Bisect module directly to find the position of the target element. The `bisect_left` method of the bisect module is used to find the index of the target element in the sorted array. If the element is already present in the array then the leftmost position where the element can be inserted within the list gets returned.
Bisect Module Recap:
➥ The purpose of the Bisect algorithms is to find the index/position of a required element within a given list where the element has to be inserted within the list. Therefore, it helps to keep the list sorted after the insertion is complete.
`bisect_left` method of the bisect module is used to find the index of the target element in the sorted list. If the element is already present in the list then the leftmost position where the element can be inserted within the list is returned.
Solution:
```from bisect import bisect_left
def search_insert(nums, target):
return bisect_left(nums, target)```
Test Case Analysis:
Let’s run this solution on our examples:
Yeah! It passed all the test cases.
Complexity Analysis:
• Time Complexity: As this method is similar to binary search, the time complexity of this method is O(log(n)).
• Space Complexity: No extra space is used. Hence, the space complexity of this method is O(1).
## Conclusion
I hope you enjoyed this coding interview question. Please stay tuned and subscribe for more interesting coding problems.
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#1
07-17-2012, 05:24 PM
nellie Junior Member Join Date: Jul 2012 Posts: 4
Probability calculation in Q8 and Q10
I correctly approximated the P(f(x) != g(x)) by generating a lot of sample points. However, how do you calculate the probability exactly? Thank you in advance for your help!
#2
07-17-2012, 08:22 PM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by nellie I correctly approximated the P(f(x) != g(x)) by generating a lot of sample points. However, how do you calculate the probability exactly? Thank you in advance for your help!
You can compute the total area of misclassification divided by 4 (the total area of the input space). The area of misclassifcation will be that of points for which and give different values. In one of the theory lectures, there will be a figure depicting this area. It is easier to compute it numerically as you did.
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#3
10-10-2012, 09:19 AM
ketchers Junior Member Join Date: Oct 2012 Posts: 5
Re: Probability calculation in Q8 and Q10
I would be curious on how to calculate - or estimate - the average number of iterations for convergence - as well as the probability - perhaps these are really the same question in disguise. It is a very simple procedure, but there is enough random elements to it so that it leaves the realm of what I am familiar with calculating:
As I understand it ...
1) choose N random points in the unit square
2) choose a random line
This gives the target
3) start with w(0) = [0;0]
4) run algorithm where in computing w(n+1) a random selection from the mis-classified points is used. (perhaps the random choice here is not-relevant)
A pointer to where a similar calculation is made would be fine. Thanks!
#4
10-10-2012, 09:49 AM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by ketchers I would be curious on how to calculate - or estimate - the average number of iterations for convergence - as well as the probability
Calculating these quantities analytically is not tractable. Estimating them using Monte Carlo methods, i.e., by running many random instances of the problem and averaging, is what we are after here. As you point out, there are many sources of randomness and some will result in significant variation. However, repeating the experiment a large number of times will overcome that variance. The numbers given for this problem were chosen to achieve that.
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#5
10-15-2012, 02:59 PM
gah44 Invited Guest Join Date: Jul 2012 Location: Seattle, WA Posts: 153
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by yaser Calculating these quantities analytically is not tractable. Estimating them using Monte Carlo methods, i.e., by running many random instances of the problem and averaging, is what we are after here. As you point out, there are many sources of randomness and some will result in significant variation. However, repeating the experiment a large number of times will overcome that variance. The numbers given for this problem were chosen to achieve that.
It should be possible to give an order of magnitude estimate, though if that was easy then you wouldn't have to write the program to answer the problem.
It looks a little similar to the relaxation algorithms for solving partial differential equations. In that case, the theory is well studied.
Even without knowing that, you could find the scaling law. How the number of iterations changes, on the average, with the number of points. But once you have the MC program, it is easy enough to change the number of points.
#6
10-15-2012, 03:04 PM
gah44 Invited Guest Join Date: Jul 2012 Location: Seattle, WA Posts: 153
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by yaser You can compute the total area of misclassification divided by 4 (the total area of the input space). The area of misclassifcation will be that of points for which and give different values. In one of the theory lectures, there will be a figure depicting this area. It is easier to compute it numerically as you did.
The area is a sum of two figures that are either triangles or quadrilaterals. You have to figure out the intersection of each of the lines with the region boundary, then remember which is in and which is out, and compute the area of the triangle or quadrilateral. Getting all the different possible cases right is what would take time. (I didn't try actually doing it, just figuring out how I might do it.)
#7
04-06-2013, 07:12 PM
mvellon Junior Member Join Date: Apr 2013 Posts: 9
Re: Probability calculation in Q8 and Q10
Your value of g, assuming it doesn't exactly match f, will be a line with a slope and intercept close, though not exactly that of f. Imagine then, two intersecting lines. The area in between the lines correspond to values in X that g misclassifies. The ratio between that area and that bounded by the region [-1,-1]x[1,1] (area 4) will be the probability of error. Alas, calculating the area between the lines is ugly. The problem would have been much simpler if the domain for X had been points inside the circle at (0,0) with radius 1.
#8
09-02-2015, 05:06 AM
henry2015 Member Join Date: Aug 2015 Posts: 31
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by yaser You can compute the total area of misclassification divided by 4 (the total area of the input space). The area of misclassifcation will be that of points for which and give different values. In one of the theory lectures, there will be a figure depicting this area. It is easier to compute it numerically as you did.
Hi, could you elaborate why we need to divide the total area of misclassification by 4?
If the input space is 2D, I can see that there are two triangles formed by f(x) and g(x), and the total area of the two triangles include all the errors. So I thought the P(f(x) != g(x)) = area of error / total area of input space...no?
Thanks in advance.
#9
09-02-2015, 10:22 PM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Probability calculation in Q8 and Q10
Quote:
Originally Posted by henry2015 Hi, could you elaborate why we need to divide the total area of misclassification by 4?
The total area of the input space is 4 and that corresponds to total probability 1, so one needs to normalize areas by 4 in order to get the corresponding probabilities.
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#10
09-02-2015, 10:48 PM
henry2015 Member Join Date: Aug 2015 Posts: 31
Re: Probability calculation in Q8 and Q10
Argh, I see.
Missed the boundaries stated in the problem. My bad.
Thanks.
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The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. Abu-Mostafa, Malik Magdon-Ismail, and Hsuan-Tien Lin, and participants in the Learning From Data MOOC by Yaser S. Abu-Mostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity. | 2,097 | 8,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.914278 |
https://math.stackexchange.com/questions/1896628/properties-of-the-a-transpose-a-matrix | 1,726,558,180,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00161.warc.gz | 346,176,309 | 39,121 | # Properties of the A-transpose-A matrix
I believe that $$A^TA$$ is a key matrix structure because of its connection to variance-covariance matrices.
In Professor Strang's linear algebra lectures, "A-transpose-A" - with this nomenclature, as opposed to $$X'X$$, for example - is the revolving axis.
Yet, it is not easy to find on a quick Google search a list of its properties. I presume that part of the reason may be that they are shared by variance-covariance matrices. But I'd like to confirm this (does it have identical properties to a var-cov matrix?), and have the list easily available from now on here at SE-Mathematics.
Just to not shy away from the initial effort, here is what I think I have so far:
1. Symmetry
2. Positive semidefinite-ness
3. Real and positive eigenvalues
4. The trace is positive (the trace is the sum of eigenvalues)
5. The determinant is positive (the determinant is the product of the eigenvalues)
6. The diagonal entries are all positive
7. Orthogonal eigenvectors (**)
8. Diagonalizable as $$Q\Lambda Q^T$$
9. It is possible to obtain a Cholesky decomposition.
10. Rank of $$A^TA$$ is the same as rank of $$A$$.
11. $$\text{ker}(A^TA)=\text{ker}(A)$$
(**) The eigenvectors of A-transpose-A form the matrix $$V$$ in singular value decomposition (SVD) of $$A,$$ while the square root of the eigenvalues of A-transpose-A are the singular values of the SVD. Similarly, the eigenvectors of A-A-transpose $$AA^\top$$ include the columns in the matrix $$U$$ of the SVD of $$A.$$ The importance of this is exemplified in the fact that SVD can be used to solve least squares regression by computing the Penrose-Moore pseudo-inverse $$A^\dagger = V\Sigma^\dagger U^*,$$ although the QR decomposition is a more expedient computational method.
There is a nice post on the topic here.
• Item 1 and 2 are straight forward. Item is a result due to symmetry. It is usually discussed before or in context of eigenvalue decomposition or Jordan normal form Commented Aug 18, 2016 at 23:22
• It is a key matrix structure because of the role it plays in orthogonal projection. Covariance matrices are just special cases. Commented Aug 18, 2016 at 23:51
• This is known as a Gramian matrix en.wikipedia.org/wiki/Gramian_matrix Commented Aug 19, 2016 at 0:44
• Equivalently, you are asking for a list of properties of symmetric positive semidefinite matrices (because every s.p.s.d. matrix can be written as $A^TA$ for some $A$). A bunch of such properties are listed on the Wikipedia page.
– user856
Commented Aug 19, 2016 at 1:33
• @symplectomorphic I have been intrigued by your comment for a while now, and would like to get some reference summarizing this issue. For instance, it is immediate to think about ordinary linear regression and PCA, both being forms of projection (over column space of the model matrix, or on the eigenvectors of the covariance). However, I am shooting for something more general, more conceptual... Commented Sep 7, 2016 at 21:25
Yes, it has all the properties of a covariance matrix because it is one. You can define a multivariate normal distribution for which $A^T A$ is the covariance matrix.
• You can't literally list all the properties, because there are infinitely many. Of course, many of those are trivial consequences of other properties, others so obscure that nobody would ever care about them. But everything follows pretty much from the orthogonal diagonalization: $A^T A = U^T D U$ where $U$ is a real orthogonal matrix and $D$ is diagonal with nonnegative diagonal elements. Commented Aug 18, 2016 at 23:35
• If $Z$ is a vector whose entries are iid standard normal random variables, $A^T Z$ has mean $0$ and covariance matrix $A^T A$. Commented Oct 2, 2023 at 1:30 | 947 | 3,740 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-38 | latest | en | 0.881474 |
https://www.physicsforums.com/threads/i-with-finding-a-voltage-across-a-capacitor.928587/ | 1,721,826,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00674.warc.gz | 785,769,500 | 17,071 | # I with finding a voltage across a Capacitor
• anonx12
In summary, the conversation is about finding the voltage across a capacitor in a circuit using the Voltage Divider Rule and Thevenin Equivalent. The correct answer is 13.61V, but the person asking for help is getting a different answer. The solution involves calculating the equivalent resistance of the parallel circuits and using the voltage divider rule. However, the person has made a mistake and needs to correct it in order to solve the problem correctly.
anonx12
## Homework Statement
Find the voltage across the capacitor given the data underneath the circuit.
## Homework Equations
Voltage Divider Rule
Thevenin Equivalent
Ohms Law
## The Attempt at a Solution
I have tried using the Voltage divider rule but I can't seem to get the right answer. The correct answer is 13.61V . I calculated the total current and overall resistance, but I don't know how to arrive where they got. Can someone help me?
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There is no current flowing in the R2 + C1 leg of the circuit so what is the voltage drop across R2 ?
Last edited:
I calculated the equivalent resistance of the ones in parallel, excluding R1. That number is 1435 ohms. I then tried to use the voltage divider rule: Vr2= Req/(R8+Req)
I think you missed the point of my question . Please show us your actual workings .
Nidum said:
I think you missed the point of my question . Please show us your actual workings .
The voltage across resistor 2 is the same as the voltage across the capacitor. Since the voltage across parallel circuits is the same, we only need to use the voltage divider rule among Resistor 1 and the resistors in parallel. Doing this however, doesn't result in the correct answer.
My worked solutions is : Vr2= (Req/(Req+R1))(35V) Vr2=5.32V Req= ((1/2)+(1/8)+(1/14))^-1
anonx12 said:
The voltage across resistor 2 is the same as the voltage across the capacitor
That is not correct . If you can work out why then solving the rest of the problem will be easy .
In this simpler problem what is V2 ?
nb : Assume that the circuit is in a settled condition .
## What is a capacitor?
A capacitor is an electronic component that stores and releases electrical energy. It is made up of two conductive plates separated by an insulating material, known as a dielectric. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which allows the capacitor to store energy.
## How does a capacitor work?
When a voltage is applied to a capacitor, it charges up until the potential difference between the plates is equal to the applied voltage. Once this happens, the capacitor stops charging and begins to store the electrical energy in the form of an electric field. When the voltage is removed, the capacitor discharges and releases the stored energy.
## What is the voltage across a capacitor?
The voltage across a capacitor is the potential difference between the two plates. Since a capacitor stores electrical energy, the voltage across it will vary depending on the amount of charge stored. The higher the charge, the higher the voltage will be.
## How do I find the voltage across a capacitor?
The voltage across a capacitor can be found by using the formula V = Q/C, where V is the voltage in volts, Q is the charge in coulombs, and C is the capacitance in farads. This formula shows that the voltage across a capacitor is directly proportional to the charge stored and inversely proportional to the capacitance.
## Why is the voltage across a capacitor important?
The voltage across a capacitor is important because it determines how much energy is stored in the capacitor. This can affect the performance of electronic circuits and devices, as well as the overall efficiency of a system. It is also important to consider the voltage across a capacitor when designing circuits to ensure that the voltage does not exceed the maximum rating of the capacitor.
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# For each order, a certain company charges a delivery fee d
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For each order, a certain company charges a delivery fee d [#permalink]
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For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and
d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500
If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499? (1) The delivery fee for one of the two orders was$3.
Sufficient
Notice that we can have a case when charges for both deliveries are calculated with the second formula, for example x1=$250 and x2=$350, in this case the total price still would be $10. _________________ Moderator Joined: 02 Jul 2012 Posts: 1231 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 117 Kudos [?]: 1394 [0], given: 116 Re: For each order [#permalink] ### Show Tags 10 Dec 2012, 00:01 1) Obviously insufficient. The other order could have any value. 2) We can easily create a situation for which the value is more than$499. So lets try to create a situation for which the value is lesser than $499 to prove insufficiency. We can see that atleast one package has to be more than$100. To minimize the value of this , the first package has to be worth $100. So, for insufficiency, the second package can be at most only$399 and so "d" can only be a maximum of nearly $6. But in such a case, the toatl will not add up to 10. Hence the second package HAS to be more than$399.
Sufficient
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For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and
d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500
If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499? Notice that the highest charge for the delivery with the second formula is for x=500, thus it equals to d=3+(500-100)/100=$7.
(1) The delivery fee for one of the orders was $3. The price of the merchandise for that order is less than or equal to 100, but we know nothing about the second order. Not sufficient. (2) The sum of the delivery fees for the two orders was$10. If the delivery charges for the two orders are $3 and$7, then the price of the second merchandise must be more than or equal to $500, which means that the total price must be greater than$499. Sufficient.
Now, if both delivery charges were calculated with the second formula, then we'd have $$(3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10$$ --> $$x_1+x_2=600>499$$.
So, we have that in both possible cases the total price of the merchandise in the two orders is greater than $499. Sufficient. Answer: B. Hope it's clear. _________________ Director Joined: 29 Nov 2012 Posts: 898 Followers: 14 Kudos [?]: 1050 [0], given: 543 Re: For each order [#permalink] ### Show Tags 18 Jan 2013, 02:51 Bunuel wrote: For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and d = 3, if 0<x<=100 d = 3 + (x-100)/100, if 100<x<=500 d = 7, if x>500 If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than$499?
Notice that the highest charge for the delivery with the second formula is for x=500, thus it equals to d=3+(500-100)/100=$7. (1) The delivery fee for one of the orders was$3. The price of the merchandise for that order is less than or equal to 100, but we know nothing about the second order. Not sufficient.
(2) The sum of the delivery fees for the two orders was $10. If the delivery charges for the two orders are$3 and $7, then the price of the second merchandise must be more than or equal to$500, which means that the total price must be greater than $499. Sufficient. Now, if both delivery charges were calculated with the second formula, then we'd have $$(3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10$$ --> $$x_1+x_2=600>499$$. So, we have that in both possible cases the total price of the merchandise in the two orders is greater than$499. Sufficient.
Hope it's clear.
Thanks this was really helpful, I missed out on the second case. where there can be 2 values within the range of d=3
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26 Mar 2013, 05:36
Bunuel wrote:
For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and
d = 3, if 0<x<=100
d = 3 + (x-100)/100, if 100<x<=500
d = 7, if x>500
If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than $499? Notice that the highest charge for the delivery with the second formula is for x=500, thus it equals to d=3+(500-100)/100=$7.
(1) The delivery fee for one of the orders was $3. The price of the merchandise for that order is less than or equal to 100, but we know nothing about the second order. Not sufficient. (2) The sum of the delivery fees for the two orders was$10. If the delivery charges for the two orders are $3 and$7, then the price of the second merchandise must be more than or equal to $500, which means that the total price must be greater than$499. Sufficient.
Now, if both delivery charges were calculated with the second formula, then we'd have $$(3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10$$ --> $$x_1+x_2=600>499$$.
So, we have that in both possible cases the total price of the merchandise in the two orders is greater than $499. Sufficient. Answer: B. Hope it's clear. Thanks Its really so helpful.I was confused of the statement 2.Now its clear Manager Joined: 12 Dec 2012 Posts: 230 Concentration: Leadership, Marketing GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33 GPA: 2.82 WE: Human Resources (Health Care) Followers: 4 Kudos [?]: 78 [0], given: 181 Re: For each order [#permalink] ### Show Tags 14 May 2013, 10:17 Bunuel wrote: [b]Now, if both delivery charges were calculated with the second formula, then we'd have $$(3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10$$ --> $$x_1+x_2=600>499$$ can you please clarify this part more? thanks in advance _________________ My RC Recipe http://gmatclub.com/forum/the-rc-recipe-149577.html My Problem Takeaway Template http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html Manager Joined: 24 Apr 2013 Posts: 54 Schools: Duke '16 Followers: 0 Kudos [?]: 10 [0], given: 76 Re: For each order, a certain company charges a delivery fee d [#permalink] ### Show Tags 14 May 2013, 14:55 WOW! I didnt even know where to start i was off and in the end i took a guess and went for B Director Joined: 29 Nov 2012 Posts: 898 Followers: 14 Kudos [?]: 1050 [0], given: 543 Re: For each order [#permalink] ### Show Tags 30 Jul 2013, 04:25 TheNona wrote: Bunuel wrote: [b]Now, if both delivery charges were calculated with the second formula, then we'd have $$(3+\frac{x_1-100}{100})+(3+\frac{x_2-100}{100})=10$$ --> $$x_1+x_2=600>499$$ can you please clarify this part more? thanks in advance Cross multiply you will get 300 + x1 - 100 + 300 + x2 - 100 = 1000 After we solve its 600... _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html Manager Joined: 12 Jan 2013 Posts: 244 Followers: 4 Kudos [?]: 69 [0], given: 47 Re: For each order, a certain company charges a delivery fee d [#permalink] ### Show Tags 12 Jan 2014, 05:12 Marcab wrote: For each order, a certain company charges a delivery fee d that depends on the total price x of the merchandise in the order, where d and x are in dollars and d = 3, if 0<x<=100 d = 3 + (x-100)/100, if 100<x<=500 d = 7, if x>500 If George placed two separate orders with the company, was the total price of the merchandise in the two orders greater than$499?
(1) The delivery fee for one of the two orders was $3. (2) The sum of the delivery fees for the two orders was$10.
Right off the bat, you need to know if: 2*x > 499
1) This one tells you that one of the orders is >= 100, it's insufficient because the other order could be any value
2) This tells you that the total delivery was > 500 + (another value between 1 and 100, inclusive), because there are no other combinations of 3 and 7 other than 3 + 7 that yield 10. So this is clearly sufficient.
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# Changes in State for Gases
One very interesting branch of physics is thermodynamics, especially for getting insight in air compressors. In this article we are talking about changes of state for gases, following up on our introduction to thermodynamics.
## The five different changes in state
Changes in state for a gas can be followed from one point to another in a p/V diagram. For real-life cases, three axes for the variables p, V and T are required. With a change in state, we are moved along a 3-dimensional curve on the surface in the p, V and T space. However, to simplify, we usually consider the projection of the curve in one of the three planes. This is usually the p/V plane. Five different changes in state can be considered:
1. Isochoric process (constant volume)
2. Isobaric process (constant pressure)
3. Isothermal process (constant temperature)
4. Isentropic process (without heat exchange with surroundings)
5. Polytropic process (complete heat exchange with the surroundings)
## What is an isochoric process?
Heating a gas in an enclosed container is an example of the isochoric process at constant volume.
## What is an isobaric process?
Heating a gas in a cylinder with a constant load on the piston is an example of the isobaric process at constant pressure.
## What is an isothermal process?
If a gas in a cylinder is compressed isothermally, a quantity of heat equal to the applied work must be gradually removed. This is unpractical, as such a slow process cannot occur.
## What is an isentropic process?
An isentropic process exists if a gas is compressed in a fully-insulated cylinder without any heat exchange with the surroundings. It may also exist if a gas is expanded through a nozzle so quickly that no heat exchange with the surroundings has time to occur.
## What is a polytropic process?
The isothermal process involves full heat exchange with the surroundings and the isotropic process involves no heat exchange whatsoever. In reality, all processes occur somewhere in between these extreme: the polytropic process.
The relation for such a process is:
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To understand the workings of compressed air, a basic introduction to physics can come a long way. Learn more about thermodynamics and how they are vital in understanding how air compressors work. | 828 | 3,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.842954 |
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1. ## u_{xy}+u_{yz}+u_{zx}-u=0
Using separation of variables
$u_{xy}+u_{yz}+u_{zx}-u=0$
$u(x,y,z)=\varphi(x)\psi(y)\omega(z)$
$\varphi'(x)\psi'(y)\omega(z)+\varphi(x)\psi'(y)\om ega'(z)+\varphi'(x)\psi(y)\omega'(z)-\varphi(x)\psi(y)\omega(z)=0$
$\varphi'(x)[\psi'(y)\omega(z)+\psi(y)\omega'(z)]+\varphi(x)[\psi'(y)\omega'(z)-\psi(y)\omega(z)]=0$
$\displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\ psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)$
Now what?
2. Well, pick a constant of separation:
$\displaystyle\frac{\varphi'(x)}{\varphi(x)}=\lambd a=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\ psi'(y)\omega'(z)-\psi(y)\omega(z)}\right).$
The first equality gives you a DE for $\varphi.$ The second equality gives you another equation that, I believe separates out thus:
$\psi'(y)\omega(z)+\psi(y)\omega'(z)=-\lambda(\psi'(y)\omega'(z)-\psi(y)\omega(z)).$
You can do the same sort of trick you just did. That is, solve for $\psi'(y)/\psi(y),$ and choose another separation constant.
You follow?
3. Originally Posted by dwsmith
$u(x,y,z)=\varphi(x)\psi(y)\omega(z)$
How do you know this?
4. $\varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda$
$\displaystyle\lambda=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\ psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)}$
$\displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\ psi(y)\omega'(z)=0$
$\Rightarrow\psi'(y)[\lambda\omega'(z)+\omega(z)}]+\psi(y)[\omega'(z)-\lambda\omega(z)]=0$
$\displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$
$\psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu$
$\displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$
$\omega'(z)-\lambda\omega(z)+\mu\lambda\omega'(z)+\mu\omega(z) =0$
$\displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}$
$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
I tried the substitution of $\displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t}$ in order to manipulate the exponential method to the separations method but it fell short.
Exponential solution:
$\displaystyle u(x,y,z;s,t)=\exp{\left(x\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$
What substitution do I need to make?
Thanks.
5. Originally Posted by chiph588@
How do you know this?
This is from Elementary PDE by Berg and McGregor 1966.
...different technique for finding a particular solutions of homogeneous linear PDE is called the method of separation of variables. The exponential solution $\exp{(rx+sy)}=\exp{(rx)}*\exp{(sy)}$, are products of functions of the separable variables.... We seek a solution of the form $u(x,y)=\varphi(x)\psi(y)$ (pg. 14).
6. Originally Posted by dwsmith
$\varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda$
$\displaystyle\lambda=-\left(\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)}{\ psi'(y)\omega'(z)-\psi(y)\omega(z)}\right)}$
$\displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\ psi(y)\omega'(z)=0$
$\Rightarrow\psi'(y)[\lambda\omega'(z)+\omega(z)}]+\psi(y)[\omega'(z)-\lambda\omega(z)]=0$
$\displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$
$\psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu$
$\displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$
$\omega'(z)-\lambda\omega(z)+\mu\lambda\omega'(z)+\mu\omega(z) =0$
$\displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}$
$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
I tried the substitution of $\displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t}$ in order to manipulate the exponential method to the separations method but it fell short.
Exponential solution:
$\displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$
What substitution do I need to make?
Thanks.
Dear dwsmith,
Think about it like this. You have the two solutions,
$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$--------(Seperation method)
$\displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$---------(Exponential method)
Consider the exponential part with y as the variable. Since the coefficient of y must be equal in both of these equations,
$\mu=s---------(1)$
Now move on to the exponential term with z as the variable. Using the same reasoning,
$t=\dfrac{\lambda-\mu}{1+\mu\lambda}\Rightarrow t=\dfrac{\lambda-s}{1+s\lambda}\Rightarrow \lambda=\dfrac{t+s}{1-ts}--------(2)$
(1) and (2) gives the substitutions necessary to convert the solution obtained from the seperation method to the solution obtained from the exponential method. Hope you understood.
7. By making that substitution, we don't obtain the same solutions.
$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
$\displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}$
$\displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda} \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
Any thoughts?
I could say $\displaystyle \frac{1}{\lambda}=\lambda_2$ but would this be ok?
8. Originally Posted by dwsmith
By making that substitution, we don't obtain the same solutions.
$\displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
$\displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}$
$\displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda} \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$
Any thoughts?
I could say $\displaystyle \frac{1}{\lambda}=\lambda_2$ but would this be ok?
Dear dwsmith,
You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,
$\displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega'(z)-\psi(y)\omega(z)}{\psi'(y)\omega(z)+\psi(y)\omega' (z)}\right)$
You will have to do it all over again.....
9. Originally Posted by Sudharaka
Dear dwsmith,
You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,
$\displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left(\frac{\psi'(y)\omega'(z)-\psi(y)\omega(z)}{\psi'(y)\omega(z)+\psi(y)\omega' (z)}\right)$
You will have to do it all over again.....
Thanks. I am not redoing the problem though haha. | 2,428 | 7,228 | {"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": 50, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.355341 |
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Angle of elevation and depression Warm Up 1. Identify the pairs of alternate interior angles. 2 and 7; 3 and 6 2. Use your calculator to find tan 30 to the nearest hundredth.
0.58 3. Solve hundredth. 1816.36 . Round to the nearest Objective Solve problems involving angles of elevation and angles of depression.
Vocabulary angle of elevation angle of depression An angle of elevation is the angle formed by a horizontal line and a line of sight to a point above the line. In the diagram, 1 is the angle of elevation from the tower T to the plane P. An angle of depression is the angle formed by a
horizontal line and a line of sight to a point below the line. 2 is the angle of depression from the plane to the tower. Since horizontal lines are parallel, 1 2 by the Alternate Interior Angles Theorem. Therefore the angle of elevation from one point is congruent to the angle of depression from the other point. Example 1A: Classifying Angles of Elevation and
Depression Classify each angle as an angle of elevation or an angle of depression. 1 1 is formed by a horizontal line and a line of sight to a point below the line. It is an angle of depression. Example 1B: Classifying Angles of Elevation and
Depression Classify each angle as an angle of elevation or an angle of depression. 4 4 is formed by a horizontal line and a line of sight to a point above the line. It is an angle of elevation. Check It Out! Example 1 Use the diagram above to
classify each angle as an angle of elevation or angle of depression. 1a. 5 5 is formed by a horizontal line and a line of sight to a point below the line. It is an angle of depression. 1b. 6 6 is formed by a horizontal line and a line of sight to a point above the line. It is an angle of elevation.
Example 2: Finding Distance by Using Angle of Elevation The Seattle Space Needle casts a 67-meter shadow. If the angle of elevation from the tip of the shadow to the top of the Space Needle is 70, how tall is the Space Needle? Round to the nearest meter. Draw a sketch to represent the given information. Let A
represent the tip of the shadow, and let B represent the top of the Space Needle. Let y be the height of the Space Needle. Example 2 Continued You are given the side adjacent to A, and y is the side opposite A. So write a tangent ratio. y = 67 tan 70
Multiply both sides by 67. y 184 m Simplify the expression. Check It Out! Example 2 What if? Suppose the plane is at an altitude of 3500 ft and the angle of elevation from the airport to
the plane is 29. What is the horizontal distance between the plane and the airport? Round to the nearest foot. You are given the side opposite A, and x is the side adjacent to A. So write a tangent ratio. Multiply both sides by x and divide by tan 29. x 6314 ft
Simplify the expression. 29 3500 ft Example 3: Finding Distance by Using Angle of Depression An ice climber stands at the edge of a crevasse that is 115 ft wide. The angle of depression from the edge where she stands to the bottom of the opposite side is 52.
How deep is the crevasse at this point? Round to the nearest foot. Example 3 Continued Draw a sketch to represent the given information. Let C represent the ice climber and let B represent the bottom of the opposite side of the
crevasse. Let y be the depth of the crevasse. Example 3 Continued By the Alternate Interior Angles Theorem, mB = 52. Write a tangent ratio. y = 115 tan 52 y 147 ft Multiply both sides by 115.
Simplify the expression. Check It Out! Example 3 What if? Suppose the ranger sees another fire and the angle of depression to the fire is 3. What is the horizontal distance to this fire? Round to the nearest foot. 3 By the Alternate Interior Angles Theorem, mF = 3.
Write a tangent ratio. x 1717 ft Multiply both sides by x and divide by tan 3. Simplify the expression. Example 4: Shipping Application An observer in a lighthouse is 69 ft above the
water. He sights two boats in the water directly in front of him. The angle of depression to the nearest boat is 48. The angle of depression to the other boat is 22. What is the distance between the two boats? Round to the nearest foot. Example 4 Application Step 1 Draw a sketch.
Let L represent the observer in the lighthouse and let A and B represent the two boats. Let x be the distance between the two boats. Example 4 Continued Step 2 Find y.
By the Alternate Interior Angles Theorem, mCAL = 58. In ALC, So . Example 4 Continued Step 3 Find z. By the Alternate Interior Angles Theorem,
mCBL = 22. In BLC, So Example 4 Continued Step 4 Find x. x=zy x 170.8 62.1 109 ft So the two boats are about 109 ft apart.
Check It Out! Example 4 A pilot flying at an altitude of 12,000 ft sights two airports directly in front of him. The angle of depression to one airport is 78, and the angle of depression to the second airport is 19. What is the distance between the two airports? Round to the nearest foot. Check It Out! Example 4 Continued
Step 1 Draw a sketch. Let P represent the pilot and let A and B represent the two airports. Let x be the distance between the two airports. 19 78
12,000 ft 78 19 Check It Out! Example 4 Continued Step 2 Find y. By the Alternate Interior Angles Theorem,
mCAP = 78. In APC, So Check It Out! Example 4 Continued Step 3 Find z. By the Alternate Interior Angles Theorem, mCBP = 19. In BPC, So
Check It Out! Example 4 Continued Step 4 Find x. x=zy x 34,851 2551 32,300 ft So the two airports are about 32,300 ft apart. Lesson Quiz: Part I Classify each angle as an angle of elevation or angle of depression.
1. 6 angle of depression 2. 9 angle of elevation Lesson Quiz: Part II
3. A plane is flying at an altitude of 14,500 ft. The angle of depression from the plane to a control tower is 15. What is the horizontal distance from the plane to the tower? Round to the 54,115 ft nearest foot. 4. A woman is standing 12 ft from a sculpture. The angle of elevation from her eye to the top of the sculpture is 30, and the angle of depression to its base is 22. How tall is the
sculpture to the nearest foot? 12 ft
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$\sin^2 x-\sin^2 y=\cos^2 y-\cos^2 x$
#### Work Step by Step
Start with the left side: $\sin^2 x-\sin^2 y$ Rewrite $\sin^2 x$ as $1-\cos^2 x$ and $\sin^2 y$ as $1-\cos^2 y$: $=(1-\cos^2 x)-(1-\cos^2 y)$ Simplify: $=1-\cos^2 x-1+\cos^2 y$ $=\cos^2y-\cos^2 x$ Since this equals the right side, the identity has been proven.
