text: 5/p. Start your preparation today with our free TEAS practice test questions. Write. Choose from 500 different sets of teas test math ati flashcards on Quizlet. Start your test prep right now with our free TEAS practice test! To solve an inequality, simply treat the inequality symbol as if it’s an equals sign: The correct answer is (D). 2005 . Choose from 500 different sets of teas test math flashcards on Quizlet. STUDY. 0/0 Steps . Only $2.99/month. Let us be your guide with a prep course that fits your schedule. Science Take the Quiz. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Write the answer from step 2 above the denomintor. Teas Math. 2005. The distance from Chicago, IL to St. Louis, MO is 300 miles. To add mixed number fractions, first convert them to improper fractions, then find a common denominator and add the numerators, placing them over the common denominator. Log in Sign up. PRACTICE QUESTIONS. If you leave this page, your progress will be lost. SETS. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Back to Practice Test Pack Free ATI TEAS 6 Science Practice Test. Write. PREP PACK -$79. Free ATI TEAS Practice Test 2021 (Reading, Math, Science, English). 4.2 (15) Preview. $−0.2,\; −\frac{1}{4},\; \frac{3}{8},\; π$, $−\frac{1}{4},\; −0.2,\; \frac{3}{8},\; π$, $−0.2,\; −\frac{1}{4},\; π,\; \frac{3}{8}$, $−\frac{1}{4},\; −0.2,\; π,\; \frac{3}{8}$. Search. Click on the book image below for more information or go to the download link below the book cover. Flashcards. Remember your order of operations: The correct answer is (A). FULL PRACTICE TESTS. Since she only has \$300 to spend, \$40$n$ must be less than or equal to \$300: The correct answer is (C). You have not finished your quiz. If you need more help, Test-Guide.com recommends BoostPrep.com. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Our free TEAS Math practice test includes 30 problems that you should be able to complete within 45 minutes. The BoostPrep TEAS review course includes: Math – Review of all TEAS math topics and 200 practice questions. This website is not endorsed, affiliated, or approved by the Assessment Technologies Institute®. The first step is to convert the mixed fraction 7$\frac{1}{4}$into an improper fraction. All of our practice questions include answers and detailed explanations. Share. In the Roman numeral system, M = 1000, D = 500, C = 100, L = 50, X = 10, V = 5, and I = 1. A millimeter is one-thousandth of a meter, which means there are 1,000 millimeters in 1 meter. Log in Sign up. The easiest way to answer this question is to round the numbers. ATI TEAS 6 MATH PRACTICE TEST; ATI TEAS 6 SCIENCE PRACTICE TEST; ATI TEAS 6 ENGLISH PRACTICE TEST; Things to Know about Your ATI TEAS 6 Exam What is the TEAS Test? PLAY. alligood1. Many universities and other educational institutions across the country require TEAS scores as a prerequisite for their nursing program. Below is an ATI math practice test. All of these sets are available to students in my comprehensive TEAS prep program. Two of the walls are 10 x 10 and two of the walls are 10 x 15: The correct answer is (B). Math - TEAS, TEAS- Math. News Home actual teas 6 test questions quizlet. Start studying Math - TEAS, TEAS- Math. First you must calculate how many hours the nurse works each week. Estimate volume of rectangular box - height is 22.1cm, length is 98cm, and width is 54 cm. SETS. Reading Take the Quiz. Math - TEAS, TEAS- Math. scottsa125. Once you are finished, click the button below. The “FOIL method” is the easiest way to remember how to multiply two-termed expressions. MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\$','\$']], processEscapes: true}}); Which of the following is the distance in meters (m) between these two cities? You can take the free TEAS math practice test below. A computer is worth$1,500 at the time of purchase. Our questions are arranged in line with the four content areas: Reading, Mathematics, Science, English & Language Arts. In order to convert a decimal to a percent you need to move the decimal point two places to the right and add the percent symbol. Order of Operations (PEMDAS) Whole Numbers. Only $2.99/month. PrenursingSmarterPro. Learn. Approach the problem strategically. Preview. No products in the cart. The correct answer is (D). STUDY. Three 8-hour shifts plus one 12-hour shift equals 36 hours per week. Log in Sign up. The ATI TEAS VI is the most commonly used nursing school entrance exam. Free ATI TEAS 6 Practice Test; Free HESI A2 Practice Test ; Free Wonderlic SLE Practice Test; Practice Tests; Online Courses; Cart. ATI TEAS 6 MATH Practice Test – PDF Download Version. The change would equal: The correct answer is (C). First you must convert the fraction to a decimal by moving the decimal point 2 places to the left, which gives you 1.815. The correct answer is (C). 5,277 5,283. Match. Home Browse. If loading fails, click here to try again. Practice Test 1 of 0. Free ATI TEAS 6 Practice Test Problems . Try a free ATI TEAS VI Math practice test with detailed answer explanations to see how ready you are for your nursing school entrance exam. Let’s say the tip is \$10 and the total was \$87. Doing well on the TEAS Nursing Exam is critical for you getting into a great nursing school. First divide both sides by 3: The correct answer is (B). To help pass the actual TEAS VI check out the 1,250 question TEAS VI Practice Test. The concepts addressed include arithmetic, algebra, measurement, and data. 4.5 (51) Preview. Start studying Teas Math. TEAS V practice tests and TEAS V Math questions do not. Preview. Teas math practice tests. Upgrade to remove ads. Please visit using a browser with javascript enabled. You will have 54 minutes to complete 36 questions. Our Recommended TEAS Study Course. Learn. (note: 1 mile = 1.6km). Create. Our free TEAS Math practice test includes 30 problems that you should be able to complete within 45 minutes. If you are serious about getting a great score on your TEAS test, try out our recommended TEAS Prep Course. Get Quizlet's official TEAS - 1 term, 1 practice question, 1 full practice test. Multiply the whole number by the denominator. The smallest number is the greatest negative value: The correct answer is (A). Upgrade to remove ads. The easiest way to make an estimate is to round each number so there is only one nonzero digit. 5,344 − 57 . TEAS. A pie chart represents percentages of a total, and would thus be the best graphic representation of a budget breakdown. A Free Professional quiz that is designed to help you with your TEAS test prep. 2. Learn. The ATI TEAS exam is made-up of a math section. Gravity. Many students dread taking the math section of the ATI exam because they are not good at math. 0% Complete. Please wait while the activity loads. PrenursingSmarterPro. Write. Steps … Flashcards. The attendance at a nursing convention was 400 members, 75% of whom voted for an increase in membership dues. ATI TEAS 6 Practice Math Tests. You will have 40 minutes to answer the 32 sample questions. The correct answer is (A). It covers such subjects as science, math, English and reading comprehension. English Take the Quiz. *Test of Essential Academic Skills (TEAS®) and ATI® Nursing Education are registered trademarks of the Assessment Technologies Institute®, LLC. TEAS® Math Practice Test. Log in Sign up. ; English and Language Usage – Complete grammar review with practice questions. Then click 'Next Question' to answer the next question. February 18, 2021 ATI TEAS 6 study guide Math. Any items you have not completed will be marked incorrect. The Science section no longer includes the Earth & Physical Science topic. 5,288 Remember that calculators are not permitted on the TEAS math test, so make sure you are able to subtract large numbers without using a calculator. FULL PRACTICE TESTS. Start your test review right now with our TEAS Math practice test. (subtraction by regrouping) PLAY AGAIN ! Multiply the. This can be solved with a calculator or subtraction by regrouping. Sign in to view your test packs Sign up. If a smaller item precedes a larger item, as in CM, it is treated as the difference of the two, so CM = 1000 − 100 = 900. Those test areas include reading, math, science, … The above graph is known as a scatterplot, which is a plot of points on$x$and$y$axes. Prepare for the ATI TEAS 6 Math section with TEAS practice tests that permit a calculator and are designed for the sixth edition of ATI Testing's nursing exam. Then add and subtract in order from left to right. Which of the following lists the values in order from least to greatest? Created by. TERMS . The correct answer is (B). Math Take the Quiz. PrenursingSmarterPro. The correct answer is (D). TERMS. After 5 years, the computer is outdated and has no monetary value. 14 airplanes minus 12 airplanes equals 2. Add the answer from step 1 to the numerator. Flashcards. The Math section of the ATI TEAS (TEAS 6) test comprises about 22% of your total testing efforts. Then multiply and divide in order from left to right. Browse. \$10/$87 = approximately 11%. Topics covered come from areas of math that you should have studied during grades 7 through 12 in school. 122 Terms. Test. Because Timmy is interrupted during each hour, he can only make four paper airplanes per hour, multiplied by 3 hours, for a total of 12 airplanes. In this case, each point would have an$x$value equal to the temperature in degrees Celsius, and a$y$value equal to the amount of money in dollars. All of our practice questions include answers and detailed explanations. ati teas practice test quizlet The test covers numbers and algebra as well as measurement and data. Plug in °C = 30 into the given formula and solve for °F: The correct answer is (B). The sample indicates that 5 out of every 300 randomly selected toys will be faulty. Gravity. The TEAS Math test includes challenging questions that focus on numbers, operations, data interpretation, algebra, and measurement. The format of this practice test is the same as that of an actual one, but the details in questions are different. Analyze the provided information. Also, this page requires javascript. Complete operations within parentheses first. The correct answer is (B). We have 150 free TEAS practice questions that match the style and format of the actual test. Our ATI TEAS questions can help you test your knowledge and improve your mental arithmetic. There are 1,000 liters in 1 kiloliter: The correct answer is (B). The correct answer is (D). Which of the following is the number of members who voted for the increase? Create. When subtracting polynomials the first step is to distribute the negative sign through the parentheses. Three comics at \$2.65 would equal a total of \$7.95. Succeed on the ATI TEAS exam with Kaplan. First, we must calculate the area of the walls Amanda wants to paint. TEAS. No products in the cart. The correct answer is (B). The TEAS VI is used as an admission exam for nursing schools. D. Each 3 inches of height adds 2 inches to the crutch length. PLAY. According to the order of operations, do what is inside the parenthesis first: The correct answer is (C). If 5 gallons of paint will cover 2,000 square feet, how many gallons of paint are needed to paint 30,000 square feet. Preview. 89 Terms. Cart. Our questions cover all four content areas: Reading, Math, Science, and English. Free ATI TEAS ® Practice Test Questions. Rounding up gives us 63%. Terms in this set (205) If Lynn can type a page in p minutes, what piece of the page can she do in 5 minutes? Which of the following expressions appropriately compares 4.67 and 4 1/3? Assuming the computer's value decreases at a constant rate, which of the following expressions represents the comps value at any time (t) in years? TEAS Math Practice Test. The correct answer is (B). I’ve created several TEAS Quizlet sets that you can use as you prepare for test day. We have 150 free TEAS practice questions that match the style and format of the actual test. Our ATI TEAS 6 Practice Test book covers all of the skills on the real ATI TEAS 6 math test. Learn teas test math ati with free interactive flashcards. Test. Absolute value equations can be rewritten in two ways to solve: The correct answer is (A). Chanelle_Clarke2. The TEAS math test consists of 36 questions to be completed in 54 minutes. This changes the sign on each term inside the parentheses: The correct answer is (C). Then multiply this by 18: The correct answer is (C). PrenursingSmarterPro. ; Reading – Reading tips and strategies with practice reading quizzes. Then you can easily add up the numbers: The correct answer is (A). If this activity does not load, try refreshing your browser. Search. Free Resources. PRACTICE QUESTIONS. Download printable PDF sample question answers. What subjects are on the TEAS test? Our TEAS Test Prep will help students applying for nursing school in the USA and prepare the Test of Essential Academic Skills (TEAS Test) online. (A) is the closest answer to our approximation. Take this free practice test to see what type of questions are on the Test of Essential Academic Skills VI (TEAS VI). STUDY. Take this practice test to become familiar with the test format and the math skills required to find the correct answers. Get Quizlet's official TEAS - 1 term, 1 practice question, 1 full practice test. So there must be 5,000 millimeters in 5 meters. Start studying TEAS TEST: MATH. Consequently, a proportion can be set up that relates the unknown number of faulty toys in the total number of toys to the ratio of faulty toys to the sample: The correct answer is (B). Learn teas test math with free interactive flashcards. Match. 5 ÷ 8 = 0.625, which equals 62.5%. The Test of Essential Academic Skills (TEAS) is an aptitude test that evaluates students’ understanding of key concepts and practical skills related to the nursing and allied health field. The TEAS Math test includes challenging questions that focus on numbers, operations, data interpretation, algebra, and measurement. TEAS test math tips. The ratio of 20 to 48 can be written as a fraction and then simplified: The correct answer is (C). teas test math. The ones, tens, hundreds and thousands must be treated as separate items. Welcome to TEAS Practice Test, a free website that is designed to help you with your test prep. The correct answer is (B). John makes seven paper airplanes during the two hours he works without chores, for a total of 14 airplanes. The correct answer is (B). Call us today: (412) 537-1229. This practice test contains sample questions from the TEAS Math section of the test. To do this: The correct answer is (C). The ATI TEAS VI consists of 4 separate test areas. Ready to prep? The correct answer is (B). You may use a calculator. UPDATE 2021: These Quizlet study sets have expanded since this post was first published! Apply the following approach to your TEAS math practice test problems as you study. Log in Sign up. If there are 48 total, and 28 study French, then 20 study Spanish. teas test math ati. When you are ready, click the “Start Test” button below. If Kyra buys$n$pairs of shoes at a cost of \$40 each, she pays \$40$n$. Here’s an update on what’s available: #1: PrenursingSmarter on Quizlet includes select sets. However, practice makes perfect for successfully passing the ATI TEAS math section. This is a practice exam for the math part of a Teas test. Compare the problem with your current base of knowledge and determine if it fits into a pattern of TEAS math questions that you’ve seen before. Post navigation. 553. PrenursingSmarterPro. The prefix “milli—” means one-thousandth. Get more TEAS Math practice questions at https://www.mometrix.com/academy/teas-math-practice-test/ - your source for TEAS study materials. Spell. PREP PACK -$79. Get Quizlet's official TEAS - 1 term, 1 practice question, 1 full practice test. Spell. 2. Given that it took 5 hours to travel 360km, we can set up a ratio equation to figure out how far the car will travel in 9 hours: The correct answer is (A). Reduce the final fraction if necessary. 5,287. 553. TEAS Test Study Guide Final Free Practice Test Instructions Choose your answer to the question and click 'Continue' to see how you did. The cost for each adult is \$11.00 and there are 7 adults: The correct answer is (D). 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# How do you solve this inhomogenous differential equation by Frobenius method/Series solution?

