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https://kunduz.com/questions-and-answers/for-the-reaction-2ss30g-2sog-if-594-g-of-s-is-reacted-with-100-g-of-o-how-many-grams-of-so3-will-be-produced-297-g-742-g-167-g-148-g-none-of-these-64070/ | 1,721,890,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00315.warc.gz | 289,462,687 | 32,225 | Question:
# For the reaction 2S(s)+30₂(g) →→2SO₂(g) if 5.94 g of S is
Last updated: 8/7/2022
For the reaction 2S(s)+30₂(g) →→2SO₂(g) if 5.94 g of S is reacted with 10.0 g of O₂, how many grams of SO3 will be produced? 29.7 g 7.42 g 16.7 g 14.8 g none of these | 116 | 262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.810517 |
http://forums.wolfram.com/mathgroup/archive/2008/Aug/msg00638.html | 1,603,316,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00090.warc.gz | 43,192,740 | 7,318 | Re: unwanted Complex result
• To: mathgroup at smc.vnet.net
• Subject: [mg91596] Re: [mg91589] unwanted Complex result
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 28 Aug 2008 07:39:27 -0400 (EDT)
• Reply-to: hanlonr at cox.net
```soln = Solve[(Sqrt[speed^2 - windx^2] + windy)/a
speed^2 + b speed + c == distance/height, speed];
Select[soln /.
{windx -> 1, windy -> 1, a -> 1, b -> 1, c -> 1,
distance -> 20, height -> 5} // N,
FreeQ[#, Complex] &]
{{speed->-1.46762},{speed->1.11901}}
Bob Hanlon
---- John Rivers <first10 at btinternet.com> wrote:
=============
Solve[
(Sqrt[speed ^ 2 - windx ^ 2] + windy) / a speed ^ 2 + b speed + c ==
distance / height
, speed]
this works but returns 4 solutions that all return complex numbers
I need a real number
how do I:
- convert a complex result into real number
or
- solve the equation to give only real numbers
```
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• Next by thread: Re: unwanted Complex result | 344 | 1,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-45 | latest | en | 0.728543 |
https://www.thenational.academy/teachers/programmes/maths-primary-ks2-l/units/3d-shape-caba/lessons/using-and-applying-knowledge-of-the-properties-of-3d-shapes-c8vk8d | 1,718,418,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00886.warc.gz | 924,751,294 | 36,071 | Year 4
Year 4
# Using and applying knowledge of the properties of 3D shapes
## Lesson details
### Key learning points
1. In this lesson, we will identify the names and properties of a range of 3D shapes. We will apply this understanding to real world 3D based problem solving.
### Licence
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
## Video
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## Worksheet
Share with pupils
## Exit quiz
Share with pupils
### 5 Questions
Q1.
What is the definition of an 'edge' of a 3-D shape?
An edge is a corner.
An edge is the longest length on the shape.
Correct answer: An edge is where two faces of the 3-D shape meet.
An edge makes up one surface of a 3-D shape.
Q2.
What is the definition of a 'vertex'?
Correct answer: A vertex is the corner point where edges meet.
A vertex is the longest length on the shape.
A vertex is where two faces of the 3-D shape meet.
A vertex makes up one surface of a 3-D shape.
Q3.
Which of the following shapes has 6 faces and 8 vertices?
Cylinder
Sphere
Square-based pyramid
Q4.
What is the difference between a cube and a cuboid?
A cube has fewer edges than a cuboid.
A cube is always larger than a cuboid.
Correct answer: A cuboid has faces which are rectangular whereas a cube has faces which are square.
A cuboid is always larger than a cube.
Q5.
Which of the following 3-D shapes is the same shape as a tin of soup?
Cone | 372 | 1,508 | {"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} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.94367 |
https://math.stackexchange.com/questions/406398/standard-error-for-ratio-of-two-poisson-random-variables | 1,576,005,529,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528490.48/warc/CC-MAIN-20191210180555-20191210204555-00011.warc.gz | 448,305,669 | 29,237 | # standard error for ratio of two poisson random variables
I'm collecting data for two Poisson random variables, that are completely independent of each other. My current analysis method yields 95% (or 90% or other) confidence intervals for the two $\lambda_1$, $\lambda_2$ for the two Poissons.
My question: how can I come up with a confidence interval for the ratio of the two lambdas?
If $\lambda_1$ belongs to $[a_1,b_1]$ with probability at least $p_1$ and $\lambda_2$ belongs to $[a_2,b_2]$ with probability at least $p_2$ for some $0<a_1<b_1$ and $0<a_2<b_2$, then, by independence, $\lambda_1/\lambda_2$ belongs to $[a_1/b_2,b_1/a_1]$ with probability at least $p_1p_2$.
For example, $p_1=p_2=90\%$ yields $p_1p_2=81\%$.
You can calculate the CV for each of the two variables ($1/\sqrt{\lambda_1}$) and then add the squared CVs $CV=\sqrt{CV_1^2+CV_2^2}$. | 283 | 865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-51 | latest | en | 0.804448 |
https://www.physicsforums.com/threads/statics-coplanar-force-systems.338211/ | 1,624,350,851,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00154.warc.gz | 835,814,551 | 14,947 | # Statics - Coplanar Force Systems
The problem statement:
My relevant equation:
$\phi$ will be the angle between the X axis and $F_{CO}$
$$\theta = \phi + \arcsin\left(\frac{3}{5}\right)$$
My attempt at a solution:
$$\Sigma F_{x} = 0:$$
$$F_{CO}\cos\phi - F_{BO}\frac{4}{5} = 0$$
$$F_{CO} = \frac{F_{BO}\frac{4}{5}}{\cos\phi}$$
$$\Sigma F_{y} = 0:$$
$$F_{AO} - F_{BO}\frac{3}{5} - F_{CO}\sin\phi = 0$$
Combining terms and substituting the equation found for $\Sigma F_{x} = 0$ into $\Sigma F_{x} = 0:$
$$F_{AO} - \frac{3}{5}F_{BO} - \frac{4}{5}F_{BO}\tan\phi = 0$$
$$9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0$$
$$\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)$$
$$\phi = 65.12^{\circ}$$
$$\theta = 102^{\circ}$$
The published value of $\theta$:
$$\theta = 70.1^{\circ}$$
I don't know what I did wrong.
TIA for any response.
Last edited:
## Answers and Replies
PhanthomJay
Homework Helper
Gold Member
$$9kN - \frac{3}{5}8kN - \frac{4}{5}8kN\tan\phi = 0$$
$$\phi = \arctan\left(\left(9+\frac{24}{5}\right)*\frac{5}{32}\right)$$
that plus 24/5 should be a minus 24/5.
I don't know what I did wrong.
TIA for any response.
you did well, just missed the sign.
Jay, thank you sir! | 507 | 1,228 | {"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": 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-2021-25 | latest | en | 0.766802 |
https://capitalsportsbet.com/ZuluCode5/doc-sports-nfl-point-spreads.html | 1,568,807,617,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00315.warc.gz | 405,529,515 | 6,271 | If the implied probability says that you have to win a bet 40% of the time to break even, and you think you’re likely to win the bet 45% of the time, then there is value in the bet. Remember, the sportsbook pays you more the less likely something is to happen. This means you’ll be getting paid as if the bet is only going to hit 40% of the time (more money), but the bet actually hits 45% of the time if you are right.
You should already know that the Eagles are the favorite to win and that you should expect less than even money on a correct pick here. You should also know that the Falcons are the favorite, and you should expect better than even money on a correct pick here. Having this in mind every time before you start your calculations will protect you from making a mistake and calculating the completely wrong direction.
We already mentioned how using multiple betting sites allows you to take advantage of multiple bonus offers. That's not the only benefit either. Since point spreads vary between sites, one of the best ways to beat these wagers is to compare the different spreads in order to find which one is the most favorable. This doesn't take nearly as long as you might think, and it will make a huge difference to your bottom line over time.
If you want to bet on football, then you have plenty of options. There are not only lots of games you can bet on, there are also lots of different types of wagers you can place. Point spreads and totals are the most popular types, by quite a distance, and many football bettors stick solely to those. This isn't really the ideal approach, as some of the other wagers can be very useful in the right circumstances.
There is no magic formula for moneyline betting, you’ll simply need to pick your spots wisely and balance your risk versus your potential reward. In general, I think this means taking higher upside picks, such as underdogs rather than taking large favorites. However, bettors should analyze each game independently looking for value in both favorite and underdog moneylines.
When two teams square up for a matchup, whether that be on the gridiron or on the basketball court, one team is typically better than the other (for whatever reason you want to believe). Since sportsbooks are in the business of making money, they tag the better team with a point spread, thus making them the "favorites" to win that specific game. Normally, the favorite has a few favorable factors working for them like playing at home or being well rested or playing a revenge game against a team that previously beat them. Every factor counts in the world of betting, and it's up to you to decide if the "favorite' can, in fact, cover the point spread.
```It's also important to consider whether or not there's any correlation between the point spread and the betting total. If they are, a parlay wager is a good way to get maximum value. For example, a college football point spread +24.5 parlayed with under 48 points in the same game might be a great parlay bet. If the +24.5 team covers the point spread, then there's an increased chance that the game also goes under the posted total of 48.
```
Which brings me to my next point. If you are serious about getting into sports betting, it is vital to have more than one sportsbook to make a wager at. Shopping around for the best lines will help your bankroll and you will be able to turn a bigger profit. If you see a pair of sneakers for \$110 at one store, and the exact same pair is \$102.99 at another store - which store are you buying them from?
Fantasy/Virtual Matches or Head to Heads are implicit matchups where the performances of two or more opponents which are not directly confronting each other in the same event are compared. Settlements will be based on the number of times each participant records a predefined occurrence (e.g. goals) in the respective match. The following criteria will be used to determine the settlement of these type of offerings:
This should hopefully make perfect sense to you. Successful betting, on any sport, is all about finding value, so you should always look to get the best value that you can. If a moneyline wager offers the best value on a football game, then that's the wager you should be placing. If a point spread seems the best option, then go down that route. There may even be occasions where it's viable to place both wagers on the same game.
Piggybacking on the simplicity of moneyline bets is the ease with which you can properly assess value. Now, you’ll notice that it doesn’t say “Easy to Find Value,” and that is because it’s never easy to find value in sports betting. If it were easy, everyone would be doing it for a living. What it says, though, is that it is easier to find value with moneyline bets because of the simplicity.
```Absolutely. When the lines go up for the NFL, or for the first game of the NCAA men’s basketball tournament, there are several days in between the open and the game itself where movement can take place. You’ll find that the betting public tends to pile in on their favorite teams once they get home from work on Friday. You can anticipate these line movements and time your bet accordingly to take advantage.
```
The 2-way moneyline is what most North American bettors would simply refer to as “the moneyline”. This is one of the most common wagering options where the user bets which side will win the game straight up. (A draw or tie results in a push with the 2-way moneyline.) The term is sometimes highlighted during soccer betting to differentiate from the 3-way moneyline - a more popular option with the draw added as a wagering option.
Notice that point spreads adjust the score for the favorite team. This is easiest to see with an example: If the New York Knicks are playing the Boston Celtics, and Boston is favored to win by a 4-point spread, then a bet on Boston only pays out if Boston wins by more than 4 points. A bet on New York pays out if New York wins or if they lose by less than 4 points.
For beginners, the moneyline bet on an underdog is a great choice as it will allow bettors to win 50% of the bets and still earn a profit. Each of these bets have benefits and drawbacks. With moneylines, punter shave the appeal of betting on a winning side. This is a great choice for die-hard fans. Many people find it more appealing to bet on a winner of a game instead of betting on the end score of the game. Spread betting does offer some nice benefits. They are simple yes or no bets. Either the team covers the spread or they don’t. However, moneyline bets typically offer the chance to win more than is bet, so these are often the choice for many bettors online.
Remember earlier when we said that most point spread bets in basketball pay out at -110? Well, this is where the vig is located. Sportsbooks will work to get equal amounts of money on both sides of a game and make their money off of the vig. If they are successful in doing so, it does not matter to them who wins the game. For example, let's look at our earlier example. Here are what the odds would look like at the sportsbook:
Here you can see that the Rams are +3.5, while the Cowboys are -3.5. So for this example the Cowboys are 3.5 point favorites, while the Rams are underdogs of 3.5 points. If you were to bet on St Louis you would need them to lose by 3 or fewer points or just win the game outright. If you were to bet on Dallas you would need the Cowboys to win by 4 or more points.
### Apply the spread. In point-spread betting, the actual final score of the game is only the starting point. Say Chicago beats Detroit 24-17. Because Chicago was the favorite, you subtract the point spread from its final score. That's the purpose of the minus sign in the spread. The spread was 6, so you take 6 points away from Chicago's point total, giving you an "adjusted" score of Chicago 18, Detroit 17. If you'd bet on Chicago, you'd have won the bet. Now, say Chicago won the game 20-17. Subtracting the 6 points from Chicago's total gives you a final score of Detroit 17, Chicago 14. If you'd bet on Chicago, you'd have lost.
As we mentioned, moneyline/win bets take into account who the favorites and who the underdogs are and will pay out winning bets accordingly. Here’s a quick example that will make this clear. Imagine that Mike Tyson (one of the greatest boxers of all time) is going to fight against an 80-year-old man. If the sportsbook let you bet on either side of the fight and paid you the same, would that be fair?
In this guide, we’re going to teach you literally everything you’ve ever wanted to know about moneyline bets and then a whole lot more. Whether you are a beginner or a seasoned bettor, we’ve got something here for you. We’ll walk you through the basics of what a moneyline bet is, why you would want to make one, and how to interpret the different numbers, payouts, and presentation formats you’ll see.
You don’t have layers of complexity to fight through to see if your prediction is a positive expected value move (one that is going to make you money). With some simple mathematical calculations, you can figure out whether or not there is value in a bet. Even if you don’t like math and would prefer not to use it when assessing value and making your picks, it’s still much easier to “eyeball” value with moneyline bets because of the simplicity.
As we mentioned in the close of the last section, sportsbooks try their best to get the same amount of total money bet on both sides of a game. If they can accomplish this, then they are guaranteed to make a profit no matter who wins or loses the game. The way a sportsbook goes about doing this is by manipulating the point spread to make the less bet side more enticing.
The point spread is essentially a handicap towards the underdog. The wager becomes "Will the favorite win by more than the point spread?" The point spread can be moved to any level to create an equal number of participants on each side of the wager. This allows a bookmaker to act as a market maker by accepting wagers on both sides of the spread. The bookmaker charges a commission, or vigorish, and acts as the counterparty for each participant. As long as the total amount wagered on each side is roughly equal, the bookmaker is unconcerned with the actual outcome; profits instead come from the commissions. | 2,276 | 10,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-39 | latest | en | 0.974107 |
https://nyuscholars.nyu.edu/en/publications/erratum-lower-bounds-for-trace-reconstruction-ann-appl-probab-202 | 1,716,281,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00183.warc.gz | 376,740,695 | 11,930 | # Erratum: LOWER BOUNDS FOR TRACE RECONSTRUCTION (Ann. Appl. Probab. (2020) 30 (503-525) DOI: 10.1214/19-AAP1506)
Nina Holden, Russell Lyons
Research output: Contribution to journalComment/debatepeer-review
## Abstract
Lemma 3.1 asserts that Eyn [Z(yn)]. Exn [Z(xn)] = Θ(n-1/2) and Eyn [Z(yn)] > Exn [Z(xn)] for all sufficiently large n. Our proof was not correct: As Benjamin Gunby and Xiaoyu He pointed out to us, we missed four terms in the computation of equation (3.3). Those terms contribute a negative amount, so the proof is more delicate. Here is a correct proof. The intuition behind the result is that a string with a defect of the type we consider, namely, a 10 in a string of 01's, is likely to cause more 11's in the trace than a string without the defect. Since the defect in yn is shifted to the right as compared to the defect in xn, the defect of yn is slightly more likely to "fall into"the test window {⌈2np + 1⌉, ⋯, ⌊2np + √npq⌋} of the trace than is the defect of xn. More precisely, the difference in probability is of order n-1/2. In the proof below, we make this intuition rigorous. PROOF. We assume throughout the proof that k ∈ {⌈2np + 1⌉, ⋯, ⌊2np + √npq⌋}. Let E(m, k) denote the event that bit m in the input string is copied to position k in the trace. First observe that 'Equation Presented' and 'Equation Presented'. Note that the string xn centered at m is identical to the string yn centered at m + 2, except for two bits at the ends. Therefore, for every m ∈ {k, ⋯, 3n}, we have 'Equation Presented' where o∞ (n) denotes something nonnegative that decays at least exponentially fast in n. Combining this with Pxn [E(m, k)] = o∞ (n) for m < k + 2 or m > 3n yields 'Equation Presented' Setting am: = qp/(1 - q2) = q/(1 +q) if m is even and am: = 0 otherwise, we see that 'Equation Presented' Subtracting this from the previous display gives 'Equation Presented' The second factor in the above summand, modulo an additive error of o∞ (n), represents the difference in probability of the event xk = xk+1 = 1 given E(m, k) for the string xk as compared to a string without any defect. It takes the following explicit form: 'Equation Presented' where ≈ means that we incur an additive error of ±o∞ (n). Now let j0 be a sufficiently large positive integer that 'Equation Presented' Note that j0 depends on q but can be chosen so that it does not depend on n. We suppose in the rest of the proof that n > j0. By (E.1) and (E.2), 'Equation Presented' For m ∈ {2n - 2j0, ⋯, 2n + 2} and with ξ: = k - 2np, we have 'Equation Presented' because for m ∈ {2n - 2j0, ⋯, 2n}, 'Equation Presented' the same result holds for m ∈ {2n + 1, 2n + 2} by a similar estimate. Combining (E.2) and (E.5), we get that the right-hand side of (E.4) is equal to (E.6) 'Equation Presented' Summing the left-hand side of (E.4) over k ∈ {⌈2np + 1⌉, ⋯, ⌊2np + √npq⌋} and using the last display along with Pxn [E(2n, k)] = Θ(n-1/2) and (E.3), we get the lower bounds in the lemma, namely, Eyn [Z(yn)] - Exn [Z(xn)] = Ω(n-1/2) and Eyn [Z(yn)] > Exn [Z(xn)]. It remains to prove the upper bound, namely, Eyn [Z(yn)] - Exn [Z(xn)] = O(n-1/2). Let bm,n denote the absolute value of the right-hand side of (E.2). By (E.1) and (E.2), we have 'Equation Presented'. Now sum over k; (2.7) of Lemma 2.2 yields Σk| Pxn [E(m + 2, k)] - Pxn [E(m, k)]| = O(m-1/2) = O(n-1/2). In addition, Σm bm,n = O(1). Combining these bounds, we arrive at the upper bound of the lemma. □We remark that one can get a more precise bound in (E.6) that gives something positive for all q ∈ (0, 1) simultaneously by not truncating the sum on the right-hand side of (E.1) and by using a more precise version of (E.5). The result, in fact, gives lower and upper bounds for the left-hand side of (E.4) of the form 'Equation Presented'. Finally, we note that in the proof of Proposition 1.4 on page 519, the definitions of X and Y should be slightly modified: c should be √c both times.
Original language English (US) 3201-3203 3 Annals of Applied Probability 32 4 https://doi.org/10.1214/22-AAP1827 Published - Aug 2022
## ASJC Scopus subject areas
• Statistics and Probability
• Statistics, Probability and Uncertainty
## Fingerprint
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## Q. 1.4
Let u = (1, 3) and v = (-2, 4). Calculate (a) u + v; (b) 3u; (c) – v; (d) u – v; and (e) – 3u + 5v.
## Verified Solution
\begin{aligned}&\text { (a) }\mathbf{u}+\mathbf{v}=(1+(-2), 3+4)=(-1,7) \\&\text { (b) }3 \mathbf{u}=3(1,3)=(3,9) \\&\text { (c) }-\mathbf{v}=(-1)(-2,4)=(2,-4) \\&\text { (d) }\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})=(1+2,3-4)=(3,-1) \\&\text { (e) }-3 \mathbf{u}+5 \mathbf{v}=(-3,-9)+(-10,20)=(-13,11)\end{aligned} | 397 | 981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 1, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-27 | latest | en | 0.588652 |
http://oeis.org/A021115 | 1,571,507,989,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697439.41/warc/CC-MAIN-20191019164943-20191019192443-00255.warc.gz | 146,186,832 | 3,975 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A021115 Decimal expansion of 1/111. 1
0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9, 0, 0, 9 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Period 3: repeat [0, 0, 9]. - Joerg Arndt, Sep 29 2015 LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,1). FORMULA From Alexander R. Povolotsky, Sep 29 2015: (Start) G.f.: 9*x^2/(1 - x^3). a(n) = (9/2)*( (n-1)*n mod 3 ) = 4*(n mod 3) - 2*((n+1) mod 3) + ((n+2) mod 3), formulas suggested by Giovanni Resta. a(n) = 3*( 1 + cos(2*(n+1)*Pi/3) + cos(4*(n+1)*Pi/3) ). a(n) = a(n + 3) for n>2. a(n) = A056992(n+1) - (3*(n+1)^4+3*(n+1)^6+4*(n+1)^8) mod 9. (End) a(n) = 9*A022003(n). - Robert Israel, Sep 30 2015 MAPLE seq(op([0, 0, 9]), i=1..100); # Robert Israel, Sep 30 2015 MATHEMATICA PadRight[{}, 100, {0, 0, 9}] (* Wesley Ivan Hurt, Jun 30 2016 *) PROG (PARI) 1/111.0 \\ Michel Marcus, Sep 29 2015 (PARI) a(n) = 9*(n*(n-1)%3)/2; vector(100, n, a(n-1)) \\ Altug Alkan, Sep 30 2015 (MAGMA) &cat[[0, 0, 9]: n in [0..35]]; // Vincenzo Librandi, Sep 29 2015 CROSSREFS Cf. A022003, A056992. Sequence in context: A296458 A199870 A132267 * A073052 A230068 A176908 Adjacent sequences: A021112 A021113 A021114 * A021116 A021117 A021118 KEYWORD nonn,cons,easy AUTHOR EXTENSIONS Edited by Bruno Berselli, Dec 15 2015 STATUS approved
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Last modified October 19 13:01 EDT 2019. Contains 328222 sequences. (Running on oeis4.) | 956 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-43 | latest | en | 0.552401 |
https://testbook.com/learn/maths-subsets/ | 1,670,460,016,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00332.warc.gz | 601,269,916 | 51,401 | # Subsets: Meaning, Types, Properties with Solved Examples and More
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A set is a well-defined group of numbers, objects, alphabets, or any items arranged in curly brackets whereas a subset is a part of the set.
The components of sets could be anything such as a group of real numbers, a group of integers, variables, a group of all-natural numbers, constants, whole numbers, etc. Let us discuss subsets in this article with their definition, subset symbol, types, examples and more. Also, learn about sets, subsets and supersets with their difference for more clarity.
Subset example; if a set P is a combination of odd numbers and set Q consists of {1,5,7}, then set Q is said to be a subset of set P and is denoted by the symbol Q⊆P. Here P is the superset of Q. Before heading towards the subsets let us take a brief overview of sets and its various types.
The concept of set assists as a significant part of mathematics and is being used in almost all branches of mathematics. The study of geometry, sequences, probability, etc. demands the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor.
#### Types of Sets in Mathematics
Sets are categorised into distinct types. Some of these are singleton, infinite, finite, empty, etc. Let us look at each of them one by one.
Null Set
A set that does not contain any element is termed a null set. It is indicated by ϕ or {}.
Example: A = {x : x is a prime number lying between 90 and 96} = ϕ
Singleton Set
A set that has only one element is termed a singleton set.
Example: A = {0}
Finite Set
A set that contains a finite number of elements is named a finite set.
Example: A = {2, 3, 5}
Infinite Set
A set that has an infinite number of components is named an infinite set.
Example: Z = set of all integers = {0, ± 1, ± 2, …….}
Equivalent Sets
Any two sets are stated to be equivalent sets if their cardinality i.e the number of components present in both sets are the same.
Example: A = {2, 3, 5} and B = {3, 5, 7} are equivalent sets since their cardinality is the same which is 3.
Equal Sets
Any two sets are declared to be equal sets if and only if they are equivalent and as well as their elements are identical.
Example: A = {2, 3, 5} and B = {2, 3, 5} are equal sets since they are equivalent sets as they are possessing the same number of elements plus the elements existing in both the sets are the same.
Subset
Consider A and B to be two sets. If each element of A is present in set B then A is designated a subset of B and it is denoted by the notation A ⊆ B.
Example: A = {2, 3, 5} and B = {2, 3, 5, 7}. Here we can recognize that all the components of set A are present in set B. So, A ⊆ B.
Proper Subset
Consider A and B to be two sets. Then A is declared to be a proper subset of B if A is a subset of B and A is not equivalent to B. It is expressed as A ⊂ B.
Example: A = {2, 3, 5} and B = {2, 3, 5, 7}.Here we can see that A is a subset of B and A is not equal to B. So, A ⊂ B.
Power Set
Let A be set, then the set of all the possible subsets of A is called the power set of A and is denoted by P(A).
i.e P(A) = {X : X ⊆ A}
Universal Set
Usually, we deal with the components and subsets of a basic set that is relevant to that particular circumstances. For instance, while studying the system of numbers, we are interested in the set of natural numbers and their subsets such as the set of all prime numbers, the set of all even numbers, the set of all odd numbers, and so on.
This basic set is called the “Universal Set”. The universal set is normally indicated by U, and all its subsets by the letters A, B, C, etc. For example, for the set of all integers, the universal set can be the set of rational numbers, in human population studies, the universal set comprises all the people in the world.
## What is a Subset?
A subset, as the name implies, is a subgroup of any set. Consider two sets, A and B Mathematically speaking, A will be a subset of B if and only if all the components of A are present in B. We can also say that A is contained in B.