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1 Stat 510 Notes: Nonparametric Tests and Confidence Intervals Charles J. Geyer April 13, 003 This handout gives a brief introduction to nonparametrics, which is what you do when you don t believe the assumptions for the stuff we ve learned so far. All frequentist procedures come in natural triples a hypothesis test, the confidence interval obtained by inverting the test, and the point estimate obtained by shrinking the confidence level to zero. This is called the Hodges-Lehmann estimator associated with the confidence interval. A familiar example is and and hypothesis test confidence interval point estimate t test t confidence interval sample mean Now we are going to learn about some competing triples hypothesis test confidence interval point estimate hypothesis test confidence interval point estimate sign test associated confidence interval sample median Wilcoxon signed rank test associated confidence interval associated Hodges-Lehmann estimator 1 The Sign Test and Associated Procedures 1.1 Assumptions Suppose we have independent and identically distributed data X 1, X,..., X n from some continuous distribution. That is, our assumptions are independence 1
2 identical distribution continuity There are no other assumptions. In particular, we do not assume any particular shape for the distribution of the data. Any continuous distribution whatsoever will do. The continuity assumption assures that ties are impossible. With probability one we have X i X j when i j. The continuity assumption is necessary for exact hypothesis tests. The continuity assumption is unnecessary for the point estimate and confidence interval. The parameter of interest is the median θ of the distribution of the X i. Because of the continuity assumption, the median is uniquely defined. If F is the distribution function of the distribution of the X i, then there exists exactly one point x such that F (x) = 1/ and that point is the median θ. 1. Hypothesis Test Define Y i = I (0, ) (X i θ). In words, Y i is the indicator of the sign of X i θ. The Y i are i. i. d. Bernoulli(1/) random variables (assuming θ is the median). Hence n T = is Binomial(n, 1/) under the assumption that θ is the median. Thus a sensible test of the hypotheses i=1 Y i H 0 : θ = θ 0 H 1 : θ θ 0 or of the corresponding one-sided hypothesis is based on the distribution of the test statistic T. For example suppose the data are and we wish to test the null hypothesis that θ is zero (this is the usual null hypothesis when the data are differences from a paired comparison). There are 1 positive data points so T = 1. Alternatively, we could consider T = 3 because there are 3 negative data points. It doesn t matter because the Binomial(n, 1/) null distribution of the test statistic is symmetric. The P -value for an upper-tailed test would be > 1 - pbinom(11, 15, 1 / ) [1]
3 because of the symmetry of the null distribution of the test statistic, the P -value of the two-tailed test is just twice that of a one-tailed test > * (1 - pbinom(11, 15, 1 / )) [1] Either test says rejects the null hypothesis at the conventional 0.05 significance level and says the median appears to be not equal to zero. The evidence, however, is not strong. P = is not much below P = There is about one chance in 8 of getting a P -value this big even if the median is zero. The only thing tricky here is why did we look up 11 rather than 1 when T = 1? Well, P (T 1) = 1 P (T 11) because T is discrete. Because the binomial distribution is discrete, you must be careful about discreteness. It s clear this is the right answer, because when we use the other tail of the distribution and avoid the complement rule, we get the same one-tailed P -value in a more obvious way > pbinom(3, 15, 1 / ) [1] and, of course, the same two-tailed P -value when we multiply by two. 1.3 Confidence Interval In order to invert the test to obtain a confidence interval, we need to consider tests of all possible null hypotheses. If the null hypothesis is some θ that is not zero, then the test is based on the signs of the Y i = X i θ. In theory, we need to consider an infinite collection of null hypotheses, a different test for each real number θ. In practice we only need to consider n + 1 null hypotheses, because the sign of X i θ changes only when θ goes past X i. Thus the n data points divide the real line into n + 1 intervals ( semi-infinite exterior intervals and n 1 interior intervals). The result of the test goes from reject to accept or vice versa as the θ specified by the null hypothesis goes past one of the data points, thus the endpoints of the confidence interval (the endpoints of the set of θ that are not rejected) are data points. By symmetry (the symmetry of the binomial null distribution of the test statistic) the endpoints will be the same number in from each end when the data values are put in sorted order This means there are only a finite set of possible confidence intervals ( 7.8, 3.5) ( 6.9,.4) ( 4.7, 0.) (3.7, 15.6) the ends, one in from each end, two in, three in, and so forth. 3
4 The only thing left to do is figure out the confidence levels that go with each such interval. The widest possible interval ( 7.8, 3.5) fails to cover θ if and only if θ is not within the range of the data. The corresponding test statistic for the sign test is T = 0 or T = n (all the X i θ have the same sign if all the X i are on the same side of θ). The probability of this happening under the null hypothesis is P (T = 0) + P (T = n) = P (T = 0). The next widest interval fails to cover θ if and only if θ is not within the range of the data except for the two extremes. The corresponding test statistic for the sign test is T 1 or T n 1. The probability of this happening under the null hypothesis is P (T 1). The general pattern should now be clear. The confidence interval whose endpoints are the data points k in from each end in sorted order has confidence level 1 P (T k). For sample size 15 the possible confidence levels are > 1 - * pbinom(k, 15, 1 / ) [1] No one is much interested in confidence intervals that have 99.9% coverage or 69% or 39%. That means the only interesting intervals are in from each end 99.% confidence interval ( 4.7, 0.) 3 in from each end 96.5% confidence interval (3.7, 15.6) 4 in from each end 88.% confidence interval (6.5, 14.4) You can t have a 95% confidence interval. You must pick one of these. On the other hand, these confidence intervals are exact. They have the exact stated confidence level. nonparametric. They have the stated confidence level under no assumptions other than that the data are i. i. d. 1.4 Point Estimate When the confidence level shrinks to zero, the interval shrinks to the middle value in sorted order (or the two middle values when n is even). This is the sample median, which by convention is defined to be the middle data value in sorted order when n is odd and the average of the two middle data values in sorted order when n is even. This general notion of the point estimator derived by sending the confidence level to zero is called a Hodges-Lehmann estimator. The Hodges-Lehmann estimator associated with the sign test is the sample median. For our example data, the median is Summary Thus the sign test gives us a complete theory about the median. The sample median is the natural point estimator of the median. The confidence interval 4
5 dual to the sign test is the natural confidence interval. And the sign test is the natural way to test hypotheses about the median. This triple of procedures is a complete competitor to the triple based on the mean (sample mean and t test and confidence interval). There s no reason you should only know about the latter. A knowledgeable statistician knows about both and uses whichever is appropriate. The Wilcoxon Signed Rank Test and Associated Procedures One way to think about what the sign test does is to replace the data X i by signs Y i (actually indicator variables for signs). We start with numbers and throw away the sizes of the numbers leaving only the signs. This allows us to get an exact nonparametric tests. But perhaps that throws away too much. Can we throw away less information and still get a nonparametric test? Yes. An idea that enables us to do this (invented in the 1940 s by Wilcoxon) is to replace the data by their signed ranks. Each data point is replaced by the integer giving its position in the sorted order by absolute value but keeping its actual sign. For the example we previously used for the sign test, the data now sorted in order of absolute value are and the corresponding signed ranks are Because we are throwing away less information, keeping some information about size as well as information about sign, we can expect to get more powerful tests, more accurate estimation, and so forth. And generally we do..1 Assumptions Suppose we have independent and identically distributed data X 1, X,..., X n from some symmetric continuous distribution. That is, our assumptions are independence identical distribution continuity symmetry 5
6 There are no other assumptions. In particular, we do not assume any particular shape for the distribution of the data. Any symmetric continuous distribution whatsoever will do. Note that the assumptions are exactly the same as for the sign test with the addition of symmetry. Note also that the assumptions are exactly the same as for the t test except that for the t we must strengthen the assumption of symmetry to an assumption of normality. Thus the assumptions that are different for the three tests are sign test signed rank test t test any continuous distribution any symmetric continuous distribution any normal distribution The other assumptions (i. i. d.) are the same for all three tests (and their associated confidence intervals and Hodges-Lehmann estimators). Clearly the sign test has the weakest assumptions (assumes the least), the t test has the strongest assumptions (assumes the most), and the Wilcoxon signed rank test is in the middle. The continuity assumption assures that tied ranks are impossible. With probability one we have X i X j when i j. In order to have exact tests, this assumption is necessary. The continuity assumption is unnecessary for the point estimate and confidence interval. The parameter of interest is the center of symmetry θ of the distribution of the X i. We know that the center of symmetry is also the median and the mean (if first moments exist). Thus this parameter is sometimes also referred to as the median or mean. How restrictive is an assumption of symmetry? Fairly restrictive. Many data distributions are obviously skewed. For such distributions the Wilcoxon test is bogus, but the sign test is valid. The symmetry assumption is sometimes justified by the following argument. When the data consist of pairs (X i, Y i ), the so-called paired comparison situation, we know the standard trick (already used with the t test) is to form the differences Z i = X i Y i and use the one sample procedure on the Z i. If we assume X i and Y i are independent and assume that X i + θ and Y i are equal in distribution (that is, the distribution of the Y i is the same as the distribution of the X i except shifted by the distance θ), then we can conclude that the distribution of the Z i is symmetric about θ. However, these assumptions about (X i, Y i ) pairs are neither necessary nor sufficient for this conclusion about the Z i. Thus this argument should be taken with a grain of salt.. Hypothesis Test Define Y i = (X i θ). (.1) and let R i denote the signed ranks corresponding to the Y i. Under the null hypothesis that the θ subtracted off in (.1) is the true parameter value, and under the rest of the assumptions made in the preceding section, the distribution of the R i is known. The absolute values of the ranks are just the numbers from 1 to n. (Ties are impossible because of the continuity assumption.) The sign of any R i is equally likely to be plus or minus 6
7 because of the symmetry assumption. Thus any function of the R i has a known distribution under the null hypothesis. The usual test statistic is the sum of the positive ranks T (+) = n R i I (0, ) (R i ) (.) i=1 equivalent test statistics linearly related to (.) so they produce the same P - values are the sum of the negative ranks T ( ) = or the sum of the of the signed ranks n R i I (,0) (R i ) (.3) i=1 T = T (+) T ( ) these statistics are related because the sum of T (+) and T ( ) is just the sum of the numbers from 1 to n Hence and T (+) + T ( ) = T ( ) = n(n + 1) T = T (+) n(n + 1) T (+) n(n + 1) The distributions under the null hypothesis of T (+) and T ( ) are the same, symmetric about n(n + 1)/4 and ranging from 0 to n(n + 1)/. The distribution of T is symmetric about zero and ranges from n(n + 1)/ to +n(n + 1)/. The distribution under the null hypothesis of T (+) or T ( ) is not a brand name distribution (at least not one in the textbook or the handout). It is officially called the null distribution of the Wilcoxon signed rank test statistic and is calculated by the R function psignrank. The whole test is done by an R function wilcox.test. If the data for the example we have been using all along are in a vector x, then > wilcox.test(x) Wilcoxon signed rank test data: x V = 109, p-value = alternative hypothesis: true mu is not equal to 0 does a two-tailed Wilcoxon signed rank test of the null hypothesis H 0 : θ = 0 (just as the printout says). 7
8 .3 Confidence Interval From the description of the Wilcoxon signed rank test it is not clear how to invert the test to get a confidence interval. Fortunately, there is an equivalent description of the Wilcoxon test for which the inversion is very analogous to the inversion of the sign test. For data X 1,..., X n the n(n + 1)/ numbers X i + X j, i j are called the Walsh averages. Let m = n(n + 1)/, and let W (1),..., W (m) denote the Walsh averages in sorted order. Then it is a (non-obvious) fact that the Wilcoxon signed rank test statistic T (+) is equal to the number of positive Walsh averages. Thus an argument exactly analogous to the argument about inverting the sign test says that possible confidence intervals associated with the signed rank test have the form (W (k+1), W (m k) ), that is, they have endpoints obtained by counting k in from each end of the Walsh averages in sorted order. For the data in our running example the Walsh averages are The possible confidence intervals are ( 7.80, 3.50) ( 7.35,.95) ( 6.90,.40) ( 6.5, 1.85) and so forth. The corresponding confidence levels are given by > n <- length(x) > m <- n * (n + 1) / > k <- 1:(m / ) > conf.lev <- 1 - * psignrank(k, n) 8
9 > names(conf.lev) <- k > round(conf.lev[0.80 < conf.lev & conf.lev < 0.995], 4) To get an actual confidence level of 95.1% we use k = 5 > w <- outer(x, x, "+") / > w <- w[lower.tri(w, diag = TRUE)] > w <- sort(w) > sum(w > 0) [1] 109 > > k <- 5 > n <- length(x) > m <- n * (n + 1) / > c(w[k + 1], w[m - k]) [1] > 1 - * psignrank(k, n) [1] So that s it. The 95.% confidence interval associated with the Wilcoxon signed rank test is (3.30, 15.15). This is all a lot of work, but fortunately there is a function in R that does it all for you. > wilcox.test(x, conf.int = TRUE) Wilcoxon signed rank test data: x V = 109, p-value = alternative hypothesis: true mu is not equal to 0 95 percent confidence interval: sample estimates: (pseudo)median 9.65 The only grumble about this function is that it doesn t give the actual confidence level, but the level requested. This matters less for the Wilcoxon than for the sign test because the discreteness is less coarse..4 Point Estimate As the confidence level goes to zero the interval shrinks to the median of the Walsh averages, so that is the Hodges-Lehmann estimator of the center of symmetry associated with the Wilcoxon signed rank test. 9
10 With the code above calculating the Walsh averages already executed > median(w) [1] 9.65 calculates the Hodges-Lehmann estimator. The wilcox.test function also calculates this point estimate, which it calls the (pseudo)median. The reason for that terminology is that the center of symmetry is the median assuming the distribution is symmetric. When the distribution of the X i is not symmetric, the median of the Walsh averages does not estimate the median but the median of the distribution of a typical Walsh average (X i + X j )/ where X i and X j are independent and identically distributed. In order to take care of that case, someone coined the term pseudomedian of a distribution F to mean the median of the distribution of (X 1 + X )/ where X 1 and X are independent random variables with distribution F. But having a name (pseudomedian) doesn t make it an interesting parameter. If you don t believe the distribution is symmetric, you probably don t want to use this estimator..5 Summary The Wilcoxon signed rank test also induces a triple of procedures, with confidence intervals and point estimate based on the Walsh averages. For comparison we give all three triples type level (%) interval P -value estimator sign 96.5 (3.70, 15.60) Wilcoxon 95. (3.30, 15.15) Student t 95 (3.80, 14.76) Robustness So why would one want to use one of these triples rather than another? There are three issues: one, assumptions, already covered, another the subject of this section, another the subject of the following section. The word robustness is purported to be a technical term of statistics, but, unfortunately, it has been used in many different ways by many different researchers. Some of the ways are quite precise, others are extremely vague. Without a nearby precise definition of the term, robust means little more than good (in the opinion of the person using the term). DeGroot and Schervish have a section (their Section 9.7) titled Robust Estimation but nowhere do they say what they mean by robustness more specific than An estimator that performs well for several different types of distributions, even though it may not be the best for any particular type of distribution, is called a robust estimator [their italics]. 10
11 3.1 Breakdown Point This section introduces one very precise robustness concept. The finite sample breakdown point of an estimator ˆθ n = g(x 1,..., x n ) (3.1) is m/n where m is the maximum number of the data points x i1,..., x im that can be changed arbitrarily while the rest remain fixed and the estimator (3.1) remain bounded. The asymptotic breakdown point is the limit of the finite sample breakdown point as n goes to infinity. Usually, the finite sample breakdown point is complicated and the asymptotic breakdown point is simpler. Hence when we say breakdown point without qualification we mean the asymptotic breakdown point. That s the precise technical definition. The sloppy motivating idea is the breakdown point is the fraction of junk data the estimator tolerates. We imagine that some fraction of the data are junk also called outliers or gross errors that may have any values at all. We don t want the estimator to be affected too much by the junk, and that s what the breakdown point concept makes precise. Example 3.1 (Breakdown Point of the Sample Mean). The sample mean is x n = x x n (3.) n if we leave x,..., x n fixed and let x 1, then x n as well. Hence the finite sample breakdown point is 0/n = 0. The m in the definition is zero, that is, no data points can be arbitrarily changed while the estimator (3.) remains bounded. Thus the finite sample and asymptotic breakdown points of the sample mean are both zero. The sample mean tolerates no junk data. It should be used only with perfect data. Example 3. (Breakdown Point of the Sample Median). Here we need to distinguish two cases. For n odd, the sample median is the middle value in sorted order. Then the finite sample breakdown point is (n 1)/(n) because so long as the majority of the data points, that is, (n + 1)/ of them, remain fixed, the other (n 1)/ can be arbitrarily changed and the median will always remain somewhere in the range of the fixed data points. For n even, the sample median is the average of the two middle values in sorted order. Then the finite sample breakdown point is (n/ 1)/n by a similar argument. In either case the asymptotic breakdown point is n/. (See why asymptotic breakdown point is the simpler concept?) The sample median tolerates up to 50% junk data. It is about as robust as an estimator can be (when robustness is measured by breakdown point). Example 3.3 (Breakdown Point of the Sample Trimmed Mean). DeGroot and Schervish define the average of the sample with the k smallest and k largest observations the kth level trimmed mean. This is their eccentric 11
12 terminology. Everyone else calls it a 100(k/n)% trimmed mean, that is, the trimming fraction k/n is expressed as a percent and the word level is not used. Note that a 10% trimmed mean trims 10% of the data at the low end and 10% of the data at the high end (0% all together). It is fairly easy to see that the asymptotic breakdown point of an 100α% trimmed mean is just α. The breakdown point is the trimming fraction. Thus trimmed means can have any desired breakdown point between 0 and 1 /. Note that the ordinary sample mean is a 0% trimmed mean and has breakdown point that is its trimming fraction. Also note that the sample median is a 50% trimmed mean, more precisely 50( n n n )% but the n becomes irrelevant as n goes to infinity, and has breakdown point that is its trimming fraction. Thus sample trimmed means form a continuum with the sample mean and the sample median at the ends. Example 3.4 (Breakdown Point of the Hodges-Lehmann Estimator associated with the Wilcoxon Signed Rank Test). Each Walsh average involves two data points. If we arbitrarily change m data points and leave n m fixed, then we leave fixed (n m)(n m + 1)/ Walsh averages (and mess up the rest). The median of the Walsh averages will stay bounded so long as the fixed Walsh averages are a majority, that is, so long as (n m)(n m + 1) > 1 n(n + 1) If the asymptotic breakdown point is α then m nα for large n and the equation above becomes n (1 α) 1 n So 1 α = 1/ and α = 1 1/ = So this Hodges-Lehmann estimator (median of the Walsh averages) is intermediate in robustness (as measured by breakdown point) between the sample mean and sample median and has about the same robustness as a 30% trimmed mean. 4 Asymptotic Relative Efficiency So if the sample mean is so bad (breakdown point zero, useful only for perfect data), why does anyone ever use it? The answer is efficiency. Generally, the more robust an estimator is, the less efficient it is. There is an unavoidable trade-off between robustness and efficiency. We say an estimator ˆθ n of a parameter θ is consistent and asymptotically normal if n(ˆθn θ) D Normal(0, σ ) for some constant σ which is called the asymptotic variance of the estimator and is not necessarily related to the population variance (all of the estimators discussed in this handout have this property). The asymptotic relative efficiency (ARE) of two such estimators is the ratio of their asymptotic variances. 1
13 The reason why ratio of variances (as opposed to, say, ratio of standard deviations) is interesting is that variances are proportional to costs in the following sense. Suppose two estimators have asymptotic variances σ 1 and σ, and suppose we want to have equal precision in estimation, say we want asymptotic 95% confidence intervals with half-width ɛ, that is, ɛ = 1.96 σ 1 n1 = 1.96 σ n where n 1 and n are the sample sizes for the two estimators. But this implies n 1 n = σ 1 σ = ARE If costs are proportional to sample size (not always true, but a reasonable general assumption), then ARE is proportional to costs. When we say the ARE of two estimators is.0, that means one estimator is twice as efficient as the other, but doesn t make clear which is better. That is unavoidable because which estimator we call estimator 1 and which estimator is arbitrary. Hence it is good practice to also say which is better. Although ARE is the proper measure of efficiency, it is very complicated because ARE depends on the true unknown distribution of the data. In order to see this we need to know more about asymptotics. The asymptotic variance of some of the estimators discussed in this handout are rather complicated. We will just state the results without proof. Consider a distribution symmetric about zero with probability density function f and cumulative density function F such that f is everywhere continuous and f(0) > 0. The asymptotic variance of the sample mean is the population variance σ mean = x f(x) dx. The asymptotic variance of the sample median is σ median = 1 4f(θ). The asymptotic variance of the Hodges-Lehmann estimator associated with the Wilcoxon signed rank test is σh-l 1 = [ ]. 1 f(x) dx The asymptotic variance of the 100α% trimmed mean is [ ] σα F 1 (1 α) = (1 α) x f(x) dx + αf 1 (1 α) 0 The asymptotic variance of the maximum likelihood estimator is inverse Fisher information σmle 1 = ( ) d log f(x) f(x) dx dx 13
14 We throw the latter in because it is the best possible estimator. The MLE has no special robustness or nonparametric properties. A sequence of distributions that go from very heavy to very light tails is the family of Student t distributions, that has the Cauchy distribution at one end (t distribution with one degree of freedom) and the normal distribution at the other (limit of t distributions as the degrees of freedom goes to infinity). For these distributions we have the following ARE for the estimators discussed. The table gives the ARE of each estimator relative to the MLE for the location family generated by the distribution, which is the best estimator. Of course, the MLE is different for each distribution. d. f. 0% 5% 10% 0% 30% 40% 50% H-L The column labeled 0% is the sample mean (the 0% trimmed mean). Note that it is not the best estimator unless the degrees of freedom are at least 5 (actually it isn t even the best estimator there, the 5% trimmed mean being slightly better though the ARE are the same when rounded to three significant figures). For any of the heavier tailed distributions (the rows of the table above 5 degrees of freedom), some other estimator is better. Of course, the fact that all of the numbers in the table are less than (except in the lower left corner) means none of these estimators are as good as the MLE. But the MLE is a parametric estimator, it cannot be calculated without knowing the likelihood, and the likelihood depends on the parametric model. The column labeled 50% is the sample median (the 50% trimmed mean). Note that it is never the best estimator (for any t distribution). Some trimmed mean or the Hodges-Lehmann estimator (the column labeled H-L) is always better. In fact, the 30% trimmed mean is better than the 50% trimmed mean for every t distribution. The Moral of the Story: What estimator is best depends on the true population distribution, which is unknown. So you never know which estimator is best. 14
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null | null | null | null | null | null | Living Reviews in Democracy, Vol 1 (2009)
Williamson and O'Neill: Property-Owning Democracy and the Demands of Justice
Property-Owning Democracy and the Demands of Justice
Thad Williamson
University of Richmond | Center for Security Studies |
Martin O'Neill
University of Manchester | Manchester Centre for Political Theory |
John Rawls is arguably the most important political philosopher of the past century. His theory of justice has set the agenda for debate in mainstream political philosophy for the past forty years, and has had an important influence in economics, law, sociology, and other disciplines. However, despite the importance and popularity of Rawls’s work, there is (rather surprisingly) no clear picture of what a society that met Rawls’s principles of justice would actually look like.
Much of the confusion arises from the frequent description of Rawls as a proponent of a redistributive welfare state regime. While Rawls’s principles of justice do provide philosophical support for the redress of existing inequalities and for the substantial redistribution of resources, it is incorrect to say that he favoured welfare state regimes in anything resembling their current form. In fact, Rawls was a strong critic of what he termed “welfare state capitalism” and an advocate of an institutional alternative which he termed “property-owning democracy.” Discussion of “property-owning democracy” occupied only a very small part of his seminal A Theory of Justice (1971), and was passed over entirely in Political Liberalism (1993). But in his final statement of his view of social justice (Justice as Fairness, 2001) Rawls provided pointed and explicit (albeit rather brief) discussion of the essential contrasts between welfare state capitalism and property-owning democracy, and explained why he believed that the welfare state could not in fact realize his two principles of justice.
Nonetheless, the concept of property-owning democracy is not well understood, and is still only rarely treated as integral to Rawls’s theory of justice. The aim of this review article is threefold. First, we review how Rawls and his leading interpreters have described the concept of property-owning democracy. Second, we examine how the notion of “property-owning democracy” has recently been appropriated by non-Rawlsian political philosophers working in the republican tradition, who have developed arguments from non-Rawlsian premises which also favour the widespread dispersion of property ownership. Third, we briefly review recent work attempting to translate the general notion of a property-owning democracy into concrete institutional and policy proposals that might be adopted by advanced industrialized nations.
A Theory of Justice and Property-Owning Democracy
The hugely ambitious aim of Rawls’s A Theory of Justice is to specify a public understanding of justice appropriate to societies committed to both individual freedom and democratic equality. Rawls develops his theory of justice explicitly in opposition to utilitarianism, understood as a public philosophy which equates both goodness and justice with the maximization of aggregate human welfare. Rawls believed that utilitarianism provided an inadequate philosophical grounding for an array of rights commonly associated with liberal democracies (such as freedom of speech and other civil liberties) and also failed to take individuals sufficiently seriously as important in their own right: a strictly utilitarian understanding of justice, for instance, could not preclude in advance depriving a minority of citizens of their liberties or denying them basic resources in order to advance the interests of the majority. As Rawls famously put it, “utilitarianism does not take seriously the distinction between persons.”1
Rawls also recognized, and was concerned to counteract, the force of traditional objections to the very idea of justice, exemplified by Marx and other skeptics dating back to the character Thrasymachus in Plato’s Republic. In this view, “justice” refers simply to the norms and rules governing a particular society—norms and rules which inevitably have the purpose and effect of justifying the status quo and benefitting the ruling class of a given society. For instance, in Marx’s view, under capitalism, there is no injustice, as such, involved in a laborer selling their labor time to a capitalist, who then exploits the laborer by appropriating the product of that labor and selling it for a profit. Under capitalist conceptions of justice, this is simply a voluntary transaction, even if the labourer’s only other choice was living in extreme penury, or starving to death. On this view, conceptions of justice can only be internal to a given society, and cannot provide an independent standard for judging a society’s institutions. Those who are in charge set the rules and then they also get to call it justice.2
Similarly, our day-to-day judgments about what is just and fair can often be shaped and distorted by our own gender, race, and class position. Highly educated college graduates may be more likely to believe that those who have the best education should be given more money and power. Men may be more likely to believe that the disproportionate numbers of men in positions of power are a result either of men’s inherent superior fitness for such roles or the result of choices made by individual women not to pursue such positions. Middle-class people may be more likely to believe that the poor are largely to blame for their own condition, and managers at capitalist firms may be more likely to believe that they should have the right to issue orders to subordinates.
Rawls’s theory of justice aims both to provide an alternative to utilitarianism and an answer to skeptics who believe that impartial justice is impossible or that justice is at bottom a sort of disingenuous ‘code language’, designed to uphold the status quo. The principal mechanism Rawls invokes to develop his conception of justice is the idea of the Original Position (OP). In the Original Position, independent individuals come together for the purpose of selecting principles of justice that will govern their entire society. In the OP, every individual will be under a “veil of ignorance” with respect to their individual identity: individuals will not have any information about their race, gender, class position, educational attainment, religious beliefs, and so on. They will know that they wish to live a self-directed life and to form and pursue a rational life plan, and they will know that they need certain resources (what Rawls calls “primary goods”) to pursue those plans; they will also be aware of certain basic principles of psychology, sociology, and economics regarding the nature of human societies and how they operate (i.e., the notion that people’s behaviour is influenced by material incentives.) At the outset of A Theory of Justice, Rawls argues that the governing principles that would be chosen in this initial position should be regarded as just.3
Rawls argues that two principles of justice would be selected: roughly speaking, a principle of liberty and a principle of equality. The liberty principle calls for providing each citizen with “a fully adequate scheme of equal liberties, which scheme is compatible with the same scheme of liberties for all.”4 What Rawls has in mind here principally are civil liberties and not (as in libertarian conceptions) the untrammelled right of individuals to profit from property holding or to enter into exchanges of any kind. The equality principle is twofold: all citizens are to have an equal opportunity to aspire to positions, offices and (more generally) social advancement (the “principle of fair equality of opportunity”); and inequalities between citizens are to be limited to those which maximally help the least well off group in society (the “difference principle”). Rawls accords absolute or “lexical” priority to the liberty principle and, within the equality principle, to the guarantee of fair equality of opportunity over the enactment of the difference principle. Thus, subject to the satisfaction of the liberty principle and the other part of the equality principle, political-economic arrangements must be organized so as to maximize the position of the least well off relative to any other possible arrangement.5
The “difference principle” therefore can be seen as having a dual function. On the one hand, it sets a limit (however vaguely) on the scope of acceptable inequalities. On the other hand, given plausible assumptions about the role of incentives in stimulating productivity, it effectively mandates inequalities, so long as such inequalities maximally benefit those at the bottom of society. Rawls, in effect, endorses an affluent society with inequality and a high standard of living for the worst off as superior to a poorer society with little inequality. Here Rawls accepts the standard economist’s view that there is a trade-off between strict equality and efficiency, and that material inequalities provide incentives for spurring the effort of economic producers, potentially to the benefit of all. Notably, Rawls also rejects the notion that inequality in itself is an overriding moral bad;6 what is bad are inequalities which cement the superior position of the most well-off, or which generate social harms, such as the domination of one part of society by another, or the loss of self-respect among the badly off.7
Given this set of principles, the task for Rawls is to specify a political economy that would be consistent with basic individual liberties (such as the liberty to choose one’s employment and important liberties of political participation); that would provide substantially equal opportunities to all citizens; and that would limit runaway inequalities that create permanent classes or that undermine the notion that society is a joint system of cooperation aimed at a common end. By the time of writing A Theory of Justice¸ Rawls had already ruled out centralized state socialism as a plausible vehicle for realizing these principles; state socialism (among its other problems) systematically violated basic liberties (such as freedom of employment and political liberties). Rawls thus assumed that a just society must, in some sense, be a market society. But in A Theory of Justice, (and even more explicitly in Justice as Fairness, to which we will turn later in this discussion) Rawls left open the question as to whether the just society would be either capitalist or socialist in character. A just society based on the private control of capital, however, should take a different form than traditional or “really existing” capitalist societies: instead of the control of capital being highly concentrated among a narrow band of citizens, it should be dispersed as widely as possible. That idea represents the core kernel of “property-owning democracy.”
Rawls on the Institutional Framework of a Just Economy
Rawls did not claim to have worked out the details of a political-economic regime corresponding to the idea of a property-owning democracy; his aim was simply to indicate the general outlines of the sort of political economy that might be fully consistent with the principles of justice as fairness. Moreover, this outline is pitched at the level of ideal-type regime analysis in Rawls’s writings, and it thereby self-consciously passes over detailed questions of “political sociology” regarding how such a regime will function in practice.8
In the broadest possible terms, a property-owning democracy will be a market economy in which holdings of capital are widely dispersed across the population. The view is that fair equality of opportunity and limited inequality can be better achieved through a more broad-based distribution of initial holdings rather than by relying on the mechanism of “after-the-fact” redistributive taxation. A property-owning democracy would be a “regime in which land and capital are widely though not presumably equally held,” in which “[s]ociety is not so divided that one fairly small sector controls the preponderance of productive resources,” and which is able to “prevent concentrations of power detrimental to the fair value of political liberty and fair equality of opportunity.”9
In many respects, the institutional structure Rawls proposes in A Theory of Justice for a property-owning democracy is familiar to citizens living under welfare state capitalism. Rawls assumes that there will be a political constitution providing basic liberties, a public sector that provides public goods (including an educational system that will provide “equal chances of education and culture for persons similarly endowed and motivated”), and a market and price system with a suitable system of regulation. Rawls goes on to specify five separate branches of government oversight, dealing with regulation of markets, macro-economic policy, social transfers (with each citizen guaranteed a social minimum), the distribution of property, and the provision of non-essential public goods. The overall picture is of a mixed economy with a judicious blend of market mechanisms and government oversight, embedded within a system of basic liberties (such as freedom of career choice). 10
What, then, makes property-owning democracy distinct from welfare state capitalism? The distinction is to be found in the relative weight accorded in importance to “after-the-fact” social transfers relative to alterations in the distribution of property in achieving a relatively egalitarian economy. Welfare state capitalism aims at providing an economic baseline as well as certain public goods (education, health care, housing) to all citizens; this is achieved primarily through redistributive taxation (what Rawls terms transfers). Property-owning democracy also aims to provide an economic baseline to the “least well off,” but it has a further goal as well: preventing large concentrations of wealth and dispersing ownership of property as widely as possible. One might say that welfare state capitalism simply wants to provide a social baseline at the bottom, whereas property-owning democracy also wants to put limits on accumulation at the top, thereby narrowing overall inequality from both directions (top and bottom). Moreover, property-owning democracy is also concerned to engage in redistribution in additional dimensions: i.e., not just the redistribution of income characteristic of welfare state capitalism, but also the redistribution of wealth and capital assets (as well as ensuring a more equitable distribution of human capital).11
In terms of how such goals might be realized, Rawls points to inheritance taxes as the best mechanism for distributing property more widely and preventing large estates from being transferred in whole from one generation to another. Here Rawls cites proposals for taxation on intergenerational transfers developed by economist James Meade; persons receiving such transfers would owe progressively higher taxes on these gifts according to how many such gifts they had received over their lifetime. Rawls does not stipulate that each person must receive an inheritance, and rejects the idea that there is an inherent injustice in some persons receiving more gifts than another (so long as this takes place within the framework of an overall system that is just).12 For Rawls, inheritance taxes have a more limited, though vital function: preventing large concentrations of wealth from being transmitted inter-generationally. This aim in turn corresponds to a social ideal in which there is no permanent class of politically privileged holders of wealth and capital sufficiently powerful to extract gains for itself that do not function to benefit the least well off.
Left Critiques of A Theory of Justice
As noted above, Rawls’s remarks describing his favoured socioeconomic regime in A Theory of Justice are rather sparse; Rawls devotes just 9 of the 514 pages of the book to discussion of “property-owning democracy,” and only mentions the term in the main text twice. Perhaps as a result of the lack of detail on this issue, a number of left-wing critics took Rawls to be advocating welfare state policies which would enable capitalist processes to produce as much as wealth as possible, while redistributive processes located in the state assured that the “least well off” received as much in the way of (compensatory) economic resources as economically feasible.13 The notion of a capitalist welfare state that could in fact maximize the position of the least well off immediately struck many critics on the left as implausible. In one of the best-developed early critiques of Rawls, Barry Clark and Herbert Gintis argued that Rawls relied on citizens holding an implausibly expansive sense of social justice, in order to facilitate the redistribution required to “correct” the inequalities generated by capitalism, so as to meet the requirements of the difference principle.14 In a related critique, Gerald Doppelt argued that Rawls failed to appreciate the impact of relative economic position, particularly in the production process, on the generation of self-respect. Consequently, Doppelt suggested that the different ways in which Rawls treated the cases of, on the one hand, civil and personal liberties (which are to be distributed equally, as a matter of assuring the conditions of self-respect for all) and, on the other hand, positions and power in the production process (which can be distributed unequally without undermining fundamental self-respect), was normatively unjustifiable when one considered the effects of inequality on status and self-respect.15 Likewise, David Schweickart argued that the logic of Rawls’s theory of justice should have led him to embrace democratic socialism as the social system most capable of realizing his favoured principles.16
In an important response to these early critiques, Arthur DiQuattro defended Rawls against the charge that he is a supporter of traditional capitalism or of a system of social classes (understood in the Marxist sense of the term). In particular, DiQuattro argued that Rawls did not envisage a society divided between owners and non-owners of capital; in short, Rawls did not endorse capitalism, and did not assume that the allowances made for socioeconomic inequality under the second principle of justice necessitated a capitalist organization of production. In defending Rawls from these challenges from the left, DiQuattro quite properly called attention to the crucial distinction Rawls made between property-owning democracy and capitalism.17
Shortly thereafter, Richard Krouse and Michael McPherson offered the first sustained effort in the literature to engage with what Rawls meant by property-owning democracy.18 Drawing on both Rawls’s writings and James Meade’s efforts to describe a property-owning democracy, Krouse and McPherson show how both a concern for the fair value of the political liberties and the difference principle point in the direction of a regime that broadens property ownership directly, rather than a welfare state dependent on large-scale ex post redistributions to limit inequality. While some redistribution via taxation will be necessary even in a property-owning democracy, the fundamental mechanism for achieving an egalitarian society must be to “[limit] the concentration of property over time.” Krouse and McPherson then went on to pose four critical questions, quoted verbatim below:
1. What institutional means are required to preserve [an] egalitarian distribution [of property] over time (should it at some time be achieved), and indeed can adequate means be described?
2. What would life in a property-owning democracy be like? Would the combination of (relatively) egalitarian property ownership and competitive markets produce a society that was acceptably ‘well-ordered’, harmonious, and stable?
3. Can a theory of justice illuminate the choice between the best private property regime—property-owning democracy—and the best socialist arrangements for providing justice?
4. How can this characterization of the ideal property-owning democracy help to guide the process of reform in existing, nonideal, private property societies?19
Justice as Fairness on Property-Owning Democracy
Well-developed answers to each of the questions posed by Krouse and McPherson are still lacking in the literature.20 Indeed, in the 1990s, most critical debate about Rawls’s system of justice followed the agenda set by his own Political Liberalism (1993), examining the question of whether a liberal egalitarian account of justice should aspire to being “comprehensive” as opposed to merely “political.” Rawls argued that liberal principles of justice can in fact be endorsed by persons with widely varying comprehensive religious and philosophical doctrines, and with varying conceptions of the good life, and need not (and, indeed, must not) involve one dominant social group imposing its own particular value commitments or comprehensive philosophical doctrine on others.
Accordingly, attention to questions of distributive justice and the idea of property-owning democracy faded to the background of the debate about political liberalism. Indeed, some observers have noted an internal connection between Rawls’s argumentation for political liberalism and the reduced prominence of distributive justice in his writings: a generation of debate among political theorists in the wake of A Theory of Justice, as well as the strong rightward turn in politics in both the United States and the UK in the 1980s, made it abundantly clear that the stringently egalitarian requirements of the difference principle were unlikely ever to command universal assent among philosophers, let alone among the broader public.21 That political fact in turn calls into doubt the broader project of developing principles of justice that both have “real teeth” and that could also be widely accepted within highly diverse modern societies.
Rawls himself at times seemed to downgrade the standing of the difference principle within his account of justice in Political Liberalism.22 For instance, Rawls argued that firm principles of distributive justice need not (and ought not) be written into the political constitutions of just societies, and that application of distributive principles should be left to legislators. This aspect of Rawls’s view can be explained by his understandable reluctance to see complex social policy questions settled in the courts, but nevertheless his proposed solution could be seen as giving the realization of distributive justice a status that is contingent on the decisions and preferences of legislators (who Rawls assumes will accept and seek to implement the difference principle).23 Given that really-existing democratic politics is rarely, if ever, characterized by consensus on fundamental principles of justice, especially in the context of countries like the United States, leaving the difference principle’s fate in the hands of democratic politics has struck some commentators as tantamount to abandoning it.24 That note of ambiguity in turn signalled a broader tension within Rawls’s theory: whether Rawls intended his theory of justice to reflect the self-understanding of existing democratic societies (an interpretation lent support by his engagement with the tradition of American constitutional law in Political Liberalism) or whether he intended the theory to be critical of existing institutional practices as well as inegalitarian social views. One can also see this tension as embodying a broader tension between some of the different roles that Rawls identifies for political philosophy: for example, between the fundamentally progressive enterprise of identifying a “realistic utopia”, as against the less radical, Hegelian task of offering a “reconciliation” to our existing social world.25
It is thus a striking fact than in his final sustained statement about justice, Justice as Fairness: A Restatement (2001), Rawls ‘lays down his cards’ so to speak: more than in any previous book, he makes it clear that he believes that contemporary capitalist societies, especially the United States, have veered far away from realizing liberal principles of justice. It is here as well that we find the most detailed contrast between, on the one hand, welfare state capitalism, which Rawls rejects, and, on the other hand, property-owning democracy and liberal socialism, both of which he is prepared to endorse. Rawls argues that either property-owning democracy or liberal socialism could in theory realize principles of justice, and argues that the choice between the two should be made on the basis of contingent historical and cultural factors. The implication seems here to be that in societies like the United States, with weak socialist traditions and a strong cultural emphasis on entrepreneurial individualism, property-owning democracy is the more likely vehicle for realizing the just society (with liberal socialism perhaps a more suitable option in societies with more collectivist political cultures or stronger socialist traditions).
In an equally striking move, Rawls in Justice as Fairness (approvingly citing Krouse and McPherson) rests the argument for property-owning democracy not primarily in terms of the demands of the second principle of justice, but rather in terms of the first principle. Rawls argues that the widespread political inequalities generated by welfare state capitalism represent a systemic violation of the “fair values of the political liberties.” Unlike other liberties, the fair value of the political liberties must be distributed equally; a society in which this is not the case cannot be considered to be either self-governing or free.26 This move is important both for its own sake and because it means that Rawls’s arguments for property-owning democracy are thereby not solely contingent on acceptance of the controversial difference principle. Nonetheless, as well as its falling short with regard to the first principle, Rawls also makes clear that a predictable consequence of the concentration of wealth and political power characteristic of welfare state capitalism is that such polities rarely if ever are able to enact redistributive policies sufficiently strong to establish and maintain intergenerational equality of opportunity, or to limit objectionable inequalities which serve no social purpose other than the enrichment of the already privileged.
According to Justice as Fairness, one of the main aims of property-owning democracy is “to prevent a small part of society from controlling the economy, and indirectly, political life as well…
Rawls goes onto describe POD as a socioeconomic system with at least the three following institutional features:
1. Wide Dispersal of Capital: The sine qua non of a POD is that it would entail the wide dispersal of the ownership of the means of production, with individual citizens controlling productive capital, both in terms of human and non-human capital (and perhaps with an opportunity to control their own working conditions).
2. Blocking the Intergenerational Transmission of Advantage: A POD would also involve the enactment of significant estate, inheritance and gift taxes, acting to limit the largest inequalities of wealth, especially from one generation to the next.
3. Safeguards against the Corruption of Politics: A POD would seek to limit the effects of private and corporate wealth on politics, through campaign finance reform, public funding of political parties, public provision of forums for political debate, and other measures to block the influence of wealth on politics (perhaps including publicly funded elections)
Policies of type (3) should be viewed as being in place with an eye on the protection of the fair value of the political liberties, and are therefore closely connected with creating a regime that is in accord with Rawls’s first principle of justice. Policies of type (1) and (2) should, in contrast, be viewed as providing the means for institutionalizing the demands of Rawls’s second principle of justice. Through a combination of all three kinds of policies, Rawls aims to specify a social system that has the capacity to overcome the structural limitations of welfare state capitalism in delivering a fully just set of socioeconomic arrangements.
Recent Commentary on Rawls’s Conception of Property-Owning Democracy
Spurred on in part by the striking arguments of Justice as Fairness, property-owning democracy has received increasing attention in recent years from liberal egalitarian political philosophers. (To be sure, in many accounts of Rawls’s social and political thought, property-owning democracy is discussed only briefly, if it is mentioned at all.) 27
Discussions of the dilemmas of contemporary liberal egalitarian politics offered by Simone Chambers and Will Kymlicka each stress Rawls’s critique of the welfare state and the implicit radicalism of property-owning democracy.28 In recent papers, both Ben Jackson and Amit Ron trace the intellectual origins of “property-owning democracy” phrase to the early 20th-century British conservative Noel Skelton; Jackson suggests that James Meade’s use of the term (subsequently picked up by Rawls) was a deliberately ironic attempt to invert the meaning of what hitherto had been a conservative idiom.29 Samuel Freeman argues that Rawls’s preference for property-owning democracy vis-à-vis the welfare state parallels his preference for the “liberalism of freedom” of Kant and J.S. Mill, in which citizens take an active role in developing their capacities, as opposed to the “liberalism of happiness” associated with classical utilitarians such as Bentham.30 Freeman’s extremely comprehensive treatment of the full range of Rawls’s thought also contains a relatively extended account of the distinction between welfare state capitalism and property-owning democracy.31 In a related vein, Nien-hê Hsieh draws on Rawls to argue the case for what he terms “workplace republicanism,” i.e., the introduction of workplace democracy and limitation of arbitrary managerial authority; in Justice as Fairness, Rawls forwarded some suggestive though noncommittal comments about the potential importance of workplace democracy in helping to realize a just society.32 This aspect of Rawls’s thought has also been picked up by Martin O’Neill, who explores, in a recent article, what he terms “three Rawlsian routes” for defending some form of economic democracy as a precondition for a just socioeconomic order.33
Most recently, papers by Hsieh, O’Neill, Waheed Hussain, and Thad Williamson have further developed both the basic idea of property-owning democracy and have also subjected the concept to critical scrutiny. Hsieh focuses on the role of work in a property-owning democracy; Hussain compares property-owning democracy to what he terms democratic corporatism; O’Neill offers a partial critique of Rawls’s arguments against the welfare state; and Williamson describes how a wide dispersal of real estate, cash, and capital might be actually institutionalized and sustained in a property-owned democracy.34 A forthcoming volume on property-owning democracy, edited by O’Neill and Williamson, will take the arguments of a number of these philosophers forward, as well as include the work of a number of other writers, each giving further elaborations and critiques of Rawls’s ideas regarding the institutional basis of a just social order.35
Non-Rawlsian Arguments for Property-Owning Democracy
Property-Owning Democracy and Market Socialism
The idea of a market economy based on a wider dispersal of capital than is characteristic of contemporary capitalist societies is not unique to Rawls or to the debate he stimulated; nor is the general search for a plausible alternative to capitalism in light of the historic failure of centralized state socialism. Indeed, since the late 1980s, political economists and philosophers have detailed a variety of proposals for a market socialist society, proposals that typically offer far more specificity and attention to institutional detail than the general comments about property-owning democracy offered by Rawls. Particularly cogent formulations are those of Joshua Cohen, David Miller, John Roemer, David Schweickart, and Gar Alperovitz.36 Typically, these proposals call for some form of community or public ownership of capital within a market model, while also allowing for broadly democratic planning of the economy as a whole. In most cases (Roemer is an exception), these proposals also call for giving workers effective democratic control of most or all enterprises. Notably, these models do not require or advocate political revolution, but assume the constitutional framework of liberal democracy; nor do they challenge the market as a mechanism of resource allocation, even when the models allow for significant degrees of government planning. What these versions of liberal democratic socialism do insist upon is changing who owns and reaps dividends from capital. To this extent, proposals for liberal democratic socialism bear a significant resemblance to Rawls’s conception of a property-owning democracy.37
Republicanism and Property-Owning Democracy
Also of relevance are recent discussions by republican political theorists concerning the content of a republican political economy (or “commercial republic”) which also point in the broad direction of dispersing capital more widely.38 The proposals of American political theorist Stephen Elkin are particular noteworthy, for two reasons: first, he specifically rejects the Rawlsian paradigm for thinking about politics and instead takes James Madison as his starting point for reasoning about the content of a commercial republic; second, he explicitly uses the language of property-owning democracy. Consequently, the following section of our discussion pays particular attention to Elkin’s non-Rawlsian arguments for a POD.
The core premise of republican approaches to political theory is that in thinking about politics, it is not enough only to specify the moral foundations of legitimate government, or the normative principles (including principles of social justice) to which government should aspire. Rather, we must think about how to construct and maintain a regime that, despite the presence of at least partially self-interested actors, succeeds in preventing the domination of any one group of citizens by any others, via either private or public means, while also allowing all citizens meaningfully to contribute, via the political process, to influencing the social conditions which shape their lives. Constructing such a regime requires paying careful attention to institutional design; to how leaders are selected and to the incentives that they are offered; to the character and engagement of the ordinary citizens who are charged with both selecting leaders and holding them accountable; and, not least, to the regime’s political economy and how it functions, including the distribution of wealth that it generates.
Civic republicans, especially those who draw their inspiration from Madison and other regime theorists such as Machiavelli and Montesquieu, characteristically argue that reasoning from the original position, in the Rawlsian style, can take us only so far in telling us what a workable and tolerably just political regime might look like. More than this, such republicans reject proposals for a “division of labour” within political theory, such that some scholars, expert in pre-institutional political philosophy and abstract reasoning, define and specify the normative principles which should guide political life, while a second group of scholars, who look more carefully at the facts of the world, work out how to put those principles into practice. From the viewpoint of republicans such as Elkin, such a division of labour is untenable: Elkin argues that we cannot fully make sense of political concepts such as “liberty” and “equality” until we have thought through, and indeed garnered some practical experience with, what it would mean to realize such values in practice, through real political institutions. Put another way, we cannot claim that we want something unless we understand what it would truly to take to get it, in practice, given reasonable assumptions about human nature.39 One of those assumptions, in turn, is that political actors often act from mixed or self-interested motives, as opposed to being motivated by the desire to realize justice.40
This approach to politics is exemplified in Stephen Elkin’s recent book Reconstructing the Commercial Republic. Elkin describes the “circumstances of politics” as involving a “large aggregation of people who 1) have conflicting purposes that engender more or less serious conflict; 2) are given to attempt to use political power to further their own purposes and those of people with whom they identify; 3) are inclined to use political power to subordinate others; and 4) are sometimes given to words and actions that suggest that they value limiting the use of political power by law and harnessing it to public purposes.” “These circumstances,” Elkin adds, are not “`the best of foreseeable conditions.’ They are simply the conditions that obtain as we Americans, like others, go about our political business.”41
Elkin argues that James Madison’s theory of a “commercial republic”—a liberal regime characterized by government that is at once popular, limited, and active—has six central elements. The first five include preventing faction, preventing a tyranny of lawmakers, ensuring that lawmakers consider the public interest, ensuring that lawmaking has a meaningful deliberative component, and ensuring a measure of civic virtue in the populace. For purposes of the present discussion, the key element is the sixth—namely, the “social basis for the regime,” or in other words, who has property and thereby political influence in the society, and thereby the capacity to shape how the regime operates in practice.42
Madison’s political theory rested heavily on the possibility of “men of property and substantial community position”—most often large landowners—coming to take a very broad view of their own interests. Suppose the public interest and the (enlightened) self-interest of these men overlapped substantially, and that these same “men of property” were in a position to have disproportionate political influence—for instance, by being the predominant class from which elected representatives would be chosen. If this were the case, and if the political institutions themselves were designed to give lawmakers, including the ambitious, strong reason to appeal to the public interest, then a deliberative politics that in fact served the public interest might be possible.43
As Elkin notes, this Madisonian account is a deeply unsatisfactory theory for contemporary liberal regimes. Broad-minded “men of landed property and standing” are no longer the dominant social class; instead we have the predominance of corporate property, and corporations are themselves legally required to have quite narrow interests. Moreover, if inequalities of wealth and income become excessive, and these translate into significant political inequalities, as they in fact often do, then we face the spectre of not, as Madison feared, factional majority rule, but rather factional minority rule by the wealthy and the well-off. So the problem remains—what is to be the social basis of a political regime based on self-rule and limited but active government?44
Elkin’s answer is fourfold. First, a commercial republic should be a regime in which the middle class is the politically predominant class, and can serve as a “pivot” in adjudicating conflicts between business elites and the poor; in particular, it is important that the middle class have enough power to force the business elites to justify their proposals in terms of the public interest. Second, excessive inequalities of wealth are inconsistent with the maintenance of a commercial republic, precisely because they translate into inequalities of political influence and make relations of mutual respect between all citizens impossible. Third, persistent poverty as well the economic insecurity of the near-poor and much of the middle class are inconsistent with the formation of independent, self-respecting citizens who recognize the value of deliberative politics. As Elkin puts it, “To worry about whether you can pay your bills wonderfully concentrates your mind—but not on political life.”45 Fourth, the interests of property-holders should be broadened to the greatest possible extent; this could be achieved by broadening the ownership of property and capital, and by fostering a politically strong middle class capable of challenging elite proposals, such that elite groups must argue on their behalf by appealing to the public interest.
Consequently, Elkin writes, “in a fully realized commercial republic, the fruits of prosperity should not be available only to a few; neither should economic production be in the service of creating an oligarchy with the status and material comforts of an aristocracy.”46 Elkin thus proposes ensuring that work is better paid, as well as advocating the more effective use of inheritance taxes, and above all widening the distribution of capital—in short, much of the agenda of what Rawls terms “property-owning democracy.”47
This focus on capital as opposed to income as the focus of redistributive efforts has in turn four further justifications: first, the link between accumulated capital and disproportionate, dangerous and factional political power; second, the observation that private relationships of domination rest on the divide between those with capital and those without, not the divide between higher-paid and lower-paid workers; third, the recognition that it is impossible to contain inequalities of income without also paying attention to inequalities in asset holdings; and fourth, the political judgment that it is all but impossible as a practical matter to allow the market to generate wide dispersions of rewards, and then to rely upon the tax system to correct the resultant inequalities to a tolerable level. The beneficiaries of socioeconomic inequality are not, and never plausibly will be, so committed to social justice that they will endorse large-scale redistributions of their own incomes on a regular basis. On the contrary, they will insist on the justice of keeping their own market-generated returns, a claim that will have strong resonance among the well-off. A regime that relies less on post-transfer taxation to achieve a tolerable measure of socio-economic equality will be more stable over the long term.48
Here we have a second set of arguments for moving in the direction of property-owning democracy. Clearly, there is overlap between some of Rawls’s reasoning and that of Elkin, particularly in Rawls’s stress on the fair value of political liberties. What is worth noting here is simply that one need not share any commitment to the Rawlsian edifice, or to a mode of political thinking that uses the ideal conditions of the original position as a starting point for reflection, to reach the judgment that persons interested in tolerably just liberal regimes should take an interest in property-owning democracy. In short, property-owning democracy need not be regarded as a uniquely Rawlsian idea, but instead may plausibly be endorsed by adherents of a range of sophisticated theories concerning the just society.