So I'm solving some practice questions and I have a differential equation $$y''-3y'+2y=\sin(x)$$ about $$x=0$$, and I'm told to solve it using the Frobenius method. I don't think that's possible because $$x=0$$ would be an ordinary point rather than a singular point, so I'm solving it by using a series solution method. However, I think I'm forgetting how you use it for a nonhomogenous equation? What do I do with the $$sin(x)$$ that I have at the RHS?

My current progress is below:

$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$

$$y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}$$

$$y''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$$

Then, putting it into the equation and then simplifying, I get:

$$\sum_{n=0}^{\infty}\bigg[(n+2)(n+1)a_{n+2} - 3(n+1)a_{n+1} + 2a_n \bigg]x^n = \sin(x)$$

I have two questions about this. Should I solve it homogenously first and get an expression for the $$a_n$$'s and stuff, or should I keep the $$\sin(x)$$ there? And if I do, what do I do with it? Like how do I get a proper expression that'll give me a solution to the DE? I've been trying to find examples online but most of them only have the RHS equal to constants or powers of x, and no other functions, which is why I'm kind of confused. Any help at all would make me super super grateful!!

• Looks like you need to use also $\sin x = x - \frac{x^3}{3!} + \cdots +$ Sep 29 '21 at 18:20
• Oh my,,, how did I not,,, think of it this way /)_- Okay so when I do this, I can just compare the coefficients? Would that work? (I'm also going to try this right now and see where this goes jbtw, but yeah, if you could answer this, it'd be helpful!) Sep 29 '21 at 18:23

Here is another method you can compare with: \begin{align*} y'' - 3y' + 2y = \sin(x) & \Longleftrightarrow (y' - 2y)' - (y' - 2y) = \sin(x)\\\\ & \Longleftrightarrow u' - u = \sin(x)\\\\ & \Longleftrightarrow [\exp(-x)u]' = \exp(-x)\sin(x)\\\\ & \Longleftrightarrow \exp(-x)u = -\exp(-x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\\\\ & \Longleftrightarrow y' - 2y = -\frac{\cos(x) + \sin(x)}{2} + k\exp(x)\\\\ & \Longleftrightarrow [\exp(-2x)y]' = -\exp(-2x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\exp(-x)\\\\ \end{align*}
The power series of the sine function is given by: \begin{align*} \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots \end{align*}
• Hi, thank you for posting this! So I would be able to do it this way, however this is a part of a series solution practice set and I genuinely can't figure out how to do it that way. It seems like a simple enough problem for other methods, but for this one, I somehow can't figure out how to make it work? My only problem is that I don't know what to do with the $sin(x)$ that I have on the right side, I'd be able to solve it if it was a power of x or a constant because I can compare it to the examples I've already seen, but with this, I'm really not sure how to do it with this method :/ Sep 29 '21 at 18:20

format: plain


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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 0 0 0 0.000%

## 문제

Alice and Bob haven't met for some time. Bob isn't very happy about this, so he urges Alice to finally make time for a meeting. Let's listen to an extract from a phone call:

Alice: ... maybe we should meet on neutral territory.

Bob: I've already heard this from you --- two years ago.

Alice: I know ; I just haven't found yet a suitable place that is roughly at the same distance from both yours and mine.

Bob: Well, the geometric place of the points that are equidistant from two given points on the surface of a sphere (and the earth is a sphere rather than a disc) is a great circle (namely the one which intersects the great circle through the given points orthogonally at the center of them). If you insist only on approximately equal distances though, we get a zone of some kilometers width and about 40000 km length. Not everything in this zone is water. Thus I think it is a feasible task to find a fitting place.

Alice: Now, if I tell you to pick any, we'll certainly land up in Honolulu.

Bob: Which is not a too bad idea. So, may I pick any ?

Alice: As long as I don't have to accept --- but I'm open to suggestions.

Bob: Honolulu ?

Alice: Is it situated on aforementioned geometric place at all ??!

Bob: Not quite ...

Nice. Now let's stop the preliminaries and come to the facts: Given two locations on the earth's surface you can find the geometric place of all equidistant points on the surface. For another given location calculate its distance on the surface to this geometric place. Assume that the earth is a sphere with a radius of 6378 km.

## 입력

The input file consists of two parts: a list of locations and a list of queries.

The location list consists of up to 100 lines, one line per location. Each contains a string and two floating-point numbers, separated by whitespace, representing the name of the location, its latitude and its longitude. Names are unique and shorter than 30 characters and do not contain whitespace. Latitudes are between -90 (South Pole) and 90 (North Pole) inclusive. Longitudes are between -180 and 180 inclusive where negative numbers denote locations west of the meridian and positive numbers denote locations east of the meridian. (The meridian passes through Greenwich, London.) The location list is terminated by a line consisting of a single "#".

Each line in the query list contains three names of locations. You can assume the first location to be Alice's home, the second location to be Bob's home and the third location to be a possible meeting point. The query list is terminated by a line consisting of a single "#".

## 출력

For each query, output a line saying "M is x km off A/B equidistance." with M,x,A,B appropriately replaced by the location names and the calculated distance rounded to the nearest integer.

If one of the locations in the query didn't occur in the list of locations print "?" instead of the distance.