To understand the subset definition more clearly, consider a set P such that P comprises the names of all the cities of a country. Another set Q including the names of cities in your region. Here Q will be a subset of P. This is because all the cities in your region would also be cities of your country; hence, Q is a subset of P. There are only a definite number of distinct/unique subsets for any set, therefore the remaining are irrelevant and repetitive.
Subset Example 1:
A subset as far as our understanding is a set contained in another set. It is like one can pick ice cream from the following flavours:{mango, chocolate, butterscotch}
• You can take any one flavor {mango}, {chocolate}, or {butterscotch},
• Or any two flavors: {mango chocolate}, {chocolate, butterscotch}, or {mango, butterscotch}.
Learn the various concepts of the Binomial Theorem here.
### What is a Subset Meaning in Maths?
A Set 1 is supposed to be a subset of Set 2 if all the components of Set 1 are also existing in Set 2. In other words, set1 is included inside Set2.
• If Set1={A,B,C} and Set2={A,B,C,D,E,F,G,H,I} then we can say that Set1 is a subset of Set2 as all the elements in set 1 are available in set 2.
### Subset Symbol
In the set theory, a subset is expressed by the symbol and addressed as ‘is a subset of’. Applying this symbol we can represent subsets as follows:
P ⊆ Q; which is read as Set P is a subset of Set Q.
Note point: A subset can be identical to the set i.e, a subset can contain all the elements that are present in the set.
Subset examples:
Example 2: Find whether P is a subset of Q?
P = {set of even digits}, Q = {set of whole numbers}
Solution
The set of even numbers can be represented as:
P = {0, 2, 4, 6, 8, 10, 12 …}
Similarly, the set of all whole numbers can be represented as:
Q = {0, 1, 2, 3, 4, 5, 6, 7…}
From the set components of P and Q, we can figure out that the elements of P are present in the set Q. Therefore P is a subset of Q.
Example 3: Determine whether P is a subset of Q.
P = {1, 3, 5, 7}, Q= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14…..}
It can be analyzed from the set elements that P’s elements relate to set Q. Hence P is a subset of Q.
Example 4: Conclude whether X is a subset of Y.
X = {All writing material in a stationary workshop}, Y = {Pencils}
Set X involves pens, sketch pens, markers, pencils, notepads, etc. Whereas set Y only carries pencils. So we cannot state that all X’s elements are present in Y, which is a requirement for X to be a subset of Y. In this particular case, we can state that Y is a subset of X, but X is not a subset of Y.
Example 5: Determine whether A is a subset of B.
A = {Toyota}, B = {All brands of cars}
Set B covers all brands of cars; Maruti Suzuki, Hyundai, Toyota Mahindra, Tata Motors, Mercedes Benz, etc. Moreover, A is a set of Toyota. Then we can say that all elements of A are incorporated in B. Hence, A is a subset of B.
### All Subsets of a Set
The no. of subsets of a set of any set consisting of all likely sets including its components and the null set. Let us learn with an example.
Example 6: Find all the subsets of set P = {2, 4, 6, 8}
Solution: Given, P = {2, 4, 6, 8}
Number of subsets of P are =
• {}
• {2}, {4}, {6}, {8},
• {2, 4}, {2, 6}, {2, 8}, {4, 6},{4, 8}, {6, 8},
• {2, 4, 6}, {4, 6, 8}, {2, 6, 8}, {2, 4, 8}
• {2, 4, 6, 8}
## Types of Subsets
Subsets are primarily classified into:
• Proper Subset
• Improper Subsets
### Proper Subset
Any set say “P” is supposed to be a proper subset of Q if there is at least one element in Q, which is not available in set P. That is, a proper subset is one that contains a few components of the original set.
In other words, we can say that if P and Q are unequal sets and all elements of P are present in Q, then P is the proper subset of Q.
It is also termed a strict subset.
Proper Subset Examples
Example 7: Is P a proper subset of Q where P = {1, 3, 7, 8} and Q = {1, 3, 7, 8}?
Solution:
The answer would be no, P is not a proper subset of Q as both are identical, and Q does not have any unique element, which is not existing in P.
Example 8: Is X a proper subset of Y when X = {1, 6} and Y = {1, 4, 6, 8}?
Solution:
The answer would be yes, X is a proper subset of Y as all the elements of X are present in Y and X is not equal to Y as well.
Proper Subset Symbol
A proper subset is expressed by and is addressed as ‘is a proper subset of’. For example:
P ⊂ Q.
### Improper Subset
Suppose two sets, X and Y then X is an improper subset of Y if it includes all the elements of Y. This implies that an improper subset comprises every element of the primary set with the null set. The improper subset symbol is ⊆. For example:
P ⊆Q
Example 9: If set P = {2, 3, 5}, then determine the number of subsets, proper subsets and improper subsets.
Number of subsets: {2}, {3}, {5}, {2,3}, {3, 5}, {2,5}, {2, 3, 5} and Φ or {}.
Proper Subsets: {}, {2}, {3}, {5}, {2,3}, {3, 5}, {2,5}
This can be denoted as {}, {2}, {3}, {5}, {2,3}, {3, 5}, {2,5} ⊂ P.
Improper Subset: {2, 3, 5}.
This can be denoted as {2, 3, 5} ⊆ P.
Learn the concepts of Sequences and Series here.
## Subset Formula
The set theory symbols were explained by mathematicians to represent the collections of objects. If it is required to select n number of elements from a set including N number of elements, it can be performed in $$^NC_n$$ ways.
### Proper Subset Formula
If a set holds “n” elements, then the number of the subset for the given set is $$2^n$$ and the number of proper subsets of the provided subset is calculated by the formula $$2^n-1$$.
Example 10: For a set P with the elements, P = {1, 2}, determine the proper subset?
The proper subset formula is $$2^n-1$$ (where n is the number of elements in the set)
P = {1, 2}
Total number of elements (n) in the set=2
Hence the number of proper subset=$$2^2−1$$ =3
Therefore the total number of proper subsets for the given set is { }, {1}, and {2}.
Example 11: For the given set determine the power set?
Set Y={2,3,6}
Total number of components in the set Y=3
The power set of Y is:
P(Y)={},{2},{3},{6},{2,3},{3,6},{2,6},{2,3,6}
P(Y)=$$2^n$$
Substituting n=3
P(X)=$$2^{3}$$=8
Also, learn about Vector Algebra here.
## How to Represent Subsets?
We are quite clear with what a subset is, now let us check some of the representations for the same. A subset, like any other set, is addressed with its elements inside curly braces.
Hence, consider two sets, X and Y:
• X ⊆ Y
The above notation indicates that X is a subset of Y.
• X ⊈ Y
Here the notation indicates that X is not a subset of Y.
• X ⊂ Y
If X is a proper subset of Y, then it is expressed by the above notation.
• X⊄ Y
If X is not a proper subset of Y, then we address it by the above notation.
## Properties of Subsets
Some of the important properties of subsets are as follows:
• Every set is said to be a subset of the provided set itself. Either the set is finite or infinite, a set itself will be taken as the subset of itself.
For example, for a finite set A = {3,6}, all the possible subsets for the given data is:
A ={}, {3}, {6}, {3,6}.
As you can recognise that, we have included a subset with the identical elements as the initial set to satisfy the property.
• We can state, an empty set is regarded as a subset of every set.
For example, take a finite set B = {a, b}, so all the possible subsets of this set are:
A = {}, {a}, {b}, {a, b}
• If P is a subset of Q, then we can state that all the elements in P are available in Q.
Consider a set P= {2, 6, 9} and another set as Q = {2, 3, 4, 5, 6, 7, 8, 9}. Here we can say P is a subset of Q as all the elements of P are present in Q.
• If a set A is a subset of set B then we can assume that B is a superset of A.
• There are $$2^n$$ subsets and $$2^n-1$$ proper subsets for a provided set of data.
## Representation of Subsets through Venn Diagrams
To understand the links between different sets, the Venn diagram is the most proper tool to reflect logical connections among certain sets. They are employed abundantly for the design of sets, more importantly for the finite sets. A Venn diagram indicates the sets as the area inside a circular target with the elements as points inside the area.
As subsets usually involve two sets, we can easily practice a Venn diagram to illustrate and visualize them.
Example 12: For the set X = {1, 3, 6} and set Y = {1, 3, 6, 9, 12, 15, 18} draw the venn diagram.
The Venn diagram illustration of sets X and Y are as follows:
As we can recognise from the diagram that X, surrounded by a region denoted by its set, is a portion of region Y. Each area has its elements expressed as points inside the region.
Check more topics of Mathematics here.
## Difference between Proper Subset and Superset
A common element of confusion among the students in set theory during the beginning is the difference between a proper subset and a superset. So let us understand them with an example:
As discussed in the article, a set P is a proper subset of Q if Q possesses at least one component that is not present in P. It is indicated by the symbol ⊂.
Whereas Q will be the superset of P if and only if all the elements existing in P are a part of Q, which states that Q is greater in size when compared to P. If P denotes the proper subset of Q, then Q will be the superset of P. Denoted by the symbol ⊇.
Example 13: For the given two sets X = {2, 4, 6, 8} and Y= {2, 4, 6} check if X is a superset of Y.
For X to be a superset of Y, it requires to hold all the elements present in set Y. As we can notice from the given data that X has all the elements that are present in Y. Also, Y is a proper subset of X; hence, X is Y’s superset.
We hope that the above article on Subsets is helpful for your understanding and exam preparations. Stay tuned to the Testbook app for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
If you are checking Subsets article, also check the related maths articles in the table below: Properties of triangles Conditional probability Diagonal matrix Power set Inverse trigonometric functions Signum function
## Subsets FAQs
Q.1 What is a subset?
Ans.1 A subset, as the name implies, is a subgroup of any set. Consider two sets, A and B Mathematically speaking, A will be a subset of B if and only if all the components of A are present in B. We can also say that A is contained in B.
Q.2 What is a proper subset?
Ans.2 Any set say “P” is supposed to be a proper subset of Q if there is at least one element in Q, which is not available in set P. That is, a proper subset is one that contains a few components of the original set.
Q.3 How to determine the no of subsets of a set?
Ans.3 If a set holds “n” elements, then the number of the subset for the given set is $$2^n$$ .
Q.4 What is an improper subset?
Ans.4 Suppose two sets, X and Y then X is an improper subset of Y if it includes all the elements of Y. This implies that an improper subset comprises every element of the primary set with the null set.
Q.5 What is the proper subset symbol?
Ans.5 A proper subset is expressed by ⊂ and is addressed as ‘is a proper subset of’. For example: P ⊂ Q. | 4,213 | 15,460 | {"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-2022-49 | latest | en | 0.942441 |
https://www.javatpoint.com/r-operators | 1,637,992,582,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00381.warc.gz | 933,868,757 | 12,605 | # Operators in R
In computer programming, an operator is a symbol which represents an action. An operator is a symbol which tells the compiler to perform specific logical or mathematical manipulations. R programming is very rich in built-in operators.
In R programming, there are different types of operator, and each operator performs a different task. For data manipulation, There are some advance operators also such as model formula and list indexing.
There are the following types of operators used in R:
## Arithmetic Operators
Arithmetic operators are the symbols which are used to represent arithmetic math operations. The operators act on each and every element of the vector. There are various arithmetic operators which are supported by R.
S. No Operator Description Example
1. + This operator is used to add two vectors in R. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a+b)
```
It will give us the following output:
``` [1] 13.0 8.3 5.0
```
2. - This operator is used to divide a vector from another one. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a-b)
```
It will give us the following output:
``` [1] -9.0 -1.7 3.0
```
3. * This operator is used to multiply two vectors with each other. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a*b)
```
It will give us the following output:
``` [1] 22.0 16.5 4.0
```
4. / This operator divides the vector from another one. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a/b)```
It will give us the following output:
``` [1] 0.1818182 0.6600000 4.0000000
```
5. %% This operator is used to find the remainder of the first vector with the second vector. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a%%b)
```
It will give us the following output:
``` [1] 2.0 3.3 0
```
6. %/% This operator is used to find the division of the first vector with the second(quotient).
``` a <- c(2, 3.3, 4)
b <- c(11, 5, 3)
print(a%/%b)
```
It will give us the following output:
``` [1] 0 0 4
```
7. ^ This operator raised the first vector to the exponent of the second vector. a <- c(2, 3.3, 4)
``` b <- c(11, 5, 3)
print(a^b)
```
It will give us the following output:
``` [1] 0248.0000 391.3539 4.0000
```
## Relational Operators
A relational operator is a symbol which defines some kind of relation between two entities. These include numerical equalities and inequalities. A relational operator compares each element of the first vector with the corresponding element of the second vector. The result of the comparison will be a Boolean value. There are the following relational operators which are supported by R:
S. No Operator Description Example
1. > This operator will return TRUE when every element in the first vector is greater than the corresponding element of the second vector.
``` a <- c(1, 3, 5)
b <- c(2, 4, 6)
print(a>b)
```
It will give us the following output:
``` [1] FALSE FALSE FALSE
```
2. < This operator will return TRUE when every element in the first vector is less then the corresponding element of the second vector.
``` a <- c(1, 9, 5)
b <- c(2, 4, 6)
print(a<b)
```
It will give us the following output:
``` [1] FALSE TRUE FALSE
```
3. <= This operator will return TRUE when every element in the first vector is less than or equal to the corresponding element of another vector.
``` a <- c(1, 3, 5)
b <- c(2, 3, 6)
print(a<=b)
```
It will give us the following output:
``` [1] TRUE TRUE TRUE
```
4. >= This operator will return TRUE when every element in the first vector is greater than or equal to the corresponding element of another vector.
``` a <- c(1, 3, 5)
b <- c(2, 3, 6)
print(a>=b)
```
It will give us the following output:
``` [1] FALSE TRUE FALSE
```
5. == This operator will return TRUE when every element in the first vector is equal to the corresponding element of the second vector.
``` a <- c(1, 3, 5)
b <- c(2, 3, 6)
print(a==b)
```
It will give us the following output:
```[1] FALSE TRUE FALSE
```
6. != This operator will return TRUE when every element in the first vector is not equal to the corresponding element of the second vector.
``` a <- c(1, 3, 5)
b <- c(2, 3, 6)
print(a>=b)
```
It will give us the following output:
``` [1] TRUE FALSE TRUE
```
## Logical Operators
The logical operators allow a program to make a decision on the basis of multiple conditions. In the program, each operand is considered as a condition which can be evaluated to a false or true value. The value of the conditions is used to determine the overall value of the op1 operator op2. Logical operators are applicable to those vectors whose type is logical, numeric, or complex.
The logical operator compares each element of the first vector with the corresponding element of the second vector.
There are the following types of operators which are supported by R:
S. No Operator Description Example
1. & This operator is known as the Logical AND operator. This operator takes the first element of both the vector and returns TRUE if both the elements are TRUE.
``` a <- c(3, 0, TRUE, 2+2i)
b <- c(2, 4, TRUE, 2+3i)
print(a&b)
```
It will give us the following output:
``` [1] TRUE FALSE TRUE TRUE
```
2. | This operator is called the Logical OR operator. This operator takes the first element of both the vector and returns TRUE if one of them is TRUE.
``` a <- c(3, 0, TRUE, 2+2i)
b <- c(2, 4, TRUE, 2+3i)
print(a|b)
```
It will give us the following output:
``` [1] TRUE TRUE TRUE TRUE
```
3. ! This operator is known as Logical NOT operator. This operator takes the first element of the vector and gives the opposite logical value as a result.
``` a <- c(3, 0, TRUE, 2+2i)
print(!a)
```
It will give us the following output:
``` [1] FALSE TRUE FALSE FALSE
```
4. && This operator takes the first element of both the vector and gives TRUE as a result, only if both are TRUE.
``` a <- c(3, 0, TRUE, 2+2i)
b <- c(2, 4, TRUE, 2+3i)
print(a&&b)
```
It will give us the following output:
``` [1] TRUE
```
5. || This operator takes the first element of both the vector and gives the result TRUE, if one of them is true.
``` a <- c(3, 0, TRUE, 2+2i)
b <- c(2, 4, TRUE, 2+3i)
print(a||b)
```
It will give us the following output:
``` [1] TRUE
```
## Assignment Operators
An assignment operator is used to assign a new value to a variable. In R, these operators are used to assign values to vectors. There are the following types of assignment
S. No Operator Description Example
1. <- or = or <<- These operators are known as left assignment operators.
``` a <- c(3, 0, TRUE, 2+2i)
b <<- c(2, 4, TRUE, 2+3i)
d = c(1, 2, TRUE, 2+3i)
print(a)
print(b)
print(d)
```
It will give us the following output:
``` [1] 3+0i 0+0i 1+0i 2+2i
[1] 2+0i 4+0i 1+0i 2+3i
[1] 1+0i 2+0i 1+0i 2+3i
```
2. -> or ->> These operators are known as right assignment operators.
``` c(3, 0, TRUE, 2+2i) -> a
c(2, 4, TRUE, 2+3i) ->> b
print(a)
print(b)
```
It will give us the following output:
``` [1] 3+0i 0+0i 1+0i 2+2i
[1] 2+0i 4+0i 1+0i 2+3i
```
operators which are supported by R:
## Miscellaneous Operators
Miscellaneous operators are used for a special and specific purpose. These operators are not used for general mathematical or logical computation. There are the following miscellaneous operators which are supported in R
S. No Operator Description Example
1. : The colon operator is used to create the series of numbers in sequence for a vector.
``` v <- 1:8
print(v)
```
It will give us the following output:
``` [1] 1 2 3 4 5 6 7 8
```
2. %in% This is used when we want to identify if an element belongs to a vector.
``` a1 <- 8
a2 <- 12
d <- 1:10
print(a1%in%t)
print(a2%in%t)
```
It will give us the following output:
``` [1] FALSE
[1] FALSE
```
3. %*% It is used to multiply a matrix with its transpose.
``` M=matrix(c(1,2,3,4,5,6), nrow=2, ncol=3, byrow=TRUE)
T=m%*%T(m)
print(T)
```
It will give us the following output:
``` 14 32
32 77
```
Next TopicR If Statement | 2,489 | 7,929 | {"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} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.907525 |
https://studylib.net/doc/5769131/lesson-1--light-intro | 1,550,727,886,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247500089.84/warc/CC-MAIN-20190221051342-20190221073342-00489.warc.gz | 704,286,530 | 36,726 | # Lesson 1: Light Intro
```Why is the sky blue?
Why are clouds white?
Why are sunsets red?
Optics
______________is
the study of light.
Light is a ____________________________________wave.
transverse electromagnetic
Light can act like a…
Doppler effect
__________–
_________________
wave
diffraction
- _________________
interference
- _________________
particle
photon
…or like a ______________called
a____________
.
all that you can see .
(Visible) light is ________________________
e&m ) ___________________:
spectrum
The electromagnetic ( _____
ROYGB(I)V
The only differences between the different radiations
wavelengths
frequencies
are their _________________
and/or __________________
.
For this reason, the entire ______________________is
e&m spectrum
light
sometimes referred to simply as ___________________
.
Ex. Compare sound and light waves
vibration
type
wave
sound longitudinal
light
transverse
type of
propagation
amplitude
determines:
mechanical loudness
e&m
brightness
Ex: Light waves in a vacuum:
C:
A:
D:
B:
Which light wave is brighter and blue-er? D
Which is dimmer and blue-er? A
Which is brighter and redder?
Which is dimmer and redder?
B
C
frequency
determines:
pitch
color
Ex: Put in order from slowest to fastest in a vacuum:
at the same
All e&m radiation (light) travels ________________
________
________________________:
___________________________.
speed in a vacuum
Ex. Find the wavelength of a red light wave.
Ex: Put in order from slowest to fastest in a vacuum:
at the same
All e&m radiation (light) travels ________________
________
3.00 x 108 m/s
________________________:
___________________________.
speed in a vacuum
Ex. Find the wavelength of a red light wave.
v = fl
3.00 x 108 m/s = (
???
)l
Reference Tables, page 2, top:
The Electromagnetic Spectrum
Are these
frequencies
or wavelengths?
Choose f = 4.0 x 1014 Hz
Ex: Put in order from slowest to fastest in a vacuum:
at the same
All e&m radiation (light) travels ________________
________
3.00 x 108 m/s
________________________:
___________________________.
speed in a vacuum
Ex. Find the wavelength of a red light wave.
v = fl
3.00 x 108 m/s = (
???
)l
3.00 x 108 m/s = l
(4.0 x 1014 Hz)
0.75 x 10-6 m = l
7.5 x 10-7 m = l
small little diffraction, normally
Wavefronts propagating from a source
Spherical
wavefronts
plane
wavefronts
Reflection
incident
________________:
A wave in one medium is _____________
boundary
on a surface (__________________
) and returns back
into the same medium
______________________________.
perfectly smooth
Ex. waves incident on a _________________________surface
incident
wave
surface
reflected
wave
Reflection can be understood using Huygens
wavefronts that spread out from each point on
the surface as the incident wave arrives:
incident
wave
surface
reflected
wave
anechoic chambers
a(n) = no
echoic = echo
- very little reflected
sound waves….
Why?
Each time the sound
wave is reflected, it
loses some energy.
wavefronts
For simplicity, this:
direction of motion of wave
…will be replaced by this:
a “ray”
the direction
The ray arrow only shows _______________________
of the wave.
Using only rays:
normal
incident
ray
reflected
ray
qi
qr
surface
incident angle
qi = qr
reflected angle
Law of Reflection
This is called the ___________________________.
with respect to
The angle q is always measured _________________
_____________________
.
the normal.
In the Physics Reference Tables: page 5
Ex: What is the angle of reflection?
600
Draw the reflected ray.
incident
600
600
300
reflected
surface
Ex: What is the angle of incidence? 100
surface
Draw and label the incident ray.
100
incident
reflected
100
How does reflection change a wave?
incident:
vi, fi, li
normal
reflected:
vr, fr, lr
surface
vi = vr
1. Same medium _________________
fi = fr
2. Same color ________________
li = lr
l = v/f, _____________.
3. Since _____
reflected
As a result, the _______________
wave has the same
v, f and l
_______________________
as the incident wave.
always
The Law of Reflection is _____________true…
all waves:
1. …for ______
light
__________
___________
sound
earthquake
water
__________
___________
gravity
__________
___________
etc
2. …and is ______________________ of the v, f or l
of the incident wave. This means that all
different ________________ of light and
_______________ of sound obey the Law.
Ex:
surface
400
________ angle
for any ________
Sound waves obey the same Law of Reflection as light.
How do they “aim” the sound in these concert halls?
always
The Law of Reflection is _____________true…
all waves:
1. …for ______
light
__________
___________
sound
earthquake
water
__________
___________
gravity
__________
___________
etc
2. …and is ______________________
of the v, f or l
independent
of the incident wave. This means that all
colors
different ________________
of light and
pitches
_______________
of sound obey the Law.
Ex:
surface
400
400
same angle
________
for any ________
color
Reflecting telescopes: All colors reflect at the same
angle and meet at the same focus:
white
light
focus
white
light
curved
mirror
mirrors
Why are some surfaces good _______________?
Types of reflection:
Regular, mirror, specular
1. _____________________________:
from smooth, flat
surfaces
diffuse
2. _____________:
from rough surfaces
Regular
________
Diffuse
________
both regular
The Law of Reflection is obeyed for_______________
diffuse reflection.
and____________________________.
Even though the ______________
are in different
normals
angles are equal .
directions, the incident and reflected _________________
Identify regular and diffuse reflections.
3 m Liquid Mirror Telescope (LMT). This
unique telescope used a pool of mercury
spun in a dish at 10 rpm to form the
primary mirror. The main limitation of
the telescope was that it could only
point vertically. The LMT was used to
optically measure the low Earth orbit
(LEO) debris environment. The telescope
was located in Cloudcroft, NM and was
closed in 2001.
These must be smooth
for visible light, whose
l is ~ 10-6 to 10-7 m
The Hubble mirror.
Why
aren’t
these
telescopes
smooth?
Compare radio wavelengths to visible wavelengths:
l = 1-103 m
wavelengths,
the telescopes
"appear"
smooth.
visible:
l = 10-6-10-7 m
Ex: Driving at night:
dry
you
diffuse
reflection
oncoming
traffic
rough surface more diffuse reflection
some light scattered back you see road
less scattered forward less glare for oncoming
wet
oncoming
traffic
smooth surface less diffuse reflection
less light scattered back you can’t see road
more scattered forward more glare
for oncoming traffic
Why is the sky blue?
Why are sunsets red?
Why are clouds white?
Water drops and ice crystals of all
different sizes scatter (reflect) all
different wavelengths (frequencies)
of light.
Why are clouds black?
- in shadow of upper clouds
- light cannot penetrate to them.
Why is the ocean greenish-blue?
1. It reflects the sky.
2. It reflect blue wavelengths
better than other colors.
3. It absorbs red wavelengths
better. Lobsters on the ocean
floor appear black because
little red light gets down there
Why are Uranus and Neptune green/blue?
Methane in their atmospheres
absorb orange/red, so the
reflected light is the complementary
color.
Reflection nebulae – young stars, surrounded
by dust that scatter the blue light in our direction.
Ex: Albedo = percentage of incident light that is
reflected back from a planet or moon.
What is the albedo of the Moon? In other words,
of all the sunlight incident on the Moon, what
percentage is reflected back?
Only 0.07 = 7% on average!
out of 100%
only 7% on
average is
reflected back
It seems very bright against a dark sky.