Policies to Broaden Property Ownership
The idea of property-owning democracy has thus enjoyed renewed attention from multiple strands of democratic political theory in recent years. Parallel to this development has been increased interest among policy scholars and some practitioners in “asset-based” policy approaches to redressing poverty. The basic thought behind asset-based approaches is that social policies should not rely only on efforts to prop up low incomes amongst the poor, but should also enable the disadvantaged to gain access to productive assets that might significantly improve their long-term life chances. Examples of such assets including savings accounts, educational funds, housing, pension funds and automobiles.49 To take one prominent example of an asset-based policy, governments might establish savings funds at birth for each child and capitalize each account with (for example) $1,000, in expectation that the value of the fund will steadily grow over time and represent a significant source of funds by early adulthood; a version of this policy (the Child Trust Fund) has been implemented in the UK. Ackerman and Alstott have offered a much more ambitious proposal in the context of the United States, calling on government to provide all citizens at age 18 with an $80,000 “stake,” on the view that having access to such significant funds would dramatically alter the life prospects and plans of many disadvantaged, working-class, and even middle-class young Americans.50
Most of the mainstream discussion of asset-based policies focuses on individual accounts and on bolstering access to cash savings or housing. Advocates of property-owning democracy should also naturally have an interest in ways of broadening ownership of productive capital. Government support (in the form of loans, technical assistance, and in some cases capital investments) for smaller businesses represents one traditional policy approach; another possibility is the provision of funds or incentives to allow individuals to buy corporate stock. More far-reaching are efforts to turn control over entire enterprises to workers or to local neighbourhood organizations. As Gar Alperovitz has documented, both employee ownership and community ownership (through vehicles such as community development corporations) of productive businesses have increased dramatically in the United States since the 1970s.51 Both approaches broaden the ownership of capital in ways consistent with property-owning democracy, as well as offering other possible benefits in addition (such as sustaining jobs in poorer or declining localities).
In short, there are numerous practical mechanisms available to policymakers to attempt to broaden access to capital; most of these mechanisms are potentially politically popular and capable of winning support from a range of ideological positions. The literature lacks, however, a sustained treatment of how such policies might be broadened and ratcheted up to scale in a manner which might realize the aims of property-owning democracy. Ackerman and Alstott’s stakeholder society proposal and Alperovitz’s arguments on behalf of a “pluralist commonwealth” come the closest in this regard, though neither proposal explicitly uses the language of “property-owning democracy.” In short, there is ample room for further work in translating the very ambitious aims of Rawlsian (or alternative republican) conceptions of property-owning democracy into concrete political proposals.
Similarly, there has been almost no serious discussion of the politics of property-owning democracy or of the question of whether and how existing “welfare state capitalist” societies might be changed into a form of market society more closely approximating property-owning democracy. Two observations are in order here: first, the social basis for a movement towards property-owning democracy is likely to be quite different than traditional left coalitions on behalf of socialism and social democracy; the aims of property-owning democracy are quite different in emphasis (though arguably not inconsistent with) the traditional goals of labour and labour organizations. Indeed, the more entrepreneurial, individualistic tenor of property-owning democracy coheres with Roberto Unger’s call for the “left” to cast its lot with the “petty bourgeoisie” rather than declining industrial working classes.52 In practice, a workable politics of property-owning democracy would need to be blended with other policies and initiatives more oriented towards traditional labour concerns about employment stability, wage levels, and labour law.
Second, while proposals to create Child Trust Funds and broaden access to homeownership are generally popular, the most crucial step towards a Rawlsian conception of property-owning democracy—more aggressive, stringent taxation on inheritance and estates—is likely to inspire much more opposition. This is especially true in the United States, where a concerted effort by conservative ideologues over the past fifteen years has succeeded in persuading many Americans and lawmakers that inheritance taxes represent an unjust “death tax”53; there is also substantial opposition to inheritance tax in countries such as the UK.54 Serious arguments for full-blown property-owning democracy thus must be combined with serious and persuasive arguments regarding the legitimacy of breaking up large estates through inheritance taxes and other forms of wealth taxation.55 The argument for the justice of taxing large-scale wealth in order to secure the fair value of political liberties, institute meaningful equal opportunity, and improve the lot of the least well off in turn mirrors the larger Rawlsian argument for understanding society as a system of social cooperation aimed at realizing a common life characterized by fairness, as opposed to a game in which the aim is to accumulate as many assets as possible within the permissible rules. The argument for breaking up large inheritances and the argument for viewing society as a fair system of cooperation are inextricably tied together. Without the prior commitment to viewing society as a fair system of social cooperation, arguments for breaking up large inheritances are dramatically weakened; without the political capacity to break up large accumulations of wealth in practice, Rawlsian aspirations for realizing a just society based on the two principles of justice will remain tantalizingly out of reach.
Ackerman, Bruce, and Anne Alstott.1999. The Stakeholder Society. New Haven: Yale University Press.
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1 Rawls (1971: 24). Page citations to A Theory of Justice in this article refer to the 1999 revised edition.
2 Whether Marx was as hostile to the idea of justice as some of his more dismissive comments suggest is a disputed question, and one Rawls examines at some length in his Lectures on the History of Political Philosophy. Rawls takes the view (drawing on the work of G.A.Cohen) that Marx does have a normative conception of justice underlying his analysis of capitalism, albeit one that is not explicitly expressed. See Rawls (2007): 335-371. See also G.A. Cohen (1989).
3 Rawls (1971): 10-19. It is important to recognize, however, that Rawls does not regard the OP as static. If it can be shown that in the OP the agents will reach principles of justice conflicting with our considered judgments, then the description of the OP is to be revised so as to yield a different result. In effect, the OP functions as a mechanism for testing intuitions about just principles. For useful discussion, see Kymlicka (2002): 63-70.
4 Rawls (2001): 42. We quote here from the revised statement of the two principles presented in Justice as Fairness.
5 As Samuel Freeman puts it, the difference principle calls for selecting that political-economic system which tends over time to maximize the position of the least well off; and it calls for maximizing the actual position of the least well off within that chosen system. So the difference principle regulates both the broad choice of institutional arrangements and the selection of specific policies (i.e., taxes, transfers, labor laws) within a given arrangement. Importantly, Rawls does not equate maximizing the position of the least well-off with maximizing their incomes and wealth, but rather maximizing an index of the broader bundle of primary goods that affect one’s sense of self-respect and overall life chances. See Freeman (2007a): 102-109 and Freeman (2007b): 111-115.
6 Derek Parfit (1991) gives the label “Telic egalitarianism” to the view that inequality is in itself bad. On the rejection of “Telic egalitarianism” and for discussion of Rawls’s views regarding the badness of inequality, see Martin O’Neill, “What Should Egalitarians Believe?” (2008a). G.A. Cohen criticizes Rawls’s theory of justice on a closely related issue, regarding Rawls’s attitude to inequality (Cohen, 2008). In Cohen’s view, the inequalities permitted by the difference principle may be sensible, but should not be regarded as just. Cohen argues in effect that Rawls wrongly elides justice as such with more pragmatic concerns in developing his account of social justice. Even if Cohen’s critique is accepted, that does not make Rawls’s ideas about property-owning democracy any less interesting or important; it simply means (to take Cohen’s view) that we should regard it as an effort to specify what a real-world political economy that balanced justice against other important considerations looks like.
7 See Rawls (2001), pp. 130-2. See also Scanlon, (1996); O’Neill (2008a).
8 The most sustained discussion of property-owning democracy offered by Rawls can found in Chapter V, of A Theory of Justice (1971), especially section 43, and again in the Preface to the French edition of A Theory of Justice (reproduced as the preface to the revised edition of TJ, see especially at pp. xiv-xvi). The most systematic discussion comes in pages 135-140 of Justice as Fairness: A Restatement (2001). (The idea is given no attention at all in Political Liberalism [1993].) Given this paucity of discussion in Rawls’s formal published writings, the discussion here is particularly informed by four further sources. These include a pair of articles in 1986 and 1987 by Krouse and McPherson that pay attention to Rawls’s notion of a POD and try to draw out some of its implications (Krouse and McPherson, 1986, 1987); various publications by the British economist James Meade in the 1960s and 1970s describing a “property-owning democracy”, from which Rawls explicitly takes the name for his preferred socioeconomic regime (Meade, 1965a, 1965b, 1975, 1993); work by one of Rawls’s prominent students, Joshua Cohen, specifying the problematic relationship between capitalism and democracy (especially Cohen, 1989); and finally, Rawls’s own recently published lectures on Karl Marx highlighting the contrast between Rawls’ conception of a just society organized as a property-owning democracy and Marx’s ideal of communism (Rawls, 2007).
9 Rawls (1971): 247, 245.
10 Rawls (1971): 242-251; quotation on education at 243.
11 As subsequent commentators have noted, such equalization of wealth and assets also has implications for how work is to be organized. See Hsieh (2009).
12 Rawls (1971): 245.
13 In Justice as Fairness, Rawls admits that the distinction between welfare state capitalism and property-owning democracy “is not sufficiently noted in Theory.” Rawls (2001): 135 n2.
14 Clark and Gintis (1978)
15 Doppelt (1981)
16 Schweickart (1979)
17 DiQuattro (1983)
18 Krouse and McPherson (1986, 1987)
19 Krouse and McPherson (1988): 99-100.
20 A forthcoming volume edited by O’Neill and Williamson (2010) is intended to redress that gap.
21 Chambers (2006)
22 Thus Simone Chambers argues that while “the difference principle lives on as Rawls’s favored interpretation of economic justice,” in Political Liberalism he no longer insists that “it is the only possible candidate for a fair principle.” Further, Chambers observes, “In addition to demoting its status within the theory, there is a more subtle fading away of the topic. Social justice is no longer front and center. His growing concern to find a view of justice that would be compatible with pluralism came to overshadow his deep commitment to egalitarianism.” Chambers (2006): 86.
23 Rawls (1993). Importantly, however, Rawls does treat provision of a social minimum “providing for the basic needs of all citizens” as a constitutional essential; see Rawls (2003), pp. 228-29.
24 Thus see Thomas (2009), who argues for constitutionally guaranteeing the fair value of the political liberties. Such a constitutional guarantee, in his view, would preclude democratic politics from sanctioning excessive inequalities, while avoiding the need to lock in place any particular institutional scheme designed to realize property-owning democracy. See also J. Cohen (2003) for a critique of Rawls’s supposition that consensus can be reached on the content of justice in actual democratic polities.
25 See Rawls, (2001), pp. 1-5.
26 In A Theory of Justice Rawls also connected arguments for inheritance taxes to the requirement of maintaining the fair value of political liberties, as well for realizing fair equality of opportunity. There Rawls states that “it seems” that “a wide dispersal of property is a necessary condition . . . if the fair values of the basic liberties are to be maintained.” Rawls (1971): 245-46. But this connection is foregrounded to a greater degree in Justice as Fairness, where he makes it the primary focus of his critique of welfare state capitalism.
27 See Pogge (2006); Audard (2007).
28 Chambers (2006); Kymlicka (2002).
29 Jackson (2005); Ron (2008)
30 Freeman (2007a).
31 Freeman (2007b).
32 Hsieh (2005).
33 O’Neill (2008b)
34 Hsieh (2009); Hussain (2009); O’Neill (2009); Williamson (2009).
35 O’Neill and Williamson (forthcoming, 2010).
36 Cohen (1989), Miller (1991), Roemer (1993), Schweickart (2000), Alperovitz (2004).
37 To be sure, prominent differences between the two proposals remain. In particular, unlike most forms of democratic socialism, Rawls’s property-owning democracy has no role for democratic oversight or planning of major capital investments. For further discussion, see Williamson (2009).
38 Dagger (2006); Elkin (2006).
39 Thus Elkin: “In the context of [institutional] practice, there aren’t two separate judgments—one about values, the other about practices that will serve them. There is only one: how much we value something given what it takes to realize it.” Elkin (2006): 77.
40 To be clear, Rawls also recognizes the importance of considering the institutional implications of a conception of justice before accepting it. Hence Rawls writes: “It is important to trace out, if only in a rough and ready way, the institutional content of the two principles of justice. We need to do this before we can endorse these principles, even provisionally. This is because the idea of reflective equilibrium involves our accepting the implications of ideals and first principles in particular cases as they arise. We cannot tell solely from the content of a political conception—from its principles and ideals—whether it is reasonable for us.” Rawls (2001): 136. This recognition on Rawls’s part is important, but is not sufficient to satisfy Elkin’s objection, on two grounds. First, Rawls limits his account of the institutional content of property-owning democracy to ideal-type regime analysis—even though he acknowledges the possibility that “a basic structure may generate interests that make it work very differently than its ideal description.” Rawls (2001): 137. Considerations of the kinds of interests a regime might generate are, according to Rawls, questions of “political sociology” and not something that accounts of ideal-type regimes need to consider. In Elkin’s view, in contrast, careful consideration of the interests a regime is likely to produce and how they might be held in balance so as to maintain the regime and achieve its stated values over time is precisely what “constitutional” thinking must take up. Second, as stated in the text, Elkin rejects Rawls’s view that the premise of a general agreement about the content of justice and the assumption of “full compliance” are useful beginning points for thinking about a workable constitutional regime.
41 Elkin (2006): 254-55. Elkin here is quoting Rawls (1993): xvii. It might be noted that Rawls, beginning in Political Liberalism, does revise his theory so as to take account of one major fact about the world, namely, the fact of reasonable pluralism. For a republican like Elkin, the question is why Rawls stops there as opposed to taking account of other prominent facts as well (such as large-scale corporate power). See Elkin (2006): 359-360, n10. For a related critique, see Sheldon Wolin’s (1996) critical review of Political Liberalism.
42 Elkin (2006): 21.
43 Ekin (2006): 38-42.
44 Ekin (2006): 51-73. Elkin’s argumentation on these themes is more complex and nuanced than this brief summary can do justice.
45 Elkin (2006): 134.
46 Elkin (2006): 134.
47 Elkin (2006): 292-96. Indeed, Elkin’s discussion goes significantly beyond Rawls’s own proposals, drawing on some of the ideas (such as universal capital grants and support for worker ownership of firms) discussed in subsequent literature (see main text, below, for a brief review).
48 To be sure, moving towards a broader distribution of capital and property also would face formidable political obstacles. The view is that if a suitably wider distribution could be achieved, this would be a more stable long-term basis for limiting inequalities and broadening prosperity than relying primarily on continual, large-scale progressive taxation.
49 See Sherraden (2005) for a thorough discussion.
50 Ackerman and Alstott (1999).
51 Alperovitz (2004, 2006).
52 Unger (2006).
53 Shapiro and Graetz (2005).
54 See O’Neill (2007); White (2008).
55 For a promising argument along these lines, see Alperovitz and Daly (2008).
Creative Commons License
This work is licensed under a Creative Commons License
Property-Owning Democracy and the Demands of Justice by Thad Williamson and Martin O'Neill
Living Reviews in Democracy, Volume 1 - 2009
ISSN: 1663-0165 |
CIS nccr | null | null | null | null | null | null | null | null | null |
https://ch.mathworks.com/matlabcentral/cody/problems/1095-circular-primes-based-on-project-euler-problem-35/solutions/219006 | 1,604,154,546,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00274.warc.gz | 237,535,872 | 17,825 | Cody
# Problem 1095. Circular Primes (based on Project Euler, problem 35)
Solution 219006
Submitted on 19 Mar 2013 by Jean-Marie Sainthillier
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% [y numbers]=circular_prime(197) assert(isequal(y,16)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197]))
y = 16 numbers = 2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197
2 Pass
%% [y numbers]=circular_prime(100) assert(isequal(y,13)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97]))
y = 13 numbers = 2 3 5 7 11 13 17 31 37 71 73 79 97
3 Pass
%% [y numbers]=circular_prime(250) assert(isequal(y,17)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 199]))
y = 17 numbers = Columns 1 through 16 2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 Column 17 199
4 Pass
%% [y numbers]=circular_prime(2000) assert(isequal(y,27)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 199 311 337 373 719 733 919 971 991 1193 1931]))
y = 27 numbers = Columns 1 through 8 2 3 5 7 11 13 17 31 Columns 9 through 16 37 71 73 79 97 113 131 197 Columns 17 through 24 199 311 337 373 719 733 919 971 Columns 25 through 27 991 1193 1931
5 Pass
%% [y numbers]=circular_prime(10000) assert(isequal(y,33)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 199 311 337 373 719 733 919 971 991 1193 1931 3119 3779 7793 7937 9311 9377]))
y = 33 numbers = Columns 1 through 8 2 3 5 7 11 13 17 31 Columns 9 through 16 37 71 73 79 97 113 131 197 Columns 17 through 24 199 311 337 373 719 733 919 971 Columns 25 through 32 991 1193 1931 3119 3779 7793 7937 9311 Column 33 9377
6 Pass
%% [y numbers]=circular_prime(54321) assert(isequal(y,38)&&isequal(numbers,[2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 197 199 311 337 373 719 733 919 971 991 1193 1931 3119 3779 7793 7937 9311 9377 11939 19391 19937 37199 39119]))
y = 38 numbers = Columns 1 through 8 2 3 5 7 11 13 17 31 Columns 9 through 16 37 71 73 79 97 113 131 197 Columns 17 through 24 199 311 337 373 719 733 919 971 Columns 25 through 32 991 1193 1931 3119 3779 7793 7937 9311 Columns 33 through 38 9377 11939 19391 19937 37199 39119
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Start Hunting! | 1,022 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-45 | latest | en | 0.307617 |
https://o1-labs.github.io/proof-systems/kimchi/overview.html?ref=blog.lambdaclass.com | 1,721,932,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00788.warc.gz | 366,588,100 | 11,611 | # Overview
Here we explain how the Kimchi protocol design is translated into the proof-systems repository, from a high level perspective, touching briefly on all the involved aspects of cryptography. The concepts that we will be introducing can be studied more thoroughly by accessing the specific sections in the book.
In brief, the Kimchi protocol requires three different types of arguments Argument:
• Custom gates: they correspond to each of the specific functions performed by the circuit, which are represented by gate constraints.
• Permutation: the equality between different cells is constrained by copy constraints, which are represented by a permutation argument. It represents the wiring between gates, the connections from/to inputs and outputs.
• Lookup tables: for efficiency reasons, some public information can be stored by both parties (prover and verifier) instead of wired in the circuit. Examples of these are boolean functions.
All of these arguments are translated into equations that must hold for a correct witness for the full relation. Equivalently, this is to say that a number of expressions need to evaluate to zero on a certain set of numbers. So there are two problems to tackle here:
1. Roots-check: Check that an equation evaluates to zero on a set of numbers.
2. Aggregation: Check that it holds for each of the equations.
### Roots-check
For the first problem, given a polynomial of degree , we are asking to check that for all , where stands for set. Of course, one could manually evaluate each of the elements in the set and make sure of the above claim. But that would take so long (i.e. it wouldn’t be succinct). Instead, we want to check that all at once. And a great way to do it is by using a vanishing polynomial. Such a polynomial will be nothing else than the smallest polynomial that vanishes on . That means, it is exactly defined as the degree polynomial formed by the product of the monomials:
And why is this so advantageous? Well, first we need to make a key observation. Since the vanishing polynomial equals zero on every , and it is the smallest such polynomial (recall it has the smallest possible degree so that this property holds), if our initial polynomial evaluates to zero on , then it must be the case that is a multiple of the vanishing polynomial . But what does this mean in practice? By polynomial division, it simply means there exists a quotient polynomial of degree such that:
And still, where’s the hype? If you can provide such a quotient polynomial, one could easily check that if for a random number \ (recall you will check in a point out of the set, otherwise you would get a ), then with very high probability that would mean that actually , meaning that vanishes on the whole set , with just one point!
Let’s take a deeper look into the “magic” going on here. First, what do we mean by high probability? Is this even good enough? And the answer to this question is: as good as you want it to be.
First we analyse the math in this check. If the polynomial form of actually holds, then of course for any possible \ the check will hold. But is there any unlucky instantiation of the point such that but ? And the answer is, yes, there are, BUT not many. But how many? How unlikely this is? You already know the answer to this: Schwartz-Zippel. Recalling this lemma:
Given two different polynomials and of degree , they can at most intersect (i.e. coincide) in points. Or what’s equivalent, let , the polynomial can only evaluate to in at most points (its roots).
Thus, if we interchange and , both of degree , there are at most unlucky points of that could trick you into thinking that was a multiple of the vanishing polynomial (and thus being equal to zero on all of ). So, how can you make this error probability negligible? By having a field size that is big enough (the formal definition says that the inverse of its size should decrease faster than any polynomial expression). Since we are working with fields of size , we are safe on this side!
Second, is this really faster than checking that for all ? At the end of the day, it seems like we need to evaluate , and since this is a degree polynomial it looks like we are still performing about the same order of computations. But here comes math again. In practice, we want to define this set to have a nice structure that allows us to perform some computations more efficiently than with arbitrary sets of numbers. Indeed, this set will normally be a multiplicative group (normally represented as or ), because in such groups the vanishing polynomial has an efficient representation , which is much faster to evaluate than the above product.
Third, we may want to understand what happens with the evaluation of instead. Since this is a degree , it may look like this will as well take a lot of effort. But here’s where cryptography comes into play, since the verifier will never get to evaluate the actual polynomial by themselves. Various reasons why. One, if the verifier had access to the full polynomial , then the prover should have sent it along with the proof, which would require coefficients to be represented (and this is no longer succinct for a SNARK). Two, this polynomial could carry some secret information, and if the verifier could recompute evaluations of it, they could learn some private data by evaluating on specific points. So instead, these evaluations will be a “mental game” thanks to polynomial commitments and proofs of evaluation sent by the prover (for whom a computation in the order of is not only acceptable, but necessary). The actual proof length will depend heavily on the type of polynomial commitments we are using. For example, in Kate-like commitments, committing to a polynomial takes a constant number of group elements (normally one), whereas in Bootleproof it is logarithmic. But in any case this will be shorter than sending elements.
### Aggregation
So far we have seen how to check that a polynomial equals zero on all of , with just a single point. This is somehow an aggregation per se. But we are left to analyse how we can prove such a thing, for many polynomials. Altogether, if they hold, this will mean that the polynomials encode a correct witness and the relation would be satisfied. These checks can be performed one by one (checking that each of the quotients are indeed correct), or using an efficient aggregation mechanism and checking only one longer equation at once.
So what is the simplest way one could think of to perform this one-time check? Perhaps one could come up with the idea of adding up all of the equations into a longer one . But by doing this, we may be cancelling out terms and we could get an incorrect statemement.
So instead, we can multiply each term in the sum by a random number. The reason why this trick works is the independence between random numbers. That is, if two different polynomials and are both equal to zero on a given , then with very high probability the same will be a root of the random combination . If applied to the whole statement, we could transform the equations into a single equation,
This sounds great so far. But we are forgetting about an important part of proof systems which is proof length. For the above claim to be sound, the random values used for aggregation should be verifier-chosen, or at least prover-independent. So if the verifier had to communicate with the prover to inform about the random values being used, we would get an overhead of field elements.
Instead, we take advantage of another technique that is called powers-of-alpha. Here, we make the assumption that powers of a random value are indistinguishable from actual random values . Then, we can twist the above claim to use only one random element to be agreed with the prover as: | 1,592 | 7,825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.9372 |
https://sciencedocbox.com/Physics/88904089-Chapter-1-electricity.html | 1,713,962,935,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00164.warc.gz | 471,340,129 | 24,925 | # CHAPTER 1 ELECTRICITY
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1 CHAPTER 1 ELECTRICITY Electric Current: The amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges. Electric Circuit: Electric circuit is a continuous and closed path of electric current. Note: Conventionally, in an electric circuit the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges Expression of Electric Current Electric current is expressed by the rate of flow of electric charges. Rate of flow means the amount of charge flowing through a particular area in unit time. If a net electric charge (Q) flows through a cross section of conductor in time t, then; Where, I - electric current, Q- net charge and t - time in second. SI unit of the electric charge is coulomb (C), SI unit of time is second (s). SI unit of electric current is ampere (A). Definition of one ampere One ampere is constituted by the flow of one coulomb of charge per second. This means if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere. Therefore; 1A = or 1A = 1Cs -1 Note: 1. Smaller unit of electric current 1mA (milliampere) = 10 3 A 1μA(microampere) =10 6 A
2 2. Ammeter: An apparatus to measure electric current in a circuit. It is always connected in series in a circuit. Que: How many electrons make one coulomb of charge? Ans: One coulomb (C), is equivalent to the charge contained in nearly electrons. (We know that an electron possesses a negative charge of C.) Numerical problems: Example 1. Find the amount of electric charge flowing through the circuit if an electric current of 5 A is drawn by an electric appliance for 5 minute. Solution: Given electric current (I) = 5 A Time (t) = 5 min = 5 X 60 = 300 s Electric charge (Q) =? We know; I = Or, Or, Q = I x t Q = 5 A x 300 s = 1500 C Example 2: If a current of 2 ampere is drawn for 1 hour through the filament of a bulb, find the amount of electric charge flowing through the circuit. Ans: Given electric current (I) = 2 A Time (t) = 1 hour = 1 x 60 x 60 s = 3600 s Electric charge (Q) =? We know that Q = I x t Therefore, Q = 2 x 3600 = 7200 C Electric Potential difference Electric potential difference is known as voltage, which is equal to the work done per unit charge to move the charge between two points.
3 Therefore; Voltage or electric potential difference is denoted by V. Therefore; Where, W = work done Q = Charge. The SI unit of potential difference is volt (V). Definition of one volt One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. Therefore, 1 volt = or 1V = Note: 1.Voltmeter : An apparatus to measure the potential difference between two points in an electric circuit. It is always connected in parallel across the points between which the potential difference is to be measured. Example 3: Calculate the potential difference between two points, if 1500 J of work is done to carry a charge of 50C from one point to other? Ans: Given Work done (W) = 1500J Charge (Q) = 50C Potential difference (V) =? We know that; V = Or, V = = 30 V
4 Symbols used in a Circuit diagram Ohm s Law Ohm's law is the relation between the potential difference applied to the ends of the conductor and current flowing through the conductor. Ohm s law states that at constant temperature the potential difference across the two end of a metallic conductor in an electric circuit is directly proportional to the current flowing through it. Ie, V α I Or, = Constant = R Or, V = IR R is a constant for the given metallic wire at a given temperature and is called its resistance.
5 Resistance: Resistance is the property of conductor which resists the flow of electric current through it. It is equal to the ratio of the potential difference applied across its ends and the current flowing through it. ie, Resistance = or, R = Note: 1. SI Unit of resistance is ohm. Ohm is denoted by Greek letter Ω 2. Definition of 1ohm : If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1Ω. That is, 1 ohm = or, 1Ω = 3. V- I graph The graph of V (potential difference) versus I (electric current) is always a straight line. A straight line plot shows that as the current through a wire increases, the potential difference across the wire increases linearly. 4. Variable Resistance: A component used to regulate current without changing the voltage source is called variable resistance. 5. Rheostat : This is a device which is used in a circuit to provide variable resistance, thus to change the resistance in the circuit
6 Factors on which resistance of a conductor depends: The factors on which the resistance of a conductor depends i) Length of the conductor: The resistance of the conductor is directly proportional to its length. Resistance increases with increase in length of the conductor. This is the cause that long electric wires create more resistance to the electric current. ie, Rα l (1) ii) Area of cross section: The resistance of the conductor is inversely proportional to its area of cross section. This means R will decrease with increase in the area of conductor and vice versa. More area of conductor decreases the resistance. This is the cause that thick copper wire creates less resistance to the electric current. ie, R α (2) iii) Nature of the material: resistance of the conductor depends on the nature of which it is made. For example a copper wire has less resistance than a nichrome wire of same length and area of cross-section. iv) Effect of temperature:- Resistance of all pure metals increases on raising the temperature and decreases on lowering the temperature. Resistance of alloys like manganin, nichrome and constantan remains unaffected by temperature. Resistivity Combining eqn (1) and (2) we can write, ie, R α R = ρ (iii) Where ρ (rho) is the constant of proportionality and is called the electrical resistivity of the material of the conductor. From equation (iii) The SI unit of resistivity is Ω m. Note: 1. If l = 1 m, A = 1 m 2 then, eqn (iv) becomes.
7 Resistivity, ρ = R ie, The resistivity of the material is equal to the resistance of the material of unit length and unit area of cross section. 2. Resistivity of the material is independent of the length and area of cross section. It only depends on the nature of the material and its temperature. 3. Electrical resistivity of the material in increasing order Resistivity of Conductors < Resistivity of alloys < Resistivity of insulator RESISTANCE OF A SYSTEM OF RESISTORS The resistances can be combined in two ways 1. In series 2. In parallel To increase the resistance individual resistances are connected in series combination and to decrease the resistance individual resistances are connected in parallel combination. 1. Resistors in Series When two or more resistances are connected end to end then they are said to be connected in series combination. Note: When the resistors are connected in series the current through them remains same but the voltage across them will be different.
8 Consider three resistors of resistance R 1, R 2 and R 3 are connected in series. The potential difference applied across the combination of resistors is V and the current through the circuit is I. Since in series combination current remains same but potential is different so, V = V 1 + V 2 + V (1) If V 1, V 2 and V 3 is the potential difference across each resistor R 1, R 2 and R 3 respectively, then according to Ohm's Law, V 1 =IR 1 V 2 =IR (2) V 3 =IR 3 Substituting eqn (2) in (1) or, V = I R 1 + I R 2 + I R 3 V = I (R 1 + R 2 + R 3 ) (3) Assuming that the three resistors R 1, R 2 and R 3 as a single resistor of resistance R s then, V = IR s (4) Substituting (4) in (3) we get, I R S = I (R 1 + R 2 + R 3 ) ie, R s = R 1 + R 2 + R 3 Thus when the resistors are connected in series, equivalent resistance of the series combination is equal to the sum of individual resistances. For n numbers of resistors connected in series equivalent resistance would be R s = R 1 +R 2 +R R n
9 2. Resistors in parallel. Resistors are said to be in parallel when the ends of the resistors are connected to a common terminal. Note: When the resistors are connected in parallel the voltage across the resistors remains same but the current flowing through each resistor is different. Consider three resistors of resistance R 1, R 2 and R 3 are connected in parallel. The potential difference applied across the combination of resistors is V and the current through the circuit is I. Since in parallel combination voltage remains same but current is different so, I = I 1 + I 2 + I (1) If I 1, I 2 and I 3 is the current through resistor R 1, R 2 and R 3 respectively, then according to Ohm's Law, Substituting the values I 1, I 2 and I 3 in eqn (1) we get (2) Assuming that the three resistors R 1, R 2 and R 3 as a single resistor of resistance R p then, I = ( 3) Substituting eqn (2) and (3) in (1) we get
10 = V Ie, = For resistors connected in parallel combination reciprocal of equivalent resistance is equal to the sum of reciprocal of individual resistances. For n numbers of resistors connected in parallel equivalent resistance would be =
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Nama :.. Kelas/No Absen : TASK 2 : CURRENT AND RESISTANCE 1. A car battery is rated at 80 A h. An ampere-hour is a unit of: A. power B. energy C. current D. charge E. force 2. Current has units: A. kilowatt-hour | 8,531 | 35,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.912006 |
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December 20, 2014 / 28 Kislev, 5775
At a Glance
Sponsored Post
Back in the USSR
Today's anti-democratic glue is apparently better than the glue of Slavic identity that formed the "Warsaw Pact" because it is a world view and a cultural perspective.
Many countries in the world compare the behavior of the bloc under Russian leadership to the conduct of the West under United States' leadership and conclude: The United States betrays her friends and abandons them, while Russia is faithful to her friends and defends them.
Those among us who are middle-aged or older will remember a song by the Beatles called “Back to the USSR”. Ever since the minor crisis regarding the Asad regime’s use of chemical weapons, this song has been stuck my head.
This minor crisis has revealed, emphasized and demonstrated what we wrote about here long ago, which is the weakening of the Western bloc, especially the United States, and the return of the opposing group to the center of international stage under Russian leadership. Putin’s article in the New York Times openly expressed his opinion about the old-new international situation, in which the world has stopped being a unipolar system, and has gone back to being a bipolar system, as it was until the end of the eighties, when the Soviet Union collapsed, and the allies in Eastern Europe left it in favor of joining with the Western, democratic world, and afterward, the European Union.
The Russian Bloc is based on non-democratic countries that are hostile to the West, whether from a cultural point of view, like China and Syria, or a religious point of view, like Iran. Countries where democracy is limping along like Venezuela and Nicaragua, also join up with Russia, who doesn’t bother them too much about marginal matters like human rights and political freedoms. North Korea also enjoys China’s and Russia’s political protection, especially in the UN Security Council.
Today’s anti-democratic glue is apparently better than the glue of Slavic identity that formed the “Warsaw Pact” because it is a world view and a cultural perspective. Back then, membership in the Soviet bloc was forced on the states (for instance, in Czechoslovakia and the Soviet invasion of 1968), while today, states freely choose to belong to the Russian bloc. It is not yet a consolidated and unified bloc, but one definitely sees that this union of anti-democratic forces is winning ever more diplomatic territory in the international sphere. There is an important military aspect to this alliance, due to the supply of Russian weaponry to Iran, Syria and Hezbollah.
Many countries in the world compare the behavior of the bloc under Russian leadership to the conduct of the West under United States’ leadership and conclude: The United States betrays her friends and abandons them, while Russia is faithful to her friends and defends them. When the world analyzes what the United States has done for states and rulers in recent years it finds Mubarak, who was abandoned by President Obama with the start of demonstrations against him; the president of Tunisia – bin Ali – who was forced to flee from the demonstrations without even one of his European friends to rescue him; the United States abandons its friends in the Gulf and in Saudi Arabia in the face of Iran’s threatening buildup; the West does not back Israel in its efforts to maintain its security and its strategic assets, and urges it to establish another Palestinian terror country in the mountains of Judea and Samaria, overlooking most of the territory of the State of Israel.
On the other hand, the world sees that Russia defends Iran and its nuclear project in the Security Council faithfully, and even supplies its reactors and the means of defending them; Russia is faithful to Asad and supplies him weaponry, ammunition and means of defense necessary for his survival; Russia supplies China with raw materials and places of employment.
In Economic matters as well, the West appears weak relative to Russia. Since six years ago, the Western economy – Europe and the United States together – has been caught in a structural crisis, not in a recession from which it is relatively easy to emerge. It seems that the unification of currency (the Euro) and production standards are not enough to make Europe into one body, so divisive forces exist there that even threaten the stability of some countries: the region of Catalonia wants to secede from Spain, and the Scots apparently will leave the United Kingdom in another year. Europe is addicted to Russian gas, and to oil that, by Iran’s “good will”, is allowed to pass through the Strait of Hormuz on its route from the Emirates to Europe.
Regarding the issue of Syrian chemical weapons, the West has seemed like a crumbling and disintegrating body, with no leader and no shared agenda. The British parliament is against war, the French is for it, and the American administration says that it’s getting ready to attack, Congress doesn’t support it, the American army is preparing for war and the State Department puts forth a compromise. The right hand does not know what the left is doing, and each one acts according to a different agenda. This is no way to build a bloc of states that is capable of executing a mission that everyone agrees is ethically justified: to defend the citizens of Syria from chemical weapons. And when ethics ceases to be the leading cause for the West, what is left of its values?
3 Responses to “Back in the USSR”
1. So Russia with the one hand tries to make a deal so the west won’t bomb Assad and while people try to get on board with them, they stab the west in the back by making a deal with Iran to supply S-300‘s. The issue is that they are forgetting that if Russia starts to send S-300‘s to Iran; Israel at the very least will likely attack the shipments. It’s not just the US that won’t like that sale. That could be the trigger that sets this whole thing off and in a big way. Russia simply should not take this action, BUT they are and I really doubt that they will pull back from it. The clock is ticking down in the Middle East. It just seems that we have no shortage of bad decisions built upon the dried bones of former bad decisions. As Iran looks to avoid getting their nuclear program bombed, they buy weapons that will almost certainly result in military conflict, and Russia who people just stated were brilliant political chess players just turned over the chess board. Perhaps the players don’t understand that this is NOT a game of chess! I wrote a small 6 page book that outlines what I believe the Bible states will take place soon as well as the potential trends I see at this time. I don’t accept donations and it’s free. It’s a short read. I encourage you to have a look: http://www.booksie.com/religion_and_spirituality/book/richard_b_barnes/after-the-rapture-whats-next
2. Eric Here says:
3. I commented once, if I remember when Kerry went to Russia to speack with Putin and Kerry waited many hours to be attented by Putin, than later Netanyahu went to Russia to speack with Putin and Putin recived Netanyahu effusively, and I have comented that it is time that Israel should approuch more to Russia, becouse the Obama goverment is not to trustworthy, I think it is time to have better relations with Russia becouse in Israel much russians with familyes tyes in Russia.
I am against that the US get involved in the syrian civil war.
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Printed from: http://www.jewishpress.com/indepth/analysis/dr-mordechai-kedar/back-in-the-ussr/2013/09/18/
Scan this QR code to visit this page online: | null | null | null | null | null | null | null | null | null |
https://www.askiitians.com/forums/Integral-Calculus/26/3231/eqn-of-curve.htm | 1,726,039,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00850.warc.gz | 606,313,196 | 42,906 | The tangent at a point P(x,y) on a curve meets the axes at M and N such that P divides MN internally in the ratio 2:1. Then find the equation of the curve?
Ramesh V
70 Points
15 years ago
let eqn of line L: x/a + y/b = 1
where a, b are X,Y intercepts of line L
so, dy/dx = -1/a
and the point P which divides MN internally in the ratio 2:1
so x=a/3 and y=2b/3
on substiting in diff. eqn , we have
dy/dx = -y/2x
i.e (1/y) dy = -(1/2x) .dx
ln y = -1/2lnx +lnC where lnC is constant
so eqn. of curve is : y=(kx)-1/2 | 197 | 524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.870512 |
https://wiki.sagemath.org/interact/calculus?action=recall&rev=28 | 1,656,892,618,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00193.warc.gz | 679,472,141 | 25,941 | # Sage Interactions - Calculus
## Root Finding Using Bisection
by William Stein
def bisect_method(f, a, b, eps):
try:
f = f._fast_float_(f.variables()[0])
except AttributeError:
pass
intervals = [(a,b)]
two = float(2); eps = float(eps)
while True:
c = (a+b)/two
fa = f(a); fb = f(b); fc = f(c)
if abs(fc) < eps: return c, intervals
if fa*fc < 0:
a, b = a, c
elif fc*fb < 0:
a, b = c, b
else:
raise ValueError, "f must have a sign change in the interval (%s,%s)"%(a,b)
intervals.append((a,b))
html("<h1>Double Precision Root Finding Using Bisection</h1>")
@interact
def _(f = cos(x) - x, a = float(0), b = float(1), eps=(-3,(-16..-1))):
eps = 10^eps
print "eps = %s"%float(eps)
try:
time c, intervals = bisect_method(f, a, b, eps)
except ValueError:
print "f must have opposite sign at the endpoints of the interval"
show(plot(f, a, b, color='red'), xmin=a, xmax=b)
else:
print "root =", c
print "f(c) = %r"%f(c)
print "iterations =", len(intervals)
P = plot(f, a, b, color='red')
h = (P.ymax() - P.ymin())/ (1.5*len(intervals))
L = sum(line([(c,h*i), (d,h*i)]) for i, (c,d) in enumerate(intervals) )
L += sum(line([(c,h*i-h/4), (c,h*i+h/4)]) for i, (c,d) in enumerate(intervals) )
L += sum(line([(d,h*i-h/4), (d,h*i+h/4)]) for i, (c,d) in enumerate(intervals) )
show(P + L, xmin=a, xmax=b)
## Newton's Method
Note that there is a more complicated Newton's method below.
by William Stein
def newton_method(f, c, eps, maxiter=100):
x = f.variables()[0]
fprime = f.derivative(x)
try:
g = f._fast_float_(x)
gprime = fprime._fast_float_(x)
except AttributeError:
g = f; gprime = fprime
iterates = [c]
for i in xrange(maxiter):
fc = g(c)
if abs(fc) < eps: return c, iterates
c = c - fc/gprime(c)
iterates.append(c)
return c, iterates
html("<h1>Double Precision Root Finding Using Newton's Method</h1>")
@interact
def _(f = x^2 - 2, c = float(0.5), eps=(-3,(-16..-1)), interval=float(0.5)):
eps = 10^(eps)
print "eps = %s"%float(eps)
time z, iterates = newton_method(f, c, eps)
print "root =", z
print "f(c) = %r"%f(z)
n = len(iterates)
print "iterations =", n
html(iterates)
P = plot(f, z-interval, z+interval, rgbcolor='blue')
h = P.ymax(); j = P.ymin()
L = sum(point((w,(n-1-float(i))/n*h), rgbcolor=(float(i)/n,0.2,0.3), pointsize=10) + \
line([(w,h),(w,j)],rgbcolor='black',thickness=0.2) for i,w in enumerate(iterates))
show(P + L, xmin=z-interval, xmax=z+interval)
## A contour map and 3d plot of two inverse distance functions
by William Stein
@interact
def _(q1=(-1,(-3,3)), q2=(-2,(-3,3)),
cmap=['autumn', 'bone', 'cool', 'copper', 'gray', 'hot', 'hsv',
'jet', 'pink', 'prism', 'spring', 'summer', 'winter']):
x,y = var('x,y')
f = q1/sqrt((x+1)^2 + y^2) + q2/sqrt((x-1)^2+(y+0.5)^2)
C = contour_plot(f, (-2,2), (-2,2), plot_points=30, contours=15, cmap=cmap)
show(C, figsize=3, aspect_ratio=1)
show(plot3d(f, (x,-2,2), (y,-2,2)), figsize=5, viewer='tachyon')
## A simple tangent line grapher
by Marshall Hampton
html('<h2>Tangent line grapher</h2>')
@interact
def tangent_line(f = input_box(default=sin(x)), xbegin = slider(0,10,1/10,0), xend = slider(0,10,1/10,10), x0 = slider(0, 1, 1/100, 1/2)):
prange = [xbegin, xend]
x0i = xbegin + x0*(xend-xbegin)
var('x')
df = diff(f)
tanf = f(x0i) + df(x0i)*(x-x0i)
fplot = plot(f, prange[0], prange[1])
print 'Tangent line is y = ' + tanf._repr_()
tanplot = plot(tanf, prange[0], prange[1], rgbcolor = (1,0,0))
fmax = f.find_maximum_on_interval(prange[0], prange[1])[0]
fmin = f.find_minimum_on_interval(prange[0], prange[1])[0]
show(fplot + tanplot, xmin = prange[0], xmax = prange[1], ymax = fmax, ymin = fmin)
## Numerical integrals with the midpoint rule
by Marshall Hampton
var('x')
@interact
def midpoint(n = slider(1,100,1,4), f = input_box(default = "x^2", type = str), start = input_box(default = "0", type = str), end = input_box(default = "1", type = str)):
a = N(start)
b = N(end)
func = sage_eval(f, locals={'x':x})
dx = (b-a)/n
midxs = [q*dx+dx/2 + a for q in range(n)]
midys = [func(x_val) for x_val in midxs]
rects = Graphics()
for q in range(n):
xm = midxs[q]
ym = midys[q]
rects = rects + line([[xm-dx/2,0],[xm-dx/2,ym],[xm+dx/2,ym],[xm+dx/2,0]], rgbcolor = (1,0,0)) + point((xm,ym), rgbcolor = (1,0,0))
min_y = find_minimum_on_interval(func,a,b)[0]
max_y = find_maximum_on_interval(func,a,b)[0]
html('<h3>Numerical integrals with the midpoint rule</h3>')
html('$\int_{a}^{b}{f(x) dx} {\\approx} \sum_i{f(x_i) \Delta x}$')
print "\n\nSage numerical answer: " + str(integral_numerical(func,a,b,max_points = 200)[0])
print "Midpoint estimated answer: " + str(RDF(dx*sum([midys[q] for q in range(n)])))
show(plot(func,a,b) + rects, xmin = a, xmax = b, ymin = min_y, ymax = max_y)
## Function tool
Enter symbolic functions f, g, and a, a range, then click the appropriate button to compute and plot some combination of f, g, and a along with f and g. This is inspired by the Matlab funtool GUI.
x = var('x')
@interact
def _(f=sin(x), g=cos(x), xrange=input_box((0,1)), yrange='auto', a=1,
action=selector(['f', 'df/dx', 'int f', 'num f', 'den f', '1/f', 'finv',
'f+a', 'f-a', 'f*a', 'f/a', 'f^a', 'f(x+a)', 'f(x*a)',
'f+g', 'f-g', 'f*g', 'f/g', 'f(g)'],
width=15, nrows=5, label="h = "),
do_plot = ("Draw Plots", True)):
try:
f = SR(f); g = SR(g); a = SR(a)
except TypeError, msg:
print msg[-200:]
print "Unable to make sense of f,g, or a as symbolic expressions."
return
if not (isinstance(xrange, tuple) and len(xrange) == 2):
xrange = (0,1)
h = 0; lbl = ''
if action == 'f':
h = f
lbl = 'f'
elif action == 'df/dx':
h = f.derivative(x)
lbl = '\\frac{df}{dx}'
elif action == 'int f':
h = f.integrate(x)
lbl = '\\int f dx'
elif action == 'num f':
h = f.numerator()
lbl = '\\text{numer(f)}'
elif action == 'den f':
h = f.denominator()
lbl = '\\text{denom(f)}'
elif action == '1/f':
h = 1/f
lbl = '\\frac{1}{f}'
elif action == 'finv':
h = solve(f == var('y'), x)[0].rhs()
lbl = 'f^{-1}(y)'
elif action == 'f+a':
h = f+a
lbl = 'f + a'
elif action == 'f-a':
h = f-a
lbl = 'f - a'
elif action == 'f*a':
h = f*a
lbl = 'f \\times a'
elif action == 'f/a':
h = f/a
lbl = '\\frac{f}{a}'
elif action == 'f^a':
h = f^a
lbl = 'f^a'
elif action == 'f^a':
h = f^a
lbl = 'f^a'
elif action == 'f(x+a)':
h = f(x+a)
lbl = 'f(x+a)'
elif action == 'f(x*a)':
h = f(x*a)
lbl = 'f(xa)'
elif action == 'f+g':
h = f+g
lbl = 'f + g'
elif action == 'f-g':
h = f-g
lbl = 'f - g'
elif action == 'f*g':
h = f*g
lbl = 'f \\times g'
elif action == 'f/g':
h = f/g
lbl = '\\frac{f}{g}'
elif action == 'f(g)':
h = f(g)
lbl = 'f(g)'
html('<center><font color="red">$f = %s$</font></center>'%latex(f))
html('<center><font color="green">$g = %s$</font></center>'%latex(g))
html('<center><font color="blue"><b>$h = %s = %s$</b></font></center>'%(lbl, latex(h)))
if do_plot:
P = plot(f, xrange, color='red', thickness=2) + \
plot(g, xrange, color='green', thickness=2) + \
plot(h, xrange, color='blue', thickness=2)
if yrange == 'auto':
show(P, xmin=xrange[0], xmax=xrange[1])
else:
yrange = sage_eval(yrange)
show(P, xmin=xrange[0], xmax=xrange[1], ymin=yrange[0], ymax=yrange[1])
## Newton-Raphson Root Finding
by Neal Holtz
This allows user to display the Newton-Raphson procedure one step at a time. It uses the heuristic that, if any of the values of the controls change, then the procedure should be re-started, else it should be continued.