## 예제 입력

Ulm             48.700 10.500
Freiburg        47.700 9.500
SanJose         37.366 -121.933
Atlanta         33     -84
Eindhoven       52     6
Orlando         28     -82
Vancouver       49     -123
Honolulu        22     -157
NorthPole       90     0
SouthPole       -90    0
#
SanJose Atlanta Eindhoven
Orlando Vancouver Honolulu
NorthPole SouthPole NorthPole
Ulm SanDiego Orlando
NorthPole SouthPole SouthPole
Ulm Honolulu SouthPole
#

## 예제 출력

Philadelphia is 690 km off Ulm/Freiburg equidistance.
Eindhoven is 3117 km off SanJose/Atlanta equidistance.
Honolulu is 4251 km off Orlando/Vancouver equidistance.
NorthPole is 10019 km off NorthPole/SouthPole equidistance.
Orlando is ? km off Ulm/SanDiego equidistance.
SouthPole is 10019 km off NorthPole/SouthPole equidistance.
SouthPole is 1494 km off Ulm/Honolulu equidistance.


text:
# Fix + percentages

Mrs. Vargas is a car sales agent who earns Php.5,850 monthly plus a 4% commision on all her sales. During a month, she sold a car worth Php.740,000. How much is her total earnings?

Correct result:

x =  35450 Php

#### Solution:

$x=5850 + \dfrac{ 4 }{ 100 } \cdot \ 740000=35450 \ \text{Php}$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors.

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• Persons
Persons surveyed:100 with result: Volleyball=15% Baseball=9% Sepak Takraw=8% Pingpong=8% Basketball=60% Find the average how many like Basketball and Volleyball. Please show your solution.
• The Chemistry test
The Chemistry test contained 8 questions, each with 3 points. Peter scored 21 points. How many percent did Peter write a test?
• Deposit is pesos
Sally deposits Php.22,000 in her savings account. If the bank pays 1.5% interest per year, how much will she receive at the end of the year?
• Profit gain
If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is:
• Percentages
52 is what percent of 93?
• Price inflation
This year the average wage grew by 4.5%. Prices rose by 2%. How much did the real wage increase in the given year?
• Summerjob
The temporary workers planted new trees. Of the total number of 500 seedlings, they managed to plant 426. How many percents did they meet the daily planting limit?
• Principal and interest
Peter put € 270 principal into bank and at the end of the year on account was € 282. To what annual interest rate that Peter had deposited money in the bank?
• Percents
How many percents is 900 greater than the number 750?


text:
# Continuity of a weird function defined as a summation including floor

## Homework Statement

Show that $f(x) = \sum_{i=1}^{\infty}\frac{2^{i}x - \lfloor 2^{i}x \rfloor}{2^{i}}$ is continuous at all real numbers, excluding integers.

## The Attempt at a Solution

I've tried going about via |f(x) - f(y)| < ε, but am having trouble with this, since first, I don't get anywhere and can't really pin down the floor parts of the function, and second, I'm a bit confused, since I also have to deal with n → ∞, in addition to, say, y → x to show continuity.

What I got this way is $f(y) - f(x) = \sum_{i=1}^{\infty}\frac{2^{i}(y - x) + \lfloor 2^{i}x \rfloor - \lfloor 2^{i}y \rfloor}{2^{i}}$, but after spending hours on this I have zero idea what to do with that.

Any help would be greatly appreciated, as I'm really struggling here.

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PAllen
Maybe a first place to get a handle on this is to answer why it's not continuous at integers. Do you understand that?

Another hint is that I don't thing combining the y and x non-floor terms is doing you any favors. Try keeping x terms and y terms together.

Well, I can see that at integers, the sum is always zero, and I assumed this function goes in waves, so that it's increasing from, say 0 to 1, but then has a value of 0 at 1. But I can't really show that it's discontinuous at integers, either (believe me, I've tried).

I've written the f(y) - f(x) without combining them, as well, but I just can't get anywhere that way, either. The reason why I combined them is that I wanted to somehow relate to the fact that y → x.

PAllen
What is f(1.001)? What is f(.999) ? What about 1.00001 and .99999 ? What do you think that says about the limit at integers?

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That's exactly what I'm working on right now, trying to work this out. But, for example, if y → x, can I say that $\lfloor2^{n}y\rfloor \rightarrow 2^{n}x$? Because 2n is increasing very rapidly, so I'm not sure it doesn't "beat" y → x.

PAllen
That's exactly what I'm working on right now, trying to work this out. But, for example, if y → x, can I say that $\lfloor2^{n}y\rfloor \rightarrow 2^{n}x$? Because 2n is increasing very rapidly, so I'm not sure it doesn't "beat" y → x.

What is .999 - floor(.999) ?

What is 1.001 - floor(1.001) ?

What is .999 - floor(.999) ?

What is 1.001 - floor(1.001) ?
Well, it's 0.999 and 0.001, respectively, but it's still the 2n thing that's bothering me, since n is not fixed, but rather n → ∞. I understand that you're going closer and closer, but, again, even as you go lower and lower from 1.001, 2n increases so rapidly that multiplying by it could still yield a higher integer after applying the floor function, couldn't it? I mean, how do I argue that it doesn't, I think that's what I'm having so much trouble with.

edit: But even if I can say that $\lfloor2^{n}y\rfloor \rightarrow 2^{n}x$, I'm still dealing with an infinite sum, right? And, say, the harmonic series also converges to zero, but the infinite sum does not.

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PAllen
Well, it's 0.999 and 0.001, respectively, but it's still the 2n thing that's bothering me, since n is not fixed, but rather n → ∞. I understand that you're going closer and closer, but, again, even as you go lower and lower from 1.001, 2n increases so rapidly that multiplying by it could still yield a higher integer after applying the floor function, couldn't it? I mean, how do I argue that it doesn't, I think that's what I'm having so much trouble with.

edit: But even if I can say that $\lfloor2^{n}y\rfloor \rightarrow 2^{n}x$, I'm still dealing with an infinite sum, right? And, say, the harmonic series also converges to zero, but the infinite sum does not.

As a first step, I'm trying to make sure you understand continuity enough to see why at integers it obviously isn't continuous. That is easier than the rest of the problem, and gives hints for the rest. I believe I've given you way more than enough to understand why it is not continuous at integers.

So: can you give a clear statement of why it is not continuous at integers?

Yeah, it's because $\lim_{y\to x^{+}}f(y) = 1$, whereas $\lim_{y\to x^{-}}f(y) = 0$, where x is an integer.

PAllen
Ok, next step. If you sum from i=1 ... k, for y sufficiently close to x not an integer, what happens to the difference in the floor values?

They become zero. I was thinking of trying induction on this, is this the right step? I'd try to show that for all n, you can find δ, such that if |x - y| < δ, the floors cancel out.

edit: OK, I guess I don't even need induction here, I'd just show this is valid for all n, and then argue that since the sum of the infinite series is the sum of a finite series as n → ∞, if the finite series converges for all n, then the infinite series converges by definition. Any thoughts?

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PAllen
ok, now forget y for a moment, and consider f(x,k), define by the same series as f(x) but summed from k+1 to ∞. What is an upper bound on this value? The same will be true for y. Then, what is an upper bound on f(y,k) - f(x,k)? Then and upper bound on f(y)-f(x) has become k(y-x) + this upper bound. Since as y -> x, k-> ∞, can you put all this together? Note that by construction, when the first k floor value differences become zero, y-x < 2^-k. So you have an upper bound of:

k(2^-k) + the upper bound I describe above.

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Hmm, well I think the upper bound for f(x, k) would be 2-k, wouldn't it? And then the upper bound for f(x, k) - f(y, k) would be 2-k+1. If that's correct, do I then just apply the squeeze theorem to show f(y) - f(x) also converges to zero as k → ∞?

edit: On second thought, how did you come up with the upper bound k(2-k)? I mean, I set up $\delta = min(|\frac{\lceil 2^{n}x \rceil}{2^{n+1}} - x|, |\frac{\lfloor 2^{n}x \rfloor}{2^{n+1}} - x|)$, although I guess I can add $\frac{1}{2^{n}}$, as well.

Last edited:
PAllen


text:
# Roof and Attic¶

HPXML allows the specification of multiple Attic elements, each of which relates to one (HPXML v2) or more (HPXML v3) Roof elements. That relation is optional in HPXML, but is required for HEScore because it is important to know which roof relates to each attic space.

## Attic/Roof Type¶

Each Attic is considered and the AtticType is mapped into a HEScore roof type according to the following mapping.

HPXML Attic Type to HEScore Roof type mapping (HPXML v2)
HPXML HEScore
cape cod cath_ceiling
cathedral ceiling cath_ceiling
flat roof cath_ceiling
unvented attic vented_attic
vented attic vented_attic
venting unknown attic vented_attic
other not translated
HPXML Attic Type to HEScore Roof type mapping (HPXML v3)
HPXML HEScore
CathedralCeiling cath_ceiling
FlatRoof cath_ceiling
Attic/CapeCod = ‘true’ cath_ceiling
Attic/Conditioned = ‘true’ cath_ceiling
Attic vented_attic
Other not translated

Note

The attic/roof type cond_attic has been removed from HEScore as of Q1 2022. Homes with conditioned attics should be described in a way that translates to a cathedral ceiling. In HPXML v3, the conditioned attic type is automatically translated to cathedral ceiling. In HPXML v2, the extension workaround to specify a conditioned attic has been removed and the user will need to identify the house as having an attic type that translates to cathedral ceiling to achieve the desired effect.

HEScore can accept up to two attic/roof constructions. If there are more than two specified in HPXML, the properties of the Attic elements with the same roof type are combined. For variables with a discrete selection the value that covers the greatest combined area is used. For R-values a calculation is performed to determine the equivalent overall R-value for the attic. This is discussed in more detail in Roof R-value.

Note

Starting from HPXML v3, HPXML allows multiple floors/roofs attached to a single attic. The properties of the floors/roofs attached to the same attic are combined into a single one.

## Attic and Roof Area¶

Home Energy Score needs to know the area of the thermal boundary between the living space and unconditioned spaces. The areas needed depend on which Attic/Roof Type is selected.