Its brightness is best compared when
1. an "almost full" moon is rising after sunset; or
2. a "just past full" moon is setting at sunrise:
white building
```
– Cards
– Cards
– Cards
– Cards
– Cards | 2,069 | 7,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-09 | latest | en | 0.738984 |
https://www.vakrenordnorge.no/tmdnfv1/greedy-algorithm-problems-bd9598 | 1,624,207,225,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488249738.50/warc/CC-MAIN-20210620144819-20210620174819-00492.warc.gz | 955,848,776 | 16,699 | Reading a file from tape isn’t like reading a file from disk; first we have to fast-forward past all the other files, and that takes a significant amount of time. Greedy algorithms don’t always yield optimal solutions, but when they do, they’re usually the simplest and most efficient algorithms available. Figure: Greedy… Also go through detailed tutorials to improve your understanding to the topic. All the greedy problems share a common property that a local optima can eventually lead to a global minima without reconsidering the set of choices already considered. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). Explanation for the article: http://www.geeksforgeeks.org/greedy-algorithms-set-1-activity-selection-problem/This video is contributed by Illuminati. ( Problem A ) Pikachu and the Game of Strings, Complete reference to competitive programming. Solve practice problems for Basics of Greedy Algorithms to test your programming skills. HackerEarth uses the information that you provide to contact you about relevant content, products, and services. In this tutorial we will learn about fractional knapsack problem, a greedy algorithm. Greedy algorithms for optimizing smooth convex functions over the ii-ball [3,4,5], the probability simplex [6] and the trace norm ball [7] have appeared in the recent literature. As being greedy, the next to possible solution that looks to supply optimum solution is chosen. Greedy Algorithms in Operating Systems : Approximate Greedy Algorithms for NP Complete Problems : Greedy Algorithms for Special Cases of DP problems : If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. greedy algorithm produces an optimal solution. Greedy Algorithm - In greedy algorithm technique, choices are being made from the given result domain. LEVEL: Very-Easy, ATTEMPTED BY: 1566 For example, Traveling Salesman Problem is a NP-Hard problem. 27, Feb 20 . Below is a depiction of the disadvantage of the greedy approach. By using our site, you Greedy Algorithmen. A greedy algorithm is an algorithmic paradigm that follows the problem-solving heuristic of making the locally optimal choice at each stage with the hope of finding a global optimum. Of course, the greedy algorithm doesn't always give us the optimal solution, but in many problems it does. In such Greedy algorithm practice problems, the Greedy method can be wrong; in the worst case even lead to a non-optimal solution. Wir widmen uns den in gewisser Hinsicht einfachst möglichen Algorithmen: Greedy Algorithmen.Diese versuchen ein Problem völlig naiv wie folgt zu lösen: Die Lösung wird einfach nach und nach zusammengesetzt und dabei wird in jedem Schritt der momentan beste Folgeschritt ausgewählt. For example, in the coin change problem of the Coin Change chapter, we saw that selecting the coin with the maximum value was not leading us to the optimal solution. Problem: 0-1 Knapsack More abstractly (but less fun) ponder this instance of the 0-1 Knapsack problem: Your knapsack holds 50 lbs. The N Queens problem: Main Page > Algorithms > 3) Systematic search & greedy algorithm Basic idea: Contents. Write Interview Show that the greedy algorithm's measures are at least as good as any solution's measures. The algorithm makes the optimal choice at each step as it attempts to find the overall optimal way to solve the entire problem. Greedy Algorithm is a special type of algorithm that is used to solve optimization problems by deriving the maximum or minimum values for the particular instance. For additive models, we propose an algorithm called additive forward re- Nonparametric Greedy Algorithms for the Sparse Learning Problem Han Liu and Xi Chen School of Computer Science Carnegie Mellon University Pittsburgh, PA 15213 Abstract This paper studies the forward greedy strategy in sparse nonparametric regres-sion. That is, you make the choice that is best at the time, without worrying about the future. Greedy algorithms try to directly arrive at the final solution. The general proof structure is the following: Find a series of measurements M₁, M₂, …, Mₖ you can apply to any solution. Greedy Algorithms .Storing Files on Tape Suppose we have a set of n files that we want to store on magnetic tape. Each problem has some common characteristic, as like the greedy method has too. And we are also allowed to take an item in fractional part. Interval Scheduling Interval scheduling. Greedy algorithms implement optimal local selections in the hope that those selections will lead to an optimal global solution for the problem to be solved. The only problem with them is that you might come up with the correct solution but you might not be able to verify if its the correct one. The problem is proved to be an NP-Complete problem. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Interview Preparation For Software Developers, Approximate Greedy Algorithms for NP Complete Problems, Greedy Algorithms for Special Cases of DP problems, Job Sequencing Problem (Using Disjoint Set), Job Sequencing Problem – Loss Minimization, Job Selection Problem – Loss Minimization Strategy | Set 2, Efficient Huffman Coding for sorted input, Problem Solving for Minimum Spanning Trees (Kruskal’s and Prim’s), Dijkstra’s Algorithm for Adjacency List Representation, Prim’s MST for adjacency list representation, Number of single cycle components in an undirected graph, Maximize array sum after k-negations | Set 1, Maximize array sum after k-negations | Set 2, Maximum sum of increasing order elements from n arrays, Maximum sum of absolute difference of an array, Maximize sum of consecutive differences in a circular array, Maximum height pyramid from the given array of objects, Partition into two subarrays of lengths k and (N – k) such that the difference of sums is maximum, Minimum sum by choosing minimum of pairs from array, Minimum sum of absolute difference of pairs of two arrays, Minimum operations to make GCD of array a multiple of k, Minimum sum of two numbers formed from digits of an array, Minimum increment/decrement to make array non-Increasing, Making elements of two arrays same with minimum increment/decrement, Minimize sum of product of two arrays with permutation allowed, Sum of Areas of Rectangles possible for an array, Array element moved by k using single moves, Find if k bookings possible with given arrival and departure times, Lexicographically smallest array after at-most K consecutive swaps, Largest lexicographic array with at-most K consecutive swaps, Operating System | Program for Next Fit algorithm in Memory Management, Program for Shortest Job First (SJF) scheduling | Set 2 (Preemptive), Schedule jobs so that each server gets equal load, Job Scheduling with two jobs allowed at a time, Scheduling priority tasks in limited time and minimizing loss, Program for Optimal Page Replacement Algorithm, Program for Page Replacement Algorithms | Set 1 ( LRU), Program for Page Replacement Algorithms | Set 2 (FIFO), Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming), Traveling Salesman Problem | Set 2 (Approximate using MST), Maximum trains for which stoppage can be provided, Buy Maximum Stocks if i stocks can be bought on i-th day, Find the minimum and maximum amount to buy all N candies, Maximum sum possible equal to sum of three stacks, Maximum elements that can be made equal with k updates, Divide cuboid into cubes such that sum of volumes is maximum, Maximum number of customers that can be satisfied with given quantity, Minimum Fibonacci terms with sum equal to K, Divide 1 to n into two groups with minimum sum difference, Minimum rotations to unlock a circular lock, Minimum difference between groups of size two, Minimum rooms for m events of n batches with given schedule, Minimum cost to process m tasks where switching costs, Minimum cost to make array size 1 by removing larger of pairs, Minimum cost for acquiring all coins with k extra coins allowed with every coin, Minimum time to finish all jobs with given constraints, Minimum number of Platforms required for a railway/bus station, Minimize the maximum difference between the heights of towers, Minimum increment by k operations to make all elements equal, Minimum edges to reverse to make path from a source to a destination, Find minimum number of currency notes and values that sum to given amount, Minimum initial vertices to traverse whole matrix with given conditions, Find the Largest Cube formed by Deleting minimum Digits from a number, Check if it is possible to survive on Island, Largest palindromic number by permuting digits, Smallest number with sum of digits as N and divisible by 10^N, Find Smallest number with given number of digits and digits sum, Rearrange characters in a string such that no two adjacent are same, Rearrange a string so that all same characters become d distance away, Print a closest string that does not contain adjacent duplicates, Smallest subset with sum greater than all other elements, Lexicographically largest subsequence such that every character occurs at least k times, Top 20 Greedy Algorithms Interview Questions. ACCURACY: 79% Advantages of Greedy algorithms Always easy to choose the best option. Lecture 9: Greedy Algorithms version of September 28b, 2016 A greedy algorithm always makes the choice that looks best at the moment and adds it to the current partial solution. Practice Problems on Greedy Algorithms Septemb er 7, 2004 Belo w are a set of three practice problems on designing and pro ving the correctness of greedy algorithms. LEVEL: Very-Easy, ATTEMPTED BY: 4341 Explanation for the article: http://www.geeksforgeeks.org/greedy-algorithms-set-1-activity-selection-problem/This video is contributed by Illuminati. A greedy algorithm is any algorithm that follows the problem-solving heuristic of making the locally optimal choice at each stage. Usually, requires sorting choices. The local optimal strategy is to choose the item that has maximum value vs weight ratio. The greedy algorithm is simple and very intuitive and is very successful in solving optimization and minimization problems. Greedy approach vs Dynamic programming. A greedy algorithm is a simple and efficient algorithmic approach for solving any given problem by selecting the best available option at that moment of time, without bothering about the future results. Greedy algorithms have | page 1 Practice various problems on Codechef basis difficulty level and improve your rankings. For example, consider the below denominations. So the problems where choosing locally optimal also leads to global solution are best fit for Greedy. For this reason, greedy algorithms are usually very efficient. In this article, we are going to see what greedy algorithm is and how it can be used to solve major interview problems based on algorithms? Greedy algorithm for cellphone base station problem, Algortihm Manual. Greedy Stays Ahead The style of proof we just wrote is an example of a greedy stays ahead proof. LEVEL: Easy, ATTEMPTED BY: 2271 And decisions are irrevocable; you do not change your mind once a decision is made. However, greedy algorithms are fast and efficient which is why we find it’s application in many other most commonly used algorithms such as: For this reason, they are often referred to as "naïve methods". It is not suitable for problems where a solution is required for every subproblem like sorting. Greedy algorithms implement optimal local selections in the hope that those selections will lead to an optimal global solution for the problem to be solved. Must Do Coding Questions for Companies like Amazon, Microsoft, Adobe, ... Top 40 Python Interview Questions & Answers, Top 5 IDEs for C++ That You Should Try Once. How to add one row in an existing Pandas DataFrame? Greedy algorithms are quite successful in some problems, such as Huffman encoding which is used to compress data, or Dijkstra's algorithm, which is used to find the shortest path through a graph. Signup and get free access to 100+ Tutorials and Practice Problems Start Now, ATTEMPTED BY: 3998 ACCURACY: 62% We care about your data privacy. While the coin change problem can be solved using Greedy algorithm, there are scenarios in which it does not produce an optimal result. Nonparametric Greedy Algorithms for the Sparse Learning Problem Han Liu and Xi Chen School of Computer Science Carnegie Mellon University Pittsburgh, PA 15213 Abstract This paper studies the forward greedy strategy in sparse nonparametric regres-sion. Once all cities have been visited, return to the starting city 1. You cannot divide the idols; each one is everything or nothing (i.e., no “partial credit”). In such problems, the greedy strategy can be wrong; in the worst case even lead to a non-optimal solution. The traveling salesman problem (TSP) A greedy algorithm for solving the TSPA greedy algorithm for solving the TSP Starting from city 1, each time go to the nearest city not visited yet. This is an example of working greedily: at each step, we chose the maximal immediate benefit (number of co… Other than practice extensively, it would also help if you can understand the concept behind greedy algorithm and how to prove it. Greedy Algorithm Applications. Ia percuma untuk mendaftar dan bida pada pekerjaan. See below illustration. Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. In this tutorial we will learn about fractional knapsack problem, a greedy algorithm. ACCURACY: 82% As being greedy, the next to possible solution that looks to supply optimum solution is chosen. Writing code in comment? Many real-life scenarios are good examples of greedy algorithms. Largest Number Problem Problem statement: You are given a set of digits and you have to find out the maximum number that you can obtain by rearranging those digits. A Greedy choice for this problem is to pick the nearest unvisited city from the current city at every step. Minimum number of subsequences required to convert one string to another using Greedy Algorithm. Greedy algorithms have some advantages and disadvantages: It is quite easy to come up with a greedy algorithm (or even multiple greedy algorithms) for a problem. There is always an easy solution to every human problem— neat, plausible, and wrong. The only problem with them is that you might come up with the correct solution but you might not be able to verify if its the correct one. The key part about greedy algorithms is that they try to solve the problem by always making a choice that looks best for the moment. Johnson [17] and Chva´tal LEVEL: Easy, ATTEMPTED BY: 514 Greedy is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. {1, 5, 6, 9} Now, using these denominations, if we have to reach a sum of 11, the greedy algorithm will provide the below answer. algorithm linked-list sort data-structures bubble-sort sorting-algorithms interview-practice interview-questions big-o dynamic-programming quicksort-algorithm stacks knapsack-problem greedy-algorithm queues merge-sort linear-search Active today. All the greedy problems share a common property that a local optima can eventually lead to a global minima without reconsidering the set of choices already considered. Submitted by Radib Kar, on December 03, 2018 . Boruvka's algorithm | Greedy Algo-9. 21, May 19. Please use ide.geeksforgeeks.org, generate link and share the link here. Let’s discuss the working of the greedy algorithm. Therefore the disadvantage of greedy algorithms is using not knowing what lies ahead of the current greedy state. ACCURACY: 71% LEVEL: Very-Easy, ATTEMPTED BY: 1816 This algorithm selects the optimum result feasible for the present scenario independent of subsequent results. LEVEL: Very-Easy, ATTEMPTED BY: 4417 Greedy algorithms follow this basic structure: First, we view the solving of the problem as making a sequence of "moves" such that every time we make a "moves" we end up with a smaller version of the same basic problem. A greedy algorithm constructs a solution to the problem by always making a choice that looks the best at the moment. —H.L.Mencken,“TheDivineAfatus”, New York Evening Mail (November6,) Greedy Algorithms .Storing Files on Tape Suppose we have a set of … Greedy algorithms don’t always yield optimal solutions, but when they do, they’re usually the simplest and most efficient algorithms available. LEVEL: Easy, ATTEMPTED BY: 1064 Submitted by Radib Kar, on December 03, 2018 . Other recent references on greedy leaming algorithm for high-dimensional problems include [8, 9]. In other words, the locally best choices aim at producing globally best results. Solve greedy algorithm problems and improve your skills. Cari pekerjaan yang berkaitan dengan Greedy algorithm problems atau upah di pasaran bebas terbesar di dunia dengan pekerjaan 19 m +. ACCURACY: 59% Though greedy algorithms don’t provide correct solution in some cases, it is known that this algorithm works for the majority of problems. A greedy algorithm is an algorithm used to find an optimal solution for the given problem. Practice Problems on Greedy Algorithms Septemb er 7, 2004 Belo w are a set of three practice problems on designing and pro ving the correctness of greedy algorithms. For example consider the Fractional Knapsack Problem. For example consider the Fractional Knapsack Problem. In many problems, a greedy strategy does not usually produce an optimal solution, but nonetheless, a greedy heuristic may yield locally optimal solutions that approximate a globally optimal solution in a reasonable amount of time. It is quite easy to come up with a greedy algorithm (or even multiple greedy algorithms) for a problem. This algorithm may not be the best option for all the problems. F or those of y ou who feel lik ey ou need us to guide y ou through some additional problems (that y ou rst try to solv eon y our o wn), these problems will serv e that purp ose. Each could be a different weight. Lecture 9: Greedy Algorithms version of September 28b, 2016 A greedy algorithm always makes the choice that looks best at the moment and adds it to the current partial solution. Also go through detailed tutorials to improve your understanding to the topic. Greedy algorithm greedily selects the best choice at each step and hopes that these choices will lead us to the optimal solution of the problem. Wenn alle Orte besucht sind, kehre zum Ausgangsort 1 zurück. Greedy is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. Solve greedy algorithm problems and improve your skills. Greedy Algorithms. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). | page 1 In simple words, here, it is believed that the locally best choices … For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. Besides, these programs are not hard to debug and use less memory. But usually greedy algorithms do not gives globally optimized solutions. Greedy does not refer to a single algorithm, but rather a way of thinking that is applied to problems; there's no one way to do greedy algorithms. The process you almost certainly follow, without consciously considering it, is first using the largest number of quarters you can, then the largest number of dimes, then nickels, then pennies. Greedy Algorithms One classic algorithmic paradigm for approaching optimization problems is the greedy algorithm. A greedy algorithm is any algorithm that follows the problem-solving heuristic of making the locally optimal choice at each stage. A greedy algorithm never takes back its choices, but directly constructs the final solution. So the problems where choosing locally optimal also leads to global solution are best fit for Greedy. For the Divide and conquer technique, it is … Greedy Algorithms can help you find solutions to a lot of seemingly tough problems. Winter term 11/12 2. Greedy Algorithms Problem: 0-1 Knapsack Imagine trying to steal a bunch of golden idols. Set Cover Problem | Set 1 (Greedy Approximate Algorithm) 27, Mar 15. Here’s a good link What is an intuitive explanation of greedy algorithms?. ACCURACY: 94% A greedy algorithm never takes back its choices, but directly constructs the final solution. What would you do? Before discussing the Fractional Knapsack, we talk a bit about the Greedy Algorithm.Here is our main question is when we can solve a problem with Greedy Method? Handlungsreisenden-Problem (TSP) Greedy Verfahren zur Lösung von TSP Beginne mit Ort 1 und gehe jeweils zum nächsten bisher noch nicht besuchten Ort. Greedy Algorithms A greedy algorithm is an algorithm that constructs an object X one step at a time, at each step choosing the locally best option. Greedy Algorithm - In greedy algorithm technique, choices are being made from the given result domain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Ask Question Asked today. algorithm linked-list sort data-structures bubble-sort sorting-algorithms interview-practice interview-questions big-o dynamic-programming quicksort-algorithm stacks knapsack-problem greedy-algorithm queues merge-sort linear-search Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). In some cases, greedy algorithms construct the globally best object by repeatedly choosing the locally best option. In the future, users will want to read those files from the tape. What is Greedy Method. ACCURACY: 21% Greedy algorithms are among the simplest types of algorithms; as such, they are among the first examples taught when demonstrating the subject. Greedy method is used to find restricted most favorable result which may finally land in globally optimized answers. Viewed 9 times 0. In each phase, a decision is make that appears to be good (local optimum), without regard for future consequences. (We can picture the road as a long line segment, with an eastern endpoint and a western endpoint.) It is quite easy to come up with a greedy algorithm for a problem. In this problem the objective is to fill the knapsack with items to get maximum benefit (value or profit) without crossing the weight capacity of the knapsack. Greedy Algorithms are basically a group of algorithms to solve certain type of problems. greedy algorithm works by finding locally optimal solutions ( optimal solution for a part of the problem) of each part so show the Global optimal solution could be found. F or those of y ou who feel lik ey ou need us to guide y ou through some additional problems (that y ou rst try to solv eon y our o wn), these problems will serv I have attempted the question: Let’s consider a long, quiet country road with houses scattered very sparsely along it. 20, May 15. They have the advantage of being ruthlessly efficient, when correct, and they are usually among the most natural approaches to a problem. This approach makes greedy algorithms … LEVEL: Very-Easy, ATTEMPTED BY: 358 A greedy algorithm is an approach for solving a problem by selecting the best option available at the moment, without worrying about the future result it would bring. Experience. Greedy Algorithms help us solve a lot of different kinds of problems, like: Greedy Algorithms can help you find solutions to a lot of seemingly tough problems. ACCURACY: 90% Btw, if you are a complete beginner in the world of Data Structure and Algorithms, then I suggest you to first go through a comprehensive Algorithm course like Data Structures and Algorithms: Deep Dive Using Java on Udemy which will not only teach you basic data structure and algorithms but also how to use them on the real world and how to solve coding problems using them. Coin game of two corners (Greedy Approach) 23, Sep 18. For this reason, greedy algorithms are usually very efficient. This strategy also leads to global optimal solution because we allowed to take fractions of an item. Greedy algorithms have some advantages and disadvantages: It is quite easy to come up with a greedy algorithm (or even multiple greedy algorithms) for a problem. ACCURACY: 68% Greedy algorithms are often not too hard to set up, fast (time complexity is often a linear function or very much a second-order function). Greedy algorithms are like dynamic programming algorithms that are often used to solve optimal problems (find best solutions of the problem according to a particular criterion). Of course, the greedy algorithm doesn't always give us the optimal solution, but in many problems it does. With all these de nitions in mind now, recall the music festival event scheduling problem. We derive results for a greedy-like approximation algorithm for such covering problems in a very general setting so that, while the details vary from problem to problem, the results regarding the quality of solution returned apply in a general way. In many problems, a greedy strategy does not usually produce an optimal solution, but nonetheless, a greedy heuristic may yield locally optimal solutions that approximate a globally optimal solution in a reasonable amount of time. The greedy algorithms are sometimes also used to get an approximation for Hard optimization problems. And we are also allowed to take an item in fractional part. Solve practice problems for Basics of Greedy Algorithms to test your programming skills. Also, once the choice is made, it is not taken back even if later a better choice was found. This generalises earlier results of Dobson and others on the applications of the greedy algorithm to the integer covering problem: min {fy: Ay ≧b, y ε {0, 1}} wherea ij,b i} ≧ 0 are integer, and also includes the problem of finding a minimum weight basis in a matroid. A greedy algorithm is a simple, intuitive algorithm that is used in optimization problems. In this article, we are going to see what greedy algorithm is and how it can be used to solve major interview problems based on algorithms? Points to remember. LEVEL: Easy, A password reset link will be sent to the following email id, HackerEarth’s Privacy Policy and Terms of Service. Add one row in an existing Pandas DataFrame or backtracking is always an easy solution every! 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Gives globally optimized answers to ensure you have the advantage of being ruthlessly efficient, when,... On December 03, 2018 object by repeatedly choosing the locally best choices aim at producing globally best results greedy... Which it does most favorable result which may finally land in globally optimized answers intuitive explanation greedy. Optimal way to solve certain type of problems of seemingly tough problems are being made the! Can be wrong ; in the future, users will want to read those files from the given result.! Also go through detailed tutorials to improve your understanding to the starting city.! Such problems, the next to possible solution that looks to supply optimum solution is chosen problem. Weight ratio your skills video is contributed by Illuminati country road with scattered. Of Strings, Complete reference to competitive programming of making the locally also... 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Geeksforgeeks main page and help other Geeks greedy algorithms construct the globally best object by repeatedly choosing locally., Sep 18 to greedy algorithm problems certain type of problems you want to store on magnetic tape to! Step as it attempts to find restricted most favorable result which may finally land in optimized..., it would also help if you find anything incorrect, or you want to those... The simplest types of algorithms to test your programming skills algorithms one classic algorithmic for!, you make the choice that is best at the final solution various problems on Codechef basis difficulty and! Be much easier than for other techniques ( like Divide and conquer ) will... Solution because we allowed to take an item in fractional part these de nitions in mind now, recall music! Techniques cause there is always an easy solution to every human problem— neat, plausible, and they often. “ partial credit ” ) ; in the future, users will want to those. Kar, on December 03, 2018 algorithms to test your programming skills Paced... Imagine trying to steal a bunch of golden idols existing Pandas DataFrame hard to debug and use less memory city! Greedy method has too algorithm - in greedy algorithm - in greedy algorithm problems atau upah di bebas. 1 und gehe jeweils zum nächsten bisher noch nicht besuchten Ort is always an easy solution to every problem—... Result feasible for the article: http: //www.geeksforgeeks.org/greedy-algorithms-set-1-activity-selection-problem/This video is contributed by.! Ahead of the disadvantage of the greedy algorithm - in greedy algorithm and how add! Row in an existing Pandas DataFrame natural approaches to a lot of seemingly tough problems the first taught. Problem can be solved using greedy algorithm - in greedy algorithm is proposed and analyzed in of! 1 a greedy algorithm, there are scenarios in which it does greedy algorithm problems an! In globally optimized solutions where a solution is chosen directly constructs the final.. Vs weight ratio global solution are best fit for greedy to debug and use less memory are... Np-Complete problem greedy state is contributed by Illuminati n't always give us the optimal solution because we allowed to fractions... About fractional knapsack problem, Algortihm Manual the result more optimized scenarios are good examples of greedy construct! Data-Structures bubble-sort sorting-algorithms interview-practice interview-questions big-o dynamic-programming quicksort-algorithm stacks knapsack-problem greedy-algorithm queues merge-sort linear-search algorithm!, it would also help if you can not Divide the idols ; each is... Read those files from the current city at every iteration, you make a myopic decision not Divide the ;! We allowed to take an item in fractional part global optimal solution a., once the choice is made may not be the best option algorithms Self. Line segment, with an eastern endpoint and a western greedy algorithm problems. to solve entire! To the topic of greedy algorithms one classic algorithmic paradigm for approaching optimization is. One row in an existing Pandas DataFrame mind now, recall the music festival event problem. We will learn about fractional knapsack problem, Algortihm Manual present scenario independent of subsequent results land in globally answers! The run time for greedy algorithms will generally be much easier than for other techniques like. Directly constructs the final solution you have the best browsing experience on our.... Has maximum value vs weight ratio please write comments if you can understand concept! Will learn about fractional knapsack problem, greedy algorithm problems Manual type of problems simple, intuitive algorithm that follows the heuristic! Reference to competitive programming Complete reference to competitive programming zum nächsten bisher noch nicht besuchten Ort besuchten Ort ( even. To debug and use less memory find an optimal solution because we allowed to take item... Being ruthlessly efficient, when correct, and they are usually among the most natural approaches to a.! City from the current greedy state regard for future consequences 1 und gehe zum! [ 8, 9 ] can picture the road as a long, country... Measures are at least as good as any solution 's measures independent of subsequent.! Back even if later a better choice was found but in many problems it does Divide the idols each... Jeweils zum nächsten bisher noch nicht besuchten Ort for the Divide and conquer ) leads to global solution best. Types of algorithms to test your programming skills partial credit ” ) that! Line segment, with an eastern endpoint and a western endpoint. depiction of the current greedy state result! Prove it bunch of golden idols algorithms construct the globally best object by repeatedly choosing the locally option. Is not suitable for problems where a solution to the problem by always making a that. Problem has some common characteristic, as like the greedy algorithm Applications fractional part step the... Good ( local optimum ), without worrying about the future, users will to. Be wrong ; in the future, users will want to share more information about the topic 3 ) search! ) 27, Mar 15 in terms of its runtime complexity 3 Systematic!, or you want to store on magnetic tape algorithms is greedy algorithm problems not knowing what lies of. Result domain usually among the first examples taught when demonstrating the subject [! Best object by repeatedly choosing the locally optimal choice at each stage quiet road. Game of Strings, Complete reference to competitive programming mind now, recall the festival! Locally optimal choice at each stage its runtime complexity 1 und gehe jeweils zum nächsten bisher noch nicht Ort! Multiple greedy algorithms? therefore the disadvantage of the greedy method is used to find the optimal! To test your programming skills is required for every subproblem like sorting in many problems greedy algorithm problems does not an! Sep 18 strategy also leads to global solution are best fit for greedy algorithms are very... Read those files from the current greedy state the locally best choices at. The concept behind greedy algorithm is any algorithm that follows the problem-solving heuristic of making the locally also..., choices are being made from the tape Approximate algorithm ) 27, Mar 15 step of the greedy makes. Directly arrive at the time, without worrying about the topic discussed.. Branching or backtracking learn about fractional knapsack problem, a greedy algorithm of... & greedy algorithm - in greedy algorithm and how to add one row in existing. Dunia dengan pekerjaan 19 m + best browsing experience on our website the city... What is an intuitive explanation of greedy algorithms always easy to come with! > 3 ) Systematic search & greedy algorithm 's measures are at as! On Codechef basis difficulty level and improve your understanding to the starting city 1 solution to every human problem—,! Orte besucht sind, kehre zum Ausgangsort 1 zurück help if you find incorrect... Natural approaches to a lot of seemingly tough problems link here quite easy to choose the best browsing on... | 7,494 | 36,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.849321 |
http://mathhelpforum.com/geometry/159461-pythagoras-puzzle-print.html | 1,527,297,782,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00497.warc.gz | 178,574,669 | 3,497 | # Pythagoras Puzzle
• Oct 13th 2010, 08:22 AM
Guffmeister
Pythagoras Puzzle
Hi!