# ideas from 'A simple tangent line grapher' by Marshall Hampton
# http://wiki.sagemath.org/interact
State = Data = None # globals to allow incremental changes in interaction data
@interact
def newtraph(f = input_box(default=8*sin(x)*exp(-x)-1, label='f(x)'),
xmin = input_box(default=0),
xmax = input_box(default=4*pi),
x0 = input_box(default=3, label='x0'),
show_calcs = ("Show Calcs",True),
step = ['Next','Prev', 'Reset'] ):
global State, Data
state = [f,xmin,xmax,x0,show_calcs]
if (state != State) or (step == 'Reset'): # when any of the controls change
Data = [ 1 ] # reset the plot
State = state
elif step == 'Next':
N, = Data
Data = [ N+1 ]
elif step == 'Prev':
N, = Data
if N > 1:
Data = [ N-1 ]
N, = Data
df = diff(f)
theplot = plot( f, xmin, xmax )
theplot += text( '\n$x_0$', (x0,0), rgbcolor=(1,0,0),
vertical_alignment="bottom" if f(x0) < 0 else "top" )
theplot += points( [(x0,0)], rgbcolor=(1,0,0) )
Trace = []
def Err( msg, Trace=Trace ):
Trace.append( '<font color="red"><b>Error: %s!!</b></font>' % (msg,) )
def Disp( s, color="blue", Trace=Trace ):
Trace.append( """<font color="%s">$%s$</font>""" % (color,s,) )
Disp( """f(x) = %s""" % (latex(f),) )
Disp( """f'(x) = %s""" % (latex(df),) )
stop = False
is_inf = False
xi = x0
for i in range(N):
fi = RR(f(xi))
fpi = RR(df(xi))
theplot += points( [(xi,fi)], rgbcolor=(1,0,0) )
theplot += line( [(xi,0),(xi,fi)], linestyle=':', rgbcolor=(1,0,0) ) # vert dotted line
Disp( """i = %d""" % (i,) )
Disp( """~~~~x_{%d} = %.4g""" % (i,xi) )
Disp( """~~~~f(x_{%d}) = %.4g""" % (i,fi) )
Disp( """~~~~f'(x_{%d}) = %.4g""" % (i,fpi) )
if fpi == 0.0:
Err( 'Derivative is 0 at iteration %d' % (i+1,) )
is_inf = True
show_calcs = True
else:
xip1 = xi - fi/fpi
Disp( r"""~~~~x_{%d} = %.4g - ({%.4g})/({%.4g}) = %.4g""" % (i+1,xi,fi,fpi,xip1) )
if abs(xip1) > 10*(xmax-xmin):
Err( 'Derivative is too close to 0!' )
is_inf = True
show_calcs = True
elif not ((xmin - 0.5*(xmax-xmin)) <= xip1 <= (xmax + 0.5*(xmax-xmin))):
Err( 'x value out of range; probable divergence!' )
stop = True
show_calcs = True
if is_inf:
xl = xi - 0.05*(xmax-xmin)
xr = xi + 0.05*(xmax-xmin)
yl = yr = fi
else:
xl = min(xi,xip1) - 0.01*(xmax-xmin)
xr = max(xi,xip1) + 0.01*(xmax-xmin)
yl = -(xip1-xl)*fpi
yr = (xr-xip1)*fpi
theplot += text( '\n$x_{%d}$' % (i+1,), (xip1,0), rgbcolor=(1,0,0),
vertical_alignment="bottom" if f(xip1) < 0 else "top" )
theplot += points( [(xip1,0)], rgbcolor=(1,0,0) )
theplot += line( [(xl,yl),(xr,yr)], rgbcolor=(1,0,0) ) # tangent
if stop or is_inf:
break
epsa = 100.0*abs((xip1-xi)/xip1)
nsf = 2 - log(2.0*epsa)/log(10.0)
Disp( r"""~~~~~~~~\epsilon_a = \left|(%.4g - %.4g)/%.4g\right|\times100\%% = %.4g \%%""" % (xip1,xi,xip1,epsa) )
Disp( r"""~~~~~~~~num.~sig.~fig. \approx %.2g""" % (nsf,) )
xi = xip1
show( theplot, xmin=xmin, xmax=xmax )
if show_calcs:
for t in Trace:
html( t )
## Coordinate Transformations
by Jason Grout
var('u v')
from sage.ext.fast_eval import fast_float
from functools import partial
@interact
def trans(x=input_box(u2-v2, label="x=",type=SR), \
y=input_box(u*v+cos(u*v), label="y=",type=SR), \
t_val=slider(0,10,0.2,6, label="Length of curves"), \
u_percent=slider(0,1,0.05,label="<font color='red'>u</font>", default=.7),
v_percent=slider(0,1,0.05,label="<font color='blue'>v</font>", default=.7),
u_range=input_box(range(-5,5,1), label="u lines"),
v_range=input_box(range(-5,5,1), label="v lines")):
thickness=4
u_val = min(u_range)+(max(u_range)-min(u_range))*u_percent
v_val = min(v_range)+(max(v_range)-min(v_range))*v_percent
t_min = -t_val
t_max = t_val
g1=sum([parametric_plot((i,v), t_min,t_max, rgbcolor=(1,0,0)) for i in u_range])
g2=sum([parametric_plot((u,i), t_min,t_max, rgbcolor=(0,0,1)) for i in v_range])
vline_straight=parametric_plot((u,v_val), t_min,t_max, rgbcolor=(0,0,1), linestyle='-',thickness=thickness)
uline_straight=parametric_plot((u_val, v), t_min,t_max,rgbcolor=(1,0,0), linestyle='-',thickness=thickness)
(g1+g2+vline_straight+uline_straight).save("uv_coord.png",aspect_ratio=1, figsize=[5,5], axes_labels=['$u$','$v$'])
xuv = fast_float(x,'u','v')
yuv = fast_float(y,'u','v')
xvu = fast_float(x,'v','u')
yvu = fast_float(y,'v','u')
g3=sum([parametric_plot((partial(xuv,i),partial(yuv,i)), t_min,t_max, rgbcolor=(1,0,0)) for i in u_range])
g4=sum([parametric_plot((partial(xvu,i),partial(yvu,i)), t_min,t_max, rgbcolor=(0,0,1)) for i in v_range])
vline=parametric_plot((partial(xvu,v_val),partial(yvu,v_val)), t_min,t_max, rgbcolor=(0,0,1), linestyle='-',thickness=thickness)
uline=parametric_plot((partial(xuv,u_val),partial(yuv,u_val)), t_min,t_max,rgbcolor=(1,0,0), linestyle='-',thickness=thickness)
(g3+g4+vline+uline).save("xy_coord.png", aspect_ratio=1, figsize=[5,5], axes_labels=['$x$','$y$'])
print jsmath("x=%s, \: y=%s"%(latex(x), latex(y)))
print "<html><table><tr><td><img src='cell://uv_coord.png'/></td><td><img src='cell://xy_coord.png'/></td></tr></table></html>"
## Taylor Series
by Harald Schilly
var('x')
x0 = 0
f = sin(x)*e^(-x)
p = plot(f,-1,5, thickness=2)
dot = point((x0,f(x0)),pointsize=80,rgbcolor=(1,0,0))
@interact
def _(order=(1..12)):
ft = f.taylor(x,x0,order)
pt = plot(ft,-1, 5, color='green', thickness=2)
html('$f(x)\;=\;%s$'%latex(f))
html('$\hat{f}(x;%s)\;=\;%s+\mathcal{O}(x^{%s})$'%(x0,latex(ft),order+1))
show(dot + p + pt, ymin = -.5, ymax = 1)
## Illustration of the precise definition of a limit
by John Perry
I'll break tradition and put the image first. Apologies if this is Not A Good Thing.
html("<h2>Limits: <i>ε-δ</i></h2>")
html("This allows you to estimate which values of <i>δ</i> guarantee that <i>f</i> is within <i>ε</i> units of a limit.")
html("<ul><li>Modify the value of <i>f</i> to choose a function.</li>")
html("<li>Modify the value of <i>a</i> to change the <i>x</i>-value where the limit is being estimated.</li>")
html("<li>Modify the value of <i>L</i> to change your guess of the limit.</li>")
html("<li>Modify the values of <i>δ</i> and <i>ε</i> to modify the rectangle.</li></ul>")
html("If the blue curve passes through the pink boxes, your values for <i>δ</i> and/or <i>ε</i> are probably wrong.")
@interact
def delta_epsilon(f = input_box(default=(x^2-x)/(x-1)), a=input_box(default=1), L = input_box(default=1), delta=input_box(label="δ",default=0.1), epsilon=input_box(label="ε",default=0.1), xm=input_box(label="<i>x</i><sub>min</sub>",default=-1), xM=input_box(label="<i>x</i><sub>max</sub>",default=4)):
f_left_plot = plot(f,xm,a-delta/3,thickness=2)
f_right_plot = plot(f,a+delta/3,xM,thickness=2)
epsilon_line_1 = line([(xm,L-epsilon),(xM,L-epsilon)], rgbcolor=(0.5,0.5,0.5),linestyle='--')
epsilon_line_2 = line([(xm,L+epsilon),(xM,L+epsilon)], rgbcolor=(0.5,0.5,0.5),linestyle='--')
ym = min(f_right_plot.ymin(),f_left_plot.ymin())
yM = max(f_right_plot.ymax(),f_left_plot.ymax())
bad_region_1 = polygon([(a-delta,L+epsilon),(a-delta,yM),(a+delta,yM),(a+delta,L+epsilon)], rgbcolor=(1,0.6,0.6))
bad_region_2 = polygon([(a-delta,L-epsilon),(a-delta,ym),(a+delta,ym),(a+delta,L-epsilon)], rgbcolor=(1,0.6,0.6))
aL_point = point((a,L),rgbcolor=(1,0,0),pointsize=20)
delta_line_1 = line([(a-delta,ym),(a-delta,yM)],rgbcolor=(0.5,0.5,0.5),linestyle='--')
delta_line_2 = line([(a+delta,ym),(a+delta,yM)],rgbcolor=(0.5,0.5,0.5),linestyle='--')
(f_left_plot +f_right_plot +epsilon_line_1 +epsilon_line_2 +delta_line_1 +delta_line_2 +aL_point +bad_region_1 +bad_region_2).show(xmin=xm,xmax=xM)
## A graphical illustration of sin(x)/x -> 1 as x-> 0
by Wai Yan Pong
x=var('x')
@interact
def _(x = slider(-7/10,7/10,1/20,1/2)):
html('<h3>A graphical illustration of $\lim_{x -> 0} \sin(x)/x =1$</h3>')
html('Below is the unit circle, so the length of the <font color=red>red line</font> is |sin(x)|')
html('and the length of the <font color=blue>blue line</font> is |tan(x)| where x is the length of the arc.')
html('From the picture, we see that |sin(x)| $\le$ |x| $\le$ |tan(x)|.')
html('It follows easily from this that cos(x) $\le$ sin(x)/x $\le$ 1 when x is near 0.')
html('As $\lim_{x ->0} \cos(x) =1$, we conclude that $\lim_{x -> 0} \sin(x)/x =1$.')
if not (x == 0):
pretty_print("sin(x)/x = "+str(sin(float(x))/float(x)))
elif x == 0:
pretty_print("The limit of sin(x)/x as x tends to 0 is 1.")
C=circle((0,0),1, rgbcolor='black')
mvp = (cos(x),sin(x));tpt = (1, tan(x))
p1 = point(mvp, pointsize=30, rgbcolor='red'); p2 = point((1,0), pointsize=30, rgbcolor='red')
line1 = line([(0,0),tpt], rgbcolor='black'); line2 = line([(cos(x),0),mvp], rgbcolor='red')
line3 = line([(0,0),(1,0)], rgbcolor='black'); line4 = line([(1,0),tpt], rgbcolor='blue')
result = C+p1+p2+line1+line2+line3+line4
result.show(aspect_ratio=1, figsize=[3,3], axes=False)
## The midpoint rule for numerically integrating a function of two variables
by Marshall Hampton
from sage.plot.plot3d.platonic import index_face_set
def cuboid(v1,v2,**kwds):
"""
Cuboid defined by corner points v1 and v2.
"""
ptlist = []
for vi in (v1,v2):
for vj in (v1,v2):
for vk in (v1,v2):
ptlist.append([vi[0],vj[1],vk[2]])
f_incs = [[0, 2, 6, 4], [0, 1, 3, 2], [0, 1, 5, 4], [1, 3, 7, 5], [2, 3, 7, 6], [4, 5, 7, 6]]
if 'aspect_ratio' not in kwds:
kwds['aspect_ratio'] = [1,1,1]
return index_face_set(f_incs,ptlist,enclosed = True, **kwds)
var('x,y')
R16 = RealField(16)
npi = RDF(pi)
sin,cos = math.sin,math.cos
html("<h1>The midpoint rule for a function of two variables</h1>")
@interact
def midpoint2d(func = input_box('y*sin(x)/x+sin(y)',type=str,label='function of x and y'), nx = slider(2,20,1,3,label='x subdivisions'), ny = slider(2,20,1,3,label='y subdivisions'), x_start = slider(-10,10,.1,0), x_end = slider(-10,10,.1,3*npi), y_start= slider(-10,10,.1,0), y_end= slider(-10,10,.1,3*npi)):
f = sage_eval('lambda x,y: ' + func)
delx = (x_end - x_start)/nx
dely = (y_end - y_start)/nx
xvals = [RDF(x_start + (i+1.0/2)*delx) for i in range(nx)]
yvals = [RDF(y_start + (i+1.0/2)*dely) for i in range(ny)]
num_approx = 0
cubs = []
darea = delx*dely
for xv in xvals:
for yv in yvals:
num_approx += f(xv,yv)*darea
cubs.append(cuboid([xv-delx/2,yv-dely/2,0],[xv+delx/2,yv+dely/2,f(xv,yv)], opacity = .5, rgbcolor = (1,0,0)))
html("$$\int_{"+str(R16(y_start))+"}^{"+str(R16(y_end))+"} "+ "\int_{"+str(R16(x_start))+"}^{"+str(R16(x_end))+"} "+func+"\ dx \ dy$$")
html('<p style="text-align: center;">Numerical approximation: ' + str(num_approx)+'</p>')
p1 = plot3d(f,(x,x_start,x_end),(y,y_start,y_end))
show(p1+sum(cubs))
## Gaussian (Legendre) quadrature
by Jason Grout
The output shows the points evaluated using Gaussian quadrature (using a weight of 1, so using Legendre polynomials). The vertical bars are shaded to represent the relative weights of the points (darker = more weight). The error in the trapezoid, Simpson, and quadrature methods is both printed out and compared through a bar graph. The "Real" error is the error returned from scipy on the definite integral.
from scipy.special.orthogonal import p_roots
from scipy.integrate import quad, trapz, simps
from sage.ext.fast_eval import fast_float
from numpy import linspace
show_weight_graph=False
# 'Hermite': {'w': e**(-x**2), 'xmin': -numpy.inf, 'xmax': numpy.inf, 'func': h_roots},
# 'Laguerre': {'w': e**(-x), 'xmin': 0, 'xmax': numpy.inf, 'func': l_roots},
methods = {'Legendre': {'w': 1, 'xmin': -1, 'xmax': 1, 'func': p_roots},
'Chebyshev': {'w': 1/sqrt(1-x**2), 'xmin': -1, 'xmax': 1, 'func': t_roots},
'Chebyshev2': {'w': sqrt(1-x**2), 'xmin': -1, 'xmax': 1, 'func': u_roots},
'Trapezoid': {'w': 1, 'xmin': -1, 'xmax': 1, 'func': lambda n: (linspace(-1r,1,n), numpy.array([1.0r]+[2.0r]*(n-2)+[1.0r])*1.0r/n)},
'Simpson': {'w': 1, 'xmin': -1, 'xmax': 1, 'func': lambda n: (linspace(-1r,1,n), numpy.array([1.0r]+[4.0r,2.0r]*int((n-3.0r)/2.0r)+[4.0r,1.0r])*2.0r/(3.0r*n))}}
var("x")
def box(center, height, area,**kwds):
width2 = 1.0*area/height/2.0
return polygon([(center-width2,0),(center+width2,0),(center+width2,height),(center-width2,height)],**kwds)
@interact
def weights(n=slider(1,30,1,default=10),f=input_box(default=3*x+cos(10*x)),show_method=["Legendre", "Chebyshev", "Chebyshev2", "Trapezoid","Simpson"]):
ff = fast_float(f,'x')
method = methods[show_method]
xcoords,w = (method['func'])(int(n))
xmin = method['xmin']
xmax = method['xmax']
plot_min = max(xmin, -10)
plot_max = min(xmax, 10)
scaled_func = f*method['w']
scaled_ff = fast_float(scaled_func)
coords = zip(xcoords,w)
max_weight = max(w)
coords_scaled = zip(xcoords,w/max_weight)
f_graph = plot(scaled_func,plot_min,plot_max)
boxes = sum(box(x,ff(x),w*ff(x),rgbcolor=(0.5,0.5,0.5),alpha=0.3) for x,w in coords)
stems = sum(line([(x,0),(x,scaled_ff(x))],rgbcolor=(1-y,1-y,1-y),thickness=2,markersize=6,alpha=y) for x,y in coords_scaled)
points = sum([point([(x,0),(x,scaled_ff(x))],rgbcolor='black',pointsize=30) for x,_ in coords])
graph = stems+points+f_graph+boxes
if show_weight_graph:
graph += line([(x,y) for x,y in coords_scaled], rgbcolor='green',alpha=0.4)
show(graph,xmin=plot_min,xmax=plot_max)
approximation = sum([w*ff(x) for x,w in coords])
integral,integral_error = scipy.integrate.quad(scaled_ff, xmin, xmax)
x_val = linspace(min(xcoords), max(xcoords),n)
y_val = map(scaled_ff,x_val)
trapezoid = integral-trapz(y_val, x_val)
simpson = integral-simps(y_val, x_val)
html("$$\sum_{i=1}^{i=%s}w_i\left(%s\\right)= %s\\approx %s =\int_{-1}^{1}%s \,dx$$"%(n,latex(f.subs(x="x_i")), approximation, integral, latex(scaled_func)))
error_data = [trapezoid, simpson, integral-approximation,integral_error]
print "Trapezoid: %s, Simpson: %s, \nMethod: %s, Real: %s"%tuple(error_data)
show(bar_chart(error_data,width=1),ymin=min(error_data), ymax=max(error_data))
## Vector Calculus, 2-D Motion
By Rob Beezer
A fast_float() version is available in a worksheet
# 2-D motion and vector calculus
# Copyright 2009, Robert A. Beezer
# Creative Commons BY-SA 3.0 US
#
# 2009/02/15 Built on Sage 3.3.rc0
# 2009/02/17 Improvements from Jason Grout
#
# variable parameter is t
# later at a particular value named t0
#
var('t')
#
# parameter range
#
start=0
stop=2*pi
#
# position vector definition
# edit here for new example
# example is wide ellipse
# adjust x, extents in final show()
#
position=vector( (4*cos(t), sin(t)) )
#
# graphic of the motion itself
#
path = parametric_plot( position(t).list(), (t, start, stop), color = "black" )
#
# derivatives of motion, lengths, unit vectors, etc
#
velocity = derivative( position(t) )
acceleration = derivative(velocity(t))
speed = velocity.norm()
speed_deriv = derivative(speed)
tangent = (1/speed)*velocity
dT = derivative(tangent(t))
normal = (1/dT.norm())*dT
#
# interact section
# slider for parameter, 24 settings
# checkboxes for various vector displays
# computations at one value of parameter, t0
#
@interact
def _(t0 = slider(float(start), float(stop), float((stop-start)/24), float(start) , label = "Parameter"),
pos_check = ("Position", True),
vel_check = ("Velocity", False),
tan_check = ("Unit Tangent", False),
nor_check = ("Unit Normal", False),
acc_check = ("Acceleration", False),
tancomp_check = ("Tangential Component", False),
norcomp_check = ("Normal Component", False)
):
#
# location of interest
#
pos_tzero = position(t0)
#
# various scalar quantities at point
#
speed_component = speed(t0)
tangent_component = speed_deriv(t0)
normal_component = sqrt( acceleration(t0).norm()^2 - tangent_component^2 )
curvature = normal_component/speed_component^2
#
# various vectors, mostly as arrows from the point
#
pos = arrow((0,0), pos_tzero, rgbcolor=(0,0,0))
tan = arrow(pos_tzero, pos_tzero + tangent(t0), rgbcolor=(0,1,0) )
vel = arrow(pos_tzero, pos_tzero + velocity(t0), rgbcolor=(0,0.5,0))
nor = arrow(pos_tzero, pos_tzero + normal(t0), rgbcolor=(0.5,0,0))
acc = arrow(pos_tzero, pos_tzero + acceleration(t0), rgbcolor=(1,0,1))
tancomp = arrow(pos_tzero, pos_tzero + tangent_component*tangent(t0), rgbcolor=(1,0,1) )
norcomp = arrow(pos_tzero, pos_tzero + normal_component*normal(t0), rgbcolor=(1,0,1))
#
# accumulate the graphic based on checkboxes
#
picture = path
if pos_check:
picture = picture + pos
if vel_check:
picture = picture + vel
if tan_check:
picture = picture+ tan
if nor_check:
picture = picture + nor
if acc_check:
picture = picture + acc
if tancomp_check:
picture = picture + tancomp
if norcomp_check:
picture = picture + norcomp
#
# print textual info
#
print "Position vector defined as r(t)=", position(t)
print "Speed is ", N(speed(t0))
print "Curvature is ", N(curvature)
#
# show accumulated graphical info
# adjust x-,y- extents to get best plot
#
show(picture, xmin=-4,xmax=4, ymin=-1.5,ymax=1.5,aspect_ratio=1)
## Vector Calculus, 3-D Motion
by Rob Beezer
Available as a worksheet
# 3-D motion and vector calculus
# Copyright 2009, Robert A. Beezer
# Creative Commons BY-SA 3.0 US
#
#
# 2009/02/15 Built on Sage 3.3.rc0
# 2009/02/17 Improvements from Jason Grout
#
# variable parameter is t
# later at a particular value named t0
#
# un-comment double hash (##) to get
# time-consuming torsion computation
#
var('t')
#
# parameter range
#
start=-4*pi
stop=8*pi
#
# position vector definition
# edit here for new example
# example is wide ellipse
# adjust figsize in final show() to get accurate aspect ratio
#
a=1/(8*pi)
c=(3/2)*a
position=vector( (exp(a*t)*cos(t), exp(a*t)*sin(t), exp(c*t)) )
#
# graphic of the motion itself
#
path = parametric_plot3d( position(t).list(), (t, start, stop), color = "black" )
#
# derivatives of motion, lengths, unit vectors, etc
#
velocity = derivative( position(t) )
acceleration = derivative(velocity(t))
speed = velocity.norm()
speed_deriv = derivative(speed)
tangent = (1/speed)*velocity
dT = derivative(tangent(t))
normal = (1/dT.norm())*dT
binormal = tangent.cross_product(normal)
## dB = derivative(binormal(t))
#
# interact section
# slider for parameter, 24 settings
# checkboxes for various vector displays
# computations at one value of parameter, t0
#
@interact
def _(t0 = slider(float(start), float(stop), float((stop-start)/24), float(start) , label = "Parameter"),
pos_check = ("Position", True),
vel_check = ("Velocity", False),
tan_check = ("Unit Tangent", False),
nor_check = ("Unit Normal", False),
bin_check = ("Unit Binormal", False),
acc_check = ("Acceleration", False),
tancomp_check = ("Tangential Component", False),
norcomp_check = ("Normal Component", False)
):
#
# location of interest
#
pos_tzero = position(t0)
#
# various scalar quantities at point
#
speed_component = speed(t0)
tangent_component = speed_deriv(t0)
normal_component = sqrt( acceleration(t0).norm()^2 - tangent_component^2 )
curvature = normal_component/speed_component^2
## torsion = (1/speed_component)*(dB(t0)).dot_product(normal(t0))
#
# various vectors, mostly as arrows from the point
#
pos = arrow3d((0,0,0), pos_tzero, rgbcolor=(0,0,0))
tan = arrow3d(pos_tzero, pos_tzero + tangent(t0), rgbcolor=(0,1,0) )
vel = arrow3d(pos_tzero, pos_tzero + velocity(t0), rgbcolor=(0,0.5,0))
nor = arrow3d(pos_tzero, pos_tzero + normal(t0), rgbcolor=(0.5,0,0))
bin = arrow3d(pos_tzero, pos_tzero + binormal(t0), rgbcolor=(0,0,0.5))
acc = arrow3d(pos_tzero, pos_tzero + acceleration(t0), rgbcolor=(1,0,1))
tancomp = arrow3d(pos_tzero, pos_tzero + tangent_component*tangent(t0), rgbcolor=(1,0,1) )
norcomp = arrow3d(pos_tzero, pos_tzero + normal_component*normal(t0), rgbcolor=(1,0,1))
#
# accumulate the graphic based on checkboxes
#
picture = path
if pos_check:
picture = picture + pos
if vel_check:
picture = picture + vel
if tan_check:
picture = picture+ tan
if nor_check:
picture = picture + nor
if bin_check:
picture = picture + bin
if acc_check:
picture = picture + acc
if tancomp_check:
picture = picture + tancomp
if norcomp_check:
picture = picture + norcomp
#
# print textual info
#
print "Position vector: r(t)=", position(t)
print "Speed is ", N(speed(t0))
print "Curvature is ", N(curvature)
## print "Torsion is ", N(torsion)
print
print "Right-click on graphic to zoom to 400%"
print "Drag graphic to rotate"
#
# show accumulated graphical info
#
show(picture, aspect_ratio=[1,1,1])
## Directional Derivatives
This interact displays graphically a tangent line to a function, illustrating a directional derivative (the slope of the tangent line).
var('x,y,t,z')
f(x,y)=sin(x)*cos(y)
pif = float(pi)
line_thickness=3
surface_color='blue'
plane_color='purple'
line_color='red'
tangent_color='green'
gradient_color='orange'
@interact
def myfun(location=input_grid(1, 2, default=[0,0], label = "Location (x,y)", width=2), angle=slider(0, 2*pif, label = "Angle"),
show_surface=("Show surface", True)):
location3d = vector(location[0]+[0])
location = location3d[0:2]
direction3d = vector(RDF, [cos(angle), sin(angle), 0])
direction=direction3d[0:2]
cos_angle = math.cos(angle)
sin_angle = math.sin(angle)
df = f.gradient()
direction_vector=line3d([location3d, location3d+direction3d], arrow_head=True, rgbcolor=line_color, thickness=line_thickness)
curve_point = (location+t*direction).list()
curve = parametric_plot(curve_point+[f(*curve_point)], (t,-3,3),color=line_color,thickness=line_thickness)
plane = parametric_plot((cos_angle*x+location[0],sin_angle*x+location[1],t), (x, -3,3), (t,-3,3),opacity=0.8, color=plane_color)
pt = point3d(location3d.list(),color='green', size=10)
tangent_line = parametric_plot((location[0]+t*cos_angle, location[1]+t*sin_angle, f(*location)+t*df(*location)*(direction)), (t, -3,3), thickness=line_thickness, color=tangent_color)
picture3d = direction_vector+curve+plane+pt+tangent_line
picture2d = contour_plot(f(x,y), (x,-3,3),(y,-3,3), plot_points=100)
picture2d += arrow(location.list(), (location+direction).list())
picture2d += point(location.list(),rgbcolor='green',pointsize=40)
if show_surface:
picture3d += plot3d(f, (x,-3,3),(y,-3,3),opacity=0.7)
dff = df(location[0], location[1])
dff3d = vector(RDF,dff.list()+[0])
picture3d += line3d([location3d, location3d+dff3d], arrow_head=True, rgbcolor=gradient_color, thickness=line_thickness)
picture2d += arrow(location.list(), (location+dff).list(), rgbcolor=gradient_color, width=line_thickness)
show(picture3d,aspect=[1,1,1], axes=True)
show(picture2d, aspect_ratio=1)
## 3D graph with points and curves
By Robert Marik
This sagelet is handy when showing local, constrained and absolute maxima and minima in two variables. Available as a worksheet
%hide
%auto
x,y, t, u, v =var('x y t u v')
INI_func='x^2-2*x+y^2-2*y'
INI_box='-1,3.2,-1,3.2'
INI_points='(1,1,\'green\'),(3/2,3/2),(0,1),(1,0),(0,0,\'black\'),(3,0,\'black\'),(0,3,\'black\')'
INI_curves='(t,0,0,3,\'red\'),(0,t,0,3,\'green\'),(t,3-t,0,3)'
@interact
def _(func=input_box(INI_func,label="f(x,y)=",type=str),\
bounds=input_box(INI_box,label="xmin,xmax,ymin,ymax",type=str),\
st_points=input_box(INI_points,\
label="points <br><small><small>(comma separated pairs, optionally with color)</small></small>", type=str),\
bnd_curves=input_box(INI_curves,label="curves on boundary<br> <small><small><i>(x(t),y(t),tmin,tmax,'opt_color')</i></small></small>", type=str),\
show_planes=("Show zero planes", False), show_axes=("Show axes", True),
show_table=("Show table", True)):
f=sage_eval('lambda x,y: ' + func)
html(r'Function $f(x,y)=%s$ '%latex(f(x,y)))
xmin,xmax,ymin,ymax=sage_eval('('+bounds+')')
A=plot3d(f(x,y),(x,xmin,xmax),(y,ymin,ymax),opacity=0.5)
if not(bool(st_points=='')):
st_p=sage_eval('('+st_points+',)')
html(r'<table border=1>')
for current in range(len(st_p)):
point_color='red'
if bool(len(st_p[current])==3):
point_color=st_p[current][2]
x0=st_p[current][0]
y0=st_p[current][1]
z0=f(x0,y0)
if show_table:
html(r'<tr><td>$\quad f(%s,%s)\quad$</td><td>$\quad %s$</td>\
</tr>'%(latex(x0),latex(y0),z0.n()))
A=A+point3d((x0,y0,z0),size=9,rgbcolor=point_color)
html(r'</table>')
bnd_cc=sage_eval('('+bnd_curves+',)',locals={'t':t})
if not(bool(st_points=='')):
for current in range(len(bnd_cc)):
bnd_c=bnd_cc[current]+('black',)
A=A+parametric_plot3d((bnd_c[0],bnd_c[1],f(bnd_c[0],bnd_c[1])),\
(t,bnd_c[2],bnd_c[3]),thickness=3,rgbcolor=bnd_c[4])
if show_planes:
A=A+plot3d(0,(x,xmin,xmax),(y,ymin,ymax),opacity=0.3,rgbcolor='gray')
zmax=A.bounding_box()[1][2]
zmin=A.bounding_box()[0][2]
A=A+parametric_plot3d((u,0,v),(u,xmin,xmax),(v,zmin,zmax),opacity=0.3,rgbcolor='gray')
A=A+parametric_plot3d((0,u,v),(u,ymin,ymax),(v,zmin,zmax),opacity=0.3,rgbcolor='gray')
if show_axes:
zmax=A.bounding_box()[1][2]
zmin=A.bounding_box()[0][2]
A=A+line3d([(xmin,0,0), (xmax,0,0)], arrow_head=True,rgbcolor='black')
A=A+line3d([(0,ymin,0), (0,ymax,0)], arrow_head=True,rgbcolor='black')
A=A+line3d([(0,0,zmin), (0,0,zmax)], arrow_head=True,rgbcolor='black')
show(A)
## Approximating function in two variables by differential
by Robert Marik
x,y=var('x y')
html('<h2>Explaining approximation of a function in two \
variables by differential</h2>')
html('Points x0 and y0 are values where the exact value of the function \
is known. Deltax and Deltay are displacements of the new point. Exact value \
and approximation by differential at shifted point are compared.')
@interact
def _(func=input_box('sqrt(x^3+y^3)',label="f(x,y)=",type=str), x0=1, y0=2, \
deltax=slider(-1,1,0.01,0.2),\
deltay=slider(-1,1,0.01,-0.4), xmin=0, xmax=2, ymin=0, ymax=3):
f=sage_eval('lambda x,y: ' + func)
derx(x,y)=diff(f(x,y),x)
dery(x,y)=diff(f(x,y),y)
tangent(x,y)=f(x0,y0)+derx(x0,y0)*(x-x0)+dery(x0,y0)*(y-y0)
A=plot3d(f(x,y),(x,xmin,xmax),(y,ymin,ymax),opacity=0.5)
B=plot3d(tangent(x,y),(x,xmin,xmax),(y,ymin,ymax),color='red',opacity=0.5)
C=point3d((x0,y0,f(x0,y0)),rgbcolor='blue',size=9)
CC=point3d((x0+deltax,y0+deltay,f(x0+deltax,y0+deltay)),rgbcolor='blue',size=9)
D=point3d((x0+deltax,y0+deltay,tangent(x0+deltax,y0+deltay)),rgbcolor='red',size=9)
exact_value_ori=f(x0,y0).n(digits=10)
exact_value=f(x0+deltax,y0+deltay)
approx_value=tangent(x0+deltax,y0+deltay).n(digits=10)
abs_error=(abs(exact_value-approx_value))
html(r'Function $f(x,y)=%s \approx %s$ '%(latex(f(x,y)),latex(tangent(x,y))))
html(r' $f %s = %s$'%(latex((x0,y0)),latex(exact_value_ori)))
html(r'Shifted point $%s$'%latex(((x0+deltax),(y0+deltay))))
html(r'Value of the function in shifted point is $%s$'%f(x0+deltax,y0+deltay))
html(r'Value on the tangent plane in shifted point is $%s$'%latex(approx_value))
html(r'Error is $%s$'%latex(abs_error))
show(A+B+C+CC+D)
## Taylor approximations in two variables
by John Palmieri
This displays the nth order Taylor approximation, for n from 1 to 10, of the function sin(x2 + y2) cos(y) exp(-(x2+y2)/2).
var('x y')
var('xx yy', ns=1)
G = sin(xx^2 + yy^2) * cos(yy) * exp(-0.5*(xx^2+yy^2))
def F(x,y):
return G.subs(xx=x).subs(yy=y)
plotF = plot3d(F, (0.4, 2), (0.4, 2), adaptive=True, color='blue')
@interact
def _(x0=(0.5,1.5), y0=(0.5, 1.5),
order=(1..10)):
F0 = float(G.subs(xx=x0).subs(yy=y0))
P = (x0, y0, F0)
dot = point3d(P, size=15, color='red')
plot = dot + plotF
approx = F0
for n in range(1, order+1):
for i in range(n+1):
if i == 0:
deriv = G.diff(yy, n)
elif i == n:
deriv = G.diff(xx, n)
else:
deriv = G.diff(xx, i).diff(yy, n-i)
deriv = float(deriv.subs(xx=x0).subs(yy=y0))
coeff = binomial(n, i)/factorial(n)
approx += coeff * deriv * (x-x0)^i * (y-y0)^(n-i)
plot += plot3d(approx, (x, 0.4, 1.6),
(y, 0.4, 1.6), color='red', opacity=0.7)
html('$F(x,y) = e^{-(x^2+y^2)/2} \\cos(y) \\sin(x^2+y^2)$')
show(plot) | 12,140 | 34,539 | {"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} | 3.78125 | 4 | CC-MAIN-2022-27 | latest | en | 0.517727 |
https://www.maplesoft.com/support/help/addons/view.aspx?path=convert%2FRootOf | 1,723,425,736,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00125.warc.gz | 663,061,490 | 23,204 | RootOf - Maple Help
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convert/RootOf
convert radicals and I to RootOf notation
Calling Sequence convert(expr, RootOf) convert(expr, RootOf, form=selector)
Parameters
expr - expression form=selector - selector could be one of the three values: index, interval, or numeric; this is an option to convert between RootOf selectors, namely range/interval selector, numeric selector, and the index selector. For information on these selectors, see RootOf.
Description
• The convert/RootOf function changes all occurrences of radicals (may be algebraic constants or functions), to indexed RootOf notation.
• Usually, the radical ${A}^{\frac{p}{m}}$, for integers p < m, is transformed into the expression RootOf(_Z^m-A,index=1)^p.
• I is replaced by RootOf(_Z^2+1,index=1)
• This function is mapped recursively over expressions. In particular, nested radicals are converted. If the input expression is an unnamed table then the conversion routine is mapped onto the elements of the table.
• This function uses radfield to convert radicals into independent RootOfs.
• To convert RootOf notation to I and radicals (where possible) use convert(expr, radical).
• To convert RootOf notation between different selectors, use form=selector to select the target selector format.
Examples
> $\mathrm{convert}\left({2}^{\frac{1}{5}},\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{-}{2}{,}{\mathrm{index}}{=}{1}\right)$ (1)
> $\mathrm{convert}\left({\left(1+{2}^{\frac{1}{2}}\right)}^{\frac{1}{3}}I,\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{+}{1}{,}{\mathrm{index}}{=}{1}\right){}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{-}{1}{-}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{-}{2}{,}{\mathrm{index}}{=}{1}\right){,}{\mathrm{index}}{=}{1}\right)$ (2)
> $\mathrm{convert}\left(\left(1+{2}^{\frac{1}{2}}\right)I,\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{+}{1}{,}{\mathrm{index}}{=}{1}\right){}\left({1}{+}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{-}{2}{,}{\mathrm{index}}{=}{1}\right)\right)$ (3)
> $\mathrm{convert}\left({2}^{\frac{1}{4}}+{2}^{\frac{1}{3}},\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{4}}{-}{2}{,}{\mathrm{index}}{=}{1}\right){+}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{-}{2}{,}{\mathrm{index}}{=}{1}\right)$ (4)
> $\mathrm{convert}\left({\left({x}^{2}+2x+1\right)}^{\frac{1}{2}},\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{-}{{x}}^{{2}}{-}{2}{}{x}{-}{1}{,}{\mathrm{index}}{=}{1}\right)$ (5)
> $\mathrm{convert}\left({\left({x}^{3}-6{x}^{2}-6x-7\right)}^{\frac{1}{2}},\mathrm{RootOf}\right)$
${\mathrm{RootOf}}{}\left({-}{{x}}^{{3}}{+}{{\mathrm{_Z}}}^{{2}}{+}{6}{}{{x}}^{{2}}{+}{6}{}{x}{+}{7}{,}{\mathrm{index}}{=}{1}\right)$ (6)
> $\mathrm{convert}\left(,\mathrm{radical}\right)$
$\sqrt{{{x}}^{{3}}{-}{6}{}{{x}}^{{2}}{-}{6}{}{x}{-}{7}}$ (7)
> $a≔\mathrm{RootOf}\left({x}^{3}-x-1,\mathrm{index}=1\right)$
${a}{≔}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{-}{\mathrm{_Z}}{-}{1}{,}{\mathrm{index}}{=}{1}\right)$ (8)
> $\mathrm{convert}\left(a,'\mathrm{RootOf}',\mathrm{form}=\mathrm{interval}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{-}{\mathrm{_Z}}{-}{1}{,}\frac{{10349359}}{{7812500}}{..}\frac{{662358981}}{{500000000}}\right)$ (9)
> $\mathrm{convert}\left(a,'\mathrm{RootOf}',\mathrm{form}=\mathrm{numeric}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{-}{\mathrm{_Z}}{-}{1}{,}{1.324717957}\right)$ (10)
> $b≔\mathrm{RootOf}\left({\mathrm{_Z}}^{5}-\mathrm{_Z}-1,1\right)$
${b}{≔}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{-}{\mathrm{_Z}}{-}{1}{,}{1}\right)$ (11)
> $\mathrm{convert}\left(b,'\mathrm{RootOf}',\mathrm{form}=\mathrm{index}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{-}{\mathrm{_Z}}{-}{1}{,}{\mathrm{index}}{=}{1}\right)$ (12)
> $\mathrm{convert}\left(b,'\mathrm{RootOf}',\mathrm{form}=\mathrm{interval}\right)$
${\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{-}{\mathrm{_Z}}{-}{1}{,}\frac{{1167303973}}{{1000000000}}{..}\frac{{1167303983}}{{1000000000}}\right)$ (13)
Compatibility
• The convert/RootOf command was updated in Maple 18.
• The form parameter was introduced in Maple 18.