It’s best practice to provide areas for all roof and attic floor surfaces regardless of Attic/Roof type.

### Vented Attic¶

The thermal boundary for a vented attic is at the floor of the attic. In HPXML v2 that area is attached to the Building/BuildingDetails/Enclosure/AtticAndRoof/Attics/Attic element:

                        <Attic>
<SystemIdentifier id="attic1"/>
<AttachedToRoof idref="roof1"/>
<AtticType>vented attic</AtticType>
<AtticFloorInsulation>
<SystemIdentifier id="attic1flrins"/>
<Layer>
<NominalRValue>49</NominalRValue>
</Layer>
</AtticFloorInsulation>
<AtticRoofInsulation>
<SystemIdentifier id="attic1roofins"/>
<Layer>
<NominalRValue>0</NominalRValue>
</Layer>
</AtticRoofInsulation>
<Area>1200</Area>
</Attic>

In HPXML v3 that area is retrieved from a referenced FrameFloor element from the Building/BuildingDetails/Enclosure/Attics/Attic element:

                    <Attic>
<SystemIdentifier id="attic1"/>
<AtticType>
<Attic>
<Vented>true</Vented>
</Attic>
</AtticType>
<AttachedToRoof idref="roof1"/>
<AttachedToFrameFloor idref="framefloor1"/>
</Attic>

                    <FrameFloor>
<SystemIdentifier id="framefloor1"/>
<Area>1200</Area>
<Insulation>
<SystemIdentifier id="attic1flrins"/>
<Layer>
<NominalRValue>49</NominalRValue>
</Layer>
</Insulation>
</FrameFloor>

If there are more than one Roof elements attached to an attic (HPXML v3 since HPXML v2 only allows one roof to be referenced per Attic), you will need to provide an area for each one so that the most common roof color and exterior finishes may be selected.

### Cathedral Ceiling¶

The thermal boundary for a cathedral ceiling is the roof deck, so the area of the roofs attached to the attic are used.

In HPXML v2 a single Roof can be referenced by the Attic and the area of that roof is used.

                <AtticAndRoof>
<Roofs>
<Roof>
<SystemIdentifier id="roof1"/>
<RoofColor>medium</RoofColor>
<RoofType>slate or tile shingles</RoofType>
<RoofArea>679</RoofArea>
</Roof>
</Roofs>
<Attics>
<Attic>
<SystemIdentifier id="attic1"/>
<AttachedToRoof idref="roof1"/>
<AtticType>cathedral ceiling</AtticType>
<AtticRoofInsulation>
<SystemIdentifier id="ceilins1"/>
<Layer>
<NominalRValue>15</NominalRValue>
</Layer>
</AtticRoofInsulation>
</Attic>
</Attics>
</AtticAndRoof>

In HPXML v3 multiple Roof elements can be referenced by the Attic and the sum of those ares is used. The properties of the roofs will be area weighted as described below.

            <Enclosure>
<Attics>
<Attic>
<SystemIdentifier id="attic1"/>
<AtticType>
<CathedralCeiling/>
</AtticType>
<AttachedToRoof idref="roof1"/>
</Attic>
</Attics>
<Roofs>
<Roof>
<SystemIdentifier id="roof1"/>
<Area>679</Area>
<RoofType>slate or tile shingles</RoofType>
<RoofColor>medium</RoofColor>
<Insulation>
<SystemIdentifier id="ceilins1"/>
<Layer>
<NominalRValue>15</NominalRValue>
</Layer>
</Insulation>
</Roof>
</Roofs>
</Enclosure>

## Roof Color¶

Roof color in HEScore is mapped from the HPXML Roof/RoofColor element according to the following mapping.

HPXML to HEScore roof color mapping
HPXML HEScore
light light
medium medium
medium dark medium_dark
dark dark
reflective white

If the Roof/SolarAbsorptance element is present, the HEScore roof color is set to “cool_color” and the recorded absorptance will be sent to HEScore under the “roof_absorptance” element.

Note

Starting from HPXML v3, if there’re more than one roof attached to the same attic, the roof color of that covers greatest area will be selected.

## Exterior Finish¶

HPXML stores the exterior finish information in the Roof/RoofType element. This is translated into the HEScore exterior finish variable according to the following mapping.

HPXML Roof Type to HEScore Exterior Finish mapping
HPXML HEScore
shingles composition shingles
slate or tile shingles concrete tile
wood shingles or shakes wood shakes
asphalt or fiberglass shingles composition shingles
metal surfacing composition shingles
expanded polystyrene sheathing not translated
plastic/rubber/synthetic sheeting tar and gravel
concrete concrete tile
cool roof not translated
green roof not translated
no one major type not translated
other not translated

Note

Items where the HEScore translation indicates not translated above will result in a translation error.

## Rigid Foam Sheathing¶

If the AtticRoofInsulation element has a Layer with the “continuous” InstallationType, InsulationMaterial/Rigid, and a NominalRValue greater than zero, the roof is determined to have rigid foam sheathing and one of the construction codes is selected accordingly. Otherwise one of the standard wood frame construction codes is selected.

### HPXML v2¶

<Attic>
<SystemIdentifier id="attic5"/>
<AttachedToRoof idref="roof3"/>
<AtticType>cathedral ceiling</AtticType>
<AtticRoofInsulation>
<SystemIdentifier id="attic5roofins"/>
<Layer>
<InstallationType>continuous</InstallationType>
<InsulationMaterial>
<Rigid>eps</Rigid>
</InsulationMaterial>
<NominalRValue>10</NominalRValue>
</Layer>
</AtticRoofInsulation>
<Area>2500</Area>
</Attic>

### HPXML v3¶

<Atics>
<Attic>
<SystemIdentifier id="attic5"/>
<AtticType>
<CathedralCeiling/>
</AtticType>
<AttachedToRoof idref="roof3"/>
</Attic>
</Attics>
<Roofs>
<Roof>
<SystemIdentifier id="roof3"/>
<Area>2500</Area>
<Insulation>
<SystemIdentifier id="attic5roofins"/>
<Layer>
<InstallationType>continuous</InstallationType>
<InsulationMaterial>
<Rigid>eps</Rigid>
</InsulationMaterial>
<NominalRValue>10</NominalRValue>
</Layer>
</Insulation>
</Roof>
<Roofs>

If the Roof/RadiantBarrier element exists and has a “true” value and roof deck insulation R-value is 0, the attic is assumed to have a radiant barrier according to the construction codes available in HEScore. If the Roof/RadiantBarrier element exists and has a “true” value but roof deck insulation R-value greater than 0, the roof will be modeled as a roof with no radiant barrier.

## Roof R-value¶

The roof R-value can be described by using NominalRValue or AssemblyRValue. If a user wishes to use a nominal R-value, NominalRValue elements for all layers need to be provided. Otherwise, AssemblyRValue elements for each layer need to be provided.

If nominal R-value is used, the R-value is summed for all insulation layers. If the roof construction was determined to have Rigid Foam Sheathing, an R-value of 5 is subtracted from the roof R-value sum to account for the R-value of the sheathing in the HEScore construction. The nearest discrete R-value from the list of possible R-values for that roof type is used to determine an assembly code. Then, the assembly R-value of the corresponding assembly code from the lookup table is used. The lookup table can be found at hescorehpxml\lookups\lu_roof_eff_rvalue.csv.

If assembly R-value is used, the discrete R-value nearest to assembly R-value from the lookup table is used. The lookup table can be found at hescorehpxml\lookups\lu_roof_eff_rvalue.csv.

If the attic has more than one Roof element and/or if multiple attics of the same type and their associated roofs are to be combined, a weighted average assembly R-value is determined by weighting the U-values by area. Then the discrete R-value nearest to the weighted average assembly R-value from the lookup table is used.

\begin{align*} U_i &= \frac{1}{R_i} \\ U_{eff,avg} &= \frac{\sum_i{U_i A_i}}{\sum_i A_i} \\ R_{eff,avg} &= \frac{1}{U_{eff,avg}} \\ \end{align*}

Then the nearest discrete R-value to the weighted average R-value from the lookup table is used. The lookup table can be found at hescorehpxml\lookups\lu_roof_eff_rvalue.csv.

## Attic R-value¶

Determining the attic floor insulation levels uses the same procedure as Roof R-value except the lookup table is different.

If nominal R-value is used, the attic floor center-of-cavity R-values are each R-0.5 greater than the nominal R-values in the enumeration list.

If assembly R-value is used, the lookup table at hescorehpxml\lookups\lu_ceiling_eff_rvalue.csv is used.

If the primary roof type is determined to be a cathedral ceiling, then an attic R-value is not calculated.

## Knee Walls¶

The Home Energy Score Q1 2022 release includes the ability to directly model knee walls. As such, the workaround of adding knee wall area to attic floor area is no longer used. Knee walls are now directly translated and passed to Home Energy Score as follows:

In HPXML v2, knee walls are specified via the Attic/AtticKneeWall element.

In HPXML v3, knee walls are specified via wall attachment in Attic/AttachedToWall. The attached wall must have AtticWallType of “knee wall”. See below an example:

<Attics>
<Attic>
<SystemIdentifier id="attic5"/>
<AtticType>
<Attic>
<Vented>true</Vented>
</Attic>
</AtticType>
<AttachedToRoof idref="roof3"/>
<AttachedToWall idref="kneewall"/>
<AttachedToFrameFloor idref="framefloor"/>
</Attic>
</Attics>
<Walls>
<Wall>
<SystemIdentifier id="kneewall"/>
<AtticWallType>knee wall</AtticWallType>
<WallType>
<WoodStud/>
</WallType>
<Area>200</Area>
<Insulation>
<SystemIdentifier id="kneewallins"/>
<Layer>
<InstallationType>cavity</InstallationType>
<NominalRValue>11</NominalRValue>
</Layer>
</Insulation>
</Wall>
<Walls>

The knee walls R-value can be described by nominal R-value or assembly R-value.