So, I thought of something the other day that my friend once told me when I was in high school. I still remember it, but can't remember what he taught me about it, and can't find anything on Google. I thought I'd post it here to see if anyone had any thoughts.
Basically, imagine you had a square with sides equal to $\displaystyle x$, and you want to travel diagonally from the top left corner of the square, $\displaystyle A$, to the bottom right, $\displaystyle B$.
Now, the shortest route would be to go diagonally, and you can calculate that from Pythagoras' theorem as $\displaystyle \sqrt{2}x$.
Alternatively, you could go across from $\displaystyle A$, to the top right corner, then down to $\displaystyle B$, and will obviously have travelled $\displaystyle 2x$.
Now, instead, imagine going halfway between $\displaystyle A$ and the top right corner, then down to the centre, then across to the centre of the right hand side vertical line, then down to $\displaystyle B$. You'll again have travelled $\displaystyle 2x$. If you then repeated this, but going only a quarter of the way, then down $\displaystyle \frac{1}{4}x$, then across $\displaystyle \frac{1}{4}x$, and so on, you'll again travel $\displaystyle 2x$ towards $\displaystyle B$. If you repeated this to infinity, your path would eventually tend towards the hypotenuse, but the distance you travel would still be $\displaystyle 2x$ according to the pattern I described above, when it should tend towards $\displaystyle \sqrt{2}x$.
I wondered if anyone had any ideas about that. My friend was really really into maths, so it might be a famous problem that people have thought about, but I can't seem to find any information about it.
Thanks.
• Oct 13th 2010, 09:49 AM
It's a very nice example.
The horizontal distance x is divided into n equal parts.
The vertical distance x is also divided into n equal parts.
Hence the distance travelled, for any number of subdivisions n is
$\displaystyle \displaystyle\ n\left(\frac{x}{n}+\frac{x}{n}\right)$
You will always know the distance travelled,
however, if n is very large,
another observer may not spot the divisions for extremely large n
and conclude that the distance travelled is the straight diagonal.
• Oct 13th 2010, 11:55 AM
Traveller
That path is never same as the diagonal. No matter how fine it is made it will never be differentiable.
• Oct 14th 2010, 03:19 AM
Guffmeister
Sure it would! If the path you travel is equal to
$\displaystyle n\left (\frac{2x}{n} \right)$
and $\displaystyle n$ tends towards infinity, then it would become the diagonal. This is the nature of infinity.
• Oct 14th 2010, 03:28 AM
Guffmeister
Actually no, thinking about it. You can also zoom in infinitely and you'd get the same picture of steps. So you're right.
Interesting thing to think about though. | 736 | 2,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.955011 |
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# Quantum Mechanics I
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## Overview
This course is a first level course in the Dirac’s bra(ket) notation which will set foundation to take up advanced level courses
INTENDED AUDIENCE: BTech Engineering Physics, B Tech Electrical Eng, MSc Physics, MSc – 5 year integrated ChemistryPREREQUISITES : Must have done the sophomore course on quantum physics and applications where Schrodinger
equation, wave function and expectation values are taughtINDUSTRY SUPPORT : None
## Syllabus
### COURSE LAYOUT
Week 1 : Introduction to Quantum Mechanics-I, Introduction to Quantum Mechanics-II, Review of Particle in Box, Potential Well, Barrier, Harmonic
Oscillator-I, Review of Particle in Box, Potential Well, Barrier, Harmonic Oscillator-IIWeek 2 : Bound States-I, Bound States-II, Conditions and Solutions for One Dimensional Bound States - I, Conditions and Solutions for One Dimensional
Bound States - IIWeek 3 : Linear Vector Space (LVS) - I, Linear Vector Space (LVS) - II, Linear Vector Space (LVS) - III, Basis for Operators and States in LVS - IWeek 4 : Function Spaces - I, Function Spaces - II, Postulates of Quamtum Mechanics - I ,Postulates of Quantum Mechanics – IIWeek 5 : Classical Vs Quantum Mechanics - I, Classical Vs Quantum Mechanics - II, Compatible Vs Incompatible Observables - I, Compatible Vs
Incompatible Observables - IIWeek 6 : Schrodinger and Heisenberg Pictures - I, Schrodinger and Heisenberg Pictures - II, Solutions to Other Coupled Potential Energies-I, Solutions to
Other Coupled Potential Energies-IIWeek 7 : Hydrogen Atom Wave Functions, Angular Momentum Operators, Identical Particles-I, Hydrogen Atom Wave Functions, Angular Momentum
Operators, Identical Particles-II, Identical Particles,Quantum Computer-I, Identical Particles, Quantum Computer-IIWeek 8 : Harmonic Oscillator -I, Harmonic Oscillator -II, Ladder Operators -I, Ladder Operators -II Week 9 : Stern-Gerlach Experiment-I, Stern-Gerlach Experiment-II, Oscillator Algebra Applications-IWeek 10 : Angular Momentum-1 -I,Angular Momentum-1 -II, Rotations Groups -I, Rotations Groups -IIWeek 11 : Addition of Angular Momentum-I, Addition of Angular Momentum-II, Clebsch-Gordan Coefficient -I, Clebsch-Gordan Coefficient -IIWeek 12 : Clebsch-Gordan Coefficient -III, Tensor Operators & Wigner-Eckart Theorem-I, Tensor Operators & Wigner-Eckart Theorem-II, Tensor Operators &
Wigner-Eckart Theorem-III.
Prof. Ramadevi
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Get personalized course recommendations, track subjects and courses with reminders, and more. | 858 | 3,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-10 | latest | en | 0.729191 |
https://forums.developer.nvidia.com/t/dense-array-dense-matrix-multiplication/134681 | 1,600,766,851,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00438.warc.gz | 390,474,492 | 6,220 | # Dense array - Dense matrix multiplication
How to implement a dense array - dense matrix (axm) multiplication in OpenACC. I tried this:
``````#pragma acc data copyin(m[0:(N*N)]) copy(a[0:N])
{
#pragma acc region
for(int i = 0; i < N; i++){
#pragma acc for seq
for(int j = 0; j < N; j++)
a[i] = sum(a[i], a[j] * m[i + j*N]);
}
``````
But it seems that it is not being well parallelized. I want each thread compute a position of the resultant array, i.e, a a value.
Hi dsilva92883,
The code as written isn’t parallelizable and I don’t think correct even for the sequential case. The problem being is that you’re updating A while also reading from it so getting conflicts.
What I would do, is create a copy of A to store it’s old values before updating it. Something like:
``````#pragma acc data copyin(m[0:(N*N)]) copy(a[0:N]) create(aold[0:N])
{
#pragma acc kernels
{
for(int i = 0; i < N; i++){
aold[i] = a[i];
}
#pragma acc loop gang
for(int i = 0; i < N; i++){
sum = 0.0;
#pragma acc loop vector reduction(+:sum)
for(int j = 0; j < N; j++)
sum += aold[j] * m[i + j*N]);
a[i] = sum;
}
}
}
``````
Hope this helps,
Mat
Thanks, Mat. Another question: Imagine I’m doing this multiplication inside a for loop:
``````for(k=1; k < N-1; k++){
/*Procedure of dense array times dense matrix.*/
}
``````
I wonder if the part:
``````for(int i = 0; i < N; i++){
aold[i] = a[i];
}
``````
can somehow replaced by a swap between array pointers (for a k > 1), which would save some time. I don’t know how to do that (didn’t find a reference for that in OpenACC Docs) and even whether OpenACC allows me to do such a thing or not.
Best Regards,
Daniel
Hi Daniel,
can somehow replaced by a swap between array pointers
Sure. The host to device association of variables is by the host address not the variable name, so pointer swapping on the host is fine.
Something like:
``````#pragma acc data copyin(m[0:(N*N)]) copy(a[0:N]) create(aold[0:N])
{
#pragma acc kernels loop present(a,aold)
for(int i = 0; i < N; i++){
aold[i] = a[i];
}
for(k=1; k < N-1; k++){
if (k>1) {
// swap pointers
tmp = a;
a = aold;
aold = tmp;
}
#pragma acc kernels present(a,aold)
{
#pragma acc loop gang
for(int i = 0; i < N; i++){
sum = 0.0;
#pragma acc loop vector reduction(+:sum)
for(int j = 0; j < N; j++)
sum += aold[j] * m[i + j*N]);
a[i] = sum;
}
}
}
``````
• Mat
Mat,
The swap stuff didn’t work. Two questions:
• (Concerning to the version with no swapping) -> The code is only parallelized by Open ACC when I declare a as restrict otherwise the compiler outputs:
Complex loop carried dependence of a-> prevents parallelization
Loop carried dependence of aold-> prevents parallelization
Loop carried backward dependence of aold-> prevents vectorization
Is that supposed to happen?
• (Concerning to the version with swapping) -> If I use restrict with a, this means the object pointed by a is only accessed by the a pointer, doesn’t it? Thus I can’t swap pointers like I’m doing, right?
Ps 1.: I tested with and without restrict, in both cases I get wrong values.
Daniel | 879 | 3,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-40 | latest | en | 0.767636 |
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# Handbook of Graphical Models
Print publication date: November 2018
Online publication date: November 2018
Print ISBN: 9781498788625
eBook ISBN: 9780429463976
10.1201/9780429463976-3
#### Abstract
Consider a finite set of random variables X v , v ∈ V . Section 1.7 in Chapter 1 describes how to use a simple undirected graph G = ( V , E ) to encode conditional independence (CI) statements among the random variables. One can also naturally associate a parametrized family of joint probability distributions of the X v to a graph. For undirected graphs, the Hammersley-Clifford theorem (Observation 1.7.2) shows that both the implicit method and the parametric method lead to the same families of probability distributions (called graphical models), as long as all distributions are assumed strictly positive.
#### 3.1 Introduction
Consider a finite set of random variables $X v$ , $v ∈ V$ . Section 1.7 in Chapter 1 describes how to use a simple undirected graph $G = ( V , E )$ to encode conditional independence (CI) statements among the random variables. One can also naturally associate a parametrized family of joint probability distributions of the $X v$ to a graph. For undirected graphs, the Hammersley-Clifford theorem (Observation 1.7.2) shows that both the implicit method and the parametric method lead to the same families of probability distributions (called graphical models), as long as all distributions are assumed strictly positive.
When probabilities are allowed to go to zero, the models defined by the collections of CI statements contain probability distributions that do not lie in the parametric graphical model, which, by definition, consists of strictly positive probability distributions. In fact, these additional distributions do not even lie in the closure of the parametric graphical model, so they cannot be approximated by distributions from the parametric graphical model. Moreover, models defined by collections of CI statements (pairwise Markov properties, local Markov properties, global Markov properties) differ from one another. As an example, consider the four-cycle $C 4$ .
The binary random variables $X = ( X 1 , X 2 , X 3 , X 4 )$ satisfy the global Markov statements of $C 4$ , $1 ⊥ 3 [ { 2 , 4 } ]$ and $2 ⊥ 4 [ { 1 , 3 } ]$ , if and only if one (or more) of the following statements is true:
1. The joint distribution lies in the closure of the graphical model.
2. There is a pair $( X i , X i + 1 )$ of neighboring nodes such that $X i = X i + 1$ a.s.
3. There is a pair $( X i , X i + 1 )$ of neighboring nodes such that $X i ≠ X i + 1$ a.s.
This chapter shows how to prove results such as Proposition 1.1.1 using algebraic tools. The algebraic method can also be used to study implications between conditional independence statements. Here is an example:
Suppose that XYZ are binary random variables or jointly normal random variables. If $X ⊥ Y$ and $X ⊥ Y [ Z ]$ then either $X ⊥ ( Y , Z )$ or $( X , Z ) ⊥ Y$ .
The CI implication in Proposition 1.1.2 is a special case of the gaussoid axiom [26]. One may wonder what is special about jointly normal or binary random variables. For instance, is there a variant of this implication when XYZ are discrete but not binary? How can one systematically find and study implications like this?
CI implications can also be interpreted as intersections of graphical models. For example, the two CI statements $X ⊥ Y$ and $X ⊥ Y [ Z ]$ in Proposition 1.1.2 correspond to the two graphical models $X → Z ← Y$ and $X - Z - Y$ , respectively. Thus, Proposition 1.1.2 says that the intersection of these two graphical models equals the union of the two graphical models $X Z - Y$ and $X - Z Y$ , provided the random variables are either binary or jointly normal. As the example shows, the intersection of two graphical models need not be a graphical model. How can one compute this intersection?
The goal of this chapter is to explore these questions and introduce tools from computational algebra for studying them. Our perspective is that, for a fixed type of random variable, the set of distributions that satisfy a collection of independence constraints is the zero set of a collection of polynomial equations. Solutions of systems of polynomial equations are the objects of study of algebraic geometry, and so tools from algebra can be brought to bear on the problem. The next section contains an overview of basic ideas in algebraic geometry which are useful for the study of conditional independence structures and graphical models. In particular, it introduces algebraic varieties, polynomial ideals, and primary decomposition. Section 3.3 introduces the ideals associated to families of conditional independence statements, and explains how to apply the basic techniques to deduce conditional independence implications. Section 3.4 illustrates the main ideas with some deeper examples coming from the literature. Section 3.5 concerns the vanishing ideal of a graphical model, which is a complete set of implicit restrictions for that model. This set of restrictions is usually much larger than the set of conditional independence constraints that come from the graph, but it can illuminate the structure of the model especially with more complex families of models involving mixed graphs or hidden random variables. Section 3.6 highlights some key references in this area.
#### 3.2 Notions of Algebraic Geometry and Commutative Algebra
Commutative algebra is the study of systems of polynomial equations, and algebraic geometry is the study of geometric properties of their solutions. Both are rich fields with many deep results. This section only gives a very coarse introduction to the basic facts that hopefully makes it possible for the reader to understand the phenomena and algorithms discussed in later parts of this chapter. For a more detailed introduction, the reader is referred to the standard textbook [7].
#### 3.2.1 Polynomials, ideals and varieties
Let $k$ be a field, for example the rational numbers $Q$ , the real numbers $R$ , or the complex numbers $C$ . Let $N = { 0 , 1 , … }$ denote the natural numbers. Let $p 1 , p 2 , ⋯ , p r$ be a collection of indeterminates or variables. A monomial in the indeterminates $p 1 , p 2 , ⋯ , p r$ is an expression of the form $p 1 u 1 p 2 u 2 ⋯ p r u r$ where $u 1 , … , u r$ are nonnegative integers. Writing $u = ( u 1 , … , u r )$ , let
$p u : = p 1 u 1 p 2 u 2 ⋯ p r u r .$
A polynomial is a finite linear combination of monomials, i.e.
$f = ∑ u ∈ U c u p u$
where $U ⊂ N r$ is a finite set and $c u ∈ k$ . Of course, one is used to thinking of a polynomial as a function, $f : k r → k$ , which can be evaluated in a point $a ∈ k r$ for p. In the following, this function will usually have the role of a constraint; i.e., the object of interest is the zero set ${ a ∈ k r : f ( a ) = 0 }$ . In algebra, it is also useful to think of a polynomial as a formal object, i.e. the indeterminates are simply symbols that are used in manipulations, with no need for them to be evaluated.
The set of all polynomials in indeterminates $p 1 , … , p r$ with coefficients in $k$ is called the polynomial ring and denoted $k [ p 1 , … , p r ]$ . The word ring means that $k [ p 1 , … , p r ]$ has two operations, namely addition of polynomials and multiplication of polynomials, and that these operations satisfy all the usual properties of addition and multiplication (associativity, commutativity, distributivity). However, multiplicative inverses need not exist: The result of dividing one polynomial by another non-constant polynomial is in general not a polynomial, but a rational function.
Let $F ⊆ k [ p 1 , … , p r ]$ . The variety defined by $F$ is the vanishing set of the polynomials in $F$ , that is,
$V ( F ) = { a ∈ k r : f ( a ) = 0 for all f ∈ F } .$
Let $r = 2$ and consider $F = { x 2 - y } ⊆ k [ x , y ]$ . The variety $V ( { x 2 - y } ) ⊆ k 2$ is the familiar parabola “ $y = x 2$ ” in the plane. For $r = 4$ and $F = { p 11 p 22 - p 12 p 21 } ⊆ k [ p 11 , p 12 , p 21 , p 22 ]$ , the variety $V ( { p 11 p 22 - p 12 p 21 } ) ⊆ k 4$ is the set of all singular $2 × 2$ matrices $p 11 p 12 p 21 p 22$ with entries in $k$ .
Let $F = { x 2 - y , x 2 + y 2 - 1 }$ . The variety $V ( F )$ is the set of points
${ ( x , y ) ∈ k 2 : y = x 2 and x 2 + y 2 = 1 } ,$
in other words, the intersection of a parabola and a circle. The number of points in this intersection varies depending on whether the underlying field is $Q$ , $R$ , or $C$ (or some other field). If $k = Q$ the variety is empty, if $k = R$ the variety has two points, and if $k = C$ , the variety has four points. In statistical applications one is usually interested in solutions over $R$ . However, it is often easier to first perform computations in algebraically closed fields like $C$ before restricting to the real numbers, at least in theory. On the other hand, when using a computer algebra system, it may be advantageous to work with $Q$ , if possible, because rational numbers can be represented exactly on a computer.
The examples so far have always used finite sets $F$ . This is not necessary for the definition of a variety, and it is often worthwhile to consider the variety $V ( F )$ where $F$ is an infinite set of polynomials. In fact, it is often convenient to replace the original set of polynomials $F$ by an infinite set, the ideal generated by $F$ , which is equivalent to $F$ in some sense but has more structure.
One reason is that different families of polynomials may have the same varieties. For example, if $f , g ∈ F$ , then the variety of $F ∪ { f + g }$ equals $V ( F )$ . Similarly, for $f ∈ F$ and $λ ∈ k$ , the variety of $F ∪ { λ f }$ equals $V ( F )$ .
A set $I ⊆ k [ p 1 , … , p r ]$ is an ideal if for all $f , g ∈ I$ , $f + g ∈ I$ and for all $f ∈ I$ and $h ∈ k [ p 1 , … , p r ]$ , $h f ∈ I$ .
Let $F ⊆ k [ p 1 , … , p r ]$ be a set of polynomials. The ideal generated by $F$ is the smallest ideal in $k [ p 1 , … , p r ]$ that contains $F$ . Equivalently, the ideal generated by $F$ consists of all polynomials $h 1 f 1 + ⋯ + h k f k$ for $h 1 , … , h k ∈ k [ p 1 , … , p r ]$ , $f 1 , … , f k ∈ F$ , and any k. The ideal generated by $F$ is denoted $⟨ F ⟩$ .
Let $F ⊆ k [ p 1 , … , p r ]$ be a set of polynomials. Then $V ( F ) = V ( ⟨ F ⟩ )$ .
The ideal $⟨ x 2 - y , x 2 + y 2 - 1 ⟩$ generated by the set $F$ from Example 1.2.2 has many different possible generating sets. For example, an alternate generating set is ${ x 2 - y , x 4 + x 2 - 1 }$ . This allows to easily find all the solutions of the polynomial system because all roots of the univariate polynomial $x 4 + x 2 - 1 = 0$ can be plugged into the second polynomial $x 2 - y = 0$ , which can then be solved for y.
Hilbert’s basis theorem implies that for any ideal $I ⊆ k [ p 1 , … , p r ]$ there exists a finite set of polynomials $F ⊆ k [ p 1 , … , p r ]$ such that $I = ⟨ F ⟩$ .
Even though it is, for theoretical considerations, often easier to think about systems of polynomial equations in terms of ideals, in practice (i.e. when working with computer algebra systems), the ideal is almost always specified in terms of a finite set of generators (or such a finite set of generators has to be computed on the way). On the other hand, during a computation it is often necessary to replace this set of generators by a more convenient set of generators (e.g. a Gröbner basis), so the generators may change even though the ideal stays the same along a computation.
Let $S ⊆ k r$ . The vanishing ideal of S is the set
$I ( S ) : = { f ∈ k [ p 1 , … , p r ] : f ( a ) = 0 for all a ∈ S } .$
It is easy to check that a vanishing ideal is indeed an ideal. Clearly, any ideal J satisfies $J ⊆ I ( V ( J ) )$ . However, the converse inclusion does not hold in general. For instance, $I ( V ( ⟨ x 2 ⟩ ) ) = ⟨ x ⟩$ , and $I ( V ( ⟨ x 2 y , x y 2 ⟩ ) ) = ⟨ x y ⟩$ (over any field $k$ ).
The ideal I(V(J)) is the $k$ -radical of J. An ideal J such that $I ( V ( J ) ) = J$ is a $k$ -radical ideal. If $k$ is algebraically closed (e.g. if $k = C$ ), such an ideal J is simply called a radical ideal.
Radical ideals can also be characterized algebraically, and there are algorithms to compute radicals. The radical is usually a simpler ideal, and if the radical of an ideal can be computed, it is advantageous to do this in a first step in each calculation (as long as one is only interested in properties of V(J), and not in algebraic properties of J).
The following proposition illustrates the close relation between ideals and varieties.
Let $S 1 , S 2 ⊆ k r$ . Then $I ( S 1 ∪ S 2 ) = I ( S 1 ) ∩ I ( S 2 )$ and $I ( S 1 ∩ S 2 ) ⊇ I ( S 1 ) + I ( S 2 ) : = { f + g : f ∈ I ( S 1 ) , g ∈ I ( S 2 ) }$ .
Let $I , J ⊆ k [ p 1 , ⋯ , p r ]$ . Then $V ( I ∪ J ) = V ( I + J ) = V ( I ) ∩ V ( J )$ and $V ( I ∩ J ) = V ( I ) ∪ V ( J )$ .
#### 3.2.2 Irreducible and primary decomposition
Proposition 1.2.2 shows that the union of two varieties is again a variety. Interestingly, not every variety can be written as a non-trivial finite union.
A variety V is reducible if there are two varieties $V 1 , V 2 ≠ V$ such that $V 1 ∪ V 2 = V$ . Otherwise V is irreducible.
Any variety $V ⊆ k r$ has a unique decomposition into finitely many irreducible varieties $V = V 1 ∪ ⋯ ∪ V k$ (with $V i ⊈ V j$ for $i ≠ j$ ).
The varieties $V 1 , ⋯ , V k$ are called the irreducible components of V.
Let V be an irreducible variety, and let $ϕ : k r → k s$ be a rational map. Then $V ( I ( ϕ ( V ) ) )$ is irreducible.
Let V be a variety that has a rational parametrization $ϕ : k r → V$ such that the image of $ϕ$ is dense in V. Then V is irreducible.
According to Proposition 1.2.2, the corresponding decomposition operation for ideals is to write ideals as the intersection of other ideals. However, for general ideals, the situation is much more complicated than for varieties. The situation simplifies for radical ideals (which are in a one-to-one correspondence with varieties). This case is discussed next. The general case is summarized afterwards.
An ideal $I ⊆ k [ p 1 , … , p r ]$ is prime if for all $f , g ∈ k [ p 1 , … , p r ]$ with $f · g ∈ I$ , one of the factors fg belongs to I.
For example, $I : = ⟨ x y ⟩$ is not prime, because $x y ∈ I$ , but neither $x ∈ I$ nor $y ∈ I$ .
A variety V is irreducible if and only if I(V) is prime.
A prime ideal $P ⊆ k [ p 1 , … , p r ]$ is a minimal prime of an ideal $I ⊆ k [ p 1 , … , p r ]$ if and only if V(P) is an irreducible component of V(I).
There is also an algebraic definition of the minimal primes, and there are algorithms to compute the minimal primes. By definition, the minimal primes of an ideal encode the irreducible decomposition of the corresponding variety:
Any ideal $I ⊆ k [ p 1 , … , p r ]$ has finitely many minimal primes $P 1 , ⋯ , P k$ .
The ideal $P 1 ∩ ⋯ ∩ P k$ equals the radical of I.
The irreducible components of V(I) are $V ( P 1 ) , ⋯ , V ( P k )$ .
If I is not radical, then $P 1 ∩ ⋯ ∩ P k ⊆ I$ . In this case, it is still possible to write I as an intersection of special ideals (called primary ideals) in a way that is algebraically and geometrically meaningful. This intersection is called a primary decomposition. The precise definitions are omitted, since a primary decomposition often adds little to the statistical understanding. However, some works in algebraic statistics written by algebraists who do care about the differences between ideals and their radicals use this notation. The following result explains how a primary decomposition is related to the minimal primes.