• For more information on Maple 18 changes, see Updates in Maple 18. | 1,579 | 4,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.444964 |
https://www.slideshare.net/MrsSanders/3e-pythagorean-theorem | 1,501,089,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426234.82/warc/CC-MAIN-20170726162158-20170726182158-00224.warc.gz | 874,103,639 | 32,576 | Upcoming SlideShare
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# 3e pythagorean theorem
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Pythagorean Theorem - Introduction - 7th grade math
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### 3e pythagorean theorem
1. 1. 3e: Create an argument using Pythagorean Theorem principles to show that a triangle is a right triangle<br />By: Deia Sanders<br />
2. 2. Vocabulary<br />Right Angle: an angle that measures 90 degrees<br />Vertex: The point where 2 sides of a polygon meet<br />Right Triangle: a triangle with one right angle<br />Legs of a Right triangle: the two sides of a right triangle that form the right angle<br />Hypotenuse: the side directly across from the right angle; always the long side of the right triangle <br />
3. 3. Pythagorean Theorem<br />a2 + b2 = c2<br />leg2 + leg2 = hypotenuse2<br />
4. 4. Let’s look at page 46<br /> | 314 | 1,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.671073 |
https://nrich.maths.org/public/topic.php?code=5039&cl=1&cldcmpid=141 | 1,571,186,488,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00177.warc.gz | 621,541,956 | 9,226 | # Search by Topic
#### Resources tagged with Interactivities similar to Four Triangles Puzzle:
Filter by: Content type:
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### There are 142 results
Broad Topics > Information and Communications Technology > Interactivities
### Four Triangles Puzzle
##### Age 5 to 11 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Inside Triangles
##### Age 5 to 7 Challenge Level:
How many different triangles can you draw on the dotty grid which each have one dot in the middle?
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Coloured Squares
##### Age 5 to 7 Challenge Level:
Use the clues to colour each square.
### Sort the Street
##### Age 5 to 7 Challenge Level:
Sort the houses in my street into different groups. Can you do it in any other ways?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Tetrafit
##### Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Turning
##### Age 5 to 7 Challenge Level:
Use your mouse to move the red and green parts of this disc. Can you make images which show the turnings described?
### Find the Difference
##### Age 5 to 7 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### Tessellate the Triominoes
##### Age 5 to 7 Challenge Level:
What happens when you try and fit the triomino pieces into these two grids?
### Cover the Camel
##### Age 5 to 7 Challenge Level:
Can you cover the camel with these pieces?
### 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.
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Combining Cuisenaire
##### Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### Difference
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Making Trains
##### Age 5 to 7 Challenge Level:
Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it?
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Getting the Balance
##### Age 5 to 7 Challenge Level:
If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced?
### Triangles All Around
##### Age 7 to 11 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
### 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?
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Fair Exchange
##### Age 5 to 7 Challenge Level:
In your bank, you have three types of coins. The number of spots shows how much they are worth. Can you choose coins to exchange with the groups given to make the same total?
### Two and One
##### Age 5 to 7 Challenge Level:
Terry and Ali are playing a game with three balls. Is it fair that Terry wins when the middle ball is red?
### Three Ball Line Up
##### Age 5 to 7 Challenge Level:
Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land.
### 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.
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Cuisenaire Counting
##### Age 5 to 7 Challenge Level:
Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Triangle Edges
##### Age 5 to 7 Challenge Level:
How many triangles can you make using sticks that are 3cm, 4cm and 5cm long?
### 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.
##### Age 7 to 11 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Triangle Animals
##### Age 5 to 7 Challenge Level:
How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all?
### Matching Triangles
##### Age 5 to 7 Challenge Level:
Can you sort these triangles into three different families and explain how you did it?
### 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?
### First Connect Three for Two
##### Age 7 to 11 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### One to Fifteen
##### Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
### Junior Frogs
##### Age 5 to 11 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### A Square of Numbers
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Winning the Lottery
##### Age 7 to 11 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### Which Symbol?
##### Age 7 to 11 Challenge Level:
Choose a symbol to put into the number sentence.
##### Age 5 to 11 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Sizing Them Up
##### Age 5 to 7 Challenge Level:
Can you put these shapes in order of size? Start with the smallest.
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Domino Numbers
##### Age 7 to 11 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Overlapping Again
##### Age 7 to 11 Challenge Level:
What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation.
### Posting Triangles
##### Age 5 to 7 Challenge Level:
If you can post the triangle with either the blue or yellow colour face up, how many ways can it be posted altogether?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Shapely Tiling
##### Age 7 to 11 Challenge Level:
Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle? | 2,105 | 9,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.879889 |
https://www.mathematicshub.edu.au/plan-teach-and-assess/teaching/lesson-plans/fruit-fractions-fruit-salad-crafty-creations/ | 1,722,672,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00057.warc.gz | 708,249,763 | 12,970 | Year level: 3 / 4
Strand: Number
Lesson length: 60 mins
Students demonstrate knowledge of fractions through creating their very own simulated ‘fruit salad’. Pose the questions: if you were to create a fruit salad what would it have in there? In what ratios/proportions? Are there any fruits you wouldn’t want in there? Which fruits go/don’t go together? Showcase your understanding of common (and decimal) fractions through this hands-on ‘Creative Craft’ learning experience.
This lesson is one of a series of lessons developed in collaboration with the Australian Association of Mathematics Teachers (AAMT).
Achievement standard
• Students represent unit fractions and their multiples in different ways.
• They recognise equivalent fractions and make connections between fraction and decimal notations. Students count and represent fractions on a number line.
Content descriptions
Students recognise and represent unit fractions including 1 2 , 1 3 , 1 4 , 1 5 and 1 10 and their multiples in different ways; combine fractions with the same denominator to complete the whole. AC9M3N02
They find equivalent representations of fractions using related denominators and make connections between fractions and decimal notation. AC9M4N03
Students count by fractions including mixed numerals; locate and represent these fractions as numbers on number lines. AC9M4N04
General capabilities
Literacy
Numeracy
Critical and Creative Thinking
• Create possibilities (Level 3)
• Draw conclusions and provide reasons (Level 3)
Cross-curricular priorities
Aboriginal and Torres Strait Islander Histories and Cultures: Country/Place (A_TSICP1)
The following formative assessment is suggested for this lesson.
Students present their ‘fruit salads’ to the class, or alternatively celebrate the lesson with a hands-on fruit fraction feast, allowing students to enjoy real fruits in fractions and showcase their learning.
Some students may:
• express fractions creatively; that is, creative application of fractions and decimals in crafting fruit salad representations may pose challenges, as it requires students to think abstractly and find innovative ways to visually represent proportions
• lack of deep understanding of fractions and may think, for example, that if they add another banana piece to their fruit salad which is currently 2 18 banana, it will now be 3 18 banana, not 3 19
• not be used to being autonomous in a task such as this, so may get stuck choosing materials, or which fruits they may wish to use.
It is expected that students have:
• basic arithmetic skills (addition, subtraction, and multiplication)
• some familiarity with common fruits and their shapes or parts is useful
• some experience with common (and decimal) fractions.
What you need:
• Lesson plan (Word)
• Teacher's slides (PowerPoint)
• Guess who fractions game (Word)
• Cardboard posters for each student
• Different coloured Unifix, MultiLink, Lego, coloured paper or other physical materials
• Real chopped fruit (optional) | 637 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.888776 |
https://www.jiskha.com/display.cgi?id=1372188707 | 1,503,380,218,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110471.85/warc/CC-MAIN-20170822050407-20170822070407-00641.warc.gz | 947,876,945 | 3,592 | # algebra
posted by .
x^2-15x+36=0
• algebra -
12+3=15
12*3=36
that should help
## Similar Questions
1. ### algebra
-15(3+x)=-6x-9 -45-15x=-6x-9 -45-15x+15x=-6x+-15x-9 -45=21x-9 -45=x-9 -45/9=5 x=5 is this right
2. ### algebra
15x^2 / 15x+10 simpfly each and exclude the values
3. ### algebra
find the pair of parallel lines 1)-12y+15x=4 2)4y=-5x-4 3)15x=12y=-4
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f(x)=13x^2 -15x-14 g(x)=15x^3+14x^2-15x+12 find f(3) × f(10) + f(-7) × g(9)
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f(x)=13x^2 -15x-14 g(x)=15x^3+14x^2-15x+12 find f(3) × f(10) + f(-7) × g(9)
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Solve by elimination 5x+y=2 3x+y=1 This is what I got : 3(5x+y=2) -5(3x+y=1) 15x+3y=6 -15x-5y=-5 -2y=1 2 2 Y=1/2 15x+3(1/2)=6 15x+3/2 =6 15x/15= 9/2 /15 X=3/10
7. ### Algebra
I need help trying to put the correct vocabulary words when solving this problem such as grouping, factor, prime factor, GCF, and perfect square, and please check and see if I am doing this correctly. 15x^2+31x+2 In my last problem …
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I need help trying to put the correct vocabulary words when solving this problem such as grouping, factor, prime factor, GCF, and perfect square, and please check and see if I am doing this correctly. 15x^2+31x+2 In my last problem …
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Ms.lynch has 21 coins in nickles and dimes.Their total value is \$1.65. How many of each coin does she have?
10. ### Algebra
Can you please advise if I worked this problem correctly: 11 + x = -15y 2x - 5y = 48 15y = 11+x y = .73 + 15x 2x (.73 + 15X) - 48 2x -3.65 -75x = 48 -73x - 3.65 = 48 x = 51.65 x = -.70
More Similar Questions | 656 | 1,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.722693 |
http://mymathforum.com/probability-statistics/340906-conditional-distribution-x-given-y-when-both-x-y-gaussian.html | 1,516,115,992,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00330.warc.gz | 239,299,032 | 9,062 | My Math Forum Conditional distribution of X given Y when both X and Y are gaussian
Probability and Statistics Basic Probability and Statistics Math Forum
June 14th, 2017, 04:16 AM #1 Newbie Joined: Jun 2017 From: Singapore Posts: 1 Thanks: 0 Conditional distribution of X given Y when both X and Y are gaussian Two Random variables X and Y are gaussian.Is the conditional distribution of X given Y that is X|Y also a gaussian when Y lies in certain range from a<=y<=b?I know its true when y has a specific value but what if it lies in a certain range as mentioned above?
June 14th, 2017, 08:48 AM #2 Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics I'm not quite sure what you mean Y being in a certain range... If Y is Gaussian, its support should be $\displaystyle \mathbb{R}$, so if you know Y lies in a proper subset of $\displaystyle \mathbb{R}$, it probably can't be Gaussian any more... Without the restriction, the conditional distribution X | Y would be $\displaystyle \frac{ \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right] \right) }{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left( -\frac{(y-\mu_Y)^2}{2\sigma_Y^2} \right)}$ which simplifies to $\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right]+\frac{(y-\mu_y)^2}{2\sigma_Y^2} \right)$ and then $\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)\sigma_X^2}\left[ (x-\mu_X)^2 + \frac{\rho^2\sigma_X^2(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\sigma_X\rho(x-\mu_X)(y-\mu_Y)}{ \sigma_Y} \right] \right)$ which finally leaves us with $\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)\sigma_X^2}\left[ x-\mu_X - \frac{\rho\sigma_X(y-\mu_Y)}{\sigma_Y} \right]^2 \right)$ which means $\displaystyle X | Y \sim N(\mu_X + \frac{\rho\sigma_X(y-\mu_Y)}{\sigma_Y}, (1-\rho^2)\sigma_X^2)$. Phew! That took me a long time to type. Last edited by 123qwerty; June 14th, 2017 at 09:04 AM.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 965 | 2,795 | {"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} | 3.578125 | 4 | CC-MAIN-2018-05 | longest | en | 0.767943 |
https://psu.pb.unizin.org/matheconcompanion/chapter/ch2/ | 1,719,301,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00622.warc.gz | 406,290,119 | 20,579 | # 2 Static Equilibrium Analysis: An Algebraic Approach
## 2.1 Equilibrium
### Practice Problems
There are no practice problems for this section.
### External Resources
There are no external resources for this section.
## 2.2 Using algebra to solve Linear systems: Two equations
### Practice Problems
Problem 2.2.1: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solutions:
Problem 2.2.2: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution: Inconsistent System (No Solutions)
Problem 2.2.3: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution:
Problem 2.2.4: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution: Inconsistent System (No Solution)
Problem 2.2.5: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution: Dependent solution (infinite number of solutions)
Problem 2.2.6: A moving company charges a flat rate of 150USD,and an additional 5USD for each box. If a taxi service would charge 20USD for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost?
Solution: Use moving company for more than 10 boxes.
Problem 2.2.7: If an investor invests 23,000USD into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns 710.00USD annual interest, how much was invested in each account?
Solution: 12,500 deposited in first bond, 10,500 deposited in second bond
### External Resources
Khan Academy: System of Equations using Elimination
Khan Academy: System of Equations using Substitution
Khan Academy: The Number of Solutions to a System of Equations
## 2.3 Using algebra to solve linear systems: Three or more variables
### Practice Problems
Problem 2.3.1: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution:
Problem 2.3.2: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution:
Problem 2.3.3: Solve the following system of equations, if possible, with whatever method you would like (other than a calculator.)
Solution: Dependent system (no solution)
Problem 2.3.4: At a carnival, 10,400USD in receipts were taken at the end of the day. The cost of a child’s ticket was 20USD, an adult ticket was 30USD, and a senior citizen ticket was 15USD. There were twice as many senior citizens as adults in attendance, and 20 more children than senior citizens. How many children, adult, and senior citizen tickets were sold?
Solution: 220 children, 100 adults, 200 seniors
Problem 2.3.5: You inherit one million dollars. You invest it all in three accounts for one year. The first account pays 3% compounded annually, the second account pays 4% compounded annually, and the third account pays 2% compounded annually. After one year, you earn 34,000USD in interest. If you invest four times the money into the account that pays 3% compared to 2%, how much did you invest in each account?
Solution: 400,000 in account 1 (3%), 500,000 in account 2 (4%), 100,000 in account 3 (2%)
### External Resources
Khan Academy: Intro to Linear Systems with Three Variables
Khan Academy: Solving Linear Systems with Three Variables | 849 | 3,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.875629 |
https://win-vector.com/2015/10/02/a-simpler-explanation-of-differential-privacy/?like_comment=636&_wpnonce=bdd05331a3 | 1,604,172,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00268.warc.gz | 602,880,813 | 25,269 | A Simpler Explanation of Differential Privacy
Differential privacy was originally developed to facilitate secure analysis over sensitive data, with mixed success. It’s back in the news again now, with exciting results from Cynthia Dwork, et. al. (see references at the end of the article) that apply results from differential privacy to machine learning.
In this article we’ll work through the definition of differential privacy and demonstrate how Dwork et.al.’s recent results can be used to improve the model fitting process.
The Voight-Kampff Test: Looking for a difference. Scene from Blade Runner
The Definition of Differential Privacy
We can define epsilon-differential privacy through the following game:
• A learner implements a summary statistic called A().
• A (notional) adversary proposes two data sets S and S’ that differ by only one row or example, and a test set Q.
• A() is called epsilon-differentially private iff:
| log( Prob[A(S) in Q] / Prob[A(S’) in Q] ) | ≤ epsilon
for all of the adversary’s choices of S, S’ and Q. The probabilities are defined over coin flips in the implementation of A(), not over the data or the adversary’s choices.
The adversary’s goal is to use A() to tell between S and S’, representing a failure of privacy. The learner wants to extract useful statistics from S and S’ without violating privacy. Identifying a unique row (the one which changed markings) violates privacy. If the adversary can tell which set (S or S’) the learner is working on by the value of A(), then privacy is violated.
Notice S and S’ are “data sets” in the machine learning sense (collections of rows carrying information). Q is a set in the mathematic sense: a subset of the possible values that A() can return. To simplify the demonstration, we will take Q to be an interval.
In the following examples, A() returns the (approximate) expected value of a set s. The adversary has chosen two sets S and S’ of size n = 100:
• S = {0,0,0,…,0} (100 zeros)
• S’ = {1,0,0,…,0} (1 one and 99 zeroes)
The set Q will be an interval [T, 1], where the adversary picks the threshold T.
The adversary’s goal is to pick T such that when he sees that A(s) ≥ T, he knows that A() has just evaluated S’. The learner has two (competing) goals:
• To pick an algorithm A() such that A(S) and A(S’) are so “close” that the adversary can’t pick a reliable T, to preserve differential privacy. “Close” is defined by epsilon.
• To have A() be a good estimate of the expectation, for performing useful analysis.
The Deterministic Case
Here, A(s) simply returns the expected value `mean(s)`, so it always returns A(S) = 0 when evaluating S, and A(S’) = 0.01 when evaluating S’. This is clearly not differentially private for any value of epsilon. If the adversary picks T = 1/200 = 0.005, then he can reliably identify when A() has evaluated the set S’, every time they play a round of the game.
In the figures, set1 in green represents the distribution of values returned by calls to A(S), and set2 in orange returns the distribution of values returned by calls to A(S’). I’ve plotted set2 upside down for clarity.
One way for the learner to obscure which set she is evaluating is to add a little random noise to the estimate, to “blur” it. Following Dwork et.al.’s methods, we’ll add noise from a Laplace distribution, which is symmetric and falls off slower than gaussian noise would. Here we show adding just a little bit of noise (of “width” sigma = 1/3n):
The shaded green region represents the chance that A(S) will return a value greater than the adversary’s threshold T — in other words, the probability that the adversary will mistake S for S’ in a round of the game. We’ve now made that probability non-zero, but it’s still much more likely that if A(s) > T, then s = S’. We need more noise.
In particular, we need the noise to be bigger than the gap A(S’)-A(S) = 1/n, or 0.01. Let’s pick sigma = 3/n = 0.03:
Now we see that the shaded green region has almost the same area as the shaded orange region — you can think of epsilon as expressing the difference between the shaded green region and the shaded orange region. In fact, the absolute value of the logratio of the two areas is epsilon. In other words, observing that A(s) > T is no longer strong evidence that s = S’, and we have achieved differential privacy, to the degree epsilon.
Of course, A(s) is also no longer a precise estimate of `mean(s)`. We’ll return to that point in a bit.
Simulating the Game
We can simulate the game I described above in R. I’ve put the code on github, here.
The code sweeps through a range of values of epsilon, and for each value selects a suitable noise width and runs 1000 rounds of the differential privacy game, returning a value for A(S) and A(S’). Each round we record whether A(S) and A(S’) are greater than T = 1/200.
From Differential Privacy to Machine Learning
Differential privacy aims to make the answers to “snooping queries” too vague to distinguish closely related sets (in this case, it makes the probability that A(S) ≥ T about the same as the probability that A(S’) ≥ T). But for machine learning, we are also interested in the output of A(). We can use the simulation above to estimate what happens to A(S) and A(S’) for different values of epsilon. Here we plot the (normalized) gap between the expected values of A(S) and A(S’) as a function of epsilon, from the simulation:
As epsilon gets smaller (implying stricter privacy), the relative gap between the expected values of A(S) and A(S’) gets smaller. The discrepancies in the plot are probably due to poor choices of the noise parameter (I picked them heuristically), but the trend is clear. This makes sense, since privacy implies that the A(S) should look a lot like A(S’).
However, as epsilon gets stricter, the estimates of `mean(S) = 0` and `mean(S') = 0.01` — which is what A() is supposed to estimate — also become poorer.
This is the trade-off: how can we preserve privacy and still perform useful analysis?
Differential Privacy and Adaptive Data Analysis
The recent papers by Dwork, et.al. apply differential privacy to the problem of adaptive data analysis, or reusable test sets. In standard machine learning practice, we use two data sets to model: a training set to fit the model, and a test set to evaluate the model. Ideally we only use the test set once; but in practice we go back to it again and again, as we try to improve our model. We already know that performance estimates of a model over its training set are upwardly biased (the model looks more accurate on training than it really is, because it “knows” the training set); if we go back to the test set too many times, then performance estimates of the model on the test set are also upwardly biased, because we also “know” the test set. This problem is exacerbated when we also use the test set to tune the model: for example to pick the variables we use, or tune modeling parameters. In other words, even if we observe the best practice of using a training set to fit models, a calibration set to tune models, and a holdout set to evaluate the model, we can also contaminate the calibration set if we look at it too many times.
Dwork, et.al., describe how (and how many times) we can go back to a test set without biasing its evaluations of model performance. To do this, we make sure that the modeling procedure only interacts with the test set in a differentially private manner. Because it has limited access to the test set, the modeling procedure can’t learn the test set, so evaluations of a model over the test set still accurately reflect a model’s performance on unseen data.
Describing the proofs and techniques in these papers is outside the scope of this article, but we can demonstrate the results in a simple example, similar to the example application shown in Dwork, et.al.’s Science paper.
Using Differential Privacy for Stepwise Regression
For our example, suppose we have 2000 rows of data with a binary outcome y (50% prevalence of the positive class), and 110 possible input variables to choose from. In our tests we used synthetic data with ten variables (`x1...x10`) that carry signal and one hundred noise variables (`n1...n100`).
We decide to use forward-stepwise logistic regression to choose a suitable model for predicting y. We split the sample data into 1000 rows for the training set, and 1000 rows for the test set. Given that we have selected k-1 variables and have fit a model, M(k-1), that uses those variables, we pick the kth variable by fitting a model Mnew on the training set using the (k-1) previously selected variables plus one new variable from the remaining candidates. We then evaluate the accuracy of M(k-1) and Mnew on the test set, and pick as the kth variable the one for which Mnew showed the most improvement. In other words, we fit candidate models on the training set, and evaluate them for improvement on the test set. This is a fairly common way of tuning models. Standard implementations of stepwise regression often use training set AIC as the criterion for picking the next model; we use accuracy improvement to keep the code straightforward, and closer to the assumptions in the paper.
Normally, the stepwise procedure would terminate when the model stops improving, but for this example we let it run to pick fifty variables. At each step we recorded the accuracy of the selected model on the training set (green circles), the test set (orange triangles) and on a fresh holdout set of 10,000 rows (purple squares). We picked this additional large holdout set (called “fresh”) to get a good estimate of the true accuracy of the model. Here’s the performance of the naive procedure:
The test set performance estimates were are more upwardly biased than the training set estimates! This is not too surprising, since we are using the test set to select variables. Despite the optimistic curve of the test scores, we can see from the true out-of-sample performance (`freshScore`) that the model performance is not much better than random; in fact, the algorithm only picked one of the ten signal carrying variables (the first one selected). The test set (and to a lesser extent, the training set) believe that performance continues to improve, when in fact performance degrades as more noise variables swamp out the single signal variable.
We want to use the test set to pick variables, while also using it to accurately estimate out-of-sample error. In their recent Science article, Dwork, et.al. propose an algorithm called Thresholdout to let us safely reuse the test set. We implemented our `diffPrivMethod` based on Thresholdout. `diffPrivMethod` evaluates proposed models on both the training and test sets. If the model’s delta accuracies on the test and the training sets are within a certain tolerance, then `diffPrivMethod` returns the training set delta accuracy with a bit of Laplacian noise as the model score. If the two delta accuracies are far apart, then `diffPrivMethod` adds a bit of Laplacian noise to the test set delta accuracy, and returns that as the model score. In R, the code would look something like this:
```# Return an estimate of model performance in a
# differentially private way
# v is the new candidate variable
# varSet is the set of already selected variables
diffPrivMethod = function(v) {
# Original model was fit on
# training data using varSet;
# Fit a new model using varSet+v
# using the training data
# and evaluate the difference
# between the new models on the test set
testScore <- modelScoreImprovement(varSet,v,
trainData,
testData)
# Do the same thing, but evaluate
# the improvement on the training set
trainScore <- modelScoreImprovement(varSet,v,
trainData,
trainData)
if(abs(testScore-trainScore)<=eps/2) {
# if the scores are close,
# return trainScore with a little noise
sc <- trainScore + rlaplace(1,eps/2)
} else {
# otherwise, return testScore
# with a little more noise
sc <- testScore + rlaplace(1,eps)
}
# make sure the score is in range [0,1]
pmax(0,pmin(1,sc))
}
```
We picked the tolerance and the widths of the Laplace noise somewhat arbitrarily.
The beauty of this scheme is that we never look directly at the test set. If the test and training set performances are similar, we see the (noisy) training performance; when we do look at the test set, we only see a noisy view of it, so it leaks information more slowly.
The original Thresholdout algorithm does not add noise to the training set performance, probably under the reasoning that we know it already. We decided to add noise anyway, on the principle that even knowing when you are using the test set performance leaks a little information, and we wanted to reduce the leakage a little more.
Here are the results of the stepwise regression using `diffPrivMethod` (we set the tolerance/noise parameter eps=0.04 for this run):
Now the test set estimates the true out-of-sample accuracies quite well. The chosen models achieve a peak accuracy of about 61%, at around 36 variables. This method did a better job of picking variables — it found all ten signal variables — though it did start picking noise variables fairly early on.
However, there is still money left on the table. What if we used a really large test set? Here, we show the results of running the naive stepwise regression (no differential privacy), with the same training set of 1000 rows and a test set of 10,000 rows.
Now the learning algorithm picks nine of the signal variables immediately, and achieves an accuracy of 62%. We can see that the model’s performance on the test set is still upwardly biased, but much less so than with the naive method. Because the test set is larger, we don’t contaminate it as much, even when looking at it directly.
This suggests that we can think of differential privacy as letting us simulate having a larger test set, though obviously we still need our actual test set to be a reasonable size, especially since stepwise regression requires a lot of queries to the test set. Our results also show that while these new differential-privacy based schemes let us do a good job of estimating out-of-sample performance, that alone is not enough to insure best possible model performance.
All code, data, and examples for this experiment can be found here.
References
Dwork, Cynthia, Vitaly Feldman, Moritz Hardt, Toniann Pitassi, Omer Reingold, Aaron Roth. “Preserving Statistical Validity in Adaptive Data Analysis”, arXiv:1411.2664, April 2015.
Dwork, Cynthia, Vitaly Feldman, Moritz Hardt, Toniann Pitassi, Omer Reingold, Aaron Roth. “The reusable holdout: Preserving validity in adaptive data analysis”, Science, vol 349, no 6248 pp 636-638, August 2015. [link to abstract]
Blum, Avrim and Moritz Hardt. “The Ladder: A Reliable Leaderboard for Machine Learning Competitions”, arXiv:1502.04585, February 2015.
nzumel
Data scientist with Win Vector LLC. I also dance, read ghost stories and folklore, and sometimes blog about it all.
6 replies ›
1. Shanx says:
Thanks for such a clear exposition!
The original papers show that the noise parameter is a function of the number of rows in the test set, number of queries, and the form of the statistic. The problem is, one would have to analytically derive the noise parameter for a statistic of choice.
These simulations suggest it might be possible to arrive at the noise parameter for a given dataset and learning algorithm by running the right simulations. Do you think that’s possible?
Like
1. Thanks for the excellent question! It so happens that for the simulations we ran in this article, we swept over a range of candidate values for a good noise parameter. I hesitate to recommend that as a general practice, because that is going “back to the well” of the dataset yet again.
In this case, we were generating the data, so we had as much of it as we wanted, and in practice I suppose one could have a separate calibration set to find good parameters, or use bootstrap simulation to generate another data set with appropriate distributions. But I haven’t thought too deeply about the implications of bootstrap sampling; my goal in this post was simply to demonstrate the technique.
It’s also worth noting that from our simulations the noise parameters that produce the model with the best *actual* out-of-sample performance are not necessarily the parameters for which the test set best *approximates* out-of-sample error. In other words, the best-performing model can still be upwardly biased with respect to the test set. Thresholdout is designed to optimize getting a good approximation of out-of-sample error, rather than directly optimizing out-of-sample model performance. While the two are related, there are still trade-offs.
Like
2. It looks like even though the paper isn’t open access the supporting material (and code) are: http://www.sciencemag.org/content/suppl/2015/08/05/349.6248.636.DC1 . Looking through the code we see the science graphs were not generated using Laplace noise as the authors seem to imply in the article, but a randomized chance of normal noise:
``` if abs(ftrain-fholdout) < threshold + np.random.normal(0,tolerance): fholdout = ftrain else: fholdout += np.random.normal(0,tolerance) ```
Though I think the noising of test and training used in your example may have similar benefit as the noised choice threshold (making it more obscure how test exactly relates to training when they are close to the threshold).
I think this very much supports that there is still room for experiment as to what are the best masking procedures (such as using a deterministic switch condition or, Bootstrapping http://www.win-vector.com/blog/2015/10/a-simple-differentially-private-procedure/ ).
And I found a couple more interesting commentaries on the Science article:
Also this is a good place to emphasize your demonstration is stepwise logistic regression (a hard problem of great utility to analysts that needs a lot of sub-queries) and the original paper is essentially demonstrating model size selection from roughly independently scored single variable models (also a good problem, but needs fewer queries as it doesn’t fully model variable to variable dependencies).
Also it looks like the science graphs appear “more stable” because they are actually the aggregate of 100 runs each- not the performance seen in single experiment.
Like
3. Very good post. It help me a lot to understand the DP algorithm.
Like
4. What I find puzzling about the Dwork et al paper is that it isn’t clear whether the problem it’s aiming to fix really exists or not. Does anyone use the TEST set error for variable selection? The forward-stepwise regression example shown here shows very clearly how bizarre this would be, since (as Nina points out) it leads to a scenario where test performance goes up much more quickly than training performance (“naive method” figs above).
Has anyone ever done this before? Has anyone seen a published paper using a method that gives smaller test error than training error? That seems fairly ludicrous to me. And as far as I can tell, Dwork et al don’t point to any papers in the literature that are guilty of the crime their method addresses.
[Technical remark: Nina notes that you would normally use AIC to pick the next regressor, but of course the penalty would be the same for all models with K+1 parameters, so really you’re just using training error to select which variable to add. And that’s what leads to the ‘standard’ scenario where training performance rises monotonically as you add regressors, but test performance behaves like an inverted U-shaped function. The REAL question should be: if you use the peak of this U-shaped function to select which model to use and report that peak as your test performance — which I think is standard if not enlightened practice — how biased is your reported test performance?].
Anyway, I don’t mean to be too negative because re-using test sets surely *is* a problem, and the idea of addressing it with differential privacy certainly seems novel and cool. I’m just curious to know how bad of a problem it is likely to be in terms of current practice, and I’m fairly surprised none of the reviewers at Science asked them to show application to a more realistic setting, i.e., where we can see how pernicious the problem is likely to be and how much of an improvement DP really provides.
Thanks Nina for an entertaining and informative post. I found this very clear and much easier to understand than the original paper!
Like
5. (John here) Thanks for your comment!
Nina and I did try some empirical runs (linked in article) and came to the conclusion that reserving some test data for step variable scoring is usually worse than the standard method of using all our training data to run stepwise regression. Also of interest was: if it was just a matter of picking how many variables (stopping) we wouldn’t have big problem- it is paying the multiple comparison penalty for all the different sets of variables you end up trying.
It is an interesting technique, but also kind of remarkable it got a science article using only one or two synthetic data sets (and also using a series heavily aggregated performance graphs). The acid test would indeed be publicly outperforming some standard technique.
Like | 4,717 | 21,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.914718 |
https://123deta.com/document/p1y93dyg-%E7%99%BA%E5%B1%95%EF%BC%92%E5%9B%9E%E7%9B%AE%E6%96%87%E5%AD%97%E5%BC%8F%E3%81%AE%E8%B6%B3%E3%81%97%E7%AE%97%E5%BC%95%E3%81%8D%E7%AE%97%E6%95%B0%E5%AD%A6%E7%AE%97%E6%95%B0%E3%81%AE%E6%95%99%E6%9D%90%E5%85%AC%E9%96%8B%E3%83%9A%E3%83%BC%E3%82%B8hatten13.html | 1,627,423,329,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00371.warc.gz | 86,754,535 | 27,547 | # 発展・２回目 文字式の足し算・引き算 数学・算数の教材公開ページ hatten0103 2test
(1)
## 1.
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−5a+ 1
Updating...
Updating... | 1,529 | 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} | 3.859375 | 4 | CC-MAIN-2021-31 | latest | en | 0.199339 |
https://math.stackexchange.com/questions/1477553/complex-integral-of-sqrtz-1 | 1,585,700,488,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370504930.16/warc/CC-MAIN-20200331212647-20200401002647-00548.warc.gz | 592,419,275 | 31,297 | # Complex Integral of $\sqrt{z-1}$
Let $\gamma$ be the half unit circle joining $1+i$ and $1-i$ clockwise. By direct parametrization compute,
$\int_{\gamma}$ $\sqrt{z-1}$ $dz$
where the principal branch of the square root is used.
So I first parametrized $\gamma$ as follows,
$\gamma(t)$ = $(1-i)+2it$, $0\leq t \leq 1$
$1-\exp(\frac{-\pi t i}{2})$, $1\leq t \leq 3$
Now I will have two integrals over two different paths,
$\int_{\gamma}$ $\sqrt{z-1}$ $dz$ = $\int_{\gamma_{1}}$$\sqrt{z-1}+\int_{\gamma_{2}}$$\sqrt{z-1}$
Now I know the principal branch of the square root of $z-1$ is defined as
$\exp(\frac{1}{2}\log(z-1))$
where log is a branch of the logarithm. Here I will use the logarithm function as defined in the slit complex plane, writing $z=r\exp(i\theta)$, by
$\text{Log}(z) = \log(r) + i\theta$
where log is the usual logarithm over the real numbers and $-\pi < \theta < \pi$. Am I on the right track for this problem?
• $\gamma$ as defined by that formula doesn't look like a half unit circle at all; it looks like the union of two line segments. – user14972 Oct 13 '15 at 3:55
Let's parameterize the curve $\gamma$ as $z=1+e^{it}$, where $t$ starts at $\pi/2$ and ends at $-\pi/2$. Then,
\begin{align} \int_{\gamma}\sqrt{z-1}\,dz&=\int_{\pi/2}^{-\pi/2}\sqrt{e^{it}}ie^{it}\,dt\\\\ &=\frac23 \left.\left(e^{it}\right)^{3/2}\right|_{\pi/2}^{-\pi/2}\\\\ &=-i\frac{2\sqrt{2}}{3} \end{align} | 505 | 1,415 | {"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": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-16 | latest | en | 0.763777 |
https://www.coursehero.com/file/5805421/s60/ | 1,496,062,686,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612283.85/warc/CC-MAIN-20170529111205-20170529131205-00441.warc.gz | 1,050,735,410 | 25,805 | # s60 - This is of course the famous Fibonacci sequence An...
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Solution to Problem 60 Congratulations to this week’s winners Nathan Pauli, Ray Kremer Further correct solutions were also received from William Webb, Robert McQuaid, Brian Laughlin, Richard Oak, Burkart Venzke, Douglas Vander Griend, Jim Wang. Numerous incorrect solutions were submitted. Let’s start out by looking to see what numbers must be on the list. I’ll assume the numbers are listed in increasing order, though this was not strictly a requirement. Clearly the number 1 must be the first number on the list, while 2 must be the second. As 3 cannot be written as the sum of smaller non-consecutive numbers on the list, it must be the third one. Now 4 = 1 + 3, but 5 must be added. Then 6 = 5 + 1, 7 = 5 + 2, but 8 must be added. Some patient experimenting will show you that the list must start with 1, 2, 3, 5, 8, 13, 21, 34, 55,. .. and you’ll see that the next number you have to add is the sum of the previous two.
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Unformatted text preview: This is, of course, the famous Fibonacci sequence. An easy proof that this list will always work was offered by Douglas Vander Griend. Take a number N . If it’s a Fibonacci number, it’s on the list. Otherwise, subtract off the largest Fibonacci number, F less that N . The result will be a number strictly less than the Fibonacci number before F . Induction finishes the argument. Burkart Venzke suggests modifying the third condition so that no n consecutive numbers on the list are allowed to appear in any one sum. With n = 3 the first few terms are 1, 2, 3, 4, 6, 9, 13, 28,. .. where each number now is the sum of the previous one and the one three steps before, that is a k = a k-1 + a k-3 . The general case leads to a k = a k-1 + a k-n . Eliminating the condition on consecutive list numbers is the same as setting n = 0, which gives a list of powers of 2....
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## This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
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s60 - This is of course the famous Fibonacci sequence An...
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# If m<n and a<b, then a. (m+a)<n-b b.
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If m[#permalink] 07 Mar 2004, 11:35
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If m<n and a<b, then
a. (m+a)<n-b
b. (m+a)<n+b
c. (ma)<nb
d. (m/a)<n/b
e.
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Pls include reasoning along with all answer posts.
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I believe B is the answer.
a) not satisfied.
Let m = 1 n = 2 a = 1 b = 2
m+a = 2 n-b = 0
b) satisfied
for all +ve values
for all -ve values
c) not satisfied
m = -2 n = 1 a = -2 b = 1
mn = 4 ab = 1
d) not satisfied
m = -2 n = 1 a = -2 b = 1
m/a = 1 n/b = 1
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anand,
What I would like to know is what made you choose the plug-in figures that you had chosen to test for validity?
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Pls include reasoning along with all answer posts.
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Display posts from previous: Sort by | 798 | 2,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2015-27 | longest | en | 0.858845 |
https://orbit-ml.readthedocs.io/en/latest/tutorials/ktr2.html | 1,637,993,728,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00191.warc.gz | 546,958,884 | 14,518 | # Kernel-based Time-varying Regression - Part II¶
The previous tutorial covered the basic syntax and structure of KTR (or so called BTVC); time-series data was fitted with a KTR model accounting for trend and seasonality. In this tutorial a KTR model is fit with trend, seasonality, and additional regressors. To summarize part 1, KTR considers a time-series as an additive combination of local-trend, seasonality, and additional regressors. The coefficients for all three components are allowed to vary over time. The time-varying of the coefficients is modeled using kernel smoothing of latent variables. This can also be an advantage of picking this model over other static regression coefficients models.
This tutorial covers:
1. KTR model structure with regression
2. syntax to initialize, fit and predict a model with regressors
3. visualization of regression coefficients
[16]:
import pandas as pd
import numpy as np
from math import pi
import matplotlib.pyplot as plt
import orbit
from orbit.models import KTR
from orbit.diagnostics.plot import plot_predicted_components
from orbit.utils.plot import get_orbit_style
from orbit.constants.palette import OrbitPalette
%matplotlib inline
pd.set_option('display.float_format', lambda x: '%.5f' % x)
orbit_style = get_orbit_style()
plt.style.use(orbit_style);
[2]:
print(orbit.__version__)
1.1.0dev
## Model Structure¶
This section gives the mathematical structure of the KTR model. In short, it considers a time-series ($$y_t$$) as the linear combination of three parts. These are the local-trend ($$l_t$$), seasonality (s_t), and regression ($$r_t$$) terms at time $$t$$. That is
$y_t = l_t + s_t + r_t + \epsilon_t, ~ t = 1,\cdots, T,$
where
• $$\epsilon_t$$s comprise a stationary random error process.
• $$r_t$$ is the regression component which can be further expressed as $$\sum_{i=1}^{I} {x_{i,t}\beta_{i, t}}$$ with covariate $$x$$ and coefficient $$\beta$$ on indexes $$i,t$$
For details of how on $$l_t$$ and $$s_t$$, please refer to Part I.
Recall in KTR, we express coefficients as
$B=K b^T$
where - coefficient matrix $$\text{B}$$ has size $$t \times P$$ with rows equal to the $$\beta_t$$ - knot matrix $$b$$ with size $$P\times J$$; each entry is a latent variable $$b_{p, j}$$. The $$b_j$$ can be viewed as the “knots” from the perspective of spline regression and $$j$$ is a time index such that $$t_j \in [1, \cdots, T]$$. - kernel matrix $$K$$ with size $$T\times J$$ where the $$i$$th row and $$j$$th element can be viewed as the normalized weight $$k(t_j, t) / \sum_{j=1}^{J} k(t_j, t)$$
In regression, we generate the matrix $$K$$ with Gaussian kernel $$k_\text{reg}$$ as such:
$$k_\text{reg}(t, t_j;\rho) = \exp ( -\frac{(t-t_j)^2}{2\rho^2} ),$$
where $$\rho$$ is the scale hyper-parameter.
## Data Simulation Module¶
In this example, we will use simulated data in order to have true regression coefficients for comparison. We propose two set of simulation data with three predictors each:
The two data sets are: - random walk - sine-cosine like
Note the data are random so it may be worthwhile to repeat the next few sets a few times to see how different data sets work.
### Random Walk Simulated Dataset¶
[5]:
def sim_data_seasonal(n, RS):
""" coefficients curve are sine-cosine like
"""
np.random.seed(RS)
# make the time varing coefs
tau = np.arange(1, n+1)/n
data = pd.DataFrame({
'tau': tau,
'date': pd.date_range(start='1/1/2018', periods=n),
'beta1': 2 * tau,
'beta2': 1.01 + np.sin(2*pi*tau),
'beta3': 1.01 + np.sin(4*pi*(tau-1/8)),
'x1': np.random.normal(0, 10, size=n),
'x2': np.random.normal(0, 10, size=n),
'x3': np.random.normal(0, 10, size=n),
'trend': np.cumsum(np.concatenate((np.array([1]), np.random.normal(0, 0.1, n-1)))),
'error': np.random.normal(0, 1, size=n) #stats.t.rvs(30, size=n),#
})
data['y'] = data.x1 * data.beta1 + data.x2 * data.beta2 + data.x3 * data.beta3 + data.error
return data
[7]:
def sim_data_rw(n, RS, p=3):
""" coefficients curve are random walk like
"""
np.random.seed(RS)
# initializing coefficients at zeros, simulate all coefficient values
lev = np.cumsum(np.concatenate((np.array([5.0]), np.random.normal(0, 0.01, n-1))))
beta = np.concatenate(
[np.random.uniform(0.05, 0.12, size=(1,p)),
np.random.normal(0.0, 0.01, size=(n-1,p))],
axis=0)
beta = np.cumsum(beta, 0)
# simulate regressors
covariates = np.random.normal(0, 10, (n, p))
# observation with noise
y = lev + (covariates * beta).sum(-1) + 0.3 * np.random.normal(0, 1, n)
regressor_col = ['x{}'.format(pp) for pp in range(1, p+1)]
data = pd.DataFrame(covariates, columns=regressor_col)
beta_col = ['beta{}'.format(pp) for pp in range(1, p+1)]
beta_data = pd.DataFrame(beta, columns=beta_col)
data = pd.concat([data, beta_data], axis=1)
data['y'] = y
data['date'] = pd.date_range(start='1/1/2018', periods=len(y))
return data
[8]:
rw_data = sim_data_rw(n=300, RS=2021, p=3)
[8]:
x1 x2 x3 beta1 beta2 beta3 y date
0 14.02970 -2.55469 4.93759 0.07288 0.06251 0.09662 6.11704 2018-01-01
1 6.23970 0.57014 -6.99700 0.06669 0.05440 0.10476 5.35784 2018-01-02
2 9.91810 -6.68728 -3.68957 0.06755 0.04487 0.11624 4.82567 2018-01-03
3 -1.17724 8.88090 -16.02765 0.05849 0.04305 0.12294 3.63605 2018-01-04
4 11.61065 1.95306 0.19901 0.06604 0.03281 0.11897 5.85913 2018-01-05
5 7.31929 3.36017 -6.09933 0.07825 0.03448 0.10836 5.08805 2018-01-06
6 0.53405 8.80412 -1.83692 0.07467 0.01847 0.10507 4.59303 2018-01-07
7 -16.03947 0.27562 -22.00964 0.06887 0.00865 0.10749 1.26651 2018-01-08
8 -17.72238 2.65195 0.22571 0.07007 0.01008 0.10432 4.10629 2018-01-09
9 -7.39895 -7.63162 3.25535 0.07715 0.01498 0.09356 4.30788 2018-01-10
### Sine-Cosine Like Simulated Dataset¶
[9]:
sc_data = sim_data_seasonal(n=80, RS=2021)
[9]:
tau date beta1 beta2 beta3 x1 x2 x3 trend error y
0 0.01250 2018-01-01 0.02500 1.08846 0.02231 14.88609 1.56556 -14.69399 1.00000 -0.73476 1.01359
1 0.02500 2018-01-02 0.05000 1.16643 0.05894 6.76011 -0.56861 4.93157 1.07746 -0.97007 -1.00463
2 0.03750 2018-01-03 0.07500 1.24345 0.11899 -4.18451 -5.38234 -13.90578 1.19201 -0.13891 -8.80009
3 0.05000 2018-01-04 0.10000 1.31902 0.20098 -8.06521 9.01387 -0.75244 1.22883 0.66550 11.59721
4 0.06250 2018-01-05 0.12500 1.39268 0.30289 5.55876 2.24944 -2.53510 1.31341 -1.58259 1.47715
5 0.07500 2018-01-06 0.15000 1.46399 0.42221 -7.05504 12.77788 14.25841 1.25911 -0.98049 22.68806
6 0.08750 2018-01-07 0.17500 1.53250 0.55601 11.30858 6.29269 7.82098 1.23484 -0.53751 15.43357
7 0.10000 2018-01-08 0.20000 1.59779 0.70098 6.45002 3.61891 16.28098 1.13237 -1.32858 17.15636
8 0.11250 2018-01-09 0.22500 1.65945 0.85357 1.06414 36.38726 8.80457 1.02834 0.87859 69.01607
9 0.12500 2018-01-10 0.25000 1.71711 1.01000 4.22155 -12.01221 8.43176 1.00649 -0.22055 -11.27534
## Fitting a Model with Regressors¶
The metadata for simulated data sets.