If nominal R-value is used, the nearest assembly code by R-value in the code is selected and the assembly R-value is looked up for that code.

If assembly R-value is used, the nearest assembly code by assembly R-value is looked up from that table.

Knee Wall Assembly Codes and R-values
doe2code U-value Eff-R-value
kwwf00 0.346 2.9
kwwf03 0.154 6.5
kwwf07 0.101 9.9
kwwf11 0.083 12.1
kwwf13 0.075 13.4
kwwf15 0.068 14.7
kwwf19 0.055 18.1
kwwf21 0.050 20.1

If an attic has more than one knee wall and/or if multiple attics of the same type need to be combined, the area weighted average assembly effective R-value is calculated from all the associated knee walls. The areas of all the knee walls are summed. The assembly code with the nearest assembly effective R-value is chosed to represent the knee walls.


text:
RO  EN

## Some properties of maximum deficiency energy of a graph

Authors: Omendra Singh, Pravin Garg, Neha Kansal
Keywords: Deficiency, maximum deficiency matrix, maxi- mum deficiency eigenvalues, maximum deficiency energy.

### Abstract

The concept of maximum deficiency matrix $M_{df}(G)$ of a simple graph $G$ is introduced in this paper. Let $G=(V,E)$ be a simple graph of order $n$ and let $df(v_{i})$ be the deficiency of a vertex $v_{i}$, $i=1,2, \ldots, n$, then the maximum deficiency matrix $M_{df}(G) = [f_{ij}]_{n \times n}$ is defined as: $f_{ij}= \begin{cases} max\{df(v_{i}), df(v_{j})\}, & \text{if v_{i}v_{j} \in E(G)} \\ 0~~~~~~~~~~~~~~~~~~~~~~~~, & \text{otherwise.} \end{cases}$ Further, some coefficients of the characteristic polynomial $\phi(G; \gamma)$ of the maximum deficiency matrix of $G$ are obtained. The maximum deficiency energy $E{M_{df}}(G)$ of a graph $G$ is also introduced. The bounds for $E{M_{df}}(G)$ are established. Moreover, maximum deficiency energy of some standard graphs is shown, and if the maximum deficiency energy of a graph is rational, then it must be an even integer.

Omendra Singh
Department of Mathematics,
University of Rajasthan,
Jaipur - 302004, India
Phone: +919521531401
E-mail:

Pravin Garg
Department of Mathematics,
University of Rajasthan,
Jaipur - 302004, India
Phone: +919982123280
E-mail:

Neha Kansal
Department of Mathematics,
University of Rajasthan,
Jaipur - 302004, India
Phone: +918952836930
E-mail:

0.14 Mb


text:
# How do you rationalize fractions

## Main Question or Discussion Point

how do you do that

Fractions are already rational. Here's what "rational" means: http://en.wikipedia.org/wiki/Rational_number" [Broken]

Last edited by a moderator:
arildno
Homework Helper
Gold Member
Dearly Missed
Fractions are already rational. Here's what "rational" means: http://en.wikipedia.org/wiki/Rational_number" [Broken]
Incorrect!
A "fraction" is a real number that is written as the product of one real number a (called the "numerator"), and the multiplicative inverse of a real (non-zero) number b (called the denominator).

To "rationalize" a fraction means to rewrite the denominator as a natural number, if possible.

Example:
$$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}*1=\frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$

Last edited by a moderator:
HallsofIvy
Homework Helper
What does this have to do with calculus? I am moving it to "General Math".

HallsofIvy
Homework Helper
I've never heard of "realizing" a fraction! What language is that translated from? However, I will agree that one does not rationalize fractions!
The example arildno gave was rationalizing the denominator of a fraction. There are also times when one would want to rationalize the numerator of a fraction. In general, it is not possible to rationalize both numerator and denominator at the same time.

I've never heard of "realizing" a fraction! What language is that translated from?
My apologies! I've made a big fool of myself once again. I was thinking about "realising" as in making the denominator of a fraction (containing complex numbers) real! (I was taught this term back in high school, here in Australia.) arildno and Hallsofivy are absolutely correct.

I'm really sorry for wasting everyone's time. I think I should go back to lurking.

arildno
Homework Helper
Gold Member
Dearly Missed
I've never heard of "realizing" a fraction! What language is that translated from? However, I will agree that one does not rationalize fractions!
The example arildno gave was rationalizing the denominator of a fraction. There are also times when one would want to rationalize the numerator of a fraction. In general, it is not possible to rationalize both numerator and denominator at the same time.
I stand corrected.

HallsofIvy
Homework Helper
My apologies! I've made a big fool of myself once again. I was thinking about "realising" as in making the denominator of a fraction (containing complex numbers) real! (I was taught this term back in high school, here in Australia.) arildno and Hallsofivy are absolutely correct.

I'm really sorry for wasting everyone's time. I think I should go back to lurking.
"realizing"! I like that. Reminds me of when, in an algebra class, we were learning about 'adding' an identity to a semigroup (to make it a monoid) and we had quite an argument about what the process should be called.

Suggestions were "unification", "one-ification" and "identification"!


text:
# Direct determination of neutron lifetime in $\beta$-decay

22 Jan 2020 Vasiliev Valerius V.

The neutron lifetime is determined for now most accurately by two methods --the so-called "beam method" and the method of neutron storage in a trap. The goal of this research is to obtain the neutron lifetime with a higher precision by eliminating the traditional systematic errors... (read more)

PDF Abstract

# Code Add Remove Mark official

No code implementations yet. Submit your code now

# Categories

• NUCLEAR EXPERIMENT
• DATA ANALYSIS, STATISTICS AND PROBABILITY


text:
OpenGLAttributes¶

Applies attributes to modify the appearance of objects in the viewport and in renders done by the OpenGLRender node.

user¶

Container for user-defined plugs. Nodes should never make their own plugs here, so users are free to do as they wish.

out¶

The processed output scene.

enabled¶

The on/off state of the node. When it is off, the node outputs the input scene unchanged.

The input scene

filter¶

The filter used to control which parts of the scene are processed. A Filter node should be connected here.

attributes¶

The attributes to be applied - arbitrary numbers of user defined attributes may be added as children of this plug via the user interface, or using the CompoundDataPlug API via python.

attributes.primitiveSolid¶

Whether or not the object is rendered solid, in which case the assigned GLSL shader will be used to perform the shading.

attributes.primitiveWireframe¶

Whether or not the object is rendered as a wireframe. Use the primitiveWireframeColor and primitiveWireframeWidth plugs for finer control of the wireframe appearance.

attributes.primitiveWireframeColor¶

The colour to use for the wireframe rendering. Only meaningful if wireframe rendering is turned on.

attributes.primitiveWireframeWidth¶

The width in pixels of the wireframe rendering. Only meaningful if wireframe rendering is turned on.

attributes.primitiveOutline¶

Whether or not an outline is drawn around the object. Use the primitiveOutlineColor and primitiveOutlineWidth plugs for finer control of the outline.

attributes.primitiveOutlineColor¶

The colour to use for the outline. Only meaningful if outline rendering is turned on.

attributes.primitiveOutlineWidth¶

The width in pixels of the outline. Only meaningful if outline rendering is turned on.

attributes.primitivePoint¶

Whether or not the individual points (vertices) of the object are drawn. Use the primitivePointColor and primitivePointWidth plugs for finer control of the point rendering.

attributes.primitivePointColor¶

The colour to use for the point rendering. Only meaningful if point rendering is turned on.

attributes.primitivePointWidth¶

The width in pixels of the points. Only meaningful if point rendering is turned on.

attributes.primitiveBound¶

Whether or not the bounding box of the object is drawn. This is in addition to any drawing of unexpanded bounding boxes that the viewer performs. Use the primitiveBoundColor plug to change the colour of the bounding box.

attributes.primitiveBoundColor¶

The colour to use for the bounding box rendering. Only meaningful if bounding box rendering is turned on.

attributes.pointsPrimitiveUseGLPoints¶

Points primitives have a render type (set by the PointsType node) which allows them to be rendered as particles, disks, spheres etc. This attribute overrides that type for OpenGL only, allowing a much faster rendering as raw OpenGL points.

attributes.pointsPrimitiveGLPointWidth¶

The width in pixels of the GL points rendered when the pointsPrimitiveUseGLPoints plug has overridden the point type.

attributes.curvesPrimitiveUseGLLines¶

Curves primitives are typically rendered as ribbons and as such have an associated width in object space. This attribute overrides that for OpenGL only, allowing a much faster rendering as raw OpenGL lines.

attributes.curvesPrimitiveGLLineWidth¶

The width in pixels of the GL lines rendered when the curvesPrimitiveUseGLLines plug has overridden the drawing to use lines.

attributes.curvesPrimitiveIgnoreBasis¶

Turns off interpolation for cubic curves, just rendering straight lines between the vertices instead.

attributes.visualiserScale¶

Scales non-geometric visualisations in the viewport to make them easier to work with.

attributes.visualiserMaxTextureResolution¶

Visualisers that load textures will respect this setting to limit their resolution.

attributes.visualiserFrustum¶

Controls whether applicable locations draw a representation of their projection or frustum.

attributes.lightDrawingMode¶

Controls how lights are presented in the Viewer.

attributes.lightFrustumScale¶

Allows light projections to be scaled to better suit the scene.

global¶

Causes the attributes to be applied to the scene globals instead of the individual locations defined by the filter.

extraAttributes¶

An additional set of attributes to be added. Arbitrary numbers of attributes may be specified within a single IECore.CompoundObject, where each key/value pair in the object defines an attribute. This is convenient when using an expression to define the attributes and the attribute count might be dynamic. It can also be used to create attributes whose type cannot be handled by the attributes CompoundDataPlug, with IECoreScene.ShaderNetwork being one example.

If the same attribute is defined by both the attributes and the extraAttributes plugs, then the value from the extraAttributes is taken.