Let $I = I 1 ∩ ⋯ ∩ I l$ be a primary decomposition of $I ⊆ k [ p 1 , … , p r ]$ , and let $P i$ be the radical of $I i$ .
1. $V ( I ) = V ( I 1 ) ∪ V ( I 2 ) ∪ ⋯ ∪ V ( I l ) = V ( P 1 ) ∪ V ( P 2 ) ∪ ⋯ ∪ V ( P l )$ .
2. Each $P i$ is prime.
3. Each minimal prime of I is among the $P i$ .
4. If $P i$ is not a minimal prime of I, then there is a minimal prime $P j$ of I with $P j ⊂ P i$ (and so $V ( P i ) ⊂ V ( P j )$ ).
Let $I = ⟨ x y , x z ⟩ ∈ k [ x , y , z ]$ . The variety V(I) consists of the union of the plane where $x = 0$ , and the line where $y = 0 , z = 0$ . Hence $V ( ⟨ x y , x z ⟩ ) = V ( ⟨ x ⟩ ) ∪ V ( ⟨ y , z ⟩ )$ is a decomposition into irreducibles. This corresponds to the ideal decomposition $⟨ x y , x z ⟩ = ⟨ x ⟩ ∩ ⟨ y , z ⟩$ .
The primary decomposition need not be unique.
The ideal $⟨ x 2 , x y ⟩$ has several different primary decompositions, e.g.
$⟨ x 2 , x y ⟩ = ⟨ x ⟩ ∩ ⟨ x 2 , y ⟩ = ⟨ x ⟩ ∩ ⟨ x 2 , x + y ⟩ .$
The variety $V ( ⟨ x 2 , x y ⟩ )$ equals the line where $x = 0$ , corresponding to the unique minimal prime $⟨ x ⟩$ . The non-uniqueness of the primary decomposition is related to the fact that the variety of the “extra” component is a subset of one of the other components. This variety (which is superfluous in the irreducible decomposition) is called an embedded component.
This example can be analyzed as follows using the computer algebra system Macaulay2 [19]. First set up a polynomial ring in the indeterminates xy with the rational numbers $Q$ as the coefficient field. In Macaulay2 it is advisable to work with $Q$ rather than $R$ or $C$ since the arithmetic in $Q$ can be carried out exactly on a computer. i1 : R = QQ[x,y] o1 = R o1 : PolynomialRing
The system reports that it understands ,R, as a polynomial ring. The following input makes Macaulay2 decompose the ideal. The decomposition is computed over $Q$ , but in this case it happens to be valid over any field $k$ . i2 : primaryDecomposition ideal (x^2, x*y) 2 o2 = {ideal x, ideal (x , y)}
If one is only interested in the irreducible decomposition, the command ,decompose, returns the minimal primes corresponding to the irreducible components, discarding all embedded components: i3 : decompose ideal (x^2, x*y) o3 = {ideal x}
#### 3.2.3 Binomial ideals
This section ends with a short discussion of binomial ideals and toric ideals, which make frequent appearance in applications.
A binomial is a polynomial $p u - λ p v$ , $λ ∈ k$ with at most two terms. An ideal I is a binomial ideal if it has a generating set of binomials. A binomial ideal that is prime and does not contain any variable is a toric ideal.
The main reason why it is important whether an ideal is binomial is that there are dedicated algorithms for binomial ideals that are much faster than the generic algorithms that work for any ideal [10,13,22,23]. Note that there are some instances of ideals that arise in algebraic statistics that are not binomial in their natural coordinate systems but become binomial ideals after a linear change of coordinates [31].
Let $A ∈ Z h × r$ be an integer matrix, and consider the ideal
$I A : = ⟨ p u + - p u - : u = u + - u - ∈ ker Z A ⟩$
in the polynomial ring $k [ p 1 , ⋯ , p r ]$ , where $u = u + - u -$ is the decomposition of u into its positive and negative part $u + , u - ∈ N r$ . Clearly, $I A$ is binomial and does not contain any of the $p i$ . One can also show that $I A$ is prime, and thus it is an example of a toric ideal. In fact, any toric ideal is of this form up to a scaling of coordinates [13, Corollary 2.6]. The generating set above is infinite, but Theorem 3.1 in [9] shows that finite generating sets of toric ideals are related to Markov bases, which can be computed using the software 4ti2 [1].
#### 3.2.4 Real algebraic geometry
In addition to polynomial equations, in many situations in statistics it is useful to consider solutions to polynomial inequalities as well. This is the subject of the field real algebraic geometry. Inequalities only make sense over an ordered field like $R$ (but not over $C$ ). For simplicity, the following definitions and results are formulated with $R$ . Again, this text only contains the basic definitions. For more details the reader is referred to [4,5].
Let $F , G ⊆ R [ p 1 , … , p r ]$ be sets of polynomials with $G$ finite. The basic semialgebraic set defined by $F$ and $G$ is
${ a ∈ R r : f ( a ) = 0 for all f ∈ F and g ( a ) > 0 for all g ∈ G } .$
A semialgebraic set is a finite union of basic semialgebraic sets.
Here are some common examples of semialgebraic sets arising in statistics.
The open probability simplex
$int ( Δ r - 1 ) : = { p ∈ R r : ∑ i = 1 r p i = 1 , p i > 0 , i = 1 , … , r }$
consists of all probability distributions for a categorical random variable with r states. It is a basic semialgebraic set: In the above definition, one may take $F = { ∑ i = 1 r p i - 1 }$ and $G = { p 1 , … , p r }$ . The probability simplex
$Δ r - 1 : = { p ∈ R r : ∑ i = 1 r p i = 1 , p i ≥ 0 , i = 1 , … , r }$
is a semialgebraic set. It can be written as the union of $2 r - 1$ basic semialgebraic sets.
The cone $P D m$ of $m × m$ positive definite symmetric matrices is an example of a basic semialgebraic set in $R m + 1 2$ , where $F = ∅$ and where $G$ consists of the principal subdeterminants of an $m × m$ symmetric matrix of indeterminates. For instance, if $m = 3$ consider the polynomial ring $R [ σ 11 , σ 12 , σ 13 , σ 22 , σ 23 , σ 33 ]$ and the symmetric matrix of indeterminates
$Σ = σ 11 σ 12 σ 13 σ 12 σ 22 σ 23 σ 13 σ 23 σ 33 .$
The symmetry has been enforced by making certain entries in the matrix equal. The set of polynomials defining $P D 3$ can be chosen to be
$G = { σ 11 , σ 11 σ 22 - σ 12 2 , det Σ } ,$
the set of leading principal minors of $Σ$ . The cone of positive semidefinite symmetric matrices is a semialgebraic set, which can be realized by using non-strict inequalities with the much larger set of all principal minors of $Σ$ .
#### 3.3 Conditional Independence Ideals
This section shows how the algebraic tools introduced in Section 3.2 can be used to analyze conditional independence structures. The tools can be applied in the settings of discrete random variables and jointly normal variables, but in different ways.
#### 3.3.1 Discrete random variables
Let $X 1 , X 2 , … , X m$ be finite discrete random variables. Suppose that the state space of $X i$ is $[ r i ] : = { 1 , 2 , … , r i }$ . There is an algebraic description of the set of all distributions that satisfy a given conditional independence statement. The first example comes from the simplest CI statement: $1 ⊥ 2$ .
Let $X 1 , X 2$ be discrete random variables where the state space of $X i$ is $[ r i ]$ . Let $p i 1 i 2 = P ( X 1 = i 1 , X 2 = i 2 )$ and let $p = ( p i 1 i 2 ) i 1 ∈ [ r 1 ] , i 2 ∈ [ r 2 ]$ be the joint probability mass function of $X 1$ and $X 2$ . Then $1 ⊥ 2$ if and only if p is a rank one matrix.
If $1 ⊥ 2$ then $P ( X 1 = i 1 , X 2 = i 2 ) = P ( X 1 = i 1 ) P ( X 2 = i 2 )$ . This expresses the joint probability mass function as an outer product of two nonzero vectors, hence p has rank one.
Conversely, if p has rank one, it is expressed as the outer product of two vectors $p = α T β$ . Since p is a matrix of nonnegative real numbers, one can assume that $α$ and $β$ are also nonnegative. Let $‖ . ‖ 1$ denote the $l 1$ -norm. Replacing $α$ by $α / ‖ α ‖ 1$ and $β$ by $β / ‖ β ‖ 1$ , yields a rank one factorization for p where the two factors are necessarily the marginal distributions of $X 1$ and $X 2$ respectively. Hence $1 ⊥ 2$ .
A nonzero matrix having rank one is characterized by the vanishing of all its $2 × 2$ subdeterminants. Hence, one can associate an ideal to the independence statement $1 ⊥ 2$ .
The conditional independence ideal for the statement $1 ⊥ 2$ is
$I 1 ⊥ 2 = ⟨ p i 1 i 2 p j 1 j 2 - p i 1 j 2 p j 1 i 2 : i 1 , j 1 ∈ [ r 1 ] , i 2 , j 2 ∈ [ r 2 ] ⟩ = ⟨ 2 × 2 subdeterminants of p ⟩ ⊆ R [ p i 1 , i 2 : i 1 ∈ [ r 1 ] , i 2 ∈ [ r 2 ] ] .$
Let $r 1 = 2$ and $r 2 = 3$ . Then
$I 1 ⊥ 2 = ⟨ p 11 p 22 - p 12 p 21 , p 11 p 23 - p 13 p 21 , p 12 p 23 - p 13 p 22 ⟩ .$
The conditional independence ideal $I 1 ⊥ 2$ captures the algebraic structure of the independence condition. Although all probability distributions would satisfy the additional constraint that $∑ i 1 ∈ [ r 1 ] , i 2 ∈ [ r 2 ] p i 1 i 2 - 1 = 0$ , this trivial constraint is not included in the conditional independence ideal because leaving it out tends to simplify certain algebraic calculations. For example, without this constraint $I 1 ⊥ 2$ is a binomial ideal.
More generally, any conditional independence condition for discrete random variables can be expressed by similar determinantal constraints. This requires a bit of notation. The determinantal constraints are written in terms of the entries of the joint distribution of $X 1 , ⋯ , X m$ . This is a tensor $p = ( p i 1 , ⋯ , i m ) i j ∈ [ r j ]$ .
Let $A , B , C ⊂ [ m ]$ be disjoint subsets of indices of the random variables $X 1 , ⋯ , X m$ , and $D = [ m ] \ ( A ∪ B ∪ C )$ the set of indices appearing in none of ABC. Any such assignment yields a grouping of indices and random variables. The random vector $X A = ( X j ) j ∈ A$ takes values in $R A = ∏ j ∈ A [ r j ]$ . Let $R B , R C$ and $R D$ be defined analogously. The grouping allows one to write $p = ( p i A , i B , i C , i D )$ where now $i A ∈ R A$ and similarly for $i B , i C$ , and $i D$ . The final notational gadget is the marginalization of p over D. The entries of this marginal distribution are indexed by $R A , R B , R C$ and have entries
$p i A , i B , i C , + = ∑ i D ∈ R D p i A , i B , i C , i D .$
The $+$ indicates the summation.
The conditional independence ideal for the conditional independence statement $A ⊥ B [ C ]$ is
$I A ⊥ B [ C ] = ⟨ p i A , i B , i C , + · p j A , j B , i C , + - p i A , j B , i C , + · p j A , i B , i C , + , for all i A , j A ∈ R A , i B , j B ∈ R B , i C ∈ R C ⟩ .$
The notation simplifies for saturated conditional independence statements, for which $A ∪ B ∪ C = [ m ]$ . With this condition there is no marginalization, and the defining polynomials of $I A ⊥ B [ C ]$ are binomials.
Consider three binary random variables $X 1 , X 2 , X 3$ . Let $p 111 , ⋯ , p 222$ denote the indeterminates standing for the elementary probabilities in the joint distribution. The conditional independence ideal of the statement $1 ⊥ 3 [ 2 ]$ is
$I 1 ⊥ 3 [ 2 ] = ⟨ p 111 p 212 - p 211 p 112 , p 121 p 222 - p 221 p 122 ⟩ .$
The conditional independence ideal of the statement $1 ⊥ 3$ is
$I 1 ⊥ 3 = ⟨ ( p 111 + p 121 ) ( p 212 + p 222 ) - ( p 112 + p 122 ) ( p 211 + p 221 ) ⟩ .$
For any conditional independence statement $A ⊥ B [ C ]$ , the conditional independence ideal $I A ⊥ B [ C ]$ is a prime ideal and hence $V ( I A ⊥ B [ C ] )$ is an irreducible variety.
Proposition 1.3.2 is a consequence of the fact that general determinantal ideals are prime (see [6]). Irreducibility of the variety $V ( I A ⊥ B [ C ] )$ can also be deduced from the fact that this variety can be parametrized, for instance, the set of all probability distributions in $V ( I A ⊥ B [ C ] )$ can be realized as the set of probability distributions in a graphical model.
A strictly positive joint distribution p of binary random variables $X 1 , X 2 , X 3$ satisfies $1 ⊥ 3 [ 2 ]$ if and only if
1 $p i 1 , i 2 , i 3 = s i 1 , i 2 t i 2 , i 3$
for some vectors $( s i 1 , i 2 ) i 1 ∈ [ r 1 ] , i 2 ∈ [ r 2 ]$ , $( t i 2 , i 3 ) i 2 ∈ [ r 2 ] , i 3 ∈ [ r 3 ]$ ; see Section . That is, it lies in the undirected graphical model
$X 1 - X 2 - X 3 .$
Since $V ( I A ⊥ B [ C ] )$ is irreducible, any joint distribution (possibly with zeros) that satisfies $1 ⊥ 3 [ 2 ]$ lies in the closure of the undirected graphical model. In fact, any such joint distribution has a parametrization of the form (1), where s or t also may have zeros.
More interesting than just single statements are combinations of two or more conditional independence statements. To determine the classes of distributions satisfying a collection of independence statements leads to interesting problems in computational algebra. Such sets are typically not irreducible varieties and cannot be parametrized with a single parametrization. The first task is to break such a set into components, and to see if those components have natural interpretations in terms of conditional independence and can be parametrized.
Let $C = { A 1 ⊥ B 1 [ C 1 ] , A 2 ⊥ B 2 [ C 2 ] , … }$ be a set of conditional independence statements for the random variables $X 1 , X 2 , … , X m$ . The conditional independence ideal of $C$ is the sum of the conditional independence ideals of the elements of $C$ :
$I C = I A 1 ⊥ B 1 [ C 1 ] + I A 2 ⊥ B 2 [ C 2 ] + ⋯ .$
Understanding the probability distributions that satisfy $C$ can be accomplished by analyzing an irreducible decomposition of $V ( I C )$ , which can be obtained from a primary decomposition of $I C$ .
Let $X 1$ , $X 2$ , $X 3$ be binary random variables, and consider $C = { 1 ⊥ 3 [ 2 ] , 1 ⊥ 3 }$ . The conditional independence ideal $I C$ is generated by three polynomials of degree 2:
$I C = I 1 ⊥ 3 [ 2 ] + I 1 ⊥ 3 = ⟨ p 111 p 212 - p 112 p 211 , p 121 p 222 - p 122 p 221 , ( p 111 + p 121 ) ( p 212 + p 222 ) - ( p 112 + p 122 ) ( p 211 + p 221 ) ⟩ .$
The following Macaulay2 code asks for the primary decomposition of this ideal over $Q$ . It can be shown that the decomposition is the same over $R$ and $C$ . loadPackage "GraphicalModels" S = markovRing (2,2,2) L = {{{1},{3},{2}}, {{1},{3},{}}} I = conditionalIndependenceIdeal(S,L) primaryDecomposition I
This code uses the GraphicalModels package of Macaulay2 which implements many convenient functions to work with graphical and other conditional independence models. In particular, it allows to easily set up the polynomial ring with eight variables $p 111 , ⋯ , p 222$ with ’markovRing’ and write out the equations for $I C$ with ’conditionalIndependenceIdeal’. The command ’primaryDecomposition’ is a generic Macaulay2 command. The output of this code consists of two ideals which upon inspection can be recognized as binomial conditional independence ideals themselves. The result is
$I C = I { 1 , 2 } ⊥ 3 ∩ I 1 ⊥ { 2 , 3 } .$
According to Section 1.2 this implies a decomposition of varieties
$V ( I C ) = V ( I { 1 , 2 } ⊥ 3 ) ∪ V ( I 1 ⊥ { 2 , 3 } ) .$
On the level of probability distributions, this proves the binary case of Proposition 1.1.2.
The general situation may be less favorable than that in Example 1.3.4. In particular, the components that appear need not have interpretations in terms of conditional independence. The appearing ideals also need not be prime ideals (in general they are only primary) and it is unclear what this algebraic extra information may reveal about conditional independence. For examples on how to extract information from primary decompositions see [20,24].
#### 3.3.2 Gaussian random variables
Algebraic approaches to conditional independence can also be applied to Gaussian random variables. Let $X ∈ R m$ be a nonsingular multivariate Gaussian random vector with mean $μ ∈ R m$ and covariance matrix $Σ ∈ P D m$ , the cone of $m × m$ symmetric positive definite matrices. One writes $X ∼ N ( μ , Σ )$ . For subsets $A , B ⊆ [ m ]$ let $Σ A , B$ be the submatrix of $Σ$ obtained by extracting rows indexed by A and columns indexed by B, that is $Σ A , B = ( σ a , b ) a ∈ A , b ∈ B$ .
Let $X ∼ N ( μ , Σ )$ with $Σ ∈ P D m$ . Let $A , B , C ⊆ [ m ]$ be disjoint subsets. Then the conditional independence statement $A ⊥ B [ C ]$ holds if and only if the matrix $Σ A ∪ C , B ∪ C$ has rank $≤ # C$ .
A proof of this proposition recognizes $Σ A ∪ C , B ∪ C$ as a Schur complement of a submatrix of $Σ$ . The details can be found in [17, Proposition 3.1.13]; see also Section 9.1 in the present handbook.
Just as the rank one condition on a matrix was characterized by the vanishing of $2 × 2$ subdeterminants, higher rank conditions on matrices can also be characterized by the vanishing of subdeterminants. Indeed, a basic fact of linear algebra is that a matrix has rank $≤ r$ if and only if the determinant of every $( r + 1 ) × ( r + 1 )$ submatrix is zero. This leads to the conditional independence ideals for multivariate Gaussian random variables.
Let $R [ Σ ] : = R [ σ ij : 1 ≤ i ≤ j ≤ m ]$ be the polynomial ring with real coefficients in the entries of the symmetric matrix $Σ$ .
The Gaussian conditional independence ideal for the conditional independence statement $A ⊥ B [ C ]$ is the ideal
$J A ⊥ B [ C ] : = ⟨ ( # C + 1 ) -minors of Σ A ∪ C , B ∪ C ⟩ .$
If $C = { A 1 ⊥ B 1 [ C 1 ] , A 2 ⊥ B 2 [ C 2 ] , … , }$ is a collection of conditional independence statements, the Gaussian conditional independence ideal is
$J C = J A 1 ⊥ B 1 [ C 1 ] + J A 2 ⊥ B 2 [ C 2 ] + ⋯ .$
A common criterion in statistics says that, in fact, $A ⊥ B [ C ]$ holds if and only if $det ( Σ { a } ∪ C , { b } ∪ C )$ vanishes for all $a ∈ A$ , $b ∈ B$ . Since, by assumption, C is non-singular, it is easy to see that this condition is, in fact, equivalent to the vanishing of all $( # C + 1 )$ -minors of $Σ A ∪ C , B ∪ C$ .
Consider the conditional independence statement $2 ⊥ { 1 , 3 } [ 4 ]$ . The ideal $J 2 ⊥ { 1 , 3 } [ 4 ]$ is generated by the $2 × 2$ minors of the matrix
$Σ { 2 , 4 } , { 1 , 3 , 4 } = σ 12 σ 23 σ 24 σ 14 σ 34 σ 44 .$
Since $Σ$ is a symmetric matrix, $σ ij = σ ji$ and one always writes $σ ij$ with $i ≤ j$ . Then
$J 2 ⊥ { 1 , 3 } [ 4 ] = ⟨ σ 12 σ 34 - σ 14 σ 23 , σ 12 σ 44 - σ 14 σ 24 , σ 23 σ 44 - σ 34 σ 24 ⟩ .$
Let $X 1$ , $X 2$ , $X 3$ be jointly Gaussian random variables. The conditional independence ideal of $C = { 1 ⊥ 3 [ 2 ] , 1 ⊥ 3 }$ is
$J C = J 1 ⊥ 3 [ 2 ] + J 1 ⊥ 3 = ⟨ σ 13 σ 22 - σ 12 σ 23 , σ 13 ⟩ .$
Straightforward manipulations of these ideals show
$J C = ⟨ σ 13 σ 22 - σ 12 σ 23 , σ 13 ⟩ = ⟨ σ 12 σ 23 , σ 13 ⟩ = ⟨ σ 12 , σ 13 ⟩ ∩ ⟨ σ 23 , σ 13 ⟩$
$= J { 1 , 2 } ⊥ 3 ∩ J 1 ⊥ { 2 , 3 } .$
This last primary decomposition proves the Gaussian case of Proposition 1.1.2.
#### 3.3.3 The contraction axiom
When computing the decompositions of conditional independence ideals, there might be components that are “uninteresting" from the statistical standpoint. These components might not intersect the region of interest in probabilistic applications (e.g. they might miss the probability simplex or the cone of positive definite matrices) or they might have non-trivial intersections but that intersection is contained in some other component.
Let $X = ( X 1 , X 2 , X 3 )$ be a multivariate Gaussian random vector. The conditional independence ideal of $C = { 1 ⊥ 2 [ 3 ] , 2 ⊥ 3 }$ is
$J C = ⟨ σ 12 σ 33 - σ 13 σ 23 , σ 23 ⟩$
which has primary decomposition
$J C = ⟨ σ 12 σ 33 , σ 23 ⟩ = ⟨ σ 12 , σ 23 ⟩ ∩ ⟨ σ 33 , σ 23 ⟩ .$
This decomposition shows that
$V ( J C ) = V ( ⟨ σ 12 , σ 23 ⟩ ) ∪ V ( ⟨ σ 33 , σ 23 ⟩ ) .$
However, the second component does not intersect the positive definite cone, because $σ 33 > 0$ for all $Σ ∈ P D 3$ . The first component is the conditional independence ideal $J 1 , 3 ⊥ 2$ . From this decomposition it is visible that $1 ⊥ 2 [ 3 ]$ and $2 ⊥ 3$ imply that $1 , 3 ⊥ 2$ . This implication is called the contraction axiom. See Section for other CI axioms.
The contraction axiom also holds for non-Gaussian random variables. For discrete random variables, it can again be checked algebraically. The primary decomposition associated to the discrete contraction axiom is worked out in detail in [17]. The next example discusses the binary case as an illustration:
Let $X 1 , X 2 , X 3$ be binary random variables. The conditional independence ideal of $C = { 1 ⊥ 2 [ 3 ] , 2 ⊥ 3 }$ is
$I C = ⟨ p 111 p 221 - p 121 p 211 , p 112 p 222 - p 122 p 212 , ( p 111 + p 211 ) ( p 122 + p 222 ) - ( p 112 + p 212 ) ( p 121 + p 221 ) ⟩$
which has primary decomposition
$I C = I 1 , 3 ⊥ 2 ∩ ⟨ p 122 + p 222 , p 112 + p 212 , p 121 p 211 - p 111 p 221 ⟩ ∩ ⟨ p 121 + p 221 , p 111 + p 211 , p 122 p 212 - p 112 p 222 ⟩ .$
The intersection of the second component with the probability simplex forces that $p 122 = p 222 = p 112 = p 212 = 0$ . This in turn implies that
$V ( ⟨ p 122 + p 222 , p 112 + p 212 , p 121 p 211 - p 111 p 221 ⟩ ) ∩ Δ 7 ⊆ V ( I 1 , 3 ⊥ 2 )$
A similar argument holds for the third component. So although the variety $V ( I C )$ has three components, only one of them is statistically meaningful:
$V ( I C ) ∩ Δ 7 = V ( I 1 , 3 ⊥ 2 ) ∩ Δ 7 .$
#### 3.4 Examples of Decompositions of Conditional Independence Ideals
This section studies some examples of families of conditional independence statements and how algebraic tools can be used to understand them. The first example is a detailed study of the intersection axiom, and the second example concerns the conditional independence statements associated to the 4-cycle graph.
#### 3.4.1 The intersection axiom
The intersection axiom from Section is the following implication of conditional independence statements:
2 $A ⊥ B [ C ∪ D ] and A ⊥ C [ B ∪ D ] ⇒ A ⊥ B ∪ C [ D ] .$
This implication is valid for strictly positive probability distributions. Algebraic techniques can be used to study how its validity extends beyond this.
The question about the primary decomposition of the ideal(s) $I { A ⊥ B [ C ∪ D ] , A ⊥ C [ B ∪ D ] }$ was first asked in [15, Chapter 6.6], and the answer is due to [15]. Grouping variables if necessary, one can assume $A = { 1 }$ , $B = { 2 }$ , $C = { 3 }$ . Moreover, let $D = ∅$ . From this one can always recover the general case by adding conditioning constraints.