[10]:
# num of predictors
p = 3
regressor_col = ['x{}'.format(pp) for pp in range(1, p + 1)]
response_col = 'y'
date_col='date'
As in Part I KTR follows sklearn model API style. First an instance of the Orbit class KTR is created. Second fit and predict methods are called for that instance. Besides providing meta data such response_col, date_col and regressor_col, there are additional args to provide to specify the estimator and the setting of the estimator. For details, please refer to other tutorials of the Orbit site.
[11]:
ktr = KTR(
response_col=response_col,
date_col=date_col,
regressor_col=regressor_col,
prediction_percentiles=[2.5, 97.5],
seed=2021,
estimator='pyro-svi',
)
Here predict has the additional argument decompose=True. This returns the compponents ($$l_t$$, $$s_t$$, and $$r_t$$) of the regression along with the prediction.
[12]:
ktr.fit(df=rw_data)
INFO:root:Guessed max_plate_nesting = 1
Initial log joint probability = -4772.9
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
19 -237.796 0.0939091 34.0127 0.4556 0.4556 23
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
39 -235.661 0.000621044 30.1787 1 1 54
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
59 -235.645 6.03892e-06 29.0296 1.335 0.1335 86
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
76 -235.644 2.24975e-06 27.6394 7.389e-08 0.001 157 LS failed, Hessian reset
79 -235.644 9.23206e-07 29.145 0.04556 1 163
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
83 -235.644 3.67315e-09 31.2878 0.03427 0.03427 171
Optimization terminated normally:
Convergence detected: absolute parameter change was below tolerance
[12]:
date prediction_2.5 prediction prediction_97.5 trend_2.5 trend trend_97.5 regression_2.5 regression regression_97.5
0 2018-01-01 5.07427 6.21155 7.47124 4.03221 5.12639 6.38328 0.67423 1.01414 1.43433
1 2018-01-02 3.26790 4.19824 5.23103 4.13565 5.15971 6.14440 -1.19338 -0.90444 -0.71797
2 2018-01-03 3.59868 4.68268 6.03930 4.05149 5.19522 6.43199 -0.84052 -0.49468 -0.21665
3 2018-01-04 1.78598 2.99446 4.28573 3.95409 5.14134 6.38946 -2.68558 -2.14146 -1.71629
4 2018-01-05 4.43589 5.49128 6.42865 4.18216 5.15453 6.10805 0.12080 0.35538 0.57076
## Visualization of Regression Coefficient Curves¶
The function get_regression_coefs to extract coefficients (they will have central credibility intervals if the argument include_ci=True is used).
[13]:
coef_mid, coef_lower, coef_upper = ktr.get_regression_coefs(include_ci=True)
[14]:
coef_mid.head(5)
[14]:
date x1 x2 x3
0 2018-01-01 0.02376 0.02569 0.15483
1 2018-01-02 0.02364 0.02506 0.15203
2 2018-01-03 0.02352 0.02442 0.14920
3 2018-01-04 0.02339 0.02377 0.14636
4 2018-01-05 0.02326 0.02310 0.14351
Because this is simulated data it is possible to overlay the estimate with the true coefficients.
[20]:
fig, axes = plt.subplots(p, 1, figsize=(12, 12), sharex=True)
x = np.arange(coef_mid.shape[0])
for idx in range(p):
axes[idx].plot(x, coef_mid['x{}'.format(idx + 1)], label='est' if idx == 0 else "", color=OrbitPalette.BLUE.value)
axes[idx].fill_between(x, coef_lower['x{}'.format(idx + 1)], coef_upper['x{}'.format(idx + 1)], alpha=0.2, color=OrbitPalette.BLUE.value)
axes[idx].scatter(x, rw_data['beta{}'.format(idx + 1)], label='truth' if idx == 0 else "", s=10, alpha=0.6, color=OrbitPalette.BLACK.value)
axes[idx].set_title('beta{}'.format(idx + 1))
fig.legend(bbox_to_anchor = (1,0.5));
To plot coefficients use the function plot_regression_coefs from the KTR class.
[21]:
ktr.plot_regression_coefs(figsize=(10, 5), include_ci=True);
These type of time-varying coefficients detection problems are not new. Bayesian approach such as the R packages Bayesian Structural Time Series (a.k.a BSTS) by Scott and Varian (2014) and tvReg Isabel Casas and Ruben Fernandez-Casal (2021). Other frequentist approach such as Wu and Chiang (2000).
For further studies on benchmarking coefficients detection, Ng, Wang and Dai (2021) provides a detailed comparison of KTR with other popular time-varying coefficients methods; KTR demonstrates superior performance in the random walk data simulation.
## Customizing Priors and Number of Knot Segments¶
To demonstrate how to specify the number of knots and priors consider the sine-cosine like simulated dataset. In this dataset, the fitting is more tricky since there could be some better way to define the number and position of the knots. There are obvious “change points” within the sine-cosine like curves. In KTR there are a few arguments that can leveraged to asign a priori knot attributes:
1. regressor_init_knot_loc is used to define the prior mean of the knot value. e.g. in this case, there is not a lot of prior knowledge so zeros are used.
2. The regressor_init_knot_scale and regressor_knot_scale are used to tune the prior sd of the global mean of the knot and the sd of each knot from the global mean respectively. These create a plausible range for the knot values.
3. The regression_segments defines the number of between knot segments (the number of knots - 1). The higher the number of segments the more change points are possible.
[22]:
ktr = KTR(
response_col=response_col,
date_col=date_col,
regressor_col=regressor_col,
regressor_init_knot_loc=[0] * len(regressor_col),
regressor_init_knot_scale=[10.0] * len(regressor_col),
regressor_knot_scale=[2.0] * len(regressor_col),
regression_segments=6,
prediction_percentiles=[2.5, 97.5],
seed=2021,
estimator='pyro-svi',
)
ktr.fit(df=sc_data)
INFO:root:Guessed max_plate_nesting = 1
Initial log joint probability = -407.708
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
19 -272.5 0.0196707 33.1763 1 1 32
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
39 -272.149 2.85937e-05 33.0207 1 1 66
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
53 -272.148 6.83364e-06 33.0766 2.071e-07 0.001 127 LS failed, Hessian reset
59 -272.148 5.66701e-07 33.3211 0.4191 0.8625 135
Iter log prob ||dx|| ||grad|| alpha alpha0 # evals Notes
66 -272.148 7.35255e-09 33.3342 0.1248 0.1248 147
Optimization terminated normally:
Convergence detected: absolute parameter change was below tolerance
[22]:
<orbit.forecaster.svi.SVIForecaster at 0x1536e0280>
[24]:
coef_mid, coef_lower, coef_upper = ktr.get_regression_coefs(include_ci=True)
fig, axes = plt.subplots(p, 1, figsize=(12, 12), sharex=True)
x = np.arange(coef_mid.shape[0])
for idx in range(p):
axes[idx].plot(x, coef_mid['x{}'.format(idx + 1)], label='est' if idx == 0 else "", color=OrbitPalette.BLUE.value)
axes[idx].fill_between(x, coef_lower['x{}'.format(idx + 1)], coef_upper['x{}'.format(idx + 1)], alpha=0.2, color=OrbitPalette.BLUE.value)
axes[idx].scatter(x, sc_data['beta{}'.format(idx + 1)], label='truth' if idx == 0 else "", s=10, alpha=0.6, color=OrbitPalette.BLACK.value)
axes[idx].set_title('beta{}'.format(idx + 1))
fig.legend(bbox_to_anchor = (1, 0.5));
Visualize the knots using the plot_regression_coefs function with with_knot=True.
[25]:
ktr.plot_regression_coefs(with_knot=True, figsize=(10, 5), include_ci=True);
There are more ways to define knots for regression as well as seasonality and trend (a.k.a levels). These are described in Part III
## References¶
1. Ng, Wang and Dai (2021). Bayesian Time Varying Coefficient Model with Applications to Marketing Mix Modeling, arXiv preprint arXiv:2106.03322
2. Isabel Casas and Ruben Fernandez-Casal (2021). tvReg: Time-Varying Coefficients Linear Regression for Single and Multi-Equations. https://CRAN.R-project.org/package=tvReg R package version 0.5.4.
3. Steven L Scott and Hal R Varian (2014). Predicting the present with bayesian structural time series. International Journal of Mathematical Modelling and Numerical Optimisation 5, 1-2 (2014), 4–23. | 5,155 | 15,249 | {"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} | 3.515625 | 4 | CC-MAIN-2021-49 | longest | en | 0.79639 |
http://www.enotes.com/homework-help/simplify-fraction-3-1-x-9-1-x-2-263775 | 1,462,534,437,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861754936.62/warc/CC-MAIN-20160428164234-00203-ip-10-239-7-51.ec2.internal.warc.gz | 493,088,259 | 11,556 | # Simplify the fraction (3+ 1/x)/(9-1/x^2)
Posted on
We have to simplify: (3+ 1/x)/(9-1/x^2)
(3+ 1/x)/(9-1/x^2)
=> (3 + 1/x)/(3 - 1/x)(3 + 1/x)
=> 1/(3 - 1/x)
=> x/(3x - 1)
The required simplified form is x/(3x - 1)
Posted on
We'll multiply both numerator and denominator by the largest denominator, namely x^2:
(3x^2 + x^2/x)/(9x^2 - x^2/x^2) = (3x^2 + x)/(9x^2 - 1)
The difference of two squares from denominator returns the product:
(9x^2 - 1) = (3x-1)(3x+1)
We'll factorize the numerator by x:
(3x^2 + x)/(9x^2 - 1) = x(3x+1)/(3x-1)(3x+1)
We'll divide the fraction by (3x+1):
x(3x+1)/(3x-1)(3x+1) = x/(3x-1)
Therefore, simplifying the given fraction, we'll get: x/(3x-1). | 299 | 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} | 4.4375 | 4 | CC-MAIN-2016-18 | longest | en | 0.816776 |
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## Methodology
Question 1
A £1,000 par value bond makes two interest payments each year of £45 each. What is the bond’s coupon rate?
Question 2
Calculate the price of a five-year, £1,000 par value bond that makes semiannual payments, has a coupon rate of eight percent, and offers a yield to maturity of seven percent. Recalculate the price assuming a nine percent YTM. What general relationship does this problem illustrates?
Question 3
Griswold Travel Inc. has issued six-year bonds that pay £30 in interest twice each year. The par value of these bonds is £1,000 and they offer a yield to maturity of 5.5 percent. How much are the bonds worth?
## Experience
Question 1
A £1,000 par value bond makes two interest payments each year of £45 each. What is the bond’s coupon rate?
Question 2
Calculate the price of a five-year, £1,000 par value bond that makes semiannual payments, has a coupon rate of eight percent, and offers a yield to maturity of seven percent. Recalculate the price assuming a nine percent YTM. What general relationship does this problem illustrates?
Question 3
Griswold Travel Inc. has issued six-year bonds that pay £30 in interest twice each year. The par value of these bonds is £1,000 and they offer a yield to maturity of 5.5 percent. How much are the bonds worth?
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Exercise
Bank PO :: Q Test No 167
Home > Bank PO > Q Test No 167 > General Questions
1 .
√5599 ÷ 3√3370 = x
2 3 4 5
2 .
[(3525 + 5567 + 9853) ÷ (3132 + 1975 + 2275)] x 7 = x
15 18 21 24
3 .
√25000 x √3.999 = x
15 18 21 24
4 .
[4/5 x (2/7 x 5/3) ÷ 7/9] of 347 = x
165 170 175 180
5 .
(13.98)2 + 272% of 75 = x
360 380 400 420
6 .
Direction (Q. 6 - 10) : Find the missing number in the following number series.
165, 195, 255, 285, 345, ?
395 405 435 485
7 .
7, 26, 63, 124, 215, 342, ?
481 511 391 421
8 .
2, 4, 12, 48, 240, ?
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Found in: Page 35
### College Physics (Urone)
Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000
# An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350. (a) What is his deceleration? (b) How long does the collision last?
a) -80.36 m/s2
b) 0.0933 s
See the step by step solution
## Given Data
• The initial velocity of the football, U = 7.50 m/s.
• The final velocity of the football, V = 0 m/s.
• The distance traveled in compressing the padding, D = 0.350 m.
## Deceleration and time period
a) The deceleration of the football can be calculated using the equation as:
${V}^{2}-{U}^{2}=2ad$ Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance traveled.
Substituting values in the above expression, we get,
$\left(0{\right)}^{2}-\left(7.5{\right)}^{2}=2×\left(0.35\right)×a\phantom{\rule{0ex}{0ex}}-56.25=2×\left(0.35\right)×a\phantom{\rule{0ex}{0ex}}a=\frac{-56.25}{2×\left(0.35\right)}\phantom{\rule{0ex}{0ex}}a=-80.357{\text{m/s}}^{\text{2}}$
The deceleration of the football is -80.357 m/s2.
b) The time required can be calculated as:
V = U + at
Here V is the final velocity, U is the initial velocity, a is the acceleration, and t is the time.
Substituting values in the above expression, we get,
$0=7.5+\left(-80.36\right)×t\phantom{\rule{0ex}{0ex}}t=\frac{7.5}{80.36}\phantom{\rule{0ex}{0ex}}t=0.0933\text{s}$
Hence the time period is 0.0933 s. | 557 | 1,651 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-14 | latest | en | 0.75044 |
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in Mathematics by Euler
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#### Description
Financial Polynomials
Read the following instructions in order to complete this assignment, and review the example of how to complete the math required for this assignment:
1.Solve problem 90 on page 304 of Elementary and Intermediate Algebra. Be sure that you show all steps of the squaring of the binomial and multiplication along with any simplification which might be required.
oEvaluate the polynomial resulting from step 1 using:
P = \$200 and r = 10%, and also with
P = \$5670 and r = 3.5%
oColmplete problem 70 on page 311 of Elementary and Intermediate Algebra. Show all steps of the division.
2.Write a two to three page paper that is formatted in APA style and according to the Math Writing Guide. Format your math work as shown in the Instructor Guidance and be concise in your reasoning. In the body of your essay, please make sure to include:
oYour solution to the above problems, making sure to include all mathematical work and a discussion of how and why this is applicable to your everyday life.
Plan the logic necessary to complete the problem before you begin writing.
Use the underline feature with single spacing to set up the division(s), and use the strikethrough font to show the canceling factors. Can you think of another way this division could be approached and worked out? If yes, briefly describe the method.
oIncorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):
FOIL
Like terms
Descending order
Dividend
Divisor
Euler
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7-78.
• Assume that each of the shaded tiles in the large rectangle at right has an area of $1$ square foot. Use this information to answer the following questions. Homework Help ✎
1. What is the total area of all of the shaded tiles?
If you know the number of shaded tiles and you know the area of each shaded tile, how can you find the total area of the shaded tiles?
$1 \text{ square foot} × 28 \text{ shaded tiles} = 28 \text{ square feet}$
1. What is the total area of the rectangle that is not shaded?
Use the tiles to determine the length and width of the large rectangle. | 166 | 648 | {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-43 | latest | en | 0.875235 |
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posted by .
6. You read in a book about bridge that the probability that each of the four players is dealt exactly one ace is about 0.11. This means that
A.
in every 100 bridge deals, each player has one ace exactly 11 times.
B.
in one million bridge deals, the number of deals on which each player has one ace will scarcely be within ±100 of 110,000.
C.
in a very large number of bridge deals, the percent of deals on which each player has one ace will be very close to 11%.
D.
in a very large number of bridge deals, the average number of aces in a hand will be very close to 0.11.
7.If the knowledge that an event A has occurred implies that a second event B cannot occur, the events A and B are said to be
A.
independent.
B.
disjoint.
C.
mutually exhaustive.
D.
the sample space.
E.
complementary.
8. If you draw an M&M candy at random from a bag of the candies, the candy you draw will have one of six colors. The probability of drawing each color depends on the proportion of each color among all candies made. Assume that the table below gives the probability that a randomly chosen M&M has each color.
Color Brown Red Yellow Green Orange Tan
Probability .3 .3 ? .1 .1 .1
R-3 Ref 6-2
The probability of drawing a yellow candy is
A.
0.
B.
.1.
C.
.2.
D.
.3.
E.
impossible to determine from the information given.
9. Suppose we roll a red die and a green die. Let A be the event that the number of spots showing on the red die is three or less and B be the event that the number of spots showing on the green die is more than three.
R-4 Ref 6-6
P(A B) =
A.
1.
B.
5/6.
C.
2/3.
D.
1/4.
E.
1/6.
10. Suppose we roll a red die and a green die. Let A be the event that the number of spots showing on the red die is three or less and B be the event that the number of spots showing on the green die is more than three.
R-4 Ref 6-6
P(A B) =
A.
1/6.
B.
1/4.
C.
1/3.
D.
5/6.
E.
none of these.
11. Suppose we roll a red die and a green die. Let A be the event that the number of spots showing on the red die is three or less and B be the event that the number of spots showing on the green die is more than three.
R-4 Ref 6-6
The events A and B are
A.
disjoint.
B.
conditional.
C.
independent.
D.
reciprocals.
E.
complementary.
• statistics -
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
• statistics -
dear,
PsyDAG
i asked for help because i didn't know how to do it not because i didn't want to do the work daaaaaaa!!!!
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More Similar Questions | 1,229 | 4,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-05 | latest | en | 0.939599 |
https://suburbanstylechallenge.com/skincare/frequent-question-how-many-grams-do-6-moles-of-o2-weigh.html | 1,660,579,013,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00766.warc.gz | 506,215,852 | 18,199 | # Frequent question: How many grams do 6 moles of O2 weigh?
Contents
## How many grams of 02 are in a mole?
The mass of oxygen equal to one mole of oxygen is 15.998 grams and the mass of one mole of hydrogen is 1.008 g.
## How many grams are there in 5 moles of O2?
Therefore, 160 g of oxygen is present in 5 moles.
## How much does a mole of 02 weigh?
One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.
## How many grams are in 1 mole of oxygen?
For one gram atomic weight of oxygen with atomic weight of 16 grams, one mole of oxygen also contains 6.022 × 1023 oxygen atoms. Similarly, for one gram atomic weight of silicon with atomic weight of 28 grams, one mole of silicon still contains 6.022 × 1023 silicon atoms.
IT\'S FUNNING: Is exfoliating with a washcloth bad?
## How many grams of co2 are produced from 5 moles O2?
The mass of carbon dioxide produced (to one significant figure) is 400 g.
## How many grams are there in 3.5 moles of oxygen?
The atomic mass of oxygen is approximately 16 grams, however o2 would make it 32 grams. Then one would multiply 32 by 3.5 moles which would give you 112 grams per 3.5 moles. So,3.5 moles of oxygen has mass=3.5×16=56grams.
## How many grams are in 3 moles of O2?
3 moles of Oxygen atoms weigh 48.00 grams.
## How many moles are in co2?
Molar Masses of Compounds
One mole of carbon dioxide molecules has a mass of 44.01g, while one mole of sodium sulfide formula units has a mass of 78.04g. The molar masses are 44.01g/mol and 78.04g/mol respectively.
## How many grams are in 3.5 moles?
The mass of 3.5 moles of Ca is 140 g to two significant figures.
## What is the mass of 5 moles of O2?
Therefore, 5 moles × 32.0 g/mol = 160 grams is the mass (m) when there are 5 moles of O2 .
## How many moles are in O2?
So one mole O2 Contains 32 grams of O2 . Hence 18 moles of O2 will contain 18 × 32 = 576 grams .
## How many moles of oxygen are in O2?
A mole of O2 molecules has 6.023E23 molecules, and each molecule is made up is two atoms of oxygen. That means that there are two moles of oxygen atoms in a mole of O2, or 1.2046E24 of them.
## How many moles are there in 50.0 g of O2?
To calculate the weight of an element in grams to moles, we should know the molar mass of an element. For example, 50 grams of oxygen is equal to 3 moles.
IT\'S FUNNING: How does Elemis Papaya Enzyme Peel work? | 754 | 2,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-33 | latest | en | 0.932705 |
https://www.qb365.in/materials/stateboard/11th-business-maths-differential-calculus-model-question-paper-2197.html | 1,621,332,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989819.92/warc/CC-MAIN-20210518094809-20210518124809-00032.warc.gz | 1,013,940,906 | 31,472 | #### Differential Calculus Model Questions
11th Standard
Reg.No. :
•
•
•
•
•
•
Time : 01:00:00 Hrs
Total Marks : 50
5 x 1 = 5
1. The graph of y = 2x2 is passing through
(a)
(0,0)
(b)
(2,1)
(c)
(2,0)
(d)
(0,2)
2. Which one of the following functions has the property f (x) = $f\left( \frac { 1 }{ x } \right)$
(a)
$f\left( x \right) =\frac { { x }^{ 2 }-1 }{ x }$
(b)
$f\left( x \right) =\frac { 1-{ x }^{ 2 } }{ x }$
(c)
f(x) = x
(d)
$f\left( x \right) =\frac { { x }^{ 2 }+1 }{ x }$
3. The range of f(x) = |x|, for all $x\epsilon R$, is
(a)
(0, $\infty$)
(b)
(0, $\infty$)
(c)
(-$\infty$$\infty$)
(d)
(1, $\infty$)
4. A function f(x) is continuous at x = a if $\lim _{ x\rightarrow a }{ f\left( x \right) }$ is equal to
(a)
f(-a)
(b)
f$(\frac{1}{a})$
(c)
2f(a)
(d)
f(a)
5. If y = log x then y2 =
(a)
$\frac{1}{x}$
(b)
$-\frac{1}{x^2}$
(c)
$-\frac{2}{x^2}$
(d)
e2
6. 5 x 2 = 10
7. If $f(x)=\frac { x+1 }{ x-1 }$ ,then prove that f(f(x))=x
8. For $f(x)=\frac { x-1 }{ 3x+1 }$ ,write the expression of $f\left( \frac { 1 }{ x } \right)$ and $\frac { 1 }{ f(x) }$
9. Find the derivative of the following functions from first principles log (x+1)
10. Differentiate ${ x }^{ \frac { 2 }{ 3 } }$ from first principles
11. Find $\frac { dy }{ dx }$ if x2 + xy + y2 = 100
12. 5 x 3 = 15
13. Find $\frac{dy}{dx}$ of the following functions: x = a (θ - sin θ),y = a(1- cosθ)
14. Differentiate sin3 x with respect to cos3x.
15. If ey (x + 1) = 1, show that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }$
16. Evaluate $\underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }+27 }{ { x }^{ 5 }+243 }$
17. Evaluate $\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-sin2x }{ { x }^{ 3 } }$
18. 4 x 5 = 20
19. Draw the graph of the following function f(x)=e-2x
20. Differentiate: $\frac { sinx+cosx }{ sinx-cosx }$with respect to 'x'
21. Find the derivate of (x3-27)from first principles.
22. Darw the graph of f(x) = logax; x > 0, a > 0 and $a\ne 1$. | 892 | 2,040 | {"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} | 3.9375 | 4 | CC-MAIN-2021-21 | latest | en | 0.300724 |
https://www.hackmath.net/en/example/1345?tag_id=150 | 1,544,771,921,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00368.warc.gz | 900,048,273 | 6,587 | # Tiles
Hall has dimensions 325 &time; 170 dm. What is the largest size of square tiles that can be entire hall tiled and how many we need them?
Result
a = 50 cm
b = 50 cm
n = 2210
#### Solution:
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#### To solve this example are needed these knowledge from mathematics:
Do you want to calculate greatest common divisor two or more numbers?
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The tractor plows the first day of 4.5ha, the second day 6.3ha and the third day 5.4ha. It worked whole hours a day, and its hourly performance did not change and was the highest of the possible. How many hectares did it plow in one hour (what is it perfor
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7. MO C–I–1 2018
An unknown number is divisible by just four numbers from the set {6, 15, 20, 21, 70}. Determine which ones.
8. Birthdate
Jane on birthday brought 30 lollipops and 24 chewing gum for their friends. How many friends has, if everyone receives the same number of lollipops and chewing gums? How much chewing gum and lollipops got any friend?
9. Diofant 2
Is equation ? solvable on the set of integers Z?
10. Diofant equation
In the set of integers (Z) solve the equation: ? Write result with integer parameter ? (parameter t = ...-2,-1,0,1,2,3... if equation has infinitely many solutions)
11. School books
At the beginning of the school year, the teacher gave out 480 books and 220 textbooks. How many students were in the class?
12. 9.A
9.A to attend more than 20 students but fewer than 40 students. A third of the pupils wrote a math test to mark 1, the sixth to mark 2, the ninth to mark 3. No one gets mark 4. How many students of class 9.A wrote a test to mark 5?
13. Segments
Line segments 67 cm and 3.1 dm long we divide into equal parts which lengths in centimeters is expressed integer. How many ways can we divide?
14. Cubes
Carol with cut bar 12 cm x 12 cm x 135 cm to the cubes. Find the sum of all the surfaces of the resulting cubes.
15. Self
A math text book is 2 2/9 inches thick. how many of these books will fit on a 120-inch self?
16. Florist
Florist has 84 red and 48 white roses. How many same bouquets can he make from them if he must use all roses?
17. Apples and pears
Mom divided 24 apples and 15 pears to children. Each child received the same number of apples and pears - same number as his siblings. How many apples (j=?) and pears (h=?) received each child? | 917 | 3,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-51 | latest | en | 0.954662 |
https://www.websiteperu.com/search/equations-of-parallel-and-perpendicular-lines | 1,642,850,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00585.warc.gz | 1,074,908,525 | 14,383 | # Keyword Analysis & Research: equations of parallel and perpendicular lines
## Keyword Research: People who searched equations of parallel and perpendicular lines also searched
How to write equation for parallel lines?
So, to find an equation of a line that is parallel to another, you have to make sure both equations have the same slope. In the general equation of a line y = mx + b , the m represents your slope value. An example of paralell lines would therefore be: (1) y = mx + b . (2) y = mx + c . With b and c being any constants.
Which pair of equations represents perpendicular lines?
Simply solve for the y in each equation, and if your slopes are opposite slopes with opposite signs and opposite reciprocal in which the numerator and the denominator is switched, the lines are perpendicular. y=-8/3x+9------> first line.
How do you find the equation of a parallel line?
So, to find an equation of a line that is parallel to another, you have to make sure both equations have the same slope. In the general equation of a line #y=mx+b# , the #m# represents your slope value. An example of paralell lines would therefore be: (1) #y=mx+b#. (2) #y=mx +c#. With #b# and #c# being any constants. | 283 | 1,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-05 | latest | en | 0.92729 |
https://daxpy.xyz/posts/xgboost/ | 1,721,258,836,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00359.warc.gz | 169,359,653 | 5,456 | # XGBoost
Extreme Gradient Boosting or XGBoost is a technique that has become quite useful for solving prediction problems. XGBoost is also quite interesting academically; for it combines quite few techniques together to give us one robust method. The technique is composed from gradient boosting, decision trees, matching pursuit and gradient descent in function space among others. In this post, we will explore and derive the inner workings of XGBoost.
## The Regression Problem
We are given a set of samples from population $\{(y_i,x_i)\}_{i=1}^{N}$ which we call a dataset. $y$’s are scalars and $x$ are vectors.
There is also some function $F^{\ast}$ on the population which perfectly determines the entire population i.e. $y = F^{\ast}(x)$. However, this function is unknown and all we have is the dataset.
Regression problem is computing some $\hat{F}$ which is the best approximation of $F^{\ast}$ that minimises some loss function $L(y,F(x))$ over all values of $(y,x)$.
Different choices of $L$ and restriction on the structrue of $F$ leads to different algorithms. For example; mean squared error and linear function on $x$ gives linear regession.
## T.H.E family of functions
Let assume (to make things easier) that $F$ has a simpler structure. The all powerful $F$ is from the family of functions which are weighted combination of simpler functions.
$$F(x) = \sum \beta_m h_m(x)$$
We do such approximations all the time. See mean field approximation for variational inference for a family which is product of simpler functions.
This makes our learning easier. It’s easier to see if we look at the explict structure of $F$.
$$F(x;\{\beta_t,\alpha_t\}) = \sum_t \beta_t h_t(x;\alpha_t)$$
All we need to do now is compute values of parameters $\beta$’s and $\alpha$’s which minimised $L$.
By the way, did we just sneak in a constraint on structure of $F$ ??? Are all the $h$’s same or different?
Well the structure allows everything. B.U.T, for XGBoost, all $h$’s are decision trees.
All we need to do is compute $\beta$’s and $\alpha$’s
This is easier said than done.
Joint optimisation will lead to a situation where adjusting $\alpha$ of one tree would require us to adjust $\alpha$ of another. And $\alpha$ of a decision tree determines where to add a split. This is not productive at all as we will have to throw away the existing tree and construct a whole new tree. In optimisation land this is classif case of a dis-continous objective.
So lets fix that problem by fixing things. We first find the best parameters for $h_1$ and then never change it. Then find best parameters for $h_2$ and so on. This idea is called “Gradient Boosting” where we restrict $F$ to the family of additive expansion is from the paper Greedy Function Approximation: A Gradient Boosting Machine. This stage wise stratergy is also very similar to matching pursuit algorithm in signal processing.
Given an loss function $l$ and parameterised objective function $f(x;\theta)$, we can find best $\theta$ which minimises $l$ using Gradient Descent. $$\theta \gets \theta - \gamma \nabla_l f$$
Let say we are following the approach of fixing things as listed above. We are in some intermediate step $m$. We have already found and fixed the parameters of $\beta_{1:m-1}$ and $\alpha_{1:m-1}$ and the current best regressor we have is
$$F_{t-1}(x) = \sum_{i=1}^{t-1} \beta_i h_i(x; \alpha_i)$$
We want to add to this a $p_t(x) = \beta_t h_t(x;\alpha_t)$ and reduce the error further.
$$l_t = \operatorname{arg min} \sum_{i=1}^{N} \operatorname{L}(y_i, F_{t-1}(x)+f_t(x))$$
In this situation we can find best parameters for $f_t$ using gradient descent on function space. But we need $\frac{\partial l_t}{\partial f_t}$. Let ignore a few things a write this situation simply as
$$l = \operatorname{L}(y, p+f(x))$$
If we squint our eyes, r.h.s looks like a function with a fixed point $p$ and a small delta $f(x)$. We can expand it around the fix point using second order taylor’s series approximation.
$$l = \operatorname{L}(y, p) + \nabla_p \operatorname{L}(y,p) f(x) + \nabla_p^2 \operatorname{L}(y,p) \frac{f(x)^2}{2}$$
This immediatly gives us an opening to get the derivative of loss w.r.to $f(x)$.
$$\frac{\partial l}{\partial f} =0 + \nabla_p \operatorname{L}(y,p) + \nabla_p^2 \operatorname{L}(y,p) f(x) = \nabla_p^2 \left( \frac{\nabla_p}{\nabla_p^2} + f(x)\right)$$
Since optimum occurs at the saddle point, the optimum $f(x)$ is the one which makes the derivative zero. So at $\frac{\partial l}{\partial f} =0$ we have $$f(x) = - \frac{\nabla_p}{\nabla_p^2}$$
Ultimately, to find the best additive function to the model at stage $t$, we simply have to fit $h_t(x)$ to predict the residuals $\{-{\nabla_p}/{\nabla_p^2} \}$. This again is another regression problem.
$$h_t \leftrightsquigarrow \left\{\left(x_i,-{\nabla_{F_{t-1}(x_i)}}\middle/{\nabla^2_{F_{t-1}(x_i)} }\right)\right\}_{i=1}^{n}$$
Addendum: What if we use first order taylor’s series approximation instead on the loss function? What will be the residuals in the last step?
## XGBoost Algorithm
• We begin by setting $h_0$ to simply predict $\mathbb{E}(y)$.
• At each step we first compute the residuals $$r_i = -\frac{\nabla_{\bar{y}_i}\operatorname{L}(y_i, \bar{y}_i)}{\nabla^2_{\bar{y}_i}\operatorname{L}(y_i, \bar{y}_i)} ; with~\bar{y}_i = F_{t-1}(x_i)$$ Note: Most loss functions have easy analytical form of first and second derivatives e.g. mse loss.
• Fit $h_t$ to predict $r_i$ given $x_i$.
• Compute $\beta_t$ using line search which will optimise $$\operatorname{arg min}_{\beta_t}\operatorname{L}(y, F_{t-1} + \beta_t h_t)$$
• Update the model as $$F_t = F_{t-1} + \beta_t h_t$$
There is a lot more to implementing XGBoost. Most of it is centered around the “fitting a new tree to residuals” part. A few scenarios that arise in this step are
• How to prevent overfitting of the intermediate tree on residuals?
• How to let user direct/control some part of the tree construction so that the complexity vs performance tradeoff can be tunable?
Chen et al. (2016) gives an excellet account of these issues among other, but the official XGBoost library’ documentation is also a great source for discussion on these topics.
## References
• Friedman (2001): Friedman, Jerome H. “Greedy function approximation: a gradient boosting machine” In Annals of statistics , (2001)
• Mallat and Zhang (1993): Mallat, Stephane G and Zhang, Zhifeng “Matching pursuits with time-frequency dictionaries” In IEEE Transactions on signal processing 41, (1993)
• Chen et al. (2016): Chen, Tianqi and Guestrin, Carlos “Xgboost: A scalable tree boosting system” In Proceedings of the 22nd acm sigkdd international conference on knowledge discovery and data mining pp. 785–794, (2016)
• Decision Tree
https://en.wikipedia.org/wiki/Decision_tree_learning
https://en.wikipedia.org/wiki/Gradient_descent
https://en.wikipedia.org/wiki/Linear_regression
https://en.wikipedia.org/wiki/Taylor_series | 1,947 | 6,982 | {"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} | 3.515625 | 4 | CC-MAIN-2024-30 | latest | en | 0.895419 |
https://forums.babypips.com/t/a-surprisingly-noob-question-are-100-pips-worth-1-of-a-currency-move/71312 | 1,713,937,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00694.warc.gz | 227,195,089 | 6,053 | # A surprisingly 'noob' question: are 100 pips worth 1% of a currency move?
I will explain!
I was listening to PowerTrading Radio the other day, and in hearing about the EUR/USD drop from May 2014
described as a “25%” drop, I wondered if I understood this concept at all: if the EUR/USD dropped 2,500 pips from its
1.40 high of last year, then we would talk about a 25% drop; if, however, we were describing what actually
happened, namely a 3,500-pip drop, then should we not call it a 35% drop?
I know that it is not so easy, because the value of pips for each currency pair is not fixed, although if someone
could explain how that applies to the pip-to-percentage question I put above, I would be grateful.
Thank you - and my apologies if this topic has been covered before (I tried a Google search but I could not
find the very direct answer that I was after).
Regards
PipMeHappy
Take current price and multiple by .25.
That’s 25%.
So 1.4 x .25 = .35
So if price went from 1.4 to 1.05 that would be a 25% drop.
I would assume that because any currency pair, or any asset for that matter, can appreciate in value to a ‘unlimited value’ then the percentage change is always calculated from either the maximum value obtained since inception of the asset, or the maximum value obtained within a predetermined time frame - such as the past 12 months.
I’ve never looked at pips in terms of “pip-to-percentage”, however to answer your question, I would assume 100pips are only equal to 1% when the currency pair has had an all time high value of 10,000 pips since inception (being measured from the origin of ‘0’, where the X and Y axis cross) , hence 100 pips IS 1% of 10,000.
To make it simple, if price moves from 1.40 (the determined high) to perhaps 1.10 (the determined low) I would simply say it’s fallen by 21.43%, irrespective of pip value. As long as you can determine the high and the low, the rest is easy.
…or just read matergunner, I obviously speak to much crap
Thank you, Mastergunner and Jezzode, you answered my question perfectly!
I also did want to ask, what is 25% in relation to? Jezzode: you answered that too!
Thank you - I will actually not use this for trading and stick to pips, but it was only to make sense of it.
Cheers!
1.4 - 1.05 = 0.35
0.35 * 100 / 1.4 = 25%
Simple math…
Yes, rindoan, that’s it!
I have not touched algebra since school! It is surprising how you can forget basic stuff like percentage operations.
Cheers.
F
In formula terms:
[(A - B) * 100] / A = C
where A= high, B= low, and C the final percentage differential.
Correct.
1.40 = 14,000 pips.
3500/14,000 = .25 (25%).
100 pips would be worth 1% of a change only from 10,000 pips which is parity. From 1.40 (14,000 pips), 100 pips is 71 basis points (0.71%).
Of course we trading low lifes don’t care about that because we use leverage and pyramiding to make a move of 3500 pips worth 100% of our account. | 802 | 2,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | latest | en | 0.965861 |
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2016-11-18, 15:35 #56 robert44444uk Jun 2003 Oxford, UK 22·3·7·23 Posts A new idea Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
2016-11-18, 16:38 #57
science_man_88
"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts
Quote:
Originally Posted by robert44444uk Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
my guess is it probably depends on gcd and a lot more to figure it out to an exact value for Y or Z.
2016-11-22, 11:43 #58
robert44444uk
Jun 2003
Oxford, UK
111100011002 Posts
Quote:
Let X = A*P#/(multiple of some primes) We know that the interval X +/- 2P has very few values that might be prime. What is to stop us finding Y = B*Q#/(multiple of some other primes) that is close in value to X, in fact very close to or equal to X+2(P+Q) or X-2(P+Q) Then we would have an interval approximately twice as long as I that would have very few values that might be prime, assuming P is similar in size to Q. By extension we could find Z=C#/(multiple of yet other primes) close to the range so that the interval is 3 times as long, etc. Simple algebra can find Y,Z I think, or am I missing something fundamental here.
I don't think the hypothesis I outlined above can be the case. The two values X and Y presumably share a large number of factors, given that P# and Q# are closely related, and hence the closest that X and Y would come would be defined by the multiple of their common factors.
Is this right?
Last fiddled with by robert44444uk on 2016-11-22 at 11:46
2017-03-21, 17:15 #59 robert44444uk Jun 2003 Oxford, UK 36148 Posts I am wondering how we might categorise divisors, in order to find those that are most likely to provide records. I'm looking at an approach using two measures, A - number of gaps >10 over a given range for a divisor B - "persistence" - the ratio of the number of gaps of size x+1 to the number of gaps of size x over a given test range for a divisor For a given divisor D, The test range I am using takes in multiple Y and P in Y*P#/D (centre points of large gaps). The specific ranges I am first looking at are D from 1 to 10,000 (squarefree only of course), Y from 1 to 5,000 and P from 97 to 229, so 130,000 tests for each D. I am not looking for all merit 10 gaps, as I am setting the delta =4. The probability of large gaps should be correlated to A and B, with B dominating, as explained below. The persistence measure is exponential. A 60% persistence suggests that 1 in 27,251 gaps of merit >10 will be a gap of merit 30 and 1 in 165 for a 20 merit gap. A 50% persistence on the other hand, 1 in every 1,048,576 gaps of merit >10 will actually be merit 30, and 1 in every 1,024 merit 20. So although the divisor 2 produces possibly the highest number of size 10 gaps (A=0.352%) its persistence ratio looks to be only in the 35% range, suggesting a conversion ratio of 1 in 1,314,132,370 to achieve a 30 merit gap and 1 in 36,251 for a merit 20. Other small Ds are showing >50% persistence, and A in excess of 0.2%. I know that my favoured divisor 46410 is closer to persistence ratio=60% with an A value relatively negligible. The plan is to find a divisor with persistence of 60% and a much higher A value. Grateful for your views. Last fiddled with by robert44444uk on 2017-03-21 at 17:15
2017-03-23, 18:23 #60 mart_r Dec 2008 you know...around... 32·71 Posts To get a persistence of more than 60%, the value for P must be higher. For P around 20,000 you get B>60% with D a primorial >= 17#, without having to cope with a small A value. For P around 50,000, you can even choose, say, D=41#, and still get a decent A~4% (maybe more) with B>60~65%. Downside is, the tests take longer... If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
2017-03-23, 19:21 #61
danaj
"Dana Jacobsen"
Feb 2011
Bangkok, TH
22·227 Posts
Quote:
Originally Posted by mart_r If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
Looking at the top-20 records:
7 D=30
7 D=210
2 D=2310
4 other (3 are maximal gaps, 1 is D=7230)
Looking at allgaps.dat for merits >= 30,
476 D=30
427 D=210
56 D=2310
20 D=6
9 D=46410
This is heavily impacted by what's being searched for, but I believe D=30 has much more searching than other divisors. For all gaps >= 10 merits, D=30 has over 4 times as many results as the next (D=210).
For gaps under 40k, percent of merits that are over 30. Again impacted by search ranges but maybe it tells us something:
7.17% D=30
7.54% D=210
7.99% D=2310
4.55% D=6
4.23% D=46410
27.78% D=9570
2017-03-23, 22:50 #62
robert44444uk
Jun 2003
Oxford, UK
22·3·7·23 Posts
Quote:
Originally Posted by mart_r To get a persistence of more than 60%, the value for P must be higher. For P around 20,000 you get B>60% with D a primorial >= 17#, without having to cope with a small A value. For P around 50,000, you can even choose, say, D=41#, and still get a decent A~4% (maybe more) with B>60~65%. Downside is, the tests take longer... If you're looking for merit >30, D=30 is most effective for the P's you're looking at.
Sorry Mart_r do you have your Ps and D's mixed up? The way I defined it, P is the prime in the primorial, D is the divisor and Y the multiplier.
2017-03-28, 18:49 #63
mart_r
Dec 2008
you know...around...
32×71 Posts
Quote:
Originally Posted by robert44444uk Sorry Mart_r do you have your Ps and D's mixed up? The way I defined it, P is the prime in the primorial, D is the divisor and Y the multiplier.
No, the P's and D's in my post are all where they have to be.
Only my results may vary a bit more or less compared to yours. I'm basically looking at the count of coprimes mod P#. In my example, 50000#/41#, i.e. D=304250263527210, showed a persistance of >60%, according to your definition.