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# About a counterexample of an inequality？

I have known how to use the compactive argument to prove the inequality (1), i.e. $1\leqslant p<n$, $\Omega\subset R^n$ is a bounded domain,$\forall \varepsilon>0$, there is $C_\varepsilon>0$, such that $\forall u\in W^{1,p}(\Omega)$, we have $$\|u\|_{L^p(\Omega)}\leqslant\varepsilon\|\nabla u\|_{L^p(\Omega)}+C_\varepsilon\|u\|_{L^1(\Omega)} \tag{1}$$

Now,I want to find out a counterexample to demonstrate that $$\|u\|_{L^{p*}(\Omega)}\leqslant\varepsilon\|\nabla u\|_{L^p(\Omega)}+C_\varepsilon\|u\|_{L^1(\Omega)} \tag{2}$$

namely if we replace $\|u\|_{L^p(\Omega)}$ in $(1)$ with $\|u\|_{L^{p*}(\Omega)}$ to get $(2)$, then $(2)$ is not true in general, where $\frac{1}{p*}=\frac{1}{p}-\frac{1}{n}$.

-

You can do it with scaling. Pick the unit ball for $\Omega$, and let $u$ be any nonzero smooth function with support contained in $\Omega$. Now for each $\lambda<1$, let $u_\lambda(x)=u(x/\lambda)$. If (2) holdes for $u_\lambda$, we find $$\lambda^{n/p^*}\lVert u\rVert_{L^{p^*}}\le\epsilon\lambda^{n/p-1}\lVert\nabla u\rVert_{L^p}+\lambda^nC_\epsilon\lVert u\rVert_{L^1}.$$ Now the first two exponents of $\lambda$ are the same, so this becomes $$\lVert u\rVert_{L^{p^*}}\le\epsilon\lVert\nabla u\rVert_{L^p}+\lambda^{p^*}C_\epsilon\lVert u\rVert_{L^1}.$$ Finally, pick $\epsilon$ small enough so that letting $\lambda\to0$ yields a contradiction.
There was a misprint in my answer: $L^p$ on the LHS of the two inequalities should have been $L^{p*}$. Oddly, this edit was suggested by an anonymous user and rejected by two reviewers. I found no way to override the two, other than to do the same edit myself. –  Harald Hanche-Olsen Jan 18 '13 at 13:32


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# Angular momentum in three dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.4.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We can now generalize our treatment of rotation, originally studied in two dimensions, to three dimensions. We’ll view a 3-d rotation as a combination of rotations about the ${x}$, ${y}$ and ${z}$ axes, each of which can be represented by a ${3\times3}$ matrix. These matrices are as follows:

 $\displaystyle R\left(\theta\hat{\mathbf{x}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle R\left(\theta\hat{\mathbf{y}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle R\left(\theta\hat{\mathbf{z}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (3)$

We’re interested in infinitesimal rotations, for which we retain terms up to first order in the rotation angle ${\varepsilon_{i}}$, so that ${\cos\varepsilon_{i}=1}$ and ${\sin\varepsilon_{i}=\varepsilon_{i}}$. This gives the infinitesimal rotation matrices as

 $\displaystyle R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & -\varepsilon_{x}\\ 0 & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & \varepsilon_{y}\\ 0 & 1 & 0\\ -\varepsilon_{y} & 0 & 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & -\varepsilon_{z} & 0\\ \varepsilon_{z} & 1 & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (6)$

We now consider the series of rotations as follows: first, a rotation by ${\varepsilon_{x}\hat{\mathbf{x}}}$, then by ${\varepsilon_{y}\hat{\mathbf{y}}}$, then by ${-\varepsilon_{x}\hat{\mathbf{x}}}$ and finally by ${-\varepsilon_{y}\hat{\mathbf{y}}}$. Because the various rotations don’t commute, we don’t end up back where we started. We can calculate the matrix products to find the final rotation.

 $\displaystyle R$ $\displaystyle =$ $\displaystyle R\left(-\varepsilon_{y}\hat{\mathbf{y}}\right)R\left(-\varepsilon_{x}\hat{\mathbf{x}}\right)R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & -\varepsilon_{y}\\ 0 & 1 & 0\\ \varepsilon_{y} & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & \varepsilon_{x}\\ 0 & -\varepsilon_{x} & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & \varepsilon_{y}\\ 0 & 1 & 0\\ -\varepsilon_{y} & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & -\varepsilon_{x}\\ 0 & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & -\varepsilon_{y}\\ 0 & 1 & \varepsilon_{x}\\ \varepsilon_{y} & -\varepsilon_{x} & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & \varepsilon_{y}\\ 0 & 1 & -\varepsilon_{x}\\ -\varepsilon_{y} & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1+\varepsilon_{y}^{2} & \varepsilon_{x}\varepsilon_{y} & -\varepsilon_{x}^{2}\varepsilon_{y}\\ -\varepsilon_{x}\varepsilon_{y} & 1+\varepsilon_{x}^{2} & 0\\ 0 & \varepsilon_{x}\varepsilon_{y}^{2} & 1+\varepsilon_{x}^{2}+\varepsilon_{y}^{2} \end{array}\right] \ \ \ \ \ (10)$

To get the third line, we multiplied the first two matrices in the second line, and the last two matrices in the second line. In the final result, we can discard terms containing ${\varepsilon_{x}^{2}}$ or ${\varepsilon_{y}^{2}}$ to get

$\displaystyle R=\left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & 0\\ -\varepsilon_{x}\varepsilon_{y} & 1 & 0\\ 0 & 0 & 1 \end{array}\right]=R\left(-\varepsilon_{x}\varepsilon_{y}\hat{\mathbf{z}}\right) \ \ \ \ \ (11)$

Thus the result of the four rotations about the ${x}$ and ${y}$ axes is a single rotation about the ${z}$ axis.

To convert this to quantum operators, we define the operator ${U\left[R\right]}$ by comparison with the procedure we used for 2-d rotations. That is, the operator ${U}$ is given by the corresponding angular momentum operator ${L_{x}}$, ${L_{y}}$ or ${L_{z}}$ as

 $\displaystyle U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{x}L_{x}}{\hbar}\ \ \ \ \ (12)$ $\displaystyle U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{y}L_{y}}{\hbar}\ \ \ \ \ (13)$ $\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (14)$

By comparing 7 and 11 we thus require these ${U}$ operators to satisfy

$\displaystyle U\left[R\left(-\varepsilon_{y}\hat{\mathbf{y}}\right)\right]U\left[R\left(-\varepsilon_{x}\hat{\mathbf{x}}\right)\right]U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]=U\left[R\left(-\varepsilon_{x}\varepsilon_{y}\hat{\mathbf{z}}\right)\right] \ \ \ \ \ (15)$

We can get the commutation relation ${\left[L_{x},L_{y}\right]}$ by matching coefficients of ${\varepsilon_{x}\varepsilon_{y}}$ on each side of this equation. On the RHS, the coefficient is ${\frac{iL_{z}}{\hbar}}$. On the LHS, we can pick out the terms involving ${\varepsilon_{x}\varepsilon_{y}}$ to get

$\displaystyle -\frac{1}{\hbar^{2}}\left(L_{y}L_{x}-L_{y}L_{x}-L_{x}L_{y}+L_{y}L_{x}\right)=\frac{1}{\hbar^{2}}\left[L_{x},L_{y}\right] \ \ \ \ \ (16)$

The first term on the LHS comes from the ${\varepsilon_{x}}$ term in the first ${U}$ in 15 multiplied by the ${\varepsilon_{y}}$ term in the second ${U}$ (with the ${I}$ term in the other two ${U}$s); the second term on the LHS comes from the ${\varepsilon_{x}}$ term in the first ${U}$ in 15 multiplied by the ${\varepsilon_{y}}$ term in the fourth ${U}$, and so on.

Matching the two sides, we get

$\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (17)$

By comparison with the classical definitions of the three components of ${\mathbf{L}}$, we can write the quantum operators in terms of position and momentum operators as

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle YP_{z}-ZP_{y}\ \ \ \ \ (18)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle ZP_{x}-XP_{z}\ \ \ \ \ (19)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle XP_{y}-YP_{x} \ \ \ \ \ (20)$

From the commutators of position and momentum ${\left[X,P_{x}\right]=i\hbar}$ and so on, we can verify 17 from these relations as well.

 $\displaystyle \left[L_{x},L_{y}\right]$ $\displaystyle =$ $\displaystyle \left[YP_{z}-ZP_{y},ZP_{x}-XP_{z}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[YP_{z},ZP_{x}-XP_{z}\right]-\left[ZP_{y},ZP_{x}-XP_{z}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar YP_{x}+i\hbar P_{y}X\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(XP_{y}-YP_{x}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar L_{z} \ \ \ \ \ (25)$

The third line follows because ${\left[YP_{z},XP_{z}\right]=\left[ZP_{y},ZP_{x}\right]=0}$. The other two commutation relations follow by cyclic permutation of ${x}$, ${y}$ and ${z}$:

 $\displaystyle \left[L_{y},L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{x}\ \ \ \ \ (26)$ $\displaystyle \left[L_{z},L_{x}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{y} \ \ \ \ \ (27)$


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### YeuDieuNhieu's blog

By YeuDieuNhieu, history, 4 years ago,

[](https://open.kattis.com/contests/asiasg18-prelim-open/standings) How to solve D and E, I tried but failed.