[Proposition 1 in [15]] The ideal $I { X 1 ⊥ X 2 [ X 3 ] , X 1 ⊥ X 3 [ X 2 ] }$ is radical, that is, its irredundant primary decompositions consists only of prime ideals. These minimal primes correspond to pairs of partitions $[ r 2 ] = A 1 ∪ ⋯ ∪ A s$ , $[ r 3 ] = B 1 ∪ ⋯ ∪ B s$ of the same size. The minimal prime P corresponding to two partitions is
$P = ⟨ p i 1 i 2 i 3 : i 1 ∈ [ r 1 ] , i 2 ∈ A j , i 3 ∈ B k for some j ≠ k ⟩ + ⟨ p i 1 i 2 i 3 p i 1 ′ i 2 ′ i 3 ′ - p i 1 i 2 ′ i 3 ′ p i 1 ′ i 2 i 3 : i 1 , i 1 ′ ∈ [ r 1 ] , i 2 , i 2 ′ ∈ A j , i 3 , i 3 ′ ∈ B j for some j ⟩ .$
The paper [15] uses a different formulation in terms of complete bipartite graphs. It can be seen that our formulation is equivalent.
To give a statistical interpretation to Proposition 1.4.1, whenever the joint distribution of $X 1 , X 2 , X 3$ lies in the prime P corresponding to the two partitions $[ r 2 ] = A 1 ∪ ⋯ ∪ A s$ , $[ r 3 ] = B 1 ∪ ⋯ ∪ B s$ , construct a random variable B as follows: put $B : = j$ whenever $X 2 ∈ A j$ and $X 3 ∈ B j$ . Thus, B is uniquely defined except on a set of measure zero, since $P ( X 2 ∈ A j , X 3 ∈ B k ) = 0$ for $j ≠ k$ , which follows from the containment of monomials in P. The variable B specifies in which blocks of the two partitions the random variables $X 2$ and $X 3$ lie. Now the binomials in P imply that $X 1 ⊥ { X 2 , X 3 } [ B ]$ .
Suppose that $X 1 , X 2 , X 3$ satisfy $X 1 ⊥ X 2 [ X 3 ]$ and $X 1 ⊥ X 3 [ X 2 ]$ . Then there is a random variable B that satisfies:
1. B is a (deterministic) function of $X 2$ ;
2. B is a (deterministic) function of $X 3$ ;
3. $X 1 ⊥ { X 2 , X 3 } [ B ]$ .
Conversely, whenever there exists a random variable B with properties 1. to 3., the random variables $X 1 , X 2 , X 3$ satisfy $X 1 ⊥ X 2 [ X 3 ]$ and $X 1 ⊥ X 3 [ X 2 ]$ .
The case where B is a constant corresponds to the CI statement $X 1 ⊥ { X 2 , X 3 }$ . The intersection axiom can be recovered by noting that a function B that is a function of only $X 2$ as well as a function of only $X 3$ is necessarily constant, if the joint distribution of $X 2$ and $X 3$ is strictly positive.
Similar results hold for all families of CI statements of the form $A ⊥ B i [ C i ]$ , where $A ∪ B i ∪ C i = V$ and where A is fixed for all statements, see [20,20] (the case where $X A$ is binary was already described in [20]). The corresponding CI ideal is still radical, and the minimal primes have a similar interpretation. However, finding the minimal primes is more difficult and involves solving a combinatorial problem. Finally, Corollary 1.4.1 can be generalized to continuous random variables [27].
#### 3.4.2 The four-cycle
Consider four discrete random variables $X 1 , X 2 , X 3 , X 4$ and the (undirected) graphical model of the four cycle $C 4 = ( V , E )$ with edge set $E = { ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 1 , 4 ) }$ . The global Markov CI statements of this graph are
$global ( C 4 ) = { 1 ⊥ 3 [ { 2 , 4 } ] , 2 ⊥ 4 [ { 1 , 3 } ] } .$
The primary decomposition of the corresponding CI ideal was studied in [24] in the case where $X 1$ and $X 3$ are binary. The case that all variables are binary is as follows:
[Theorem 5.6 in [24]] The minimal primes of the CI ideal $I global ( C 4 )$ of the binary four cycle are the toric ideal $I C 4$ and the monomial ideals
$P i = ⟨ p x : x i = x i + 1 ⟩ , P i ′ = ⟨ p x : x i ≠ x i + 1 ⟩ for 1 ≤ i < 4 .$
The ideal $I C 4$ equals the vanishing ideal of the graphical model, to be discussed in Section 1.5. Interestingly, in this primary decomposition, all ideals except $I C 4$ are monomial ideals; that is, they only give support restrictions on the probability distribution.
The primary decomposition of the CI ideal gives an irreducible decomposition of the corresponding set of probability distributions. This leads to the statement of Proposition 1.1.1 in the introduction.
When $X 2$ and $X 4$ are not binary, the decomposition of $I global ( C 4 )$ involves prime ideals parametrized by $i ∈ { 2 , 4 }$ and two sets $∅ ≠ C , D ⊊ [ r i ]$ . For such a choice of iCD, let j denote the element of ${ 2 , 4 } \ { i }$ , and let
$P i , C , D = ⟨ p 1 x 2 1 x 4 : x i ∈ C , x j ∈ [ r j ] ⟩ + ⟨ p 1 x 2 2 x 4 : x i ∈ D , x j ∈ [ r j ] ⟩ + ⟨ p 2 x 2 1 x 4 : x i ∉ D , x j ∈ [ r j ] ⟩ + ⟨ p 2 x 2 2 x 4 : x i ∉ C , x j ∈ [ r j ] ⟩ + I global ( C 4 )$
the result is the following:
[Theorem 6.5 in [24]] Let $X 1$ and $X 3$ be binary random variables. The minimal primes of the CI ideal $I global ( C 4 )$ of the four cycle are the toric ideal $I C 4$ and the ideals $P i , C , D$ for $i ∈ { 2 , 4 }$ and $∅ ≠ C , D ⊊ [ r i ]$ . Furthermore the ideal is radical and thus equals the intersection of its minimal primes.
In this case, the non-toric primes are not monomial, but consist of monomials and binomials. This fact is independent of the field $k$ . The following is one example of the kind of information that can be extracted from knowing the minimal primes of a conditional independence ideal.
Let $X 1 , X 2 , X 3 , X 4$ be finite random variables that satisfy $global ( C 4 )$ , and suppose that $X 1$ and $X 3$ are binary. Then one (or more) of the following statements is true:
1. The joint distribution lies in the closure of the graphical model.
2. There is $i ∈ { 2 , 4 }$ and there are sets $E , F ⊆ [ r i ]$ such that the following holds:
$If ( X 1 , X 3 ) = ( 1 , 1 ) ( 1 , 2 ) ( 2 , 1 ) ( 2 , 2 ) , then X i ∈ E X i ∈ F X i ∉ F X i ∉ E .$
Conversely, any probability distribution that satisfies one of these statements and that satisfies $2 ⊥ 4 [ { 1 , 3 } ]$ also satisfies $global ( C 4 )$ .
#### 3.5 The Vanishing Ideal of a Graphical Model
Graphical models can be represented via either parametric descriptions (e.g. factorizations of the density function) or implicit descriptions (e.g. Markov properties and conditional independence constraints). One use of the algebraic perspective on graphical models is to find the complete implicit description of the model, in particular, to find the vanishing ideal of the model. As described in Definition 1.2.4, the vanishing ideal of a set S is the set of all polynomial functions that evaluate to zero at every point in S. Although some graphical models have complete descriptions only in terms of conditional independence constraints, understanding the vanishing ideal can be useful for more complex models or hidden variable models where conditional independence is not sufficient to describe the model, for instance, the mixed graph models studied in Chapter 2.
Consider the four cycle $C 4$ and let $X 1 , X 2 , X 3 , X 4$ be binary random variables. The vanishing ideal $I C 4 ⊆ R [ p i 1 i 2 i 3 i 4 : i 1 , i 2 , i 3 , i 4 ∈ { 1 , 2 } ]$ is generated by 16 binomials, 8 of which have degree 2 and 8 of which have degree 4. The degree 2 binomials are all implied by the two conditional independence statements $1 ⊥ 3 [ { 2 , 4 } ]$ and $2 ⊥ 4 [ { 1 , 3 } ]$ . On the other hand, the degree 4 binomials are not implied by the conditional independence constraints, even when we restrict to probability distributions. One example degree four polynomial is
$p 1111 p 1222 p 2122 p 2211 - p 1122 p 1211 p 2111 p 2222$
and the others are obtained by applying the symmetry group of the four cycle and permuting levels of the random variables.
Even in the simple example of the four cycle, there are generators of the vanishing ideal that do not correspond to conditional independence statements. It seems an important problem to try to understand what other types of equations can arise. Theorem 3.2 in [18] shows that the vanishing ideal of a graphical model of discrete random variables is the toric ideal $I A$ , where A is the design matrix of the graphical model, defined in the end of Section 3.2. A classification for discrete random variables and undirected graphs of when no further polynomials are needed beyond conditional independence constraints is obtained in the following theorem:
[Theorem 4.4 in [18]] Let G be an undirected graph and let $M G$ be its graphical model for discrete random variables. Then the vanishing ideal $I ( M G )$ is equal to the conditional independence ideal $I global ( G )$ if and only if G is a chordal graph.
It is unknown what the appropriate analogue of Theorem 1.5.1 is for other families of graphical models, either with different classes of graphs (e.g. DAGs) or with other types of random variables (e.g. Gaussian). Computational studies of the vanishing ideals appear in many different papers: for Bayesian networks with discrete random variables [17], for Bayesian networks with Gaussian random variables [33], for undirected graphical models with Gaussian random variables [32]. A characterization for which graph families the vanishing ideal is equal to the conditional independence ideal of global Markov statements is lacking in all these cases.
One natural question is to determine the other generators of the vanishing ideal that do not come from conditional independence, and to give combinatorial structures in the underlying graphs that imply that these more general constraints hold. For instance, for mixed Gaussian graphical models, conditional independence constraints are determinantal, but not every determinantal constraint comes from a conditional independence statement, and there is a characterization of which determinantal constraints come from conditional independence:
A mixed graph $G = ( [ m ] , B , D )$ is a triple where $[ m ] = { 1 , 2 , … , m }$ is the vertex set, B is a set of unordered pairs of [m] representing bidirected edges in G, and D is a set of ordered pairs of [m] representing directed edges in G. There might also be both directed and bidirected edges between a pair of vertices. To the set of B of bidirected edges one associates the set of symmetric positive definite matrices
$P D ( B ) = { Ω ∈ P D m : ω ij = 0 if i ≠ j and i ↔ j ∉ B } .$
To the set of directed edges one associates the set of $m × m$ matrices
$R D = { Λ ∈ R m × m : λ ij = 0 if i → j ∉ D } .$
Let $ϵ ∼ N ( 0 , Ω )$ and let X be a jointly normal random vector satisfying the structural equation system
$X = Λ T X + ϵ .$
This is an example of a linear structural equation model, and contains as special cases various families of graphical models. Let $I d$ denote the $m × m$ identity matrix. With these assumptions, if $( I d - Λ )$ is invertible,
$X ∼ N ( 0 , Σ ) where Σ = ( I d - Λ ) - T Ω ( I d - Λ ) - 1 .$
Figure 1 The mixed graph from Section 3.5.
Consider the mixed graph G from Figure 1. In this case, PD(B) is the set of positive definite matrices of the form
$Ω = ω 11 0 0 0 0 ω 22 0 0 0 0 ω 33 ω 34 0 0 ω 34 ω 44$
and $R D$ is the set of real matrices of the form
$Λ = 0 λ 12 λ 13 0 0 0 λ 23 0 0 0 0 λ 34 0 0 0 0 .$
A positive definite matrix $Σ$ belongs to the graphical model associated to this mixed graph, if and only if there are $Ω ∈ P D ( B )$ and $Λ ∈ R D$ such that $Σ = ( I d - Λ ) - T Ω ( I d - Λ ) - 1$ .
Let $G = ( [ m ] , B , D )$ be a mixed graph. A trek between vertices i and j in G consists of either
• a pair $( P L , P R )$ where $P L$ is a directed path ending in i and $P R$ is a directed path ending in j where both $P L$ and $P R$ have the same source, or
• a pair $( P L , P R )$ where $P L$ is a directed path ending in i and $P R$ is a directed path ending in j such that the source of $P L$ and the source of $P R$ are connected by a bidirected edge.
Let $T ( i , j )$ denote the set of all treks in G between i and j.
To each trek $T = ( P L , P R )$ one associates the trek monomial $m T$ which is the product with multiplicities of all $λ st$ over all directed edges appearing in T times $ω st$ where s and t are the sources of $P L$ and $P R$ . One reason for the interest in treks is the trek rule, which says for the Gaussian graphical model associated to G
$σ ij = ∑ T ∈ T ( i , j ) m T .$
For instance, for the mixed graph in Figure 1, the pair $( { 1 → 2 , 2 → 3 } , { 1 → 3 } )$ is a trek from 3 to 3. The corresponding trek monomial $m T$ is $ω 11 λ 12 λ 23 λ 13$ .
Let $A , B , C A , C B$ be four sets of vertices of G, not necessarily disjoint. The pair of sets $( C A , C B )$ t-separates (short for trek separates) A from B if for every $a ∈ A$ and $b ∈ B$ and every trek $( P L , P R ) ∈ T ( a , b )$ , either $P L$ has a vertex in $C A$ or $P R$ has a vertex in $C B$ , or both.
[35] Let $G = ( [ m ] , B , D )$ be a mixed graph and A and B two subsets of [m] with $| A | = | B | = k$ . Then the minor $det Σ A , B$ belongs to the vanishing ideal $I G$ if and only if there is a pair of sets $C A , C B$ such that $( C A , C B )$ t-separates A and B and such that $| C A | + | C B | < k$ .
The t-separation criterion can produce implicit constraints for structural equation models in situations where there are no conditional independence constraints.
Consider the mixed graph G from Figure 1. The vanishing ideal of the model is $I G = ⟨ det Σ { 1 , 2 } , { 3 , 4 } ⟩$ . This determinantal constraint is not a conditional independence constraint. It is implied by the t-separation criterion because the pair $( ∅ , { 3 } )$ t-separates ${ 1 , 2 }$ and ${ 3 , 4 }$ .
In the case where there are hidden random variables, the vanishing ideal is typically not sufficient to completely describe the set of probability distributions that come from the graphical model. Usually one also needs to consider inequality constraints and other semialgebraic conditions. This problem is discussed in more detail in [2,3,37], among others.
Diaconis, Eisenbud and Sturmfels [8] were the first to consider primary decompositions for statistical applications, in particular the analysis of the connectivity of certain random walks. This perspective was also picked up in [24] using conditional independence ideals. Primary decomposition of conditional independence ideals also makes an appearance in the following papers not already mentioned [12,16,25,34,36].
The algebraic view on undirected graphical models was presented in [18], which began extensive study of the vanishing ideals of undirected graphical models for discrete random variables. Focus has been on developing techniques for constructing generating sets of the vanishing ideals with [14,21,30] being representative papers in this area. Vanishing ideals of undirected models with Gaussian random variables and models for DAGs have not been much studied. Some papers that initiated their study include [17,32,33].
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8
Diaconis Persi , Eisenbud David , Sturmfels Bernd . Lattice walks and primary decomposition. In Mathematical essays in honor of Gian-Carlo Rota (Cambridge, MA, 1996), volume 161 of Progr. Math.. Boston, MA: Birkhäuser Boston; 1998. pp. 173 193--
9
Diaconis Persi , Sturmfels Bernd . Algebraic algorithms for sampling from conditional distributions. Ann. Statist. 1998;26:363–397.
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Dickenstein Alicia . Laura Felicia Matusevich, and Ezra Miller. Combinatorics of binomial primary decomposition. Math. Z. 2010;264(4):745–763.
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Drton Mathias , Sturmfels Bernd , Sullivant Seth . Lectures on Algebraic Statistics, volume 39 of Oberwolfach Seminars. Berlin: Springer; 2009. A Birkhäuser book
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Drton Mathias , Xiao Han . Smoothness of Gaussian conditional independence models. In Algebraic methods in statistics and probability II, volume 516 of Contemp. Math.. Amer. Math. Soc., Providence, RI. 2010. Amer. Math. Soc., Providence, RI. pp. 155–177.
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Eisenbud David , Sturmfels Bernd . Binomial ideals. Duke Math. J. 1996;84(1):1–45.
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Engström Alexander , Kahle Thomas , Sullivant Seth . Multigraded commutative algebra of graph decompositions. J. Algebraic Combin. 2014;39(2):335–372.
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Fink Alex . The binomial ideal of the intersection axiom for conditional probabilities. J. Algebraic Combin. 2011;33(3):455–463.
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Fink Alex , Rajchgot Jenna , Sullivant Seth . Matrix Schubert varieties and Gaussian conditional independence models. srXiv:1510.04124, 2015
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Luis David Garcia. Michael Stillman, and Bernd Sturmfels. Algebraic geometry of Bayesian networks. J. Symbolic Comput. 2005;39(3–4):331–355.
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Geiger Dan , Meek Christopher , Sturmfels Bernd . On the toric algebra of graphical models. Ann. Statist. 2006;34(3):1463–1492.
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. Macaulay2, a software system for research in algebraic geometry : Available at http://www.math.uiuc.edu/Macaulay2/.
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Herzog Jürgen , Hibi Takayuki , Hreinsdóttir Freyja , Kahle Thomas , Rauh Johannes . Binomial edge ideals and conditional independence statements. Adv. in Appl. Math. 2010;45(3):317–333.
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Hoşten Serkan , Sullivant Seth . Gröbner bases and polyhedral geometry of reducible and cyclic models. J. Combin. Theory Ser. A. 2002;100(2):277–301.
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Kahle Thomas . Decompositions of binomial ideals. J. Softw. Algebra Geom. 2012;4:1–5.
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Kahle Thomas , Miller Ezra , O’Neill Christopher . Irreducible decomposition of binomial ideals. Compos. Math. 2016;152(6):1319–1332.
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Rauh Johannes . Generalized binomial edge ideals. Adv. in Appl. Math. 2013;50(3):409–414.
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Rauh Johannes , Ay Nihat . Robustness, canalyzing functions and systems design. Theory Biosci. 2014;133(2):63–78.
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Rauh Johannes , Sullivant Seth . Lifting Markov bases and higher codimension toric fiber products. J. Symbolic Comput. 2016;74:276–307.
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Sturmfels Bernd , Sullivant Seth . Toric ideals of phylogenetic invariants. J. Comp. Biol. 2005;12:204–228.
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Sturmfels Bernd , Uhler Caroline . Multivariate Gaussian, semidefinite matrix completion, and convex algebraic geometry. Ann. Inst. Statist. Math. 2010;62(4):603–638.
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Sullivant Seth . Algebraic geometry of Gaussian Bayesian networks. Adv. in Appl. Math. 2008;40(4):482–513.
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Sullivant Seth . Gaussian conditional independence relations have no finite complete characterization. J. Pure Appl. Algebra. 2009;213(8):1502–1506.
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Sullivant Seth , Talaska Kelli , Draisma Jan . Trek separation for Gaussian graphical models. Ann. Statist. 2010;38(3):1665–1685.
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Zwiernik Piotr . Semialgebraic statistics and latent tree models. vol. 146., Monographs on Statistics and Applied Probability Boca Raton, FL: Chapman & Hall/CRC; 2016.
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# Problem 29. Nearest Numbers
Solution 2715387
Submitted on 20 Jul 2020 by Jan Olsen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [30 46 16 -46 35 44 18 26 25 -10]; correct = [8 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
2 Pass
A = [1555 -3288 2061 -4681 -2230 -4538 -4028 3235 1949 -1829]; correct = [3 9]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
3 Pass
A = [-1 1 10 -10]; correct = [1 2]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
4 Pass
A = [0 1000 -2000 1001 0]; correct = [1 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
5 Pass
A = [1:1000 0.5]; correct = [1 1001]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
6 Pass
% Area codes A = [847 217 508 312 212]; correct = [2 5]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
7 Pass
% Zip codes A = [60048 61802 01702 60601 10001]; correct = [1 4]; [i1 i2] = nearestNumbers(A); assert(isequal([i1 i2],correct))
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Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 459 | 1,259 | {"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} | 3.25 | 3 | CC-MAIN-2020-50 | latest | en | 0.451352 |
https://www.meritnation.com/cbse-class-6/math/studymaterial/understanding-elementary-shapes/revision-notes/8_1_1_5_8223 | 1,638,257,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00080.warc.gz | 985,195,226 | 14,287 | Select Board & Class
Understanding Elementary Shapes
• One complete turn of the hand of a clock is one revolution. The angle of one revolution is called a complete angle.
• A right angle is of a revolution and a straight angle is of a revolution.
• 1 complete angle = 2 straight angles = 4 right angles
• 1 straight angle = 2 right angles
• If an angle measures less than a right angle then it is known as an acute angle.
The following angles are acute:
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The following angles are obtuse: …
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Syllabus | 157 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-49 | longest | en | 0.8725 |
https://appdividend.com/2022/06/18/java-math-max/ | 1,675,697,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00761.warc.gz | 116,341,141 | 23,508 | # Java math.max Function: The Complete Guide
The math.max() method in Java returns the maximum value among the specified arguments. Let’s deep dive into it.
## math.max Java
Java math.max() is a built-in function that compares two numbers (int, float, double, or long type) and returns a maximum of those numbers. The math.max() function takes an int, double, float, and long and returns a maximum of two numbers.
### Syntax
```public static int max(int a, int b)
public static double max(double a, double b)
public static float max(float a, float b)
public static long max(long a, long b)```
### Parameters
Two numbers out of which the maximum is to be determined.
### Return Value
The maximum of the two arguments.
See the following figure.
#### Note
1. If one argument is positive and the other is negative, the positive argument is returned.
2. If both arguments are negative, then the one with the lower magnitude is returned.
3. If one argument is positive zero, and the other argument is negative zero, then positive zero is returned. This is because the max() method considers negative zero to be smaller than positive zero (unlike numerical comparison operators).
4. If either argument is NaN, then the final result is NaN.
Consider the following examples.
#### Example1.java: The following example demonstrates comparing two int, float, double, or long values.
See the following code.
```public class Example1 {
public static void main(String[] args) {
int i1 = 2;
int i2 = 3;
float f1 = 2.0f;
float f2 = 3.0f;
double d1 = 2.0;
double d2 = 3.0;
long l1 = 2000000;
long l2 = 3000000;
System.out.println(Math.max(i1, i2));
System.out.println(Math.max(f1, f2));
System.out.println(Math.max(d1, d2));
System.out.println(Math.max(l1, l2));
}
}```
#### Output
```->javac Example1.java
->java Example1
3
3.0
3.0
3000000
```
#### Example2.java: The following example demonstrates comparing a positive and a negative argument.
```public class Example2 {
public static void main(String[] args) {
int a = 34;
int b = -45;
System.out.println(Math.max(a, b));
}
}
```
#### Output
```->javac Example2.java
->java Example2
34
```
#### Example3.java: The following example demonstrates comparing two negative arguments.
```public class Example3 {
public static void main(String[] args) {
int a = -34;
int b = -45;
System.out.println(Math.max(a, b));
}
}
```
#### Output
```->javac Example3.java
->java Example3
-34
```
#### Example4.java: The following example demonstrates comparing positive zero and negative zero.
```public class Example4 {
public static void main(String[] args) {
float a = -0.0f;
float b = 0.0f;
System.out.println(Math.max(a, b));
}
}
```
#### Output
```->javac Example4.java
->java Example4
0.0
```
#### Example5.java: The following example demonstrates a situation involving a NaN argument.
```public class Example5 {
public static void main(String[] args) {
float a = 2f;
System.out.println(Math.max(a, 2.0 % 0));
}
}
```
#### Output
```->javac Example5.java
->java Example5
NaN
``` | 780 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.577242 |
https://www.nano-motor.co.uk/youngs-modulus-aluminium-2/ | 1,660,188,722,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00503.warc.gz | 788,600,825 | 11,509 | # young’s modulus aluminium
Actually, Elasticity Modulus of Aluminium is around 69 GPa (10.0106psi).. Young's modulus of ceramic particle reinforced aluminium: Measurement by the.
Young’s Modulus is a mechanical property of the material where it can be called as modulus of Elasticity/Elastic Modulus. An English physician and physicist named Thomas Young described the elastic properties of materials. Young’s modulus can be defined as simply the stiffness of a solid material.
aluminum alloy young’s modulus . If you have any questions or good suggestions on our products and site, or if you want to know more information about our products, please write them and send to us, we will contact you within one business day.
T_2 = 33^o \ C {/eq} Linear thermal expansion coefficient of Aluminum {eq}\alpha = 24.0 \times 10^{-6} \ m / ^o C {/eq} Young’s modulus of Aluminum {eq}Y = 68.0 \times 10^9 \ Pa {/eq} Let {eq}L, \ \ A.
Strains from 1 to 10% were used to calculate Young’s modulus, as the stress-strain response in this region was linear.
aluminum young’s modulus . If you have any questions or good suggestions on our products and site, or if you want to know more information about our products, please write them and send to us, we will contact you within one business day.
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The values of Y determined for Aluminium, Copper, brass, and iron are in close agreement with the available standard values. These results indicate DEHI technique can be used to determine the standard values of Young’s modulus of elastic material. This result confirms that the values of Y determined for samples K 1, K 2, and K 3 are correct.
the young’s modulus of aluminium is 0.675 * 105 this can also be called as the modulus of elasticity Asked in Physics , Mechanical Engineering What is the dimension of youngs modulus of elasticity ?
The graph bars on the material properties cards below compare 2024-T4 aluminum to: 2000-series alloys (top), all aluminum alloys (middle), and the entire database (bottom). A full bar means this is the highest value in the relevant set.
M6 Bolt 12Mm and 6.5 to 12 mm in diameter. Any diameter can be made as a custom part. precision-grade shoulder diameters have a tolerance of 0.0005 to 0.0015 in. below the nominal size for inch-size screws and.motorcycle bar end weights Bar End Weights & Covers, Handlebars, Grips & Levers, Motorcycle Parts, Parts & Accessories, eBay Motors. Shop the Largest Selection, Click to See! Search eBay faster with PicClick. Money Back Guarantee ensures YOU receive the item you ordered or get your money back.bmw r1150rt review honda vfr800 Related: honda vfr 800 1998-2001 honda interceptor honda interceptor 800 honda vfr 800 vtec honda vfr 800 exhaust honda vfr 800 2002 honda vfr 800 98-01 honda vfr 800 2006 yamaha fjr1300 honda vfr 1200 buell honda vfr 800 chainThis website is for BMW enthusiasts, and owners of the new R1150R specifically. The site’s main attraction is its discussion board where users can ask questions, post.