2017-07-19, 22:27 #64 ATH Einyen Dec 2003 Denmark 17·181 Posts Regarding the "new" gap formula by Maynard, Tao and others https://www.youtube.com/watch?v=BH1GMGDYndo https://arxiv.org/abs/1408.4505 G(X) >= C * log X * loglog X * loglogloglog X / (logloglog X)^2 They found out that C can get arbitrarily large as X->infinity. I was curious about the value of this "Maynard-Tao" constant for known gaps: C= Gn * (logloglog Pn)^2 / (log Pn * loglog Pn * loglogloglog Pn) It seems to follow the "Cramér–Shanks–Granville ratio" somewhat and is largest at the smaller gaps. Code: Gn Merit Gn/(ln(Pn)^2) Maynard-Tao Pn 15900 39.62015365 0.09872683 36.37716367 1.936933265397289504398811903696*10^174 18306 38.06696007 0.07915948 34.09959273 7.041097148478282668812106731813*10^208 10716 36.85828850 0.12677617 35.50074155 1.839377720243795270729953508768*10^126 13692 36.59018324 0.09778276 33.93062260 3.254185929142547441117000456865*10^162 26892 36.42056789 0.04932537 30.92865059 4.696226774889053656642126142794*10^320 66520 35.42445941 0.01886489 27.27312299 3.292808201042179724620296543360*10^815 1476 35.31030807 0.84472754 59.57000119 1425172824437699411 1442 34.97568651 0.84833471 59.49933905 804212830686677669 1454 34.11893253 0.80062005 56.90602177 3219107182492871783 1370 33.76518602 0.83218087 58.00949419 418032645936712127 1132 32.28254764 0.92063859 61.34832684 1693182318746371 6582144 13.18288411 0.00002640 7.04128546 8.465069837806447347636518542879*10^216840 5103138 10.22031845 0.00002047 5.45890046 7.695421151871542659687327631743*10^216848 Last fiddled with by ATH on 2017-07-19 at 22:30
2017-07-20, 13:39 #65
CRGreathouse
Aug 2006
10111011000012 Posts
Quote:
Originally Posted by ATH I was curious about the value of this "Maynard-Tao" constant for known gaps: C= Gn * (logloglog Pn)^2 / (log Pn * loglog Pn * loglogloglog Pn) It seems to follow the "Cramér–Shanks–Granville ratio" somewhat and is largest at the smaller gaps.
It's much larger than the Cramér-Shanks-Granville ratio -- which we expect to be bounded, or 'nearly' bounded, unlike the one you mention (Ford-Green-Konyagin-Tao).
The state of the art today is Ford-Green-Konyagin-Maynard-Tao:
C1 = Gn * (log log log Pn) / (log Pn * log log Pn * log log log log Pn).
2017-07-20, 17:10 #66 robert44444uk Jun 2003 Oxford, UK 36148 Posts Using Antonio's examples and the FGKMT formula provides C as follows: Code: Gn C 15900 20.31243206 18306 18.72974166 10716 20.45427849 13692 19.07131193 26892 16.38392518 66520 13.501963 1476 45.22507308 1442 45.29864017 1454 43.03407437 1370 44.3097939 1132 48.34468117 6582144 2.735318749 5103138 2.12060946 Now I have to get my head around what this means I think we can say that we are nowhere near getting maximal gaps outside of the range we are searching suggesting that the merit is going to increase a lot. Last fiddled with by robert44444uk on 2017-07-20 at 17:13
Similar Threads Thread Thread Starter Forum Replies Last Post Nick Number Theory Discussion Group 6 2016-10-14 19:38 firejuggler Math 0 2016-07-11 23:09 Ricie Miscellaneous Math 24 2009-08-14 15:31 jasong Math 3 2005-05-15 04:01 clowns789 Miscellaneous Math 5 2004-01-08 17:09
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So, Voltron has EIGHT Seasons, at the very least.
8 seasons. seasons= 7 letters. 8 + 7 = 15. And we are meant to have 5 paladins. 15 + 5 = 20
If we follow the letters in the word ‘paladin’ and each letter corresponds to a number (A=1, B=2, etc.)
P=16, A=1, L=12, A=1, D=4, I=9, N=14
16+1+12+1+4+9+14=57
And if we add these to the previous number… 57+20= 77
Coran and Allura were in that pod for 10,000 years. If we divide 10,000 by 77, we get 129.87 aka 130.
If we subtract the 7 main characters, as well as Zarkon and Haggar- which makes 9 characters in total- from the 130 we have, we will get 121.
As you can guess, a phrase can be made from those numbers, if we follow the sequence accordingly.
K=11, L=12, A=1, N=14, C=3, E=5
I=9, S=19
C=3, A=1, N=14, O=15, N=14
11+12+1+14+3+5+9+19+3+1+14+15+14 = 121
## so what i’m saying is
4
GUYS THE NUMBERS FOR THE CASE FILE, IF YOU CORRESPOND WITH THE ALPHABET
S=19
A=1
T=20
A=1
N=14
i’m screaming
Numerology: Calculating Your Life Lesson Number
## Numerology Basics
Each one of us has four personal numbers that carry a specific vibration. They are the Life Lesson Number, the Soul Number, the Outer Personality Number, and the Path of Destiny Number. In this post, we’ll be focusing on the Life Lesson number.
The Life Lesson Number represents lessons one must learn in this lifetime and is most significant to career choices. It is derived from your full birthdate.
Though this post will be focused on calculating one’s number using the numbers in your birthday, I’m including information on how to calculate the numeric vibration of letters as well, because it’s intrinsically related.
To work with the value of letters beyond “I”, which is 9, we reduce the value to a single digit. For example, “O” is 15, so we reduce the number by adding the first digit to the second: 1+5=6. The number 15 is then written as 15/6. The letter “T” is 20. We reduce the 20 by adding the first digit to the second: 2+0=2. The number 20 is then written as 20/2.
Number Values of the Alphabet
A 1 J 10/1 S 19/1
B 2 K 11/2 T 20/2
C 3 L 12/3 U 21/3
D 4 M 13/4 V 22/4
E 5 N 14/5 W 23/5
F 6 O 15/6 X 24/6
G 7 P 16/7 Y 25/7
H 8 Q 17/8 Z 26/8
I 9 R 18/9
We can reduce any word to a numeric vibration by using the above table. Before we move on to calculating one’s Life Lesson number, lets experiment with a few words:
3 + 9 + 6 + 5 = 23/5
C R O W
We reduce 23 by adding 2+3=5, which is written 23/5
5 + 9 + 2 + 3 + 8 = 21/3
W I T C H
We reduce 21 by adding 2+1=3, which is written 21/3
3 + 9 + 7 + 8 + 2 = 29/11
L I G H T
We reduce 29 by adding 2+9=11, which is written 29/11
There are four double digits which generally are not reduced. These are referred to as master numbers. Master numbers offer more opportunity for expression, yet demand more effort from the person or thing in question. The master numbers are 11, 22, 33, and 44. Whenever you arrive at one of these numbers in your calculations; retain the master value.
Because the word “light” vibrates to master number 11, it’s written as 29/11 instead of 29/2, which would be the case if reduced entirely. The number 2 is still acknowledged as the base number as the word will fluctuate between the vibrations of 2 and 11.
An individual will fluctuate between any master number and its base digit because the vibration of the master number is too intense to live under continuously. Living under the vibration of one’s base number offers respite while preparing to work with the master number vibration again.
## Calculating Your Life Lesson Number
To calculate your Life Lesson Number, add all the digits of your birthdate together and reduce the sum. Here is an example of how someone with the birthdate of 24 October 1948 would calculate their Life Lesson Number:
10 + 24 + 1948
10 + 24 + 22 (1 + 9 + 4 + 8)
10 + 24 + 22 = 56/11
Since the reduced sum of their birthday is a master number, the sum is not reduced further.
Now that you know how to calculate your Life Lesson Number, read on to see the description of each numeric vibration. All number descriptions come directly from Numerology and The Divine Triangle (1979) by Faith and Dusty Bunker, the reference material for this post.
## Life Lesson Number Descriptions
1: You must learn to be original, strong willing, creative, and innovative. You should have the courage and drive to go ahead into new fields of expression and be a pioneer. You should always go forward, never turn back. At times you may be dictatorial and stubborn because you do not like to be restricted or directed. You are a good executive and work best alone. You are usually efficient and well organized. You are not naturally domestic but can manage well in any situation. You usually like sports and athletics and enjoy the thrill of winning. You are sophisticated, not emotionally romantic, and always appear at the head of social and commercial groups. By learning the lessons of number 1, you become intimately familiar with the universal energy; that probing, seeking, independent spark that moves all creation. You are creative on the physical plane because your pioneer spirit precedes all others and expresses your unique individuality.
2: You are here to learn to become a good mixer. You are a good salesperson, more persuasive than forceful. You should be a support for those in leadership roles, help them to find their goals in life, and remain behind the scenes if necessary. This quality can be a help to you in business because those who benefit from your talents will in turn help you make use of your abilities. In partnerships and groups, you will encounter the lessons you came to earn in this lifetime. Success is then very possible. You must have consideration for others and should bring people together for a common cause. Various professions are open to you as you learn to be adaptable to most things that need to be done. You could select a career in finance, music, medicine, religion, or statistical analysis and research.
3: You are best in intellectual, artistic, or creative endeavors. You need to be expressive to manifest and to see the results of your work. Beauty, fruitfulness, luxury, and pleasure are your keywords. You should have ambition and pride. You must become conscious of the law, and by being an excellent disciplinarian, you will achieve a position of authority over others. 3 combines the daring of 1 with the caution of 2. It is a number of self-expression and freedom. You must guard against becoming a jack-of-all-trades; rather, you should specialize. You could then be successful in artistic, religious, or inventive pursuits. You should avoid routine work because you dislike restriction. You should work alone for the best results. Business partnerships become too disciplined for your freedom-loving nature. You could write, lecture, teach, or find your niche in journalism. Whatever you decide to specialize in, you must use your creative and inspirational talents.
4: You must build a solid foundation on which to base your life. This demands a well ordered system of conduct and morals. Administration or some sort of management would be the best type of employment for you. You want your home life to conform to the culture in which you live. You will provide well for those within your care, and you expect them to respond with respect and dignity. You should become a diligent worker and honestly earn your success. By being thrifty, you will have an adequate savings account as security against any possible losses. You should learn not to take a chance unless it is a sure bet. You should seek high goals. You might want to achieve concrete results quickly, and therefore should strive for patience and perseverance. Learn to face reality and base all your efforts on sound practical reasoning.
5: Your keyword is freedom. If you have “free rein”, you can accomplish wonders, but if you feel bound or limited, you lose your enthusiasm and accomplish little. You would be a good explorer or Peace Corps volunteer, as you learn well by travel and experience. You are a diligent student if interested in the subject, but you may fail in subjects for which you see no useful ends. You should be eager for new experience, and shun monotony. In your quest for knowledge, you will become interested in discovering answers in books and magazines. An avid reader, a fluent talker, and a versatile doer, you are the witty conversationalist and brighten any group by your mere presence. You are here to learn and experience the value of freedom and should not tie yourself down too severely. Your talents once learned, prepare you for a literary career or a a position in sales and dealing with the public.
6: You are here to learn a sense of responsibility for your family and community. 6 is the love and domestic vibration and requires that you be responsive to the social needs of others. A fine sense of balance must be acquired so that you can equalize injustices. This keen sense bestows artistic abilities as well as judgmental talents which can be utilized in the legal system. You should develop compassion and understanding necessary to ease the burdens of those who will naturally be drawn to you. You are among those who serve, teach, and bring comfort to humanity. A wide choice of professions is yours including nursing, teaching, welfare work, ministry, medicine, restaurant enterprises, a legal profession, and possibly veterinary or animal husbandry. You may also choose to end a career in the arts, interior decorating, or hairdressing.
8: This is the number of power and ambition, the number of the executive, the boss, who lives by brain and brawn. You will learn to work and will want to see everyone else working. You can push people to become successful in their own right. You should lead and show by example how to profit in business. You are here to learn to handle power, authority, and money. You can build a business empire and should work to that end. You want success for your family and for the family name as a matter of pride. You want your offspring to carry your name with honor and dignity. Sports is another field open to you, as this number vibration bestows great strength and endurance. Many famous athletes operate under an 8.
9: You should be the universal lover of humanity, patient, kind, and understanding. You are at the peak of life’s expression and must turn and show others the way. You seem to receive wisdom from above; thus you know that the true way of happiness is in service to others. You are the marrying type, strong in passion and compassion. You easily acquire money or wealth, and know how to preserve it. You are never petty, but deal in broad concepts and can attain success in the face of difficulties. You are here to show others the way, through your breadth of thinking. You can choose from many professions; education and medicine are the most usual. You may become an orator, writer, or lecture with equal ease. Communication, foreign service, statesmanship, and leadership positions are easily within your capacity.
11: The keywords here are altruism and community. You came into a unique and testing incarnation. You must practice “love thy neighbor as thyself” and use it as your foundation. Your strong intuitions are of value in gaining wisdom and inspiration. 11 is one of the most difficult vibrations because the demand for high standards is constant. You must learn patience and at the same time be able to make quick decisions. Seek for balance between the material, physical life which has to be considered, and the inspiration, spiritual life which underlies your self-understanding. You can succeed in the field of science because all new inventions and discoveries such as laser rays, research in fields of anti-gravity or kirlian photography, or any area of electronics would appeal to you. You could choose to be an astronomer or an astrologer, or a bible researcher and interpreter. You may become a teacher or writer in the field of philosophy. You are original and creative and could become and inspirational speaker. 11 is an esoteric master number of spiritual import. It bestows courage, power, and talent with strong feelings of leadership. You must not let this power go to your head, since frame and recognition are likely; instead realize that true mastership is service.
22: You must express a basic building urge, accomplish things in a big way, and work with large groups or business concerns. You would enjoy the import-export business which could demand long distance travel and meetings with persons of authority. You like to take an inspirational idea and put it to practical use. Self-knowledge is very valuable to you. 22 gives the promise of success. You know how to use your ability to adjust the physical laws of life and living to demonstrate esoteric wisdom. You could become an executive in banking or financial affairs in a national capacity, or help organize businesses for others as an efficiency expert or the like. As an ambassador to foreign countries, you would demonstrate statesmanship. You like to be occupied in some large enterprise to challenge your power to achieve. Your lesson is to learn to take charge of large organizations and corporations and to handle money efficiently and usefully for the benefit of large groups of people.
33: You should be steady and reliable and develop a strong desire to protect others. You would like to live close to nature, and this urge may influence you to choose a life in agriculture. Your goal would be to produce food on a large scale to provide sustenance for the hungry of the world. You would never be found in any profession that could act destructively to humanity. Your talent may lie along the line of the arts: music to bring harmony, painting to bring beauty, or literature to promote education. Service in the field of medicine an healing could also attract you. Possibly, you would choose the law as a way to protect others through justice. Since the 33 consciousness is almost beyond that of humanity, a place within ministry or priesthood could lead you to the realm of your dreams as world savior. You may be required to sacrifice your own desires for the need of others in order to fulfill your Life Lesson vibration.
44: This number stands for strength and complete mental control over your life while on earth. It requires discipline in every department of life so that you may be instrumental in promoting the material advancement of the world. Your mind must be trained to let the higher forces work within it, and you must keep your body and environment in order so that you are ready for any opportunity to achieve this same results from others. Your high energy potential is meant to further evolution by helping others set their world in order. You should try to promote better ethics and justice in the world of business. You must recognize reality, then use what you learn to alleviate the physical burdens of others. You are the instrument by which this alteration takes place. By displaying bravery, resourcefulness, courage, and discipline; you serve as an example for others. Edgar Cayce is an example of the 44 vibration.
Please remember that these are just general descriptions of the numeric vibrations pertaining to the Life Lesson Number. Not everything may resonate with you, and that is perfectly fine. Numerology is meant to be tool to guide, not a stringent set of rules to live by.
Stay tuned for follow up posts on the Soul Number, the Outer Personality Number, and the Path of Destiny Number
((source))
so i redrew my fave scene from consequences because felix has a tiny waist and locus is gent l e
4
9
N°14 in the Jibcon 2017 edit spam - Sunday Panels: Jared 🦁
## the M A R K of O X I N
created by 14 y/o @amazingphil (+ played by @danisnotonfire )
If you ever wanna remake that game… I’d happily do the art for it ;D
But seriously, that video and game was just so lovely to watch… Dan excitedly playing it and Phil reminiscing was really cute!
Thank you for sharing it with us!
phan doodles | all art | retweet
redbubble shop | commissions (open)
10
NCT in Busan
The next time BTS is on American News
• Interviewer: welcome Rat Monster, Gin, Sugar, I-Hop, Bee, Jimmy, and Jungcook!
• BTS: ...
• ARMY: ....
• BTS: ...so when are we going back to Korea?
You’ve Been Warned
This was written for @thing-you-do-with-that-thing SPN Hiatus Writing Challenge Week #14 (14 already????) and the prompt is “I never wanted to hurt you.”
Master List
Characters: Sam Winchester, Reader, Dean Winchester
It was quiet in the bunker without Dean around. He had gone off to help a hunter friend of his, (female of course) take care of a Rugaru case in Texas. You and Sam didn’t have a case at the moment, so you took the chance to hang out and recharge.
Dean was going to be gone for at least a week, so the two of you planned on watching movies, doing research, and sorting through the bunkers many rooms of treasures. You loved spending time with Sam. If circumstances were different….
You had a long-standing policy not to get involved with other hunters. Mixing business and pleasure never seemed to work out well for you. Especially since you not only worked with Sam, but you lived with him too. It was a recipe for freaking disaster.
But damn if Sam didn’t make it difficult. Those eyes, those dimples, and lord that ass! Not to mention his towering height, his muscles, and his hair. (Not that you were looking or anything !) But then roll in his kindness, his empathy, and his determination to always do the right thing, and it all made up for one very hard to resist guy.
2
no one tells you how to mourn.
All Black Everything (M)
Author’s Note: i have seen the devil and his name is kim jongin. someone get me a fan. this is a continuation of the universe for Did You See? you can read both separately, however the impact is a little stronger at the end if you read the original story first.
Pairing: Kai x Reader (oc; female)
Genre: smut
Rating: NC-17
Warnings: explicit sex; explicit language
Word Count: 3,881
Nini[10:03 PM]: i can’t stop thinking about the other night…
Y/N[10:05 PM]: which night? lmao i’ve seen you every night this week
Nini[10:08 PM]: don’t be like that, duchess. you know exactly which night i’m talking about.
Y/N[10:09 PM]: no, nini.
Y/N[10:09 PM]: youll have to be specific.
Nini[10:09 PM]: tuesday
Y/N[10:10 PM]: what happened tuesday? we did a lot of things that night~
Nini[10:13 PM]: are you gonna make me say it?
Y/N[10:14 PM]: yes.
3
2
Growing up, what was your favorite TV show/star and did they make you want to become an actor?
4
P O K E M O N G E N E R A T I O N S
DRAYDEN — Episode 14 - The Frozen World
Jealous Texts
Anon requests: Can you do one where the reader is dating Jughead but they’re trying to keep their relationship low-key. Reggie flirts with the reader and bby juggie just gets a little insecure and it ends in fluff and cuddles and fluff did I mention fluff?
Description: Texts exchanged between Jughead and (Y/N) over the course of a week
Warnings: none
Word count: 311
A/N: I’ve never written in this format before, so let me know if you like it!
Wednesday, October 12
2:14 pm
To: (Y/N)
Still on for Pop’s tonight?
2:15 pm
Of course Jug
Friday, October 14
6:34 pm
To: (Y/N)
Why was Reggie talking to you?
7:12 pm
Why does it matter?
7:13 pm
To: (Y/N)
Because I worry
8:01 pm
Don’t
8:02 pm
To: (Y/N)
But I do
He wasn’t flirting with you, was he?
8:49 pm
To: (Y/N)
(Y/N)?
9:07 pm
To: (Y/N)
9:36 pm
To: (Y/N)
9:48 pm
To: (Y/N)
(Y/N) I’m sorry
10:14 pm
Not mad. Just a bit frustrated
Talk to you tomorrow
10:15 pm
To: (Y/N)
I’m sorry. I love you
10:18 pm
Love you too
Saturday, October 15
9:56 am
To: (Y/N)
Pop’s?
10:04 am
Yes. Noon.
10:05 am
To: (Y/N)
Great
1:47 pm
To: (Y/N)
I miss you and I love you
1:50 pm
I just left silly. I love you too
Monday, October 17
8:03 am
Just a reminder that I love you :)
8:05 am
To: (Y/N)
:)
Tuesday, October 18
10:13 am
To: (Y/N)
He’s doing it again
10:17 am
Reggie’s just a friend, Jug
10:19 am
To: (Y/N)
Yeah but does he know that?
10:21 am
I’ll make sure he does
10:30 am
To: (Y/N)
Ok
Wednesday, October 19
1:02 pm
1:14 pm
To: (Y/N)
Why would I be mad at you?
1:16 pm
Because you’re not talking to me
1:19 pm
To: (Y/N)
I’m sorry
There’s a lot on my mind
1:22 pm
My house.
9
Be there
1:25 pm
To: (Y/N)
Ok
11:03 pm
To: (Y/N)
Thank you for that cuddle session
I love you
11:04 pm
I love you too
The Language Of Love // Carlos De Vil Imagine
a/n: *has 14 requests to get to but writes Carlos for fun instead*
Your knuckles rapped at the door for what seemed like 5 minutes before finally it opened, revealing Carlos’s beaming face.
“What’d you get?” you asked excitedly, practically bouncing with anticipation.
Without giving you a response, Carlos turned around and walked over to the drawer next to his bed. You followed him, shutting the door behind you.
Carlos pulled a piece of paper out of the drawer and handed it to you quickly.
You read the paper frantically and yelled when you saw the grade on the top.
“A+! Oh my god Carlos you did better than I did!” You yelled excitedly as you pulled him into a tight hug.
The smile did not leave his face as he happily accepted the hug and wrapped his arms around you. | 5,378 | 21,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-47 | latest | en | 0.857976 |
https://fr.slideserve.com/hawa/chapter-6-5758428-powerpoint-ppt-presentation | 1,685,558,358,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00486.warc.gz | 313,084,012 | 19,460 | # CHAPTER 6
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## CHAPTER 6
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##### Presentation Transcript
1. CHAPTER 6 Algebra: Equations and Inequalities
2. 6.3 Applications of Linear Equations
3. Objectives Use linear equations to solve problems. Solve a formula for a variable.
4. Strategy for Solving Word Problems Step 1 Read the problem carefully several times until you can state in your own words what is given and what the problem is looking for. Let x (or any variable) represent one of the quantities in the problem. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x. Step 3 Write an equation in x that models the verbal conditions of the problem. Step 4 Solve the equation and answer the problem’s question. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words.
5. Algebraic Translations of English Phrases
6. Example 1: Education Pays Off This graph shows the average yearly earnings in the United States by highest educational attainment.The average yearly salary of a man with an associate degree exceeds that of a man with some college by \$3 thousand. The average yearly salary of a man with a bachelor’s degree or more exceeds that of a man with some college by \$41 thousand. Combined, three men with these educational attainments earn \$188 thousand. Find the average yearly salary of men with each of these levels of education.
7. Example 2: continued Step 1: Let x represent one of the unknown quantities. Let x = the average yearly salary of a man with some college. Step 2: Represent the other unknown quantities in terms of x. x + 3 = the average yearly salary of a man with an associate degree x + 41 = the average yearly salary of a man with a bachelor’s degree or more.
8. Example 2: continued Step 3: Write an equation in x that models the conditions. x + (x + 3) + (x + 41) = 188 Step 4: Solve the equation and answer the question.
9. Example 2: continued The average salary with some college = 48 The average salary with an associate degree = x + 3 = 48 + 3 = 51 The average salary with a bachelor’s degree or more = x + 41 = 48 + 41 = 89. Some college = \$48 thousand per year Associate degree = \$51 thousand Bachelor’s degree = \$89 thousand Step 5: Check the proposed solution in the wording of the problem. The solution checks.
10. The total price of an article purchased on a monthly deferred payment plan is described by the following formula: T is the total price, D is the down payment, p is the monthly payment, and m is the number of months one pays. Solve the formula for p. T – D = D – D + pm T – D = pm T – D = pm m m T – D = p m Example 6: Solving a Formula for One of its Variables | 697 | 2,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-23 | latest | en | 0.900371 |
http://mymathforum.com/number-theory/14142-clock-hands.html | 1,555,597,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517682.16/warc/CC-MAIN-20190418141430-20190418163430-00011.warc.gz | 120,190,275 | 9,576 | My Math Forum the clock hands
Number Theory Number Theory Math Forum
August 15th, 2010, 12:05 PM #1 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 the clock hands I don't know if it's the right place to ask this question so I'm sorry if it's not. I'm a rather layperson in mathematics but there's one puzzle I've been given by my friend recently and which i have to solve in order to borrow something from her. Can you help me with this, please? It goes like this: "When I checked my watch this morning the hour hand was exactly where the minute hand is now, and the minute hand was one minute division before where the hour hand is now. And both hands are exactly at minute divisions now. What was the time when I checked it in the morning?" I went like: x=the place of the hour hand now, y=the place of the minute hand now, likewise z=the place of the hour hand earlier, w=the place of the minute hand earlier So: (60z + w)/12=y and (60x + y)/12 - 1=w It made the final equation 720z + 60x - 143y = 12 What should I do next? I don't know how to do the equations with three unknowns (though the unknowns are all integers) Anyone can please explain how I should proceed? I would appreciate that.
August 15th, 2010, 09:34 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: the clock hands 7:12. At 2:36 the hour hand is 13 ticks past 12 and the minute hand is 36 ticks past twelve. In the morning the minute hand was 12 ticks past 12 and the hour hand was 36 ticks past 12.
August 16th, 2010, 01:52 AM #3 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 Re: the clock hands Thanks a lot! I got what I wanted. How did you do that? you did any calculations or you just knew it straight away? I'm just curious...
August 16th, 2010, 09:25 AM #4 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: the clock hands I wrote p = b - 1 q = a a b and 12 24 36 48 00 then the answer became apparent.
August 16th, 2010, 10:47 AM #5 Newbie Joined: Aug 2010 Posts: 3 Thanks: 0 Re: the clock hands Right...foolish me give me some of your brains Greg please...Thanks once again.
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null | null | null | null | null | null | Sean Hannity
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8-21 Beach and Company Hour 3
Aug 21, 2014|
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Automatically Generated Transcript (may not be 100% accurate)
Wow -- Hamburg they're hiring them assistant not a system a temporary a school superintendent. And after they looked at his credentials as academic credentials they asked them to -- car. I -- a dark driveway without hitting a telephone pole that that's never got. Had been prior superintendent. And a better car widths -- a backup camera needs that we superintendent. Plus the guys 77 years also won one would think that maybe doesn't need any. Often grounds meetings. We have other administrators. Get my drift. So that's a good choice good choice there. Al but if the sword dual bathtubs to be installed in his office side by side parallel bathtubs. Be careful be very careful. You're taking your -- to see what's that -- the fresh -- band OK you know all the little -- kids are going to be there while -- screaming I'm sweating and yet basically any -- -- that watch Nickelodeon will probably be there and I think. Just approved I'll graded that you are on the way home he should stuff -- Chucky cheese eaten and I. Well and that's that's the tests. That's better than lassie saving -- from following in the well I'm just telling you. Taking your kid at chuck. Well he still likes -- ugly I don't think the employees want me back when it's funny if you were dating girls there for the first time and you Broderick refugees. How did you get a second date I think also that young married couples who haven't decided yet. On how many children they want or they want children at all. They should go to -- injuries they should I might give them some guidance generals thought it. -- Are we talking about what's going on taking a breather air and Ferguson Missouri is this a lull before the storm. It's funny because I mentioned a couple of times earlier. There or not a lot of people on street last night. Because it was raining. The other thing about it yeah SunGard yes I'm going to protest. It might get different golf they're gonna be fire bombs there policeman a bare armed. Were also are going to have the possibility of tear gas but I'm going oh it's raining like -- maybe maybe tomorrow when it's a little nicer. So people are are more are. More afraid of getting wet than they are. Of it dealing with the stuff I just told him. So is this a while before the storm yes I think it is and the reporter -- blood from -- Malloy -- think the same thing. Will be policeman Darren Wilson. Be indicted. I I want. What is legitimate injustice I don't want vigilante justice. I don't want mob justice I don't want racial quote unquote justice I want justice period. There and that is just is why the most debate with somebody somebody make it just to see again it was a some kind of Americo. -- justice supposedly blind but Eric Holder says I'm not only the attorney general -- a black man. Maybe maybe you know what I'm not giving him. The benefit of the doubt the exact quote I am the attorney general but I am also black man you know what I think. I'm gonna take the positive side I think he was speaking at a school for the blind. So that people wouldn't know he was a black man until he told him maybe he was maybe was that a Dave Chappelle concert. I'm not really sure a crack at it. Any regrets saying I am the attorney general what I'm also black -- and people only on its. -- He's a black man. So I mean it's it's an educational process notice how he's not a cell wall by a band guys like bank cards and Arianna I exactly I understand that job. Accomplished black people who don't -- their lives they've all sold out in order to do that but not not holder. -- -- -- Let's go to Carol in Cheektowaga Cara -- on WB yeah. Our two day time on girl what do you have on your mind. Well -- have a flat he urged okay. There were and why. He won't -- see it the it's harder for you walk. Last eight -- they -- it won't walked -- -- cannot -- yet why don't. Eat well -- -- are you know I Mike and Ike and I can understand. Well thank you -- by a -- lot of guilt all these. -- Main rate debt is debt is -- the main rate they don't like but a now -- Barrett that -- all other bank. I feel start in this but not there I don't care what are -- Because. I think yet everything stacked against and even now in it I'm not big rock. You know I mean that it needed so I'm pretty beat a dead but I think that he worked in our that this duty. But -- I'll I'll believe that it chill. The black population who don't agree. With Eric Holder and as Alabama. And -- racial stop. -- -- -- -- Very loudly. At your make it look like -- What I had been a governor you said you feel sorry -- policeman Darrin walls united so busy -- governor said we must get justice for a group of video Michael Brown family. When indeed he didn't mention -- Wilson in his -- address and he mentioned justice for the Michael Brown family several times so that gives you an idea where his mindset is. All right all right and and you know. I -- I I don't ever hit it again and got my eye out. It's going to be. And even if he gets a fair trial it's. Where city it's going to act and even if they know about -- I guess -- yet -- I praying are scared that I really am I'm praying that the justice -- some. Are available right way but the only one that he out of that little. -- just. At -- Judge Judy if you noticed a commercial street at eight. Well you know it's true if if he does go to trial you're right in -- a change of venue. Via city is predominantly black now normally I wouldn't even bring that up -- and every lawyer in the world. -- complains that innovators -- many blacks on the jury or not enough blacks on the jury. But the -- they bring up the racial side of it all the time and I think that's abhorrent to be honest -- that makes the assumption that. If if you are one color you -- going to vote a certain way no matter what the evidence says and I find that to be wrong but they always do it and so that's what we have. As a population makeup -- enough Ferguson. Well like I said before but I I believe that gadgets that the population color shouldn't matter but and that case. And I think it's going to. Will it be a problem. I'm with Leo and you know I think I write for the like for our Carol it's all hope for the best that you have very much yeah I never use that you know because I always found. How -- that the second line is already. Another words no group has everybody in the group thinking the same thing acting the same way we're all individuals and I totally agree -- that I I do I do believe that. And yet you talk to defense attorneys. And it publicly update them and hide its. If if you'll have to say you have a black defendant. The how many times have you heard of defense attorneys say there aren't enough blacks on the jury. Or you'll hear somebody's go the other way say OK we got a good mix on the jury of blackened and why. -- I find all of that. Discussion. Objectionable. Because it does treat everybody like a monolith. So -- you -- both sides of a straight are you saying that every black jurors going to favor every black defendant no matter what the evidence says. Are you saying every white jurors to convict every black defendant no matter what the evidence says. Well then they shouldn't be jurors nine -- being naive. And I and I'm looking at -- through rose colored glasses. But today -- listen to the evidence you make your decision and you vote accordingly. If you could be blindfolded in the jurors' box that would be ideal. -- but that isn't the way it works but defense attorneys and jury -- specialist. Know what they're doing -- -- make their living and they see it all the time and certainly the worst example we have ever seen is the OJ Simpson jury. I'm no question about that OJ was guilty you know what I know -- can say whatever you want to we know it. But. He was a free manner of course and -- again and trading cards order -- invaders we'll be back should -- state in Vegas will be back with beach -- company. Along with that and then along with the number of arrests that the deplete their hat at people that are just trying to make trouble. It really is a lot calmer atmosphere. When you found out there was going to be some rain -- you put on a big frowns and Aldridge. I'm going to be said then it's going to rain now I don't wanna embarrass you but when you go to the beach. Don't you usually get wet. And -- vs them -- get wet from the water next of this and or wet from the rain from a house the bikinis won't -- is -- and all of the women yeah. The women don't go out in the rain now. Does that put the cake out of the rain repairs there if they did a lot of slice -- MacArthur is a chocolate cake. I can't you know I understand that this change is gonna give kind of. You know. Disgusting with it's it's it's well that's a long walk through that as you know you're probably you're probably via -- it up and down with your Childs. Yeah as you know a -- child as a chick magnet that option isn't he Q road I don't curly hair and -- yes I'm his you know that kind of stuff so that wouldn't work probably not. -- you don't wanna get wet when you're going to a place to get wet right to get that -- Plus I wanna sit out I'm the patio in young -- in the song that's -- you are sun goddess you know where one of those collars like they've put around Japan at the vets. You also reflects a certain. You know -- polls. The couple hole maybe you can win a mystery ball maybe I am in there. Tonight I just want to have a good time Saturday girls just wanna have fun that's the rights over the water he's a you wanna file with the boys and all of address governor. Owning up visiting run with -- -- besides -- so while police statements that picture and Elijah. -- As I don't I Israel 1806 with six guys who -- six and I'm thirty let's go to or asking the basic questions. Regarding Ferguson is this the lull before the storm. Will be policeman. Daryn -- also want he has his name he's not just the policeman -- a person. And he was there to protect and serve the public. And he has broken eye socket to it's a show for it. He is is he going to be indicted what do you think is just this blind if so Eric Holder wanted to make sure that everybody and so he was black. He said I'm not only the attorney general I am a black man. For those -- you ever seen in my guess that's clarification. Let's go to Emily in north buffalo heavily on WB yeah. Are they duplicate Mike -- -- I I knew a year cold heart and using your theory entry changed its app. Candy I did get a couple of comments -- about. After I heard. Eric holder's comments. The only way it would be justified if you were talking tool apply audience. That's right school -- lines -- I you know I am at every child might hear like killing -- murder no matter where it is located. What they have yet I just can't. A radical suggestion. Would be. -- -- -- -- -- Future problems. In predominately. Minority communities. Case in point Ferguson. I don't think jet that only minority. Patrol officers. Patrol those areas. This is an important wouldn't have escalated. -- would black I'm black we wouldn't have Al Sharpton we wouldn't have -- Jesse Jackson and others. There. You know protests being -- this is not going to be resolved peacefully. Because it broke a police officer is following. You know not guilty there's going to be more right. Big time right because they're gonna feel it's there was a cover up the you know most fix the rig was there and they want -- -- they want is -- -- that there's -- they're not going to be have a Julia. And you know not if they have if they hire more minority police officers. Then there'll be wiped out from minority let them patrolled their own people. That's the only way bet they're going to be happy. And Jesse Jackson and Eric Holder and everything will be able to complain that something happened. -- okay well said thank you very much and for those that Leo. And there are people of their. Who'll look at this -- was a fair guy you've got to think that what part does race play as far as the attorney general. Coming into town. To head up the investigation. With the FBI and with the criminal division of the civil rights part. Section of -- of of the Department of Justice. I mean if if you wanna be fair it should be treated like any other case but obviously it's been set aside a special. Because of vote the racial angle of this as the last caller just said if this were a black cop that shot. The Michael Brown it would -- be an issue. It would not and and most of a crime against blacks are committed by blacks make no mistake about it. So -- is bill for the racial angle is on the far and away. The most the most visible of all the angles here I just hope that this guy this cup. The guy who took an oath to load the two. Protect and serve the community are hoping its partnership. Because if not what can you expect just -- whatever the mob wants give it to them. Why don't we get our rob us out there and eskimo what he feels we'll be back -- would beach company and whose radio and I thirty WB. You're hearing the voice of buffalo. WGN. That call us now an 8030930. Cell calls are free and start 930 -- our toll free line is 180616. 9236. We -- back where Visio they have -- to be -- talking about was everybody's talking about it that would be Ferguson Missouri is this a lull before the storm no question about that. But I have a -- now portable. These protesters are. There's a lot of them are not from the area -- from out of town. And I don't know about you as I mentioned earlier if suddenly I had a mad desire to be in another city to protest something. That allegedly -- ought to me. I don't offered to take five or six weeks off. I don't know if I could afford to be you know hotel room motel room for that long and eat out -- India and but not not go to my job. -- take care might hats. I mean how these people sold mobile. I'd like to know it's like a movable feast for them they just get up there must have a bag -- near the door is like denial of Lucy episode when she was pregnant -- little Ricky. Dora -- bags packed ready to write out the door and go to the nearest protest. At four of you who say and you know says this can Hamilton says this Tony. Can -- that says is that I disagree with them what we've talked about why oh why dude who knows certain groups. Have people speak for them that are morons. And like Al Sharpton. And he says. We meaning of the black community did not make them our spokesperson. That they have a -- to people for the media so you let's go home that they're they're speaking for the population when indeed they're not. Ever those who tried to minimize Al Sharpton by saying he's a charlatan. And I would be among a group. He's an idiot idea among the group he'd have to study to be in he'd have to -- I have been taking night courses to be -- -- right and he's dangerous. And -- problematic is no question about that. But for those of you say well the media made him -- really. Oh really there is going to be a funeral on Monday. Four Michael Brown. You know speaking out Al Sharpton. And Al Sharpton is hijacking this thing he is doing this some kind of big march going on in New York City on Sunday. And he is going to be leaving that march. I don't think the march was originally designed for this event was for something else. But he's taking this and and blending it with that. On Sunday and then Monday he's gonna speak at the funeral so don't tell me that the medium made him. A lot of people in the media that I don't think he's a moron. But that's the deal so yeah this is what you get. This is what you did you get Al Sharpton speaking. All right immediately services some people see him or Jesse Jackson the shades were down over our eyes because this just another blah blah blah moment. I -- I agree the canopy the media hands. Made him. To the point where he's acceptable. To anybody that's holding a rally like that because he's gonna say things that they're going to want to each year. He's got every -- yeah yeah I think they'll follow for that by. I do agree that for the most part the media has made Al Sharpton and Jesse Jack. Well how about to have immediate Kansas makes somebody about Ben Carson OK he's an articulate spokesperson bodies to the right by the guy he's for the ride and -- Ruling and figure that a lot more match point on is being that the that the media being -- being -- winning the but the whole point is that this is what you get for if you want. To better your position in life. He got to get better than Al Sharpton he really do. Because you've got to get to some somebody when that person speaks everybody goes you know. That's somewhat you know it doesn't that's something I'm about to give that some thought. He's not a guy. And neither is Jesse Jackson and to prove -- I mean Jesse Jackson's up there try to raise funds try to do fund raising you got booed off the stage. This is what he thinks about he thinks about an opportunity to further his agenda and further his his his wallet. And that's what it's about and sharpton's even worse I think. Because of the -- he is he's not going to bring anybody together so the only person I know that brings their regulars buy hourly. Because he says that's his goal that's his mission did you know that really I'm trying to get them. Get me together with some of the promotion woman and I don't know if that works and seeing nuclear fourteen of the hook -- but got the shaft and doesn't bring anybody there. He just he just immediately divides people. And as soon as you see them it's one of those things you know a billboard advertising rates are you changeable -- because once somebody sees it. -- and next time they go buy it they don't even -- -- because they've already registered as having CNET -- and it's the same thing with Sharpton. As soon as you see Sharpton you immediately. -- I do you immediately disk now what jurors seeing what you're hearing as meaning. This sharply about equivalent to our standing in the situation. As a spokesman Mike please tell us Mike the situation. And now for the Italian American society -- Mike the situation. Yeah that's good that really works. And I who's there. I think Gary is there from -- -- one of our -- table I Guerrier on W via. They -- -- city Tony. -- I'm still concerned about. -- dog in the rest of society being trained. That. Any officer cannot shoot a black suspect. -- The one thing they're really I find troubling on this whole event. Is the officer has been tried convicted. And sentence. -- dubious witness statements hearsay innuendo. They don't want the -- They -- -- -- they go on trial they don't want a grand jury they want him to go directly to electric -- Does that mean. That we are returning to. Create constitutional. And in some cases. Pre magna cart along were going back to. -- For every -- I like I like your attitude on this because I felt the same way it's a lack of respect for law and order. And that starts at the top it starts with the Oval Office. If the people on the street see the president disregarding laws or not funding laws of the Campion for us. Or just picking and choosing which laws -- bay. And it's quite obvious that he and his a right hand man. But the attorney general have the attitudes they have. I think that without that respect for law you're gonna do things you normally wouldn't do I think there -- emboldened. And their enables through doing this kind of stuff and it's gonna have a bad end result. Didn't we fight a civil war partially for the reason. That people were being improperly. Single now. And punished and in a lot of cases kill. Because of their color of their skin of course absolutely. I have a solution out there skip work or go back to meet able justice the next I want these things happen. Let's bring domestic. Or sacrifice. Always -- and then them distinction re entrails and say guilty or not guilty -- that are gonna go. I think you're on the something and Gerri thank you thank you very much is that what he's saying really we're gonna sacrifice -- -- Personally have a lot of town opens fire here but I'm just saying you know you can argue the change of venue immediately. It's our you're on and we like to sacrifice -- version of the scene having will be back -- with Beijing company under Israeli and I'm thirty WBA. Then -- could be a lot of. You know this. These mobs out about free speech or anything like that. First of all we said that all the things a president that there should be president that are constitutionally. Protected. Free speech they have a right to assemble. And they have a right to a free press -- they have almost. But free speech apparently only extends to the mob itself. Because they showed video last night of a woman I guessing because there was a lot of commotion should probably a middle aged woman. Who had a small sign. Rose colored kind of supporting the policeman. Supporting Darren Wilson. And she rushed by the mob the police had to really kind of rescue her and get her inside a police cruiser and get out of -- way so. Enough about free speech it's free speech only if they agree with the -- of -- don't agree with you let's arrest them and maybe even injure them -- -- them. -- that's the way it as and that's what's going on there meanwhile I think to give into it. It would be a dramatic mistake and yet the idiot governor they have as tried to walk back what he said earlier. What he said was not at a press conference where he just casually said something off the top of his head what he said it was a pre recorded message. And Dana Perino used to be the bush. A press secretary. Of the white house Press Secretary for George Bush. I said that you know that speech was written. It was approved. It was edited and that was recorded before it was played. It went through all of those steps so obviously going through all of those steps we have all of those people involved with -- that's what he meant. But he got a lot of blowback on that even from people who might be on his side. Because the call for vigorous prosecution. Of the policeman. Where and no evidence had even been gathered. No conclusions have been made no grand jury testimony at the point he he release that and so he apparently got enough blow back. That he tried to revise that and a spokesperson. Said no what he meant was. The whole process will be vigorous while that isn't what he says. As a matter of fact -- he said was we we have to seek justice for the Michael Brown family. Now Michael Brown was the one losses wife we understand that but Michael Brown moments before was committing a robbery. And there the question is cool broke the orbital bones of a cop. -- Obi wan side says it was Michael ground. And I don't know what the other side says maybe he fell down -- something maybe. He accidentally slipped on banana peel some ridiculous thought like that. Okay so you got the man you got video governor there the mayor has been eclipsed. Of the city he says he's hardly been involved with anything impact he needed to get information on how to get all of -- governor one time. And so it's that the governor and now the head of a circus yes Eric Holder. As as Eric knows he probably learned this a law school that justice is blind except for him. When he says I am the attorney general but I am also a black man. Oh really tell me what has that do with anything. I has nothing to do anything. Except for Nazis tried to accomplish. The fact that you're gonna get a fair -- because I'm in town that's fine. But I don't think I don't think this mob that we've seen on the streets or into innuendo. Or subtlety. I I don't think -- -- shading. I think I think it's meat and potatoes this is what you get this is what you see and that's it. I think. By reinforcing the fact that he's a black -- and I guess he's been a black man Wallace live from nothing though -- That he black man's -- like I think he graduated a few years ago graduated early because he's not -- really really black he's now I'd be age. So he Wednesday got a set I'm a beige man and that that would've you know may be -- half a crowd. What I what gets me angry as he's using race just like the mob is using race. There's no question in my mind and is this all before the storm take back to the bank absolutely positively. As a reporter from KM -- Alison no blood said. That if indeed the video policeman is not indicted. They'll be problems. If he is indicted and a charge if the chargers aren't severe enough they'll be problems. If he is indicted in the charges are severe enough and he is acquitted -- problems. If the charges are severe enough and he is indicted and he is convicted. And sentence at the sentence isn't harsh enough. They'll be trouble so in most of these scenarios that I can come up with they'll be trouble. You take that in the bank I wish I wish I didn't feel that way about that is the way guys I see this thing going. And I'm I'm just asking. A whether this is real justice or this is show business. When asked why they sent Eric Holder. Wannabe commentators -- optics. It looks good to send Eric Holder it shows a degree of here's how important this event is we're sending the attorney general. Let's go to Allah in west Seneca LA area going -- sue Ferguson Missouri to protest. No. -- Sandberg and eat candy I can't believe what's going in our country. First of all let me in major bank and the year thank you people -- up for -- did a wonderful job and -- they use some common sense. And it too which I find so that they. I mean when I look back. My generation. At what's happening now I'm scared. Yeah I think we're losing our country there's no question I'm with the -- on this. Mean we we have the greatest country in the world. And god gave it says quite a hat and people use the brains they -- given immunity -- and that rightly. I mean with -- sensible think that. Now the demonstrated number one on the fact that they're not even from that area. Yet and don't know. Number so instead of all this demonstrating why don't they -- down on their knees and start pretty. That is what we need. A return and that he. And number three I remember and this it really deeply. Controlling a police officer was to respect that I remember it started tourney he couldn't. Hope -- -- he was the all week meant they -- I've seen and been around the wrong and he has seen things. But nobody wants to -- that older people well that's what we don't we don't have any rain -- Well the problem is that everybody. Who is of a newer generation thinks that the older generation didn't have anything to do with the country that they've invented the country since they were born whether their athletes politicians or anything else. But I have the same feeling you do well I'm not won two of panic. However I am want to see that. With the lack of respect for law and order and the people who are in top positions in our government. I don't see a way out of this thing at least for a few years okay got it around and thank you alive. I really don't I think it's I think we're on -- on on a bad road right now. And it's gonna get worse analyses and common sense and get back to some respect for the law respect for the constitution about some respect for ourselves. About that. All right that about wraps it up good luck on -- moment tonight we'll see you tomorrow at 9 AM Monday Israeli and I started -- BBN. Well it's they never have to these beasts.