• +14

 » 4 years ago, # |   0 D: Find all bridge in the graph, then DFS to see how many vertices reachable from vertex 0 after removing all the bridge.E: Binary search the answer. Suppose we need to check whether it is possible to find a combination of N numbers with the minimum not less than X. Build a bipartite graph where vertices on the left represent row, and vertices on the right represent column. For each cell ai, j, if ai, j ≥ X, connect an edge from i-th vertices on the left to j-th vertices on the right. If the maximum matching of the bipartite graph consist N edges, it is possible to find a combination of N numbers satisfying the condition above (we choose all cell (i, j) such that the edge connecting i-th vertices on the left and j-th vertices on the right is in the maximum matching).
•  » » 4 years ago, # ^ |   0 Thank you ! got ACed on D. I'm trying to solve E.
 » 4 years ago, # |   0 Anyway, can anyone explain G and J?
•  » » 4 years ago, # ^ |   0 G: precalculate all N such that Bob will win. There are only less than 400 such values. You need an efficient enough precalculate program. Ours runs in less than 10 minutes.
•  » » » 4 years ago, # ^ |   0 How did you precalculate these N efficiently?
•  » » » » 4 years ago, # ^ |   +8 For each i<=10^9, determine if there is any N found so far such that i-N+1 is prime. If none of them is, i is a new N we found.
•  » » » » » 4 years ago, # ^ |   0 Can you prove an upper bound on how many i from 1 to N that Bob will win?
•  » » » » » » 4 years ago, # ^ |   0 No, I guess it's related to distribution of primes which is really complicated. In contest, I discovered there are really few N that Bob will win only by experiment.
•  » » » » » 4 years ago, # ^ |   0 If someone is interested. My offline sol to find all possible nos. And AC soln with all no hardcoded. But I as of now I don't have any proof on the upper bound.
•  » » » » » 4 years ago, # ^ |   0 Unable to solve A with DFS? I tried but got TLE, so I changed to an algorithm O(9*n^4).
•  » » » » » » 4 years ago, # ^ |   0 My Soln of A. Can you share yours?
•  » » » » » » » 4 years ago, # ^ |   +3
•  » » » » » » » » 4 years ago, # ^ |   0 Thanks. You can reduce your time complexity to O(9*8*n^2). By using this piece of code. const lli r[10]={-2,-2,-1,-1,1,1,2,2}; const lli c[10]={-1,1,-2,2,-2,2,-1,1}; for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) for (int k=0;k<8;k++) { p=x+r[k]; q=y+c[k]; if (1<=p&&p<=n&&1<=q&&q<=n&&a[i][j]=='I' && a[p][q]=='C') dd[p][q]=1; }
•  » » 4 years ago, # ^ |   +21 J: Enumerate all the possible cycles For a cycle, contract all the vertices into one vertex then, we can calculate the number of spanning tree after contracting with Kirchhoff theorem The first part can be computed by a O(2VV2) TSP DP. The overall time complexity is O(2VV3) since you need to compute at most 2V times Gaussian elimination.
•  » » » 4 years ago, # ^ |   0 Do you know any tutorial on how to compute determinant of a N * N matrix in O(N3)?
•  » » » » 4 years ago, # ^ |   0 After applying Gaussian elimination in O(N3), the determinant of the matrix is equal to the product of all the element on the main diagonal.
 » 4 years ago, # |   0 How to solve B?
•  » » 4 years ago, # ^ |   0 B can be solved by using DSU or DFS to check whether (i,n-i+1) is in the same connected conponents. ( i<=1<=(n+1)/2). Here is my sol (using DSU): Here
 » 4 years ago, # |   0 how to solve F?
•  » » 4 years ago, # ^ |   +8 My soln for F -Chose a const SIZE ~ sqrt(n). (500 gives AC, and 1500 gives TLE on last TC) Now divide all B into two half one less than SIZE and other greater than or equal to SIZE.For Type 1 queries - if B >  = SIZE. Directly update all values. //O(n/SIZE) if B < SIZE, //O(1) Keep an array inc[SIZE][SIZE] i.e. (B,A) And update as inc[b][a] +  = cstFor Type 2 queries - cnt=a[D]; for(i=1;i
•  » » » 4 years ago, # ^ |   0 Can I see your full code? send it for me to this email thanks. phamnhi8910@gmail.com
•  » » » » 4 years ago, # ^ |   0 Link of code is already present in the comment. Just click on soln in first line. :P
 » 4 years ago, # |   0 How to solve C, H, I?
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 C: Dynamic Programming. Consider building the big string (I'll call it T) character by character. Because it has to be a palindrome, when we assign T[0], we also assign T[2N - 1]. Keep a pointer to the head and tail of S. When we assign positions in T, we may need to move the head and tail pointers of S. When S is a subsequence of the prefix / suffix of T that we've built so far, just multiply by 26remaining. The state is just how many characters are remaining and the positions of the pointers in S for a O(n3) solution.H: Just BFS from outside the grid and count how many walls you hit. If you enter an "active" cell don't continue the BFS from there.I: It looks like finding the coefficient of xk in (1 + x + x2 + ... + xm)n. No idea how to do that efficiently.
•  » » » 4 years ago, # ^ | ← Rev. 3 →   +13 I is a pretty standard problem. For example, see 451E - Devu and Flowers for the application of the same idea.Firstly, consider the case when m (the number of objects of each type) is infinite. In this case the answer is simply (you need to count ways of splitting k in n non-negative integer summands; that is the same as splitting n + k in n positive integer summands; that is the same as splitting n + k balls lying in a row in n non-empty groups by placing n - 1 "walls" that separate the groups in n + k - 1 positions (between each pair of balls)).But we overcounted something: we counted the partitions of k in n non-negative summands where some summands are larger than m, though we should have not done that. Let's use inclusion-exclusion to handle the overcounting. How many partitions of k in n non-negative summands there are, if we are required to make some i fixed summands larger than m? Let's note that there are as many of them as there are partitions of k - (m + 1)i in n non-negative summands (simply subtract m + 1 from each "definitely overflowing" summand, after that there will be no restrictions on said summand).In the end, by inclusion-exclusion, the answer is equal to (we need to choose i definitely overflowing summands out of n and count the partitions of k into n summands such that i of them are definitely overflowing).
•  » » » » 4 years ago, # ^ |   +8 Does there any way to solve this problem with polynomial multiplication?
•  » » » 4 years ago, # ^ |   +20 For I, we can rewrite the expression as (1 - xm + 1)n(1 - x) - n. Just use the binomial expansion on both and you have to evaluate the following:
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +8 I do not really understand the solution for C."Because it has to be a palindrome, when we assign T[0], we also assign T[0]." Do you mean T[0] and T[2*N-1]?How do you handle S being in the middle (i.e. it does not lie entirely in the first N letters or the last N letters) for the DP state? I do not see how the transitions work for this case.EDIT: I misread the question as substring instead of subsequence, so the solution makes a lot more sense now...
 » 4 years ago, # |   -8 how to solve problem C An alphabetical string is a string consisting of 0 or more capital letters (i.e. [‘A’..‘Z’]). Given an alphabetical string S[1..N], determine the number of palindromic alphabetical strings of length 2N that contains S as a subsequence (not necessarily contiguous). A string is palindromic if it is equal to its reverse.


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# Basic concept of utility

I'm new to the concept of utility and I'm struggling to understand an important idea. Say we have some bet. I don't understand how the utility of the expected value of the bet differs from the expected value of the utility of the bet. In my mind, both correspond to what I would expect to get out of the bet on average, scaled to take in account my preferences for different monetary values.

Perhaps a simple example showing how exactly the two ideas are different would help. Thanks!

• Your example is not completely clear, but typically the context would be something like this. – GeoMatt22 Apr 28 '17 at 0:42

Let's pick a simple utility function and a simple gamble. Our utility function will be $$U(X)=\sqrt{X}$$ Our lottery will pay 1 unit with 50% probability and 4 units with 50% probability. The expected return is: $$E(X)=.5*\times1+.5\times4=2.5.$$ The utility of the expected value is: $$U(E(X))=\sqrt{2.5}\approx{1.581139}.$$

The utility of each payoff is: $$U(1)=1\wedge{U}(4)=2.$$ The expected utility of the lottery is: $$E(U(X))=.5\times{U}(1)+.5\times{U}(4)=.5\times{1}+.5\times{2}=1.5.$$

The expected utility is the expectation of the satisfaction that would be received over the sample space of possible outcomes. The utility of the expected value may not happen unless one of the prizes happens to match it. The utility of the expected value is the satisfaction that would be received if the expected utility were actually received. It won't be received.

Wins are experienced through the emotion they generate, losses too. The expected utility is your expected satisfaction.

• re: your final sentence, which I like. For a while, lotteries in Australia were marketed as a 'ticket to dream', tacitly acknowledging the chance of winning being negligible. Hence, the utility of a loss was the emotion you experienced between paying for the ticket and the dream being over when you discovered you didn't win, and utility of a win was irrelevant in calculating the utility of the lottery ticket. – Robert de Graaf Apr 28 '17 at 4:26

The idea of utility is very simple. It converts the numerical outcome into the utility to you. Say, you get 10 apples or 5 oranges, or some combination of two. How would you compare apples to oranges? Easy, I offer you to take either of two choices, and you pick one. So, theory goes that there's some kind of a function $U(x_1,x_2,\dots,x_n)$ such that when you feed it a basket of goods $1,2,\dots,n$ with quantities $x_i$ of each good, this function returns a scalar value, which reflects the utility of this basket to you. You pick the one with a greater utility.

So, if I feed it $U(10,0)$ or $U(0,5)$, where the good 1 is apples, and the good 2 is oranges, it'll return two different numbers, the greater one will indicate which basket has more utility to you. YOu can combine goods in the same basket $U(1,1)$ - one apple and one orange.

So, in a nutshell, the utility function is a concept that allows us to compare apples to oranges. The theory states that you pcik the baskets based on their utility, which is calculated by plugging the basket descriptions into this utility function.


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# SharePoint 2013 - Page Status Bar

I want page status bar to be displayed on top of Main Content Place holder, not with the ribbon. What changes I need to do?