Shear Modulus, 25, 34, GPa, 3.62594, 4.93128, 106 psi. Tensile Strength, 75, 360, MPa, 10.8778, 52.2136, ksi. Young's Modulus, 68, 88.5, GPa, 9.86256.
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Young’s Modulus, Elastic Modulus Or Modulus of Elasticity takes the values for stress and strain to predict the performance of the material in many other scenarios, such as Beam Deflection. and is calculated using the formula below:
about town bolts Usain St Leo Bolt OJ CD (/ ju s e n /; born 21 August 1986) is a Jamaican retired sprinter.He is a world record holder in the 100 metres, 200 metres and 4 100 metres relay.owing to his achievements and dominance in sprint competition, he is widely considered to be the greatest sprinter of all time. | 1,054 | 4,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.905939 |
https://fovconsulting.com/and-book/329-discrete-structures-and-graph-theory-books-638-819.php | 1,601,396,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00021.warc.gz | 391,695,970 | 10,124 | # Discrete structures and graph theory books
7.88 · 6,951 ratings · 894 reviews
Posted on by
## PHI Learning - Discrete Mathematics and Graph Theo
A network has points, connected by lines. In a graph, we have special names for these. We call these points vertices sometimes also called nodes , and the lines, edges. There are several roughly equivalent definitions of a graph. Set theory is frequently used to define graphs.
File Name: discrete structures and graph theory books.zip
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Published 18.05.2019
## Graph theory
Graphs are one of the prime objects of study in discrete mathematics. Main article: Mathematical logic. All that discrets is which vertices are connected to which others by how many edges and not the exact layout. A long-standing topic in discrete geometry is tiling of the plane.
Technically, suitable as a textbook for advanced undergraduate and beginning graduate students in mathematics and computer science, but our definition thwory for simple graphs. The unification of two argument graphs is defined as the most general graph or the computation thereof that is consistent with i. The primary aim of this book is to present a coherent introduction to graph theory. Coding Theory.
## Top Authors
The study of graphs, mathematical structures used to model pairwise relations between objects from a certain collection. Algorithmic Graph Theory and Sage. An Introduction to Combinatorics and Graph Theory. Applied Combinatorics. Digraphs Theory, Algorithms and Applications. Explorations in Algebraic Graph Theory with Sage.
### Updated
Note also it is a cycle, the last vertex is joined to the first. Linux and Unix. All categories Follow Books under this sub-category 15 books. A similar problem is finding induced subgraphs in a given graph.
Bibcode : EPJB Algebraic varieties also have a well-defined notion boosk tangent space called the Zariski tangent spacethe last vertex is joined to the first. Note also it is a cycle, making many features of calculus applicable even in finite settings. Random House.
## 5 thoughts on “Discrete Mathematics/Graph theory - Wikibooks, open books for an open world”
1. In mathematics , graph theory is the study of graphs , which are mathematical structures used to model pairwise relations between objects. A graph in this context is made up of vertices also called nodes or points which are connected by edges also called links or lines. A distinction is made between undirected graphs , where edges link two vertices symmetrically, and directed graphs , where edges link two vertices asymmetrically; see Graph discrete mathematics for more detailed definitions and for other variations in the types of graph that are commonly considered. 😣
2. Interpreter Middleware Virtual machine Operating system Software quality. Social choice theory is about voting. Industrial and applied mathematics. Cambridge University Press.
3. Requiring only high school algebra as mathematical background, Pla. Software Engineering. Bibcode : JAP. Included within theoretical computer science is the study of algorithms and data structures.💂
4. Approximation theory Numerical analysis Differential equations Dynamical systems Control theory Variational calculus. Related Titles? Cambridge University Press? Graphs without multiple edges or loops are known as simple graphs. | 667 | 3,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | latest | en | 0.936476 |
http://stuff.dewsoftoverseas.com/bquestion329.htm | 1,516,110,074,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00148.warc.gz | 321,874,279 | 2,254 | In America, the way we talk about paper is amazingly convoluted. The short answer to your question is that 500 sheets of bond paper with a size of 17" by 22" have a weight of 20 pounds. The manufacturer would cut a sheet that big into four letter-size sheets, so a 500-sheet ream of 20 pound bond paper weighs 5 pounds.
If you had said it was something other than bond paper, then the size of a standard sheet used to determine the weight might be different. See the second link below for details. For example, Bristol paper is heavier and stiffer (like the paper in a manila file folder). Its standard sheet size is 22.5" by 28.5".
In general, the more a sheet of a certain grade of paper weighs, the thicker it is. You can purchase 20 and 24 pound bond paper at an office supply store. 24 pound bond paper is thicker, heavier and more opaque than 20 pound bond.
The metric system has a much better way of measuring paper. A0 paper is a square meter of paper. One side is 84.1 centimeters long and the other is 118.9 centimeters (the longer side's length is the square root of 2 longer than the shorter side's length). A1 paper is half of an A0 sheet (preserving the square root of 2 rule). A2 is half of an A1 sheet, and so on. And paper weight is measured in grams per square meter, so it is very easy to figure out what is going on! | 331 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-05 | latest | en | 0.947804 |
http://www.includehelp.com/algorithms/priority-scheduling-pre-emptive-algorithm.aspx | 1,519,417,141,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814833.62/warc/CC-MAIN-20180223194145-20180223214145-00292.warc.gz | 459,131,708 | 9,257 | # Implementation of Priority Scheduling (Pre-emptive) algorithm using C++
In this article, we are going to learn about priority scheduling algorithm (pre-emptive) and implementing this algorithm using C++ program.
Submitted by Aleesha Ali, on January 29, 2018
Pre-emptive: If a process of higher priority comes then first CPU will be assign to the Process with higher priority first.
Scheduling criteria tells us that any algorithm is how much efficient, the main criteria of scheduling are given below:
• CPU Utilization
• Throughput
• Arrival time
• Turnaround time
• Waiting time
• Completion time
• Burst time
*Ready Queue is a queue where all the processes wait to get CPU for its execution.
CPU Utilization: The amount of time CPU is busy.
Throughput: The number of process computed per unit time.
Arrival time: The time at which the process enters into ready queue.
Turn around time: The interval between the time of submission of a process to the time of completion.
Waiting time: The total amount of the time a process spends in ready queue.
Completion time: The time at which process completes its execution.
Burst time: The time needed by CPU to completes its execution.
## Priority Scheduling Algorithm
In this algorithm priority is defined by manufacture of operating system, sometimes we assume minimum number has higher priority or vice a versa.
In my algorithm I use higher number has higher priority means process having higher priority will be schedule first.
## C++ Program for Priority Algorithm
```//Implementation of Priority(Preeemptive) Using C++
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
typedef struct proccess
{
int at,bt,ct,ta,wt,btt,pr;
string pro_id;
/*
artime = Arrival time,
bt = Burst time,
ct = Completion time,
ta = Turn around time,
wt = Waiting time
*/
}schedule;
bool compare(schedule a,schedule b)
{
return a.at<b.at;
/* This schedule will always return TRUE
if above condition comes*/
}
bool compare2(schedule a,schedule b)
{
return a.pr>b.pr;
/* This schedule will always return TRUE
if above condition comes*/
}
int main()
{
schedule pro[10];
int n,i,j,pcom;
cout<<"Enter the number of process::";
cin>>n;
cout<<"Enter the Process id arrival time burst time and priority :::";
for(i=0;i<n;i++)
{
cin>>pro[i].pro_id;
cin>>pro[i].at;
cin>>pro[i].bt;
pro[i].btt=pro[i].bt;
cin>>pro[i].pr;
}
sort(pro,pro+n,compare);
/*sort is a predefined funcion defined in
algorithm.h header file, it will sort the
schedulees according to their arrival time*/
i=0;
pcom=0;
while(pcom<n)
{
for(j=0;j<n;j++)
{
if(pro[j].at>i)
break;
}
sort(pro,pro+j,compare2);
if(j>0)
{
for(j=0;j<n;j++)
{
if(pro[j].bt!=0)
break;
}
if(pro[j].at>i)
i+=pro[j].at-i;
pro[j].ct=i+1;
pro[j].bt--;
}
i++;
pcom=0;
for(j=0;j<n;j++)
{
if(pro[j].bt==0)
pcom++;
}
}
for(i=0;i<n;i++)
{
pro[i].ta=pro[i].ct-pro[i].at;
pro[i].wt=pro[i].ta-pro[i].btt;
//before executing make it in one statement
cout<<pro[i].pro_id<<"\t"<<pro[i].at<<"\t"<<pro[i].btt<<"\t"<<pro[i].ct<<"\t"<<pro[i].ta<<"\t"<<pro[i].wt<<"\t"<<pro[i].pr;
cout<<endl;
}
return 0;
}
```
Output | 830 | 3,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-09 | latest | en | 0.862251 |
https://www.mersenneforum.org/showthread.php?s=56d36602d9441c48a92c7f58316ac23b&p=114666 | 1,620,604,367,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00246.warc.gz | 916,474,628 | 9,738 | mersenneforum.org A Kind of Solitaire
Register FAQ Search Today's Posts Mark Forums Read
2007-09-18, 20:57 #1 davar55 May 2004 New York City 3·17·83 Posts A Kind of Solitaire I recently came across this and thought it might be interesting here: Start with a deck of T cards where T=n*(n+1)/2 is triangular. Form a collection of piles of arbitrary sizes using up all T cards. At each step, take one card away from every pile and with these removed cards form a new pile. Then: (a) Eventually, the piles will have sizes {1,2,3,...,n} after which each step cycles on this pattern. (b) This pattern occurs within n*(n-1) steps. (This is called Bulgarian solitaire and was introduced by Martin Gardner.) Proofs of (a) and (b) exist, but you're welcome to try.
2007-09-19, 10:51 #2 retina Undefined "The unspeakable one" Jun 2006 My evil lair 17FC16 Posts It is a dull type of solitaire game. Since we know the outcome in advance and there is no strategy at all. What actually is your puzzle here?
2007-09-19, 11:44 #3
davieddy
"Lucan"
Dec 2006
England
194A16 Posts
Quote:
Originally Posted by retina What actually is your puzzle here?
To prove a) and b)
2007-09-19, 13:21 #4
Wacky
Jun 2003
The Texas Hill Country
32×112 Posts
Quote:
Originally Posted by retina What actually is your puzzle here?
I have to agree with this sentiment. It seems to me that there have been posted here a number of "mathematical exercises" and/or "conjectures" that we are being ask to solve in the name of a "puzzle".
These contrast highly from those examples such as the "hole in the sphere" or the logic problems that are characteristic of the problems posed by the "Mathematical Recreations" Greats of the last centuries.
2007-09-19, 14:09 #5
davieddy
"Lucan"
Dec 2006
England
11001010010102 Posts
Quote:
Originally Posted by Wacky I have to agree with this sentiment. It seems to me that there have been posted here a number of "mathematical exercises" and/or "conjectures" that we are being ask to solve in the name of a "puzzle". These contrast highly from those examples such as the "hole in the sphere" or the logic problems that are characteristic of the problems posed by the "Mathematical Recreations" Greats of the last centuries.
Well I am working on it with enthusiasm ATM.
I can see why 1,2,3,,,,,n stays as it is.
If Martin Gardner posed it, it qualifies for this thread IMHO.
What do you think of Nim?
David
2007-09-19, 15:42 #6
Wacky
Jun 2003
The Texas Hill Country
32·112 Posts
Quote:
Originally Posted by davieddy If Martin Gardner posed it, it qualifies for this thread IMHO. What do you think of Nim?
I will note that Martin Gardner presented many mathematically related topics which were not "puzzles" in the sense of Henry Dudeney or Sam Loyd.
In particular, I recall that he popularized John Conway's cellular automaton "Game of Life", as well as many other interesting topics.
As for Nim, I worked out the winning strategy decades ago. As a result, I find it of little more interest than Tic Tac Toe.
Last fiddled with by Wacky on 2007-09-19 at 15:44
2007-09-19, 15:49 #7
davieddy
"Lucan"
Dec 2006
England
2×3×13×83 Posts
Quote:
Originally Posted by Wacky As for Nim, I worked out the winning strategy decades ago. As a result, I find it of little more interest than Tic Tac Toe.
Do you want a "Bronze medal"
2007-09-21, 11:24 #8 davieddy "Lucan" Dec 2006 England 11001010010102 Posts Nim Well I deem my "Nim" thread a success! Is a proof of this solitaire puzzle beyond me? It makes a nice little simulation program. David BTW in the light of Mally's demise, I regret my reference to a Bronze medal. Last fiddled with by davieddy on 2007-09-21 at 11:38
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Sun May 9 23:52:47 UTC 2021 up 31 days, 18:33, 0 users, load averages: 3.56, 3.53, 3.22 | 1,184 | 4,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-21 | latest | en | 0.943808 |
http://mathhelpforum.com/algebra/149514-simultaneous-equations.html | 1,527,275,279,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867173.31/warc/CC-MAIN-20180525180646-20180525200646-00245.warc.gz | 186,533,648 | 9,432 | 1. ## SImultaneous equations
Can these simultaneous equations be solved?
e+4f=1
3d+2e+4f=1
2. Originally Posted by Stuck Man
Can these simultaneous equations be solved?
e+4f=1
3d+2e+4f=1
Hi
2 linear equations with 3 unknowns cannot be solved
What you can do is express two variables in terms of one
For instance
$\displaystyle f = \frac14 - \frac{e}{4}$
$\displaystyle d = -\frac{e}{3}$ | 129 | 392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-22 | latest | en | 0.814232 |
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# did anyone do ccea maths c1 today
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1. i got everthing right apart from maybe question 8, i got
- 4 - (2 root 2) < X < - 4 + (2 root 2). is that right
2. Yea i did it today! cant remember what i got but it was something like that!
3. What answers did use get in each question? What did use think of the paper overall?
4. Hi yeah I did it,
I got something similar to that answer for question 8 as well. What did you get for question 7?
5. (Original post by Pyro2000x)
i got everthing right apart from maybe question 8, i got
- 4 - (2 root 2) < X < - 4 + (2 root 2). is that right
This is right.
6. (Original post by Alfie1993)
Hi yeah I did it,
I got something similar to that answer for question 8 as well. What did you get for question 7?
what was it about, we had to give up our question book
7. (Original post by Pyro2000x)
what was it about, we had to give up our question book
The cost function which you had to say was increasing when t>2
8. Here are my answers can anyone give any feedback?
1. Simplifying an equation - 5
b) graph functions, f(x-1) moved one to the right
f(-x) reflected in the y axis
2. x=20 y=45 z=30
3a)i) gradient of AB = 2
ii) equation of line perpendic to AB going through (-1,1)
2y=-x-1
b) differentiating y=4x^3 - x^-1/3
i got - 12x^2 + 1/3x^-4/3
4. curve and line meeting at points C and D
i) find co ordinates - C (2,-1) D (4,3)
ii) exact length - sq root of 20/ 2root 5
5a) factor theorem ii) factories fully - (2x-1)(x+2)(x+1)
iii)solve equation so x=1/2, x=-2, x=-1
b) triangle q
6root 3 + 4
couldn't do i)
ii) 17/2
7a)i) dy/dx of x^4 + 32x
4x^3 + 32
ii) find x co ord of turning point and its nature
x= -2, minimum
b) cost
t > cube root of 6 ( i know thats wrong)
8. curves that didn't meet
i got the discriminant to be m^2 + 8m + 8<0 and instead of using the quad formula i did m^2 + 8m < - 8
and ended up with like -8<m<-16 or something stupid
anyone get any of the same answers?
9. (Original post by B18)
Here are my answers can anyone give any feedback?
1. Simplifying an equation - 5
b) graph functions, f(x-1) moved one to the right
f(-x) reflected in the y axis
2. x=20 y=45 z=30
3a)i) gradient of AB = 2
ii) equation of line perpendic to AB going through (-1,1)
2y=-x-1
b) differentiating y=4x^3 - x^-1/3
i got - 12x^2 + 1/3x^-4/3
4. curve and line meeting at points C and D
i) find co ordinates - C (2,-1) D (4,3)
ii) exact length - sq root of 20/ 2root 5
5a) factor theorem ii) factories fully - (2x-1)(x+2)(x+1)
iii)solve equation so x=1/2, x=-2, x=-1
b) triangle q
6root 3 + 4
couldn't do i)
ii) 17/2
7a)i) dy/dx of x^4 + 32x
4x^3 + 32
ii) find x co ord of turning point and its nature
x= -2, minimum
b) cost
t > cube root of 6 ( i know thats wrong)
8. curves that didn't meet
i got the discriminant to be m^2 + 8m + 8<0 and instead of using the quad formula i did m^2 + 8m < - 8
and ended up with like -8<m<-16 or something stupid
anyone get any of the same answers?
The cube root of 6 was wrong, it was supposed to be t>2. Number 8, you did wrong, but you would have gained some marks for getting as far as the quadratic inequality. I think your factorising in the long division may be wrong. Also, the equation of the perpendicular line; I think you may have got a sign or two wrong. Apart from that, all good
10. unreal i am almost certain i got 100%
11. How did you think you had 100% when you didnt even attempt 6)i.
Sorry to sound harsh!
12. It was a decent paper. I messed up in a few spots but I think I'll get a high B/ an A.
13. (Original post by acrawford957)
How did you think you had 100% when you didnt even attempt 6)i.
Sorry to sound harsh!
It wasn't him who said that.
(Original post by xDanny 117)
It was a decent paper. I messed up in a few spots but I think I'll get a high B/ an A.
What did you mess up on?
14. (Original post by acrawford957)
How did you think you had 100% when you didnt even attempt 6)i.
Sorry to sound harsh!
that is not me
15. (Original post by acrawford957)
How did you think you had 100% when you didnt even attempt 6)i.
Sorry to sound harsh!
lol yeah... i never said i thought i got 100% that was someone else
16. (Original post by CD315)
The cube root of 6 was wrong, it was supposed to be t>2. Number 8, you did wrong, but you would have gained some marks for getting as far as the quadratic inequality. I think your factorising in the long division may be wrong. Also, the equation of the perpendicular line; I think you may have got a sign or two wrong. Apart from that, all good
would you talk me through 6i) 7b) and 8? also what was the perpendicular equation meant to be?
17. (Original post by CD315)
What did you mess up on?
Used the wrong 'length of a line' formula in a question, couldn't get a final value for when the cost is increasing in Q7, only got to the quadratic inequality in Q8.
18. (Original post by B18)
would you talk me through 6i) 7b) and 8? also what was the perpendicular equation meant to be?
You had a negative 1 which I think was supposed to be a positive.
6i; wasn't this the speed one? You just had something along the lines of 17/x + 13/(x-2) if I remember correctly? Then just add them and set equal to 4. (They're the expressions for time taken.)
7b; you differentiated the function and said it was greater than 0. You ended up with it being 2.
8; set the curves equal to each other and you'll form a quadratic. Since there is no solutions to this, due to them not meeting, you take the discriminant and say it's less than 0. I ended up with a quadratic inequality which I then drew a graph of.
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Useful resources | 1,933 | 6,303 | {"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} | 3.1875 | 3 | CC-MAIN-2016-44 | latest | en | 0.957005 |
https://eliteresearchpapers.com/calculate-the-direct-labor-rate-variance-for-june-if-the-variance-is-favorable-enter-an-f-after-your-number-with-a-space-between-the-number/ | 1,725,739,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00137.warc.gz | 211,697,887 | 14,609 | # Calculate the direct labor rate variance for June. If the variance is favorable, enter an F after your number with a space between the number
ABC Company began operations on June 1, 2019. For its first month of operations, ABC
Company established the following standards for unit of its single product:
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standard quantity standard price
direct materials 8 pounds \$3.50 per pound
direct labor 7 hours \$15.00 per hour
variable overhead 7 hours \$7.50 per hour
The following information is available for surge for the month of June:
1. 65,000 pounds of direct materials were purchased at a total cost of \$188,500.
2. 45,000 direct labor hours were worked at a total cost of \$731,250.
3. The variable overhead cost for the month totaled \$325,000.
4. 6,500 units were produced.
5. At June 30, ABC Company had 8,000 pounds of direct materials on hand.
Calculate the direct labor rate variance for June. If the variance is favorable, enter
an F after your number with a space between the number and the F (i.e., 10,000 F). If
the variance is unfavorable, enter a U after your number with a space between the
number and the U (i.e., 10,000 U).
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The post Calculate the direct labor rate variance for June. If the variance is favorable, enter an F after your number with a space between the number appeared first on BEST NURSING TUTORS . | 410 | 1,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.934062 |
https://www.cfd-online.com/Forums/cfx/21958-ansys-flotran-3d.html | 1,518,925,461,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811352.60/warc/CC-MAIN-20180218023321-20180218043321-00497.warc.gz | 867,630,047 | 16,902 | # Ansys flotran-3D
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December 19, 2005, 15:46 Ansys flotran-3D #1 omer Guest Posts: n/a I am new to Ansys. Would anyone know how i can get hold of a tutorial on 3d flow problem, lets say, flow through a circular pipe. And how can i use the solution from this , to apply it as load data to the structure for vibration analysis? Any help would be appreciated! Regards. hamidgarusi likes this.
December 23, 2005, 13:02 Re: Ansys flotran-3D #2 Rishi Guest Posts: n/a The 3D problem analysis for a circular pipe as u say is pretty easy....i cant give you a link for a tutorial though...if you are using ICEM-CFD....the icem-cfd tutorials have such a problem.....about your second question...you have to use the results file of the previous solution as load data for the new analysis...browse to the file and provide it....hope this helps....
January 3, 2006, 13:25 Re: Ansys flotran-3D #3 vasudev Guest Posts: n/a please send me the complete steps of cfd analysis.
January 3, 2006, 13:27 Re: Ansys flotran-3D #4 vasudev Guest Posts: n/a i wnt to learn cfd flotran
January 3, 2006, 17:00 Re: Ansys flotran-3D #5 Rishi Guest Posts: n/a if u r using icem-cfd..there r 2 types of meshing techniques;hexa and tetra...while tetra is a lot easier..it falls short in boundary condition analysis..where hexa is much accurate and a wise option...for hexa the steps are... 1.create the geometry(using points,curves and surfaces) create the cylindrical surface and the inlets and outlets for the cylinder. 2.then initialise a block 3.associate the edges of the block to the respective curves. 4.form the o-grid and then change sizes as u want by rescaling the o-grid. 5.mesh and check quality. get icem-cfd and u will have the tutorial i told u about
February 3, 2006, 23:41 Re: Ansys flotran-3D #6 Shailesh Rao A Guest Posts: n/a hi I want to know the details of modeling and parameters to be considered for fluid flow in circular pipe when the pipe is spinning
May 28, 2014, 01:10
#7
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hamid
Join Date: May 2014
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Quote:
Originally Posted by omer ;74286 I am new to Ansys. Would anyone know how i can get hold of a tutorial on 3d flow problem, lets say, flow through a circular pipe. And how can i use the solution from this , to apply it as load data to the structure for vibration analysis? Any help would be appreciated! Regards.
hi i want to know 3d flow analysis in ansys 15.
May 28, 2014, 08:24 #8 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 14,309 Rep Power: 110 Try the CFX tutorials. They are provided in the help menu.
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Similar Threads Thread Thread Starter Forum Replies Last Post Jim CFX 14 February 29, 2016 07:34 pablo--- ANSYS 2 December 11, 2009 05:35 Cirion0000 ANSYS 1 December 11, 2009 02:39 Paul Atreides CFX 1 December 29, 2006 19:02 Paul Atreides Main CFD Forum 0 December 27, 2006 11:23
All times are GMT -4. The time now is 23:44. | 942 | 3,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-09 | latest | en | 0.89268 |
https://www.hpmuseum.org/forum/showthread.php?tid=11775&pid=107441&mode=threaded | 1,702,236,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00519.warc.gz | 872,220,426 | 5,831 | (41C) Area of Triangle (SSS)
11-11-2018, 07:53 AM (This post was last modified: 11-11-2018 08:51 AM by Dieter.)
Post: #5
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: (41C) Area of Triangle (SSS)
(11-11-2018 05:04 AM)Gamo Wrote: My program only work when
A ≤ B < C
A = B = C
A ≠ B ≠ C
This means that your example A=8, B=8, C=3 would not work.
In general, according to these rules no isosceles triangle can be entered where the two equal sides are larger than the third one.
You can modify the program so that it will always work when A ≤ B ≤ C.
Still the LBL 02 part should be removed. Take a look at the code:
The program first checks if A=B...
Code:
RCL 01 RCL 02 X=Y GTO 02
...and if true, it jumps to LBL 02 where the same test is done once again:
Code:
LBL 02 RCL 01 RCL 02 X=Y GTO 03
Since this always tests true again, the program always continues at LBL 03 (where the program checks if B and C are equal as well):
Code:
LBL 03 RCL 02 RCL 03 X=Y GTO 04 "ISOSCELES" PROMPT
That's why "ISOSCELES" appears twice in the program.
So you should remove the complete LBL 02 part and replace the "GTO 02" line with "GTO 03".
BTW, in an HP-41 program quotation marks are used for text. You you should not use them for commands, and vice versa.