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How Hollywood’s biggest politicos leaned right, not left
by Jonathan Kirsch
March 14, 2012 | 10:54 am
Ronald Reagan, Shirley Temple, Sony Bono, George Murphy and Arnold Schwarzenegger are all entertainers who launched their political careers in California, and they are all Republicans. Indeed, aside from Al Franken, no prominent Democratic officeholder on the scene today started out in the entertainment industry. Yet, ironically, a myth that began in the McCarthy era — and persists today — holds that Hollywood celebrities on the left play a powerful role in American politics.
“Quite candidly, when Hollywood speaks, the world listens,” Sen. Arlen Specter once observed. “Sometimes when Washington speaks, the world snoozes.”
The myth is misbegotten, or so argues film historian and USC professor Steven J. Ross in “Hollywood Left and Right: How Movie Stars Shaped American Politics” (Oxford University Press, $29.95), a benchmark study of the role that Hollywood stars and moguls have played in American politics. Like Neal Gabler’s classic “An Empire of Their Own,” Ross’ book allows us look behind the curtain and to glimpse the inner workings of the entertainment industry.
Hollywood began to figure in politics as early as 1918, when federal agents reported that movie stars were playing “an active part in the Red movement.” But, from the start and throughout its history, activists on the left have always been less successful than those on the right. “It was the Republican Party, not the Democratic Party, that established the first political beachhead in Hollywood,” Ross explains. “The Hollywood left has the political glitz, but the Hollywood right sought, won, and exercised electoral power.”
Ross surveys nearly a century of Hollywood history through the lens of politics. Of necessity, he drills down into the nuance and detail of corporate and union politics in the movie business. But he also comes into tight focus on a few of the more famous faces. Charlie Chaplin, for example, is singled out as the first star to strike a political stance — an explicitly anti-fascist stance. “No silent star,” Ross writes, “brought political messages to the mass public more effectively than the man millions of moviegoers affectionately called ‘Charlie.’ ”
But Ross also reminds us that Chaplin was hounded by right-wing activists, both in Hollywood and in federal law enforcement, throughout his long career, and he was ultimately driven into exile as much for his politics as for his supposed promiscuity. “You are the one artist of the theatre,” observed the writer Lion Feuchtwanger, a Jewish refugee from Nazi Germany, “who will go down in American history as having aroused the political antagonism of a whole nation.”
By contrast, studio mogul Louis B. Mayer, one of the founders of MGM, is singled out as the archetype of Hollywood Republicanism. He was hailed by Rabbi Edgar Magnin as “an ardent enemy of pseudo-liberals, Reds, and pinks,” and Ross himself credits Mayer with teaching the Republican Party “how to use radio, film, and movie stars to sell candidates and ideas to a mass public.” At a time when Fanny Brice, Eddie Cantor and George Jessel were campaigning for Democratic presidential candidate Al Smith, for example, Mayer served as executive director of the Southern California campaign committee for Herbert Hoover.
“Hollywood Right and Left” does not overlook the McCarthy era, although it is only one episode in a much grander saga. But Ross approaches the subject from a new and illuminating angle by focusing on the plight of Edward G. Robinson, an early and committed anti-fascist at a time when Irving Thalberg was comforting his boss, Louis B. Mayer, with a rosy report from Nazi Germany: “Hitler and Hitlerism will pass,” Thalberg said. “[T]he Jews will still be there.” Robinson, by contrast, worked with other stars to organize a boycott of Nazi Germany in 1938, an effort that was not popular among isolationists in America.
“[The] movie colony may root for the Jews all they wish, but don’t think that the people of the United States are going to fall in with your plans,” one estranged movie fan wrote. “Those of us who know World History and the Bible know that the Jews have always been in trouble up to their ears.”
Robinson, who was condemned as “Yiddish riff raff” by another letter writer, was repaid for his activism with surveillance by the FBI during the war, a place on the blacklist, and repeated appearances before the House Un-American Activities Committee when it targeted Hollywood in the late 1940s and early 1950s. “Outraged by the smear campaign against him,” Ross writes, “Robinson spent the next three years of his life, and over $100,000 of his own money, trying to clear his name and resume his career.” Ultimately, he was reduced to abasing himself as “an unsuspecting agent of the Communist conspiracy,” although he refused to name names. Ironically, he was “restored to semi-respectability” only when Cecil B. DeMille, “one of Hollywood’s most prominent anti-communists,” cast Robinson in “The Ten Commandments” in 1956.
The excesses of the McCarthy era eventually subsided, but Ross makes the point that the balance of power in Hollywood remained on the right as Murphy and Reagan used the denunciation of supposed “Red Menace” in Hollywood to launch their own political careers. Reagan, of course, has been credited with nothing less than a revolution in American politics, while Jane Fonda, an activist on the left in the same era, crashed and burned. Her counterpart on the right, at least in terms of the visibility and intensity of his role in politics, is Charlton Heston, whom Ross describes as “the first prominent practitioner of image politics,” if only because Heston played not only Moses, but also “three saints, three presidents, and two geniuses.”
Fonda “demonstrated that celebrities could use their star power to draw attention to controversial political issues,” Ross explains. “Her subsequent vilification revealed how the public often view such activism, especially left activism, with suspicion and cynicism.” The woman who came to be known as “Hanoi Jane,” Ross points out, “paid a high price for her activism.”
The bottom line, according to Ross, is that one wing of the entertainment industry seems to have connected with the hearts and minds of the American electorate, and the other has not. “From Louis B. Mayer to Arnold Schwarzenegger, the Hollywood right has told a simple but compelling story of American triumphalism: America is the greatest nation in the world. What else do you need to know?” The Hollywood left, by contrast, has been undercut by its willingness to look behind the façade. “Few citizens want to hear a Jane Fonda, Warren Beatty, or Sean Penn point out what is wrong with the United States.” In that sense, Ross’s even-handed but eye-opening book serves as a corrective to some very famous entertainers who simply failed to understand how they come across to their audience.
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# Week 6 06.12.21-10.12.21
Literacy:
Our Story of the Week is 'Dear Santa' by Rod Campbell. Watch the video below and while watching ask your child to predict what might be hiding underneath the presents. Highlight the language - too small, too big and too scary. Can they predict what the final gift could be.
Can you write a letter to Santa? Choose some nice paper or card and tell a grown up what you would like Santa to bring you this year. Is there anything else you would like? Maybe you could have a go at writing your name at the end. You could even post your letter off to Santa when you are finished.
## Dear Santa by Rod Campbell - With fun questions!
'Dear Santa' by Rod Campbell.
Maths:
Our number song this week is 'One ,two Buckle my shoe. By singing number rhymes children will learn how to count. Have fun joining in with the rhyme.
## One Two Buckle My Shoe
This week in maths we are learning to separate a group of three or four of objects in different ways and we are beginning to to recognise that the total is still the same.
Separating objects means a variety of different ways to split and arrange a set of objects. The total does not change. Watch online, click on the video below.
## Separates a group of objects in different ways and understand the total is still the same
Challenge:
1. Gather a collection of four soft toys and two pillows or cushions.
2. Explain that it is time for the toys to go to bed.
3. Encourage the child to arrange the toys on the pillows in different ways and then count the total number of toys. This will encourage them to recognise that the total number stays the same, no matter how the toys are arranged.
How to Get Your Child Thinking:
Try asking questions, such as: How many toys are on this pillow? How many toys altogether? If we move the toys all on to this pillow, how many are there? What can you tell me about the number of toys? Have a go at using three pillows or cushions instead of two. Again, encourage the child to arrange the four toys between the three pillows and count the total. Talk about what they notice.
Phonics:
Children in nursery are beginning to learn some phase 2 sounds and are practicing blending sounds together. Watch the video below and practice saying the sounds.
Top | 516 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-33 | latest | en | 0.923569 |
https://www.physicsforums.com/threads/harmonic-oscillator.912578/ | 1,508,426,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00625.warc.gz | 958,367,425 | 16,931 | Harmonic Oscillator
1. Apr 25, 2017
alex91alex91alex
1. The problem statement, all variables and given/known data
I am having issues with d) and would like to know if I did the a, b, and c correctly. I have tried to look online for explanation but with no success.
A harmonic oscillator executes motion according to the equation x=(12.4cm)cos( (34.4 rad /s)t+ π/5 ) .
a) Determine the amplitude of the oscillation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
d) Determine when the object is at its equilibrium position.
2. Relevant equations
max v=Aw
T=1/f
3. The attempt at a solution
a) Determine the amplitude of the oscillation.
Amplitude would just be 12.4cm, we can take it straight out from the equation.
b) Determine the maximum velocity of the oscillation.
c) Determine the period of the oscillation.
Here we know that 2π rad is one turn so 34.4 rad is 5.48 turns.
So we have 5.48 turns per second.
T=1/f=1/5.48=0.183secs
d) Determine when the object is at its equilibrium position.
I know that the object is in equilibrium when x=0, so
0=(12.4cm)cos( (34.4 rad /s)t+ π/5 )
I may be missing some algebra skills, but I even try to compute it online and it wields no solution. What am I doing wrong?
2. Apr 25, 2017
Staff: Mentor
Your work on parts (a) through (c) looks fine.
For part (d), consider what the argument of the cosine function needs to be for the cosine to be zero. (Hint: there are many such angles)
3. Apr 25, 2017
alex91alex91alex
So cos( (34.4 rad /s)t+ π/5 ) need to equal 0?
(34.4rad/s)t+π/5 has to be equal to π/2 or 3π/2?
4. Apr 25, 2017
alex91alex91alex
That is going to wield t= 0.02739s and 0.1187s, does not feel right though because a period takes 0.183secs.
5. Apr 25, 2017
Staff: Mentor
Yes.
In fact, the cosine will be zero every time its argument is equivalent to π/2 or 3π/2. You should be able to write it as a function of n, where n = 0,1,2,.... Or you can solve for the first instance (n = 0 so that the argument is π/2) and then it will happen every half period after that. | 626 | 2,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.896694 |
https://us.metamath.org/mpeuni/cygabl.html | 1,709,591,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00286.warc.gz | 589,124,961 | 8,062 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cygabl Structured version Visualization version GIF version
Theorem cygabl 18946
Description: A cyclic group is abelian. (Contributed by Mario Carneiro, 21-Apr-2016.) (Proof shortened by AV, 20-Jan-2024.)
Assertion
Ref Expression
cygabl (𝐺 ∈ CycGrp → 𝐺 ∈ Abel)
Proof of Theorem cygabl
Dummy variables 𝑛 𝑥 𝑎 𝑏 𝑖 𝑚 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2826 . . 3 (Base‘𝐺) = (Base‘𝐺)
2 eqid 2826 . . 3 (.g𝐺) = (.g𝐺)
31, 2iscyg3 18941 . 2 (𝐺 ∈ CycGrp ↔ (𝐺 ∈ Grp ∧ ∃𝑥 ∈ (Base‘𝐺)∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)))
4 eqidd 2827 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → (Base‘𝐺) = (Base‘𝐺))
5 eqidd 2827 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → (+g𝐺) = (+g𝐺))
6 simpll 763 . . . 4 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Grp)
7 oveq1 7157 . . . . . . . . . . 11 (𝑛 = 𝑖 → (𝑛(.g𝐺)𝑥) = (𝑖(.g𝐺)𝑥))
87eqeq2d 2837 . . . . . . . . . 10 (𝑛 = 𝑖 → (𝑦 = (𝑛(.g𝐺)𝑥) ↔ 𝑦 = (𝑖(.g𝐺)𝑥)))
98cbvrexvw 3456 . . . . . . . . 9 (∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) ↔ ∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
109biimpi 217 . . . . . . . 8 (∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) → ∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
1110ralimi 3165 . . . . . . 7 (∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
1211adantl 482 . . . . . 6 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
13123ad2ant1 1127 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ∀𝑦 ∈ (Base‘𝐺)∃𝑖 ∈ ℤ 𝑦 = (𝑖(.g𝐺)𝑥))
14 simpll 763 . . . . . . . . 9 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → 𝐺 ∈ Grp)
15 simpr 485 . . . . . . . . . . 11 ((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) → 𝑥 ∈ (Base‘𝐺))
1615anim1ci 615 . . . . . . . . . 10 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ) ∧ 𝑥 ∈ (Base‘𝐺)))
17 df-3an 1083 . . . . . . . . . 10 ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺)) ↔ ((𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ) ∧ 𝑥 ∈ (Base‘𝐺)))
1816, 17sylibr 235 . . . . . . . . 9 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺)))
19 eqid 2826 . . . . . . . . . 10 (+g𝐺) = (+g𝐺)
201, 2, 19mulgdir 18204 . . . . . . . . 9 ((𝐺 ∈ Grp ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ ∧ 𝑥 ∈ (Base‘𝐺))) → ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2114, 18, 20syl2anc 584 . . . . . . . 8 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ (𝑚 ∈ ℤ ∧ 𝑛 ∈ ℤ)) → ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2221ralrimivva 3196 . . . . . . 7 ((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
2322adantr 481 . . . . . 6 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
24233ad2ant1 1127 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ∀𝑚 ∈ ℤ ∀𝑛 ∈ ℤ ((𝑚 + 𝑛)(.g𝐺)𝑥) = ((𝑚(.g𝐺)𝑥)(+g𝐺)(𝑛(.g𝐺)𝑥)))
25 simp2 1131 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → 𝑎 ∈ (Base‘𝐺))
26 simp3 1132 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → 𝑏 ∈ (Base‘𝐺))
27 zsscn 11983 . . . . . 6 ℤ ⊆ ℂ
2827a1i 11 . . . . 5 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → ℤ ⊆ ℂ)
2913, 24, 25, 26, 28cyccom 18291 . . . 4 ((((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) ∧ 𝑎 ∈ (Base‘𝐺) ∧ 𝑏 ∈ (Base‘𝐺)) → (𝑎(+g𝐺)𝑏) = (𝑏(+g𝐺)𝑎))
304, 5, 6, 29isabld 18856 . . 3 (((𝐺 ∈ Grp ∧ 𝑥 ∈ (Base‘𝐺)) ∧ ∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Abel)
3130r19.29an 3293 . 2 ((𝐺 ∈ Grp ∧ ∃𝑥 ∈ (Base‘𝐺)∀𝑦 ∈ (Base‘𝐺)∃𝑛 ∈ ℤ 𝑦 = (𝑛(.g𝐺)𝑥)) → 𝐺 ∈ Abel)
323, 31sylbi 218 1 (𝐺 ∈ CycGrp → 𝐺 ∈ Abel)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 396 ∧ w3a 1081 = wceq 1530 ∈ wcel 2107 ∀wral 3143 ∃wrex 3144 ⊆ wss 3940 ‘cfv 6354 (class class class)co 7150 ℂcc 10529 + caddc 10534 ℤcz 11975 Basecbs 16478 +gcplusg 16560 Grpcgrp 18048 .gcmg 18169 Abelcabl 18843 CycGrpccyg 18932 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1904 ax-6 1963 ax-7 2008 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2153 ax-12 2169 ax-ext 2798 ax-sep 5200 ax-nul 5207 ax-pow 5263 ax-pr 5326 ax-un 7455 ax-cnex 10587 ax-resscn 10588 ax-1cn 10589 ax-icn 10590 ax-addcl 10591 ax-addrcl 10592 ax-mulcl 10593 ax-mulrcl 10594 ax-mulcom 10595 ax-addass 10596 ax-mulass 10597 ax-distr 10598 ax-i2m1 10599 ax-1ne0 10600 ax-1rid 10601 ax-rnegex 10602 ax-rrecex 10603 ax-cnre 10604 ax-pre-lttri 10605 ax-pre-lttrn 10606 ax-pre-ltadd 10607 ax-pre-mulgt0 10608 This theorem depends on definitions: df-bi 208 df-an 397 df-or 844 df-3or 1082 df-3an 1083 df-tru 1533 df-ex 1774 df-nf 1778 df-sb 2063 df-mo 2620 df-eu 2652 df-clab 2805 df-cleq 2819 df-clel 2898 df-nfc 2968 df-ne 3022 df-nel 3129 df-ral 3148 df-rex 3149 df-reu 3150 df-rmo 3151 df-rab 3152 df-v 3502 df-sbc 3777 df-csb 3888 df-dif 3943 df-un 3945 df-in 3947 df-ss 3956 df-pss 3958 df-nul 4296 df-if 4471 df-pw 4544 df-sn 4565 df-pr 4567 df-tp 4569 df-op 4571 df-uni 4838 df-iun 4919 df-br 5064 df-opab 5126 df-mpt 5144 df-tr 5170 df-id 5459 df-eprel 5464 df-po 5473 df-so 5474 df-fr 5513 df-we 5515 df-xp 5560 df-rel 5561 df-cnv 5562 df-co 5563 df-dm 5564 df-rn 5565 df-res 5566 df-ima 5567 df-pred 6147 df-ord 6193 df-on 6194 df-lim 6195 df-suc 6196 df-iota 6313 df-fun 6356 df-fn 6357 df-f 6358 df-f1 6359 df-fo 6360 df-f1o 6361 df-fv 6362 df-riota 7108 df-ov 7153 df-oprab 7154 df-mpo 7155 df-om 7574 df-1st 7685 df-2nd 7686 df-wrecs 7943 df-recs 8004 df-rdg 8042 df-er 8284 df-en 8504 df-dom 8505 df-sdom 8506 df-pnf 10671 df-mnf 10672 df-xr 10673 df-ltxr 10674 df-le 10675 df-sub 10866 df-neg 10867 df-nn 11633 df-n0 11892 df-z 11976 df-uz 12238 df-fz 12888 df-seq 13365 df-0g 16710 df-mgm 17847 df-sgrp 17896 df-mnd 17907 df-grp 18051 df-minusg 18052 df-mulg 18170 df-cmn 18844 df-abl 18845 df-cyg 18933 This theorem is referenced by: lt6abl 18951 frgpcyg 20655
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http://www.docstoc.com/docs/51990803/Expectation-Maximization-Algorithm | 1,436,171,313,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375098071.98/warc/CC-MAIN-20150627031818-00136-ip-10-179-60-89.ec2.internal.warc.gz | 472,321,349 | 47,236 | # Expectation Maximization Algorithm by khn19658
VIEWS: 13 PAGES: 61
• pg 1
``` Expectation Maximization Algorithm
Rong Jin
A Mixture Model Problem
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Apparently, the dataset consists of two modes
How can we automatically identify the two modes?
Gaussian Mixture Model (GMM)
Assume that the dataset is generated by two
mixed Gaussian distributions
Gaussian model 1: 1 1 , 1 ; p1
Gaussian model 2: 2 2 , 2 ; p2
If we know the memberships for each bin,
estimating the two Gaussian models is easy.
How to estimate the two Gaussian models
without knowing the memberships of bins?
EM Algorithm for GMM
Let memberships to be hidden variables
{x1 , x2 ,..., xn } x1 , m1 , x2 , m2 ,..., xn , mn
EM algorithm for Gaussian mixture model
Unknown memberships: x1 , m1 , x2 , m2 ,..., xn , mn
1 1 , 1 ; p1
Unknown Gaussian models:
2 2 , 2 ; p2
Learn these two sets of parameters iteratively
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Random assign the
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10
memberships to
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each bin
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2
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Random assign the
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10
memberships to
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6
each bin
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2 Estimate the means
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1
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and variance of
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0.6 each Gaussian
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model
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E-step
Fixed the two Gaussian models
Estimate the posterior for each data point
p( x, m 1) p( x,1 ) p( x | 1 , 1 ) p1
p(m 1| x)
p ( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
p( x, m 2) p ( x, 2 ) p( x | 2 , 2 ) p2
p (m 2 | x)
p ( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
EM Algorithm for GMM
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18
16
Re-estimate the
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12 memberships for
10
8 each bin
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4
2
0
1
0 5 10 15 20 25
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M-Step
Fixed the memberships
Weighted by posteriors
Re-estimate the two model Gaussian
n
l p(mi 1| xi ) log p( xi ,1 ) p(mi 2 | xi ) log p( xi , 2 )
ˆ ˆ
i 1
n
p(mi 1| xi ) log p1 log p( xi | 1 , 1 ) p(mi 2 | xi ) log p2 log p( xi | 2 , 2 )
ˆ ˆ
i 1
Weighted by posteriors
i 1 p(mi 1| xi ) , i 1 p(mi 1| xi ) xi , 2 i 1 p(mi 1| xi ) xi2 2
n n n
ˆ ˆ ˆ
p1 1 1 1
i 1 p ˆ (mi 1| xi ) i 1 p
ˆ (mi 1| xi )
n n
n
i 1 p (mi 2 | xi ) i 1 p(mi 2 | xi ) xi 2 i 1 p (mi 2 | xi ) xi2
n n n
ˆ ˆ ˆ
p2 , 2 , 2 22
i 1 p(mi 2 | xi ) i 1 p(mi 2 | xi )
n n
n ˆ ˆ
EM Algorithm for GMM
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18
16 Re-estimate the
memberships for
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12
10
8 each bin
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4
2
Re-estimate the
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1
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0 5 10 15 20 25 models
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At the 5-th Iteration
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Red Gaussian
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component slowly
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shifts toward the left
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end of the x axis
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At the10-th Iteration
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Red Gaussian
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8
component still
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slowly shifts
2
0
0.9
toward the left end
0 5 10 15 20 25
0.8
of the x axis
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0
0 5 10 15 20 25
At the 20-th Iteration
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Red Gaussian
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component make
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more noticeable shift
toward the left end of
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4
2
0
1
0 5 10 15 20 25
the x axis
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0
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At the 50-th Iteration
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Red Gaussian
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component is close
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to the desirable
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location
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At the 100-th Iteration
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16 The results are
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12 almost identical to
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8 the ones for the
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4 50-th iteration
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EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
EM as A Bound Optimization
EM algorithm in fact maximizes the log-likelihood function
of training data
Likelihood for a data point x
p( x) p( x,1 ) p( x, 2 ) p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
Log-likelihood of training data
l 1 , 2 i 1 log p( xi ) i 1 log p( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
Logarithm Bound Algorithm
0
l (1 ,2 )
10 , 2
0
Logarithm Bound Algorithm
Point
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
1 , 2
0 0
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (11 , 2 ) Q(1 , 2 )
1
Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2 1 , 2
2 2
• Repeat the procedure
Logarithm Bound Algorithm
Optimal
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2 1 , 2
2 2
,... • Repeat the procedure
• Converge to the local optimal
EM as A Bound Optimization
Parameter for previous iteration: 1' , 2
'
Parameter for current iteration: 1 , 2
Compute Q(1 , 2 )
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
Concave property 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p ( xi | 1 , 1 ) p1
p ( xi | of
' logarithm function
' ' ' ' ' ' ' '
i 1 log
n
log( p (1 p) ) p log (1 ' p) log '
p( x | , ) p '
p ( xi | 2 , 2 ) p2
0 p , 0 ' ' i ' 2 2 2 ' ' '
p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p ( xi | 2 , 2 ) p2
1, ' ' '
p ( xi | 1 , 1 ) p1
' ' '
p( xi | 1 , 1 ) p1
log
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p ( x | ' , ' ) p ' p ( x | ' , ' ) p ' log p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
n p ( xi | 1 , 1 ) p1 p( xi | 2 , 2 ) p2
i 1
p (1' | xi ) log p ( 2
'
| xi ) log '
p ( xi | 1' , 1' ) p1
'
p( xi | 2 , 2 ) p2
' '
Q(1 , 2 ) l (1 , 2 ) l (1' , 2 )
'
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
n
i 1
log '
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
' ' ' ' '
p( xi | 1 , 1 ) p1
' ' '
p ( xi | 1 , 1 ) p1
p( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2 p( xi | 1 , 1 ) p1
' ' ' ' ' ' ' ' '
i 1 log
n
p ( xi | 2 , 2 ) p2
' ' '
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p' p( x | ' , ' ) p ' p( x | ' , ' ) p '
i 1 1 1 i 2 2 2 i 2 2 2
' Definition of posterior
p( xi | 1 , 1 ) p1
' '
p( xi | 1 , 1 ) p1
p( xi | '1 , 1 ) p1 ' log
' ' '
n p ( xi | 1 , 1 ) p1 'p ( xi | 2 , 2 ) p2' i m 1 p
' ' ' ' ' ' '
p( x | p(1 , 1 )1| 1xi ;1' , 2 )
'
i 1 p( xi | 1 , '1 ) p1' p( xi | 2 , 2 ) p2
' ' ' '
p( xi | 2 , 2 ) p2
'
p ( xi | 2 , 2 ) p2
p( x | ' , ' ) p ' p( x | ' , ' ) p ' log
i 1 1 1 i 2 2 2 p( xi | 2 , 2 ) p2
' ' '
n p ( xi | 1 , 1 ) p1 p ( xi | 2 , 2 ) p2
i 1
p (m1 1| xi ;1' , 2 ) log
'
p(m1 2 | xi ;1' , 2 ) log
'
'
p ( xi | 1 , 1 ) p1
' ' '
p ( xi | 2 , 2 ) p2
' '
Log-Likelihood of EM Alg.
-375
-380
-385
Loglikelhood
-390
-395
-400
-405
-410
0 10 20 30 40 50 60 70 80 90 100
Iteration
Maximize GMM Model
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
What is the global optimal solution to GMM?
n
x
i 1 i
1 x1 , 1 0, 1 , 2 1, p1 p2 0.5
n
Maximizing the objective function of GMM is ill-
posed problem
Maximize GMM Model
l i 1 log p ( xi ) i 1 log p ( x | 1 , 1 ) p1 p( x | 2 , 2 ) p2
n n
1 x 2 1 x 2
p( x | 1 , 1 ) exp , p( x | 1 , 1 ) exp
1 2
2 1
2 2 12 2 2
2 2 2
2
What is the global optimal solution to GMM?
n
x
i 1 i
1 x1 , 1 0, 1 , 2 1, p1 p2 0.5
n
Maximizing the objective function of GMM is ill-
posed problem
Identify Hidden Variables
For certain learning problems, identifying hidden variables is
Consider a simple translation model
For a pair of English and Chinese sentences:
e : (e1 , e2 ,..., es ) c : (c1, c2 ,..., cl )
A simple translation model is
Pr(e | c ) j 1 Pr(e j | c ) j 1
s s
t
k 1
Pr(e j | ck )
The log-likelihood of training corpus e1 , c1 ,..., en , cn
l i 1 log Pr(ei | ci ) i 1 j i1 log
n n e
ci
k 1
Pr(ei , j | ci ,k )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
Identify Hidden Variables
Consider a simple case e : (e1 e2 )
Pr(e | c ) j 1
2
2
k 1
Pr(e j | ck )
Pr(e1 | c1 ) Pr(e2 | c1 ) Pr(e1 | c2 ) Pr(e2 | c2 )
Pr(e1 | c1 ) Pr(e2 | c2 ) Pr(e1 | c2 ) Pr(e2 | c1 ) c : (c1 c2 )
Alignment variable a(i)
a:
map a position in English sentence to a position in Chinese sentence
Rewrite Pr(e | c ) a Pr(e1 | ca(1) ) Pr(e2 | ca(2) )
EM Algorithm for A Translation Model
Introduce an alignment variable for each translation pair
e1 , c1 , a1 , e2 , c2 , a2 ,..., en , cn ,, an
EM algorithm for the translation model
E-step: compute the posterior for each alignment variable Pr(a j | e j , c j )
M-step: estimate the translation probability Pr(e|c)
|e j | |e j |
Pr(a, e j , c j ) Pr(e j ,k | c j ,a (k ) ) Pr(e j ,k | c j ,a (k ) )
k 1 k 1
Pr(a | e j , c j )
a ' Pr(a ', e j , c j ) |e j | |e j |
a ' Pr(e j ,k | c j ,a '(k ) ) t 1 Pr(e j ,s | c j ,t )
ci
k 1 k 1
EM Algorithm for A Translation Model
Introduce an alignment variable for each translation pair
e1 , c1 , a1 , e2 , c2 , a2 ,..., en , cn ,, an
EM algorithm for the translation model
E-step: compute the posterior for each alignment variable Pr(a j | e j , c j )
M-step: estimate the translation probability Pr(e|c)
|e j | |e j |
Pr(a, e j , c j ) Pr(e j ,k | c j ,a (k ) ) Pr(e j ,k | c j ,a (k ) )
k 1 k 1
Pr(a | e j , c j )
a ' Pr(a ', e j , c j ) |e j | |e j |
a ' Pr(e j ,k | c j ,a '(k ) ) t 1 Pr(e j ,s | c j ,t )
ci
k 1 k 1
We are luck here. In general, this step can be extremely
difficult and usually requires approximate approaches
Compute Pr(e|c)
First compute Pr(e | c; ei , ci )
Pr(e | c; ei , ci ) (e ei ) (c ci ) a Pr(a | ei , ci ) (a(e) c)
(e e ) (c c )
a Pr(a, ei , ci ) (a(e) c)
i i
Pr(ei , ci )
|e j |
t 1 Pr(e j , s | c j ,t )
ci
Pr(e | c)
k 1^ e j ,k e
(e ei ) (c ci ) |e j |
t 1 Pr(e j ,s | c j ,t )
ci
k 1
Pr(e | c)
(e ei ) (c ci )
t 1 Pr(e | c j ,t )
ci
Compute Pr(e|c)
First compute Pr(e | c; ei , ci )
Pr(e | c; ei , ci ) (e ei ) (c ci ) a Pr(a | ei , ci ) (a(e) c)
(e e ) (c c )
a Pr(a, ei , ci ) (a(e) c)
i i
Pr(ei , ci )
|e j |
t 1 Pr(e j , s | c j ,t )
ci
Pr(e | c)
k 1^ e j ,k e
(e ei ) (c ci ) |e j |
t 1 Pr(e j ,s | c j ,t )
ci
k 1
Pr(e | c)
(e ei ) (c ci )
t 1 Pr(e | c j ,t )
ci
Pr(e | c) i 1 Pr(e | c; ei , ci )
n
Bound Optimization for A Translation Model
θ Pr(e | c) for the current iteration
θ ' Pr'(e | c) for the previous iteration
l (θ) i 1 log Pr(ei | ci ; θ) i 1 j i1 log
n n e
ci
k 1
Pr(ei , j | ci ,k )
l (θ ') i 1 log Pr(ei | ci ; θ ') i 1 j i1
n n e
log
ci
k 1
Pr'(ei , j | ci ,k )
ci Pr(e | c )
Q(θ, θ ') l (θ) l (θ ') i 1 j i1 log k 1
n e i, j i ,k
ci Pr'(e | c )
l 1 i, j i ,l
Bound Optimization for A Translation Model
ci Pr(e | c )
Q(θ, θ ') l (θ) l (θ ') i 1 j i1 log k 1
n e i, j i ,k
ci Pr'(e | c )
l 1 i, j i ,l
ci Pr'(ei , j | ci ,k ) Pr(ei , j | ci ,k )
log
n ei
i 1 j 1
k 1 ci Pr'(e | ci ,l ) Pr'(ei , j | ci ,k )
l 1 i, j
Pr'(ei , j | ci ,k ) Pr(ei , j | ci ,k )
n ei ci
log
i 1 j 1 k 1 Pr'(ei , j | ci ,k )
ci
l 1
Pr'(ei , j | ci ,l )
Pr'(e | c)
Pr(e | c) i 1 (e ei ) (c ci )
n
ci
t 1
Pr'(e | c j ,t )
Iterative Scaling
Maximum entropy model
exp( x wy ) exp( xi wyi )
, l ( Dtrain )
N
p ( y | x ; ) log
y exp( x wy ) i 1
y exp( xi wy )
Iterative scaling
All features xi , j 0
Sum of features are constant j 1 xi, j g
d
Iterative Scaling
Compute the empirical mean for each feature of every class,
i.e., ey, j N xi, j ( y, yi ) N for every j and every class y
i 1
Start w1 ,w2 …, wc = 0
Repeat
Compute p(y|x) for each training data point (xi, yi) using w from the
previous iteration
Compute the mean of each feature of every class using the estimated
probabilities, i.e., my, j N xi, j p( y | xi ) N for every j and every y
i 1
Compute w j , y
1
g
log e j , y log m j , y for every j and every y
Update w as w j , y w j , y w j , y
Iterative Scaling
w1 , w2 ,..., wc : parameters for the current iteration
' w1 , w2 ,..., wc : parameters for the last iteration
' ' '
exp( x wy )
p ( y | x ; )
y exp( x wy )
exp( xi wyi )
l ( ) p ( y | x ; )
N N
log log
i 1 i 1
y exp( xi wy )
exp( xi w'yi )
l ( ') p( y | x; ')
N N
log log
i 1 i 1
y exp( xi w'y )
exp( x w ) exp( xi w'yi )
l ( ) l ( ')
N i yi
log '
y exp( xi wy ) y exp( xi wy )
i 1
Iterative Scaling
exp( x w ) exp( xi w'yi )
l ( ) l ( ')
N i yi
log
y exp( xi wy ) y
i 1
exp( xi w'y )
i 1 xi wyi w'yi log
N
y
exp( xi w'y ) log exp( x w )
y i y
Can we use the concave property of logarithm function?
No, we can’t because we need a lower bound
Iterative Scaling
log x x 1 log exp(x w ) exp(x w ) 1
y i y y i y
l ( ) l ( ')
i 1 xi wyi w'yi log
N
y exp( x w )
exp( xi w'y ) log y i y
x w log exp( x w ) exp( x w ) 1
N
i 1 i yi w'yi y i
'
y y i y
• Weights w y still couple with each other
• Still need further decomposition
Iterative Scaling
exp q p p exp q for i, p 0, p 1
i i i i i i i i i
exp( xi wy ) exp d
x w
j 1 i , j y , j
d
exp j 1 d
xi , j
k 1 xi,k
wy , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
d d xi , j
exp wy , j k 1 xi , k j 1
g
exp gwy , j
k 1
l ( ) l ( ')
N
i 1 xi wyi w'yi log y
exp( xi w'y ) y exp( xi wy ) 1
N
j xi, j log
xi , j
i 1
wyi , j w'yi , j exp( xi w'y ) exp( gwy , j ) 1
y y j g
Iterative Scaling
N
xi , j
Q( , ') i 1 j xi , j wyi , j w'yi , j log y exp( xi w'y ) y j exp( gwy , j ) 1
g
N
log
i 1 y
exp( xi w'y ) 1 N
i 1
y j xi, j wy, j w'y, j
y, yi
xi , j
g
exp( gwy , j )
Q( , ')
wy , j
i 1 y j xi , j y, yi xi , j exp( gwy , j ) 0
N
y j xi, j y, yi
N
i 1
wy , j log
i 1 y j xi, j
N
Wait a minute, this can not be right! What happens?
Logarithm Bound Algorithm
• Come up with a lower bounded
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 )
l (1 ,2 )
Q(1 , 2 ) is a concave function
l (1 , 2 ) l (10 , 20 ) Q(1 , 2 ) Touch point: Q(1 10 , 2 20 ) 0
• Search the optimal solution
that maximizes Q(1 , 2 )
1 , 2
0 0
,
1
1
1
2
Iterative Scaling
Q( , ')
N
i 1 log y
exp( xi w'y ) 1 N
i 1
y j
'
xi , j wy , j wy , j y, yi
xi , j
g
exp( gwy , j )
Q( ', ')
N
i 1
log y
1 y j xi , j wy , j wy , j y, yi
exp( xi w'y )
N
i 1
' '
xi , j
g
'
exp( gwy , j )
i 1
N
log y exp( xi wy ) 1 i 1 y j
'
N xi , j
g
exp( gw'y , j )
0
Where does it go wrong?
Iterative Scaling
log x x 1 log exp(x w ) exp(x w ) 1
y i y y i y
l ( ) l ( ')
N
i 1 xi wyi w'yi log y exp( x w )
exp( xi w'y ) log y i y
i 1
N
x w
i yi w'yi log y
exp( x w ) exp( x w ) 1
i
'
y y i y
Not zero when = ’
exp( xi wy ) exp( xi wy )
log x x 1 log y 1
exp( xi wy )
' y
y exp( xi wy )
'
y
Iterative Scaling
exp( xi wy ) exp( xi wy )
log x x 1 log y 1
exp( xi wy )
' y
y exp( xi wy )
'
y
l ( ) l ( ')
exp( xi wy )
N
i 1 i
x
wyi w'yi y
y
1
exp( xi w'y )
y Definition y ' conditional
wy w of
N exp( xi w'y ) exp( xi y )
i 1 xi yi y
exponential model 1
y exp( xi w'y )
N
i 1 x
i yi
y p( y | xi ; ') exp( xi y ) 1
Iterative Scaling
exp( xi y ) exp d
x
j 1 i , j y , j
exp j 1 d
d xi , j
k 1 xi,k
y , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
d
d xi , j
exp y , j k 1 xi , k j 1
g
exp g y , j
k 1
l ( ) l ( ')
N
i 1 x
i yi y p( y | xi ; ') exp( xi y ) 1
N xi , j
i 1
xi , j yi , j y p( y | xi ; ') j exp( g y , j ) 1
j
g
xi , j
i 1 j y xi , j y , j ( y, yi ) p( y | xi ; ')
N
exp( g y , j ) 1
g
Iterative Scaling
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( g y , j ) 1
g
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') xi , j exp( g y , j ) 0
N
N
1 x ( y, yi )
i 1 i , j y , j
y , j wy , j w'y , j log
N
g
i 1
p( y | xi ; ') xi , j
Iterative Scaling
How about d1 xi, j gi constant ?
j
exp( xi y ) exp d
x
j 1 i , j y , j
exp j 1 d
d xi , j
k 1 xi,k
y , j k 1 xi ,k
d
j 1 d
d xi , j
xi,k
exp y , j k 1 xi , k j 1
d
d xi , j
gi
exp gi y , j
k 1
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( gi y , j ) 1
gi
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') xi, j exp( gi y , j ) 0
N
Is this solution unique?
Iterative Scaling
exp( xi y ) exp j 1 xi, j y, j
d
d 1
d
d xi , j
exp j 1 d y , j xi , k j 1
d
exp d y , j xi , k
1
Q( , ') i 1 j y xi , j y , j ( y, yi ) p( y | xi ; ') exp( xi , j y , j d ) 1
N
d
Q( , ')
y , j
i 1 xi , j y , j ( y, yi ) p( y | xi ; ') exp(d y , j xi , j ) 0
N
Faster Iterative Scaling
The lower bound may not be tight given all the
coupling between weights is removed
xi , j
Q( , ') jy
N
i 1
xi , j y , j ( y, yi ) p( y | xi ; ') exp( gi y , j ) 1
gi
i 1 j y q( y , j )
N
Univariate functions!
A tighter bound can be derived by not fully
decoupling the correlation between weights
xi, j g
Q( , ') y, yi xi , j y , j log p( y | x)e y , j i
j i, y i gi
y
Faster Iterative Scaling
Log-likelihood
You may feel great after the struggle of the derivation.
However, is iterative scaling a true great idea?
Given there have been so many studies in optimization, we
should try out existing methods.
Comparing Improved Iterative Scaling to
Newton’s Method
Dataset Iterations Time (s)
Dataset Instances Features
Rule 823 42.48
Rule 29,602 246
81 1.13
Lex 42,509 135,182
Try out the standard numerical 241
Lex 102.18
Summary 24,044 198,467
methods before you get excited 176 20.02
Shallow 8,625,782 264,142 | 18,549 | 37,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | longest | en | 0.590513 |
https://www.physicsforums.com/threads/spaningset-theorem.79024/ | 1,519,545,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00612.warc.gz | 925,563,553 | 15,629 | # Spaningset theorem
1. Jun 14, 2005
### Mathman23
Spaning set theorem (new Question)
Hi
I have a couple of questions regarding an assignment which deals with the spanning set theorem.
Hope You can help
The matrix $$A = [a1 \ a2 \ a3 \ a4 \ a5] = \left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 & -1 \\ 0 & 1 & 1 & 2 & 0 \\ 1 & 0 & 0 & 1 & 2 \\ -1 & 2 & 2 & 3 & 1 \end{array} \right]$$
a) First I determin the rank of A rank(A) = 3
The dimension of Null A: dim (Null A) = 2
b) Determin a basis for A's column space.
I do this using the spanning set theorem.
Since a3 = a2, a4 = 2a2 + a1 then
B(ColA) = sp{a1,a2, a5} According to the theorem.
c) Next the basis for the B(Null A).
First I row reduce A and then up and write up the set of solutions for A which results in the set B(Null A) = span{(0,-1,1,0) , (1,-2,0,1)}
Is that the correct approach ??
d) There is a vector x = a1 + a2 + a3 + a4 + a5. I'm tasked with showing that this vector belongs to Col A. Finally I'm tasked with finding the vector x with respect to the basis B.
I need some assistance is solving c) and d) therefore I hope there is somebody out there who can guide me :-)
Sincerely and Best Regards,
Fred
/Fred
Last edited: Jun 14, 2005 | 412 | 1,223 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-09 | longest | en | 0.864837 |
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