• check out this article – Mark L Jan 13 '16 at 9:39
• I don't think the status bar is part of the ribbon. Do you mean the status bar part of

? – moe Jan 13 '16 at 9:41
• Actually by default , it sticks to Ribbon, but I have custom master page, so I want this to be displayed in above content block, rather than with ribbon, because sometimes it gets hidden behind ribbon. – Ovais Khatri Jan 13 '16 at 9:48
• @Mark, I dont want any custom message to be displayed, neither I want to use any code behind, there should a way to do this by positioning it in masterpage layout. – Ovais Khatri Jan 13 '16 at 9:49
• @OvaisKhatri I haven't tested this, but since DeltaPageStatusBar is a <SharePoint:AjaxDelta> region, you could move that block inside the <SharePoint:AjaxDelta id="DeltaPlaceHolderMain"></SharePoint:AjaxDelta> block to force it to be part of the DeltaPlaceHolderMain. That might require some .css tuning afterwards. – moe Jan 13 '16 at 11:03

var element = $('#DeltaPageStatusBar').detach();$('#DeltaPlaceHolderMain').prepend(element);


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## Variables

### Learning Objectives

• Define and identify variables

Greg and Alex have the same birthday, but they were born in different years. This year Greg is $20$ years old and Alex is $23$, so Alex is $3$ years older than Greg. When Greg was $12$, Alex was $15$. When Greg is $35$, Alex will be $38$. No matter what Greg’s age is, Alex’s age will always be $3$ years more, right?

In the language of algebra, we say that Greg’s age and Alex’s age are variable and the three is a constant. The ages change, or vary, so age is a variable. The $3$ years between them always stays the same, so the age difference is the constant.

In mathematics, letters of the alphabet are used to represent variables. Suppose we call Greg’s age $g$. Then we could use $g+3$ to represent Alex’s age. See the table below.

Greg’s age Alex’s age
$12$ $15$
$20$ $23$
$35$ $38$
$g$ $g+3$

Letters are used to represent variables. Letters often used for variables are $x,y,a,b,\text{ and }c$.

### Variables and Constants

A variable is a letter that represents a number or quantity whose value may change (ex. $x, y, z, a, t, k$ etc.).

A constant is a number whose value always stays the same.

### EXAMPLE

Identify the variable(s) in each expression or equation

1. $x+2$
2. $5-3y$
3. $7+5b-z=9$

### TRY IT

To write algebraically, we need some symbols as well as numbers and variables. There are several types of symbols we will be using. There are multiple symbols and phrases to represent the four basic arithmetic operations: addition, subtraction, multiplication, and division. We will summarize them here:

Operation Notation Say: The result is…
Addition $a+b$ $a\text{ plus }b$ the sum of $a$ and $b$
Subtraction $a-b$ $a\text{ minus }b$ the difference of $a$ and $b$
Multiplication $a\cdot b,\left(a\right)\left(b\right),\left(a\right)b,a\left(b\right)$ $a\text{ times }b$ The product of $a$ and $b$
Division $a\div b,a/b,\frac{a}{b},b\overline{)a}$ $a$ divided by $b$ The quotient of $a$ and $b$

In algebra, the cross symbol, $\times$, is not used to show multiplication because that symbol may cause confusion. Does $3xy$ mean $3\times y$ (three times $y$ ) or $3\cdot x\cdot y$ (three times $x\text{ times }y$ )? To make it clear, use • or parentheses for multiplication.

Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in written language. They indicate which expressions are to be kept together and separate from other expressions. The table below lists three of the most commonly used grouping symbols in algebra.

Common Grouping Symbols
parentheses ( )
brackets [ ]
braces { }

Here are some examples of expressions that include grouping symbols.

$\begin{array}{cc}8\left(14 - 8\right)21 - 3\\\left[2+4\left(9 - 8\right)\right]\\24\div \left\{13 - 2\left[1\left(6 - 5\right)+4\right]\right\}\end{array}$

## Contribute!

Did you have an idea for improving this content? We’d love your input.


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# OpenGL Code that should work.

This topic is 3930 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Can someone tell me why this code doesn't do what i must do? Supposdly, when i press "a" the rectangle should go further, but it just desappears.

typedef float dpi_vertex[3];
typedef float dpi_matrix44[16];

void cDPI_Frame::getMatrix44(dpi_matrix44 m)
{
dpi_matrix44 x;
dpi_crossProduct(x, front, up);

m[0]=x[0];
m[1]=x[1];
m[2]=x[2];
m[3]=0;

m[4]=up[0];
m[5]=up[1];
m[6]=up[2];
m[7]=0;

m[8]=front[0];
m[9]=front[1];
m[10]=front[2];
m[11]=0;

m[12]=origin[0];
m[13]=origin[1];
m[14]=origin[2];
m[15]=1;

}

void cDPI_Frame::getOrigin(dpi_vertex v)
{
v[0]=origin[0];
v[1]=origin[1];
v[2]=origin[2];
}

void cDPI_Frame::MoveFoward(float p)
{
origin[0]+=front[0]*p;
origin[1]+=front[1]*p;
origin[2]+=front[2]*p;
}

void dpi_crossProduct(dpi_vertex result, dpi_vertex vectora, dpi_vertex vectorb)
{
result[0]=(vectora[1]*vectorb[2])-(vectorb[1]*vectora[2]);
result[1]=-(vectora[0]*vectorb[2])+(vectorb[0]*vectora[2]);
result[2]=(vectora[0]*vectorb[1])-(vectorb[0]*vectora[1]);
}

cDPI_Frame camera;
dpi_matrix44 temp;
dpi_vertex v;

int result=0;
result=SDL_Init(SDL_INIT_VIDEO);
if(result!=0)
return false;
result=(int)SDL_SetVideoMode(width, height, 32, SDL_OPENGL);
if(result!=0)
return false;

glViewport(0, 0, 640, 480);
glMatrixMode(GL_PROJECTION);

glOrtho(-100, 100, -100, 100, -100, 100);

glMatrixMode(GL_MODELVIEW);

for(;;)
{
glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT);
input->update();

if(input->isKeyPressed(SDLK_ESCAPE))
break;

if(input->isKeyPressed(SDLK_a))
{
printf("origin=(%f, %f, %f)\n", v[0], v[1], v[2]);
camera.MoveFoward(0.01);
}

camera.getMatrix44(temp);
camera.getOrigin(v);
glPushMatrix();
glMultMatrixf(temp);
glTranslatef(-v[0], -v[1], -v[2]);

glBegin(GL_TRIANGLES);
glColor3f(1.0, 0.0, 0.0);
glVertex3f(-0.5, 0.5, 0.0);
glVertex3f(-0.5, -0.5, 0.0);
glVertex3f(0.5, -0.5, 0.0);
glVertex3f(0.5, 0.5, 0.0);
glEnd();

glPopMatrix();
render->update();
}

There are only relevant pieces of code. Any doubt about it tell me. I have used gluLookat too and some other tricks but all of them do the same stuff, make the cube desappear instead of move foward.

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I think that your issue is that you do not take into account the framerate your box is running at, the code

camera.MoveFoward(0.01);

moves your cube forward by 0.01 units per _frame_, but you do see the cube before you hit 'a' right? use the SDL call SDL_GetTicks() {it returns how much time in thousandth's of a second has passed since SDL was started up} to see how much time has passed between frames and use that value to scale your 0.01f.

Best Regards

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Well, the problem is that i have used:

camera.MoveFoward(0.001);
and
camera.MoveFoward(0.0001);

both with the saame effect.
At that rate there should be something visible.

Also, after debugging, i saw that the square desappear when the counter reachers -1.00
Before that it stays.

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Might not be helpful, but

You're moving your camera, not the box.

You're using glOrtho(), which I've only ever used for 2d and masking. I'd use gluPerspective(). Make sure your box isn't outside of the clipping planes.

If I'm thinking straight, the glOrtho() thing doesn't show perspective (objects don't get smaller the further they are from the camera). Think of it as a cuboid as opposed to a pyramid with the top cut off.

so that's possibly it.

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Hmm... if I may ask, what are you initialising "front"'s values to?

Other than that, I would tend to agree with kRogue - it may simply be that your code is executing so fast that even using a step of 0.1*front or 0.0001*front is moving the object out of your viewing space too quickly.

I would second his suggestion to base the degree of forward movement on the time between frames - even if this is not your current problem, it should produce movement that better matches what I imagine you expect.

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Sorry, i forgot the constructor:

cDPI_Frame::cDPI_Frame(){	origin[0]=0.0;	origin[1]=0.0;	origin[2]=0.0;		front[0]=0.0;	front[1]=0.0;	front[2]=-1.0;		up[0]=0.0;	up[1]=1.0;	up[2]=0.0;		}

By the way, i want to say that the FPS i not an issue. As you can see there is a printf just above the MoveFoward member. And believe me, it doen't go THAT fast.

I can agree with the theory of the ortho-problem.
I will try moving to the sides instead and after i have what i want i may change it to perspective.

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Hmm... You do make a good point about the frame rate.

However, I do notice something else:

You appear to be moving the object along the z-axis, if I'm not much mistaken, in which case, since you are viewing the scene orthographically, it seems to me now that TheUmpteenth makes a good point. I would expect, given this, that you should see the object sit still for a moment, and then disappear as it exits the viewing volume. Note that, as TheUmpteenth says, an orthographic projection should not show any perspective.

If it disappears immediately, try changing the third value of "front" to "1" - if it remains on-screen for a longer period and then disappears, then you would seem to be moving the object behind the camera on the first step (since it would seem that you have the camera and object placed at the same point in space).

If you want the object to move across the screen, then I believe that you should be moving along x or y (or both, of course), instead of z, as you appear to be doing now.

If you want the object to become smaller as it moves "into" the screen, then I would recommend either a perspective projection or simply scaling the object.

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Well, actually what i wanted to make is to move with the arrows foward and backward, right and left, so i can "explore" the polygons.
Of course, this are just some tests before the real thing.
I'll try moving to other sides and changing to perspective.
thanks

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Well, yep the problem was that, the orthographic projection. I tried moving to the sides and it worked.

Thanks for your help.

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np :) glad you got it fixed

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text:
# Set Intersection/Examples/Overlapping Integer Sets

## Example of Set Intersection

Let:

 $\displaystyle A$ $=$ $\displaystyle \set {x \in \Z: 2 \le x}$ $\displaystyle B$ $=$ $\displaystyle \set {x \in \Z: x \le 5}$

Then:

$A \cap B = \set {2, 3, 4, 5}$

and so is finite.