ISOSCELES
"PROMPT"
you better write
"ISOSCELES"
PROMPT
Dieter
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Messages In This Thread (41C) Area of Triangle (SSS) - Gamo - 11-10-2018, 12:20 PM RE: (41C) Area of Triangle (SSS) - Albert Chan - 11-10-2018, 03:22 PM RE: (41C) Area of Triangle (SSS) - Dieter - 11-10-2018, 08:53 PM RE: (41C) Area of Triangle (SSS) - Albert Chan - 11-12-2018, 02:55 PM RE: (41C) Area of Triangle (SSS) - Dieter - 11-12-2018, 08:28 PM RE: (41C) Area of Triangle (SSS) - Albert Chan - 11-12-2018, 10:33 PM RE: (41C) Area of Triangle (SSS) - Gamo - 11-11-2018, 05:04 AM RE: (41C) Area of Triangle (SSS) - Dieter - 11-11-2018 07:53 AM RE: (41C) Area of Triangle (SSS) - Gamo - 11-11-2018, 12:23 PM RE: (41C) Area of Triangle (SSS) - Dieter - 11-11-2018, 04:25 PM RE: (41C) Area of Triangle (SSS) - Albert Chan - 11-12-2018, 01:32 AM RE: (41C) Area of Triangle (SSS) - Albert Chan - 11-16-2018, 03:33 AM RE: (41C) Area of Triangle (SSS) - Albert Chan - 12-04-2019, 03:03 PM
User(s) browsing this thread: 1 Guest(s) | 844 | 2,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.802419 |
http://mathhelpforum.com/algebra/111708-help-solving-y.html | 1,480,703,717,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00378-ip-10-31-129-80.ec2.internal.warc.gz | 178,425,287 | 10,345 | 1. help solving for y
(y/7)^2 + ((x-1)/3)^2 = 1
How do I go about doing this?
2. Originally Posted by clambottomjewels
(y/7)^2 + ((x-1)/3)^2 = 1
How do I go about doing this?
This is the equation of an ellipse:
$\left(\dfrac y7\right)^2+\left(\dfrac{x-3}3\right)^2=1~\implies~\dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1$
So what exactly do you want to do with it?
$\dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1 ~\implies~y^2=49-\dfrac{49}9 \cdot (x-3)^2$
Can you take it from here?
3. Originally Posted by clambottomjewels
(y/7)^2 + ((x-1)/3)^2 = 1
How do I go about doing this?
• Take $\left(\frac{1}{3}(x-1)\right)^2$ from both sides
• Take the square root of both sides (and don't forget the $\pm$
• Multiply both sides by 7
You should get an answer of $y = \pm7\sqrt{1-\left(\frac{1}{3}(x-1)\right)^2}$. This is equivalent to
$\pm \frac{7}{3} \sqrt{-x^2+2x+8}$ | 349 | 860 | {"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": 6, "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} | 4.21875 | 4 | CC-MAIN-2016-50 | longest | en | 0.857263 |
https://www.numbersaplenty.com/2050120 | 1,653,375,312,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00450.warc.gz | 821,487,111 | 3,546 | Search a number
2050120 = 235107479
BaseRepresentation
bin111110100100001001000
310212011020101
413310201020
51011100440
6111535144
723266012
oct7644110
93764211
102050120
111180316
1282a4b4
1356a1b7
143b51b2
152a769a
hex1f4848
2050120 has 32 divisors (see below), whose sum is σ = 4665600. Its totient is φ = 810688.
The previous prime is 2050109. The next prime is 2050141. The reversal of 2050120 is 210502.
It is a Harshad number since it is a multiple of its sum of digits (10).
It is a nialpdrome in base 8.
It is a junction number, because it is equal to n+sod(n) for n = 2050097 and 2050106.
It is an unprimeable number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 4041 + ... + 4519.
It is an arithmetic number, because the mean of its divisors is an integer number (145800).
22050120 is an apocalyptic number.
2050120 is a gapful number since it is divisible by the number (20) formed by its first and last digit.
It is an amenable number.
2050120 is an abundant number, since it is smaller than the sum of its proper divisors (2615480).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
2050120 is a wasteful number, since it uses less digits than its factorization.
2050120 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 597 (or 593 counting only the distinct ones).
The product of its (nonzero) digits is 20, while the sum is 10.
The square root of 2050120 is about 1431.8240115321. The cubic root of 2050120 is about 127.0358879296.
Adding to 2050120 its reverse (210502), we get a palindrome (2260622).
The spelling of 2050120 in words is "two million, fifty thousand, one hundred twenty". | 532 | 1,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.88735 |
https://www.mathstunners.org/problems/1160.12 | 1,686,248,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00114.warc.gz | 961,663,474 | 3,519 | ## 1160.12 – ALAS LASS part 2
The mother wrote back:
$\begin{array}{cccc} A & L & A & S \\ L & A & S & S \\ & & N & O \\ M & O & R & E \\ \hline C & A & S & H \end{array}$
As in 1160.11, each letter stands for a numeral (0 1 2 3 4 5 6 7 8 or 9). The same letter always stands for the same numeral and two different letters never stand for the same numeral. Your job is to figure out what each letter stands for.
Solution
There are two solutions that we know of.
1: A = 5, C = 9, E = 4, H = 6, L = 1, M = 2, N = 3, O = 8, R = 0, and S = 7.
\begin{aligned} 51&57 \\ 15&77 \\ &38 \\ \underline{28}&\underline{04} \\ 95&76 \end{aligned}
2: A = 1, C = 7, E = 5, H = 9, L = 3, M = 2, N = 8, O = 6, R = 0, and S = 4.
\begin{aligned} 13&14 \\ 31&44 \\ &86 \\ \underline{26}&\underline{05} \\ 71&49 \end{aligned}
(There are rumors of a 3rd solution with C = 8. If you find it, let us know.)
This problem naturally follows after no. 1160.11, SEND MORE MONEY. There is a very lengthy explanation of the solution to that problem, which you could use to develop the solution to this one. | 402 | 1,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-23 | latest | en | 0.84999 |
https://im.openupresources.org/6/students/2/14.html | 1,550,317,711,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480272.15/warc/CC-MAIN-20190216105514-20190216131514-00590.warc.gz | 579,871,975 | 8,176 | # Lesson 14: Solving Equivalent Ratio Problems
Let's practice getting information from our partner.
## 14.1: What Do You Want to Know?
Here is a problem: A red car and a blue car enter the highway at the same time and travel at a constant speed. How far apart are they after 4 hours?
What information would you need to solve the problem?
## 14.2: Info Gap: Hot Chocolate and Potatoes
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
3. Explain how you are using the information to solve the problem.
If your teacher gives you the data card:
2. Ask your partner “What specific information do you need?” and wait for them to ask for information.
If your partner asks for information that is not on the card, do not do the calculations for them. Tell them you don’t have that information.
3. Have them explain “Why do you need that information?” before telling them the information.
4. After your partner solves the problem, ask them to explain their reasoning, even if you understand what they have done.
Both you and your partner should record a solution to each problem.
• Lin read the first 54 pages from a 270-page book in the last 3 days.
• Diego read the first 100 pages from a 320-page book in the last 4 days.
• Elena read the first 160 pages from a 480-page book in the last 5 days.
If they continue to read every day at these rates, who will finish first, second, and third? Explain or show your reasoning.
## Summary
To solve problems about something happening at the same rate, we often need:
• Two pieces of information that allow us to write a ratio that describes the situation.
• A third piece of information that gives us one number of an equivalent ratio. Solving the problem often involves finding the other number in the equivalent ratio.
Suppose we are making a large batch of fizzy juice and the recipe says, “Mix 5 cups of cranberry juice with 2 cups of soda water.” We know that the ratio of cranberry juice to soda water is $5:2$, and that we need 2.5 cups of cranberry juice per cup of soda water.
We still need to know something about the size of the large batch. If we use 16 cups of soda water, what number goes with 16 to make a ratio that is equivalent to $5:2$?
To make this large batch taste the same as the original recipe, we would need to use 40 cups of cranberry juice.
cranberry juice (cups) soda water (cups)
row 1 5 2
row 2 2.5 1
row 3 40 16 | 604 | 2,510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-09 | latest | en | 0.923289 |
https://discuss.pytorch.org/t/complex-indexing-to-avoid-for-loop-on-gpu/178977 | 1,685,446,813,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645595.10/warc/CC-MAIN-20230530095645-20230530125645-00124.warc.gz | 255,101,612 | 4,044 | Complex indexing to avoid for-loop on GPU
Hello,
I am trying to perform complex indexing to avoid using for-loops on GPU. My problem is as follows, I have a large matrix X = [B, N, H], a smaller matrix H = [B, N, M, H] (which is initialized with as zeros), and a list containing indices L, where each list has the coordinates l = [b, n, m], where B: batch size, N: number of tokens in sequence, M: number of sub_tokens - M is a subset of N, H: hidden dimension.
Specifically, I want to populate H with samples from X; however, not every N in H samples M times. Also there is no sampling across the batch B. Let me illustrate with an example.
X = tensor(
`````` [[[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]]]
``````
)
Index = [[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 2, 0], [0, 2, 1], [1, 0, 0], [1, 1, 1], [1, 2, 1], [1, 2, 2]]
H = tensor(
`````` [[[[0, 1, 2, 3],
[4, 5, 6, 7]],
[[0, 1, 2, 3],
[0, 0, 0, 0]],
[[0, 1, 2, 3],
[4, 5, 6, 7]]],
[[[12, 13, 14, 15],
[0, 0, 0, 0]],
[[16, 17, 18, 19],
[0, 0, 0, 0]],
[[16, 17, 18, 19],
[20, 21, 22, 23]]]])
``````
If anyone has an idea how to solve this that would be amazing!
Thank you,
Christoph
Hi,
I found a different solution that tackles the problem from a different angle and does not require complex indexing as mentioned in the post.
Thanks. | 559 | 1,401 | {"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} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.903558 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/5625/2/a/q/ | 1,679,543,704,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00767.warc.gz | 977,432,666 | 60,351 | # Properties
Label 5625.2.a.q Level $5625$ Weight $2$ Character orbit 5625.a Self dual yes Analytic conductor $44.916$ Analytic rank $0$ Dimension $6$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [5625,2,Mod(1,5625)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(5625, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("5625.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$5625 = 3^{2} \cdot 5^{4}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 5625.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$44.9158511370$$ Analytic rank: $$0$$ Dimension: $$6$$ Coefficient field: 6.6.44400625.1 comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{6} - 11x^{4} - x^{3} + 29x^{2} + 3x - 1$$ x^6 - 11*x^4 - x^3 + 29*x^2 + 3*x - 1 Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 75) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{5}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q - \beta_1 q^{2} + (\beta_{4} - \beta_{3} + 1) q^{4} + ( - \beta_1 + 1) q^{7} + (\beta_{3} - \beta_{2} - 2 \beta_1) q^{8}+O(q^{10})$$ q - b1 * q^2 + (b4 - b3 + 1) * q^4 + (-b1 + 1) * q^7 + (b3 - b2 - 2*b1) * q^8 $$q - \beta_1 q^{2} + (\beta_{4} - \beta_{3} + 1) q^{4} + ( - \beta_1 + 1) q^{7} + (\beta_{3} - \beta_{2} - 2 \beta_1) q^{8} + (\beta_{5} - \beta_1) q^{11} + ( - \beta_{5} - \beta_{3} - \beta_1) q^{13} + (\beta_{4} - \beta_{3} - \beta_1 + 3) q^{14} + (\beta_{5} - 3 \beta_{3} + \beta_{2} + \beta_1 + 2) q^{16} + ( - \beta_{5} - \beta_{4} - \beta_{3} + \beta_{2} - 3) q^{17} + ( - \beta_{4} + 2 \beta_{3} + \beta_{2} + 3) q^{19} + (\beta_{4} - \beta_{3} - 2 \beta_{2} + 2) q^{22} + (\beta_{5} - \beta_{4} - \beta_{3} - \beta_1 - 2) q^{23} + (\beta_{4} - \beta_{3} + \beta_{2} - \beta_1 + 4) q^{26} + (\beta_{4} - \beta_{2} - 4 \beta_1 + 1) q^{28} + ( - \beta_{5} - \beta_{2} - \beta_1) q^{29} + (\beta_{5} + \beta_{4} + \beta_{3} - 2 \beta_{2} - 2 \beta_1 - 1) q^{31} + ( - \beta_{5} - \beta_{4} + 2 \beta_{3} - 3 \beta_{2} - \beta_1 - 2) q^{32} + ( - \beta_{5} + 2 \beta_{3} + \beta_{2} + 4 \beta_1 + 3) q^{34} + (\beta_{5} + \beta_{2} - \beta_1 + 4) q^{37} + ( - \beta_{5} + 2 \beta_{3} + 2 \beta_{2} + \beta_1 + 2) q^{38} + (\beta_{5} + \beta_{4} + \beta_{3} + 3 \beta_{2} + \beta_1 + 1) q^{41} + ( - \beta_{4} + \beta_{3} + 2 \beta_{2} - \beta_1 + 1) q^{43} + ( - 5 \beta_{3} - \beta_{2} - 3 \beta_1 - 4) q^{44} + (\beta_{4} - 2 \beta_{3} - 3 \beta_{2} + 3 \beta_1 + 2) q^{46} + (\beta_{4} - 3 \beta_{3} - 2 \beta_1 - 4) q^{47} + (\beta_{4} - \beta_{3} - 2 \beta_1 - 3) q^{49} + (\beta_{5} + \beta_{4} + 5 \beta_{3} - \beta_{2} - 5 \beta_1 + 5) q^{52} + (\beta_{4} + 4 \beta_{3} + 2 \beta_{2} - 2) q^{53} + (\beta_{5} + 2 \beta_{4} - 4 \beta_{3} - \beta_1 + 4) q^{56} + (\beta_{5} + \beta_{4} - 4 \beta_{3} + 2 \beta_{2} + 2) q^{58} + ( - 2 \beta_{5} - 3 \beta_{3} - \beta_{2} + \beta_1 - 4) q^{59} + (\beta_{5} - \beta_{4} + \beta_{3} - 3 \beta_{2} - \beta_1 + 3) q^{61} + (2 \beta_{5} + 2 \beta_{4} - 7 \beta_{3} - \beta_{2} + 1) q^{62} + (\beta_{5} + \beta_{4} - 5 \beta_{3} + 2 \beta_{2} + 4 \beta_1 - 6) q^{64} + ( - 2 \beta_{5} - \beta_{4} + \beta_{3} - 2 \beta_{2} - \beta_1 + 1) q^{67} + (\beta_{5} - 2 \beta_{4} + 9 \beta_{3} + 2 \beta_{2} - \beta_1 - 3) q^{68} + (\beta_{5} + 2 \beta_{4} + 3 \beta_{3} - \beta_{2} + \beta_1 + 3) q^{71} + (\beta_{5} - 2 \beta_{4} - 2 \beta_{3} + \beta_{2} + 2) q^{73} + ( - \beta_{5} + \beta_{4} + 2 \beta_{3} - 2 \beta_{2} - 4 \beta_1 + 4) q^{74} + ( - 2 \beta_{5} + \beta_{4} + 3 \beta_{3} + 2 \beta_{2} - 4) q^{76} + (\beta_{5} + \beta_{4} - \beta_{3} - 2 \beta_{2} - \beta_1 + 2) q^{77} + ( - 2 \beta_{5} - 2 \beta_{4} + \beta_{3} + \beta_{2} + 2 \beta_1 - 1) q^{79} + ( - 3 \beta_{5} - \beta_{4} + 11 \beta_{3} - \beta_{2} - 2 \beta_1 + 2) q^{82} + ( - \beta_{5} + 2 \beta_{4} - \beta_{3} + \beta_{2} + 2) q^{83} + ( - 2 \beta_{5} + \beta_{4} + 4 \beta_{3} + \beta_{2} + 2 \beta_1 + 7) q^{86} + (\beta_{5} + \beta_{4} - 4 \beta_{3} - \beta_{2} - \beta_1 + 3) q^{88} + ( - 2 \beta_{4} + 8 \beta_{3} - 3 \beta_{2} + \beta_1 + 5) q^{89} + ( - \beta_{5} + \beta_{4} - 2 \beta_{3} + \beta_{2} - 2 \beta_1 + 4) q^{91} + (\beta_{5} - \beta_{4} - 3 \beta_{3} - 2 \beta_{2} - 4 \beta_1 - 11) q^{92} + (2 \beta_{4} - \beta_{3} - 3 \beta_{2} - \beta_1 + 6) q^{94} + (2 \beta_{5} - \beta_{4} - 2 \beta_{2} - 3 \beta_1) q^{97} + (2 \beta_{4} - \beta_{3} - \beta_{2} + 6) q^{98}+O(q^{100})$$ q - b1 * q^2 + (b4 - b3 + 1) * q^4 + (-b1 + 1) * q^7 + (b3 - b2 - 2*b1) * q^8 + (b5 - b1) * q^11 + (-b5 - b3 - b1) * q^13 + (b4 - b3 - b1 + 3) * q^14 + (b5 - 3*b3 + b2 + b1 + 2) * q^16 + (-b5 - b4 - b3 + b2 - 3) * q^17 + (-b4 + 2*b3 + b2 + 3) * q^19 + (b4 - b3 - 2*b2 + 2) * q^22 + (b5 - b4 - b3 - b1 - 2) * q^23 + (b4 - b3 + b2 - b1 + 4) * q^26 + (b4 - b2 - 4*b1 + 1) * q^28 + (-b5 - b2 - b1) * q^29 + (b5 + b4 + b3 - 2*b2 - 2*b1 - 1) * q^31 + (-b5 - b4 + 2*b3 - 3*b2 - b1 - 2) * q^32 + (-b5 + 2*b3 + b2 + 4*b1 + 3) * q^34 + (b5 + b2 - b1 + 4) * q^37 + (-b5 + 2*b3 + 2*b2 + b1 + 2) * q^38 + (b5 + b4 + b3 + 3*b2 + b1 + 1) * q^41 + (-b4 + b3 + 2*b2 - b1 + 1) * q^43 + (-5*b3 - b2 - 3*b1 - 4) * q^44 + (b4 - 2*b3 - 3*b2 + 3*b1 + 2) * q^46 + (b4 - 3*b3 - 2*b1 - 4) * q^47 + (b4 - b3 - 2*b1 - 3) * q^49 + (b5 + b4 + 5*b3 - b2 - 5*b1 + 5) * q^52 + (b4 + 4*b3 + 2*b2 - 2) * q^53 + (b5 + 2*b4 - 4*b3 - b1 + 4) * q^56 + (b5 + b4 - 4*b3 + 2*b2 + 2) * q^58 + (-2*b5 - 3*b3 - b2 + b1 - 4) * q^59 + (b5 - b4 + b3 - 3*b2 - b1 + 3) * q^61 + (2*b5 + 2*b4 - 7*b3 - b2 + 1) * q^62 + (b5 + b4 - 5*b3 + 2*b2 + 4*b1 - 6) * q^64 + (-2*b5 - b4 + b3 - 2*b2 - b1 + 1) * q^67 + (b5 - 2*b4 + 9*b3 + 2*b2 - b1 - 3) * q^68 + (b5 + 2*b4 + 3*b3 - b2 + b1 + 3) * q^71 + (b5 - 2*b4 - 2*b3 + b2 + 2) * q^73 + (-b5 + b4 + 2*b3 - 2*b2 - 4*b1 + 4) * q^74 + (-2*b5 + b4 + 3*b3 + 2*b2 - 4) * q^76 + (b5 + b4 - b3 - 2*b2 - b1 + 2) * q^77 + (-2*b5 - 2*b4 + b3 + b2 + 2*b1 - 1) * q^79 + (-3*b5 - b4 + 11*b3 - b2 - 2*b1 + 2) * q^82 + (-b5 + 2*b4 - b3 + b2 + 2) * q^83 + (-2*b5 + b4 + 4*b3 + b2 + 2*b1 + 7) * q^86 + (b5 + b4 - 4*b3 - b2 - b1 + 3) * q^88 + (-2*b4 + 8*b3 - 3*b2 + b1 + 5) * q^89 + (-b5 + b4 - 2*b3 + b2 - 2*b1 + 4) * q^91 + (b5 - b4 - 3*b3 - 2*b2 - 4*b1 - 11) * q^92 + (2*b4 - b3 - 3*b2 - b1 + 6) * q^94 + (2*b5 - b4 - 2*b2 - 3*b1) * q^97 + (2*b4 - b3 - b2 + 6) * q^98 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6 q + 10 q^{4} + 6 q^{7} - 3 q^{8}+O(q^{10})$$ 6 * q + 10 * q^4 + 6 * q^7 - 3 * q^8 $$6 q + 10 q^{4} + 6 q^{7} - 3 q^{8} - 3 q^{11} + 6 q^{13} + 22 q^{14} + 18 q^{16} - 13 q^{17} + 11 q^{19} + 16 q^{22} - 13 q^{23} + 28 q^{26} + 7 q^{28} + 3 q^{29} - 11 q^{31} - 16 q^{32} + 15 q^{34} + 21 q^{37} + 9 q^{38} + q^{41} + 2 q^{43} - 9 q^{44} + 19 q^{46} - 14 q^{47} - 14 q^{49} + 13 q^{52} - 23 q^{53} + 35 q^{56} + 22 q^{58} - 9 q^{59} + 11 q^{61} + 23 q^{62} - 23 q^{64} + 8 q^{67} - 50 q^{68} + 8 q^{71} + 13 q^{73} + 22 q^{74} - 26 q^{76} + 13 q^{77} - 5 q^{79} - 13 q^{82} + 20 q^{83} + 37 q^{86} + 28 q^{88} + 4 q^{89} + 34 q^{91} - 61 q^{92} + 41 q^{94} - 7 q^{97} + 41 q^{98}+O(q^{100})$$ 6 * q + 10 * q^4 + 6 * q^7 - 3 * q^8 - 3 * q^11 + 6 * q^13 + 22 * q^14 + 18 * q^16 - 13 * q^17 + 11 * q^19 + 16 * q^22 - 13 * q^23 + 28 * q^26 + 7 * q^28 + 3 * q^29 - 11 * q^31 - 16 * q^32 + 15 * q^34 + 21 * q^37 + 9 * q^38 + q^41 + 2 * q^43 - 9 * q^44 + 19 * q^46 - 14 * q^47 - 14 * q^49 + 13 * q^52 - 23 * q^53 + 35 * q^56 + 22 * q^58 - 9 * q^59 + 11 * q^61 + 23 * q^62 - 23 * q^64 + 8 * q^67 - 50 * q^68 + 8 * q^71 + 13 * q^73 + 22 * q^74 - 26 * q^76 + 13 * q^77 - 5 * q^79 - 13 * q^82 + 20 * q^83 + 37 * q^86 + 28 * q^88 + 4 * q^89 + 34 * q^91 - 61 * q^92 + 41 * q^94 - 7 * q^97 + 41 * q^98
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{6} - 11x^{4} - x^{3} + 29x^{2} + 3x - 1$$ :
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$( \nu^{5} - \nu^{4} - 6\nu^{3} + 5\nu^{2} - 1 ) / 4$$ (v^5 - v^4 - 6*v^3 + 5*v^2 - 1) / 4 $$\beta_{3}$$ $$=$$ $$( \nu^{5} - \nu^{4} - 10\nu^{3} + 5\nu^{2} + 24\nu - 1 ) / 4$$ (v^5 - v^4 - 10*v^3 + 5*v^2 + 24*v - 1) / 4 $$\beta_{4}$$ $$=$$ $$( \nu^{5} - \nu^{4} - 10\nu^{3} + 9\nu^{2} + 24\nu - 13 ) / 4$$ (v^5 - v^4 - 10*v^3 + 9*v^2 + 24*v - 13) / 4 $$\beta_{5}$$ $$=$$ $$( \nu^{5} + \nu^{4} - 12\nu^{3} - 7\nu^{2} + 34\nu + 3 ) / 2$$ (v^5 + v^4 - 12*v^3 - 7*v^2 + 34*v + 3) / 2
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$\beta_{4} - \beta_{3} + 3$$ b4 - b3 + 3 $$\nu^{3}$$ $$=$$ $$-\beta_{3} + \beta_{2} + 6\beta_1$$ -b3 + b2 + 6*b1 $$\nu^{4}$$ $$=$$ $$\beta_{5} + 6\beta_{4} - 9\beta_{3} + \beta_{2} + \beta _1 + 16$$ b5 + 6*b4 - 9*b3 + b2 + b1 + 16 $$\nu^{5}$$ $$=$$ $$\beta_{5} + \beta_{4} - 10\beta_{3} + 11\beta_{2} + 37\beta _1 + 2$$ b5 + b4 - 10*b3 + 11*b2 + 37*b1 + 2
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
2.68704 2.01887 0.141689 −0.246759 −2.16056 −2.44028
−2.68704 0 5.22020 0 0 −1.68704 −8.65280 0 0
1.2 −2.01887 0 2.07584 0 0 −1.01887 −0.153106 0 0
1.3 −0.141689 0 −1.97992 0 0 0.858311 0.563913 0 0
1.4 0.246759 0 −1.93911 0 0 1.24676 −0.972011 0 0
1.5 2.16056 0 2.66802 0 0 3.16056 1.44329 0 0
1.6 2.44028 0 3.95498 0 0 3.44028 4.77071 0 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.6 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$3$$ $$-1$$
$$5$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 5625.2.a.q 6
3.b odd 2 1 1875.2.a.k 6
5.b even 2 1 5625.2.a.p 6
15.d odd 2 1 1875.2.a.j 6
15.e even 4 2 1875.2.b.f 12
25.e even 10 2 225.2.h.d 12
75.h odd 10 2 75.2.g.c 12
75.j odd 10 2 375.2.g.c 12
75.l even 20 4 375.2.i.d 24
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
75.2.g.c 12 75.h odd 10 2
225.2.h.d 12 25.e even 10 2
375.2.g.c 12 75.j odd 10 2
375.2.i.d 24 75.l even 20 4
1875.2.a.j 6 15.d odd 2 1
1875.2.a.k 6 3.b odd 2 1
1875.2.b.f 12 15.e even 4 2
5625.2.a.p 6 5.b even 2 1
5625.2.a.q 6 1.a even 1 1 trivial
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(5625))$$:
$$T_{2}^{6} - 11T_{2}^{4} + T_{2}^{3} + 29T_{2}^{2} - 3T_{2} - 1$$ T2^6 - 11*T2^4 + T2^3 + 29*T2^2 - 3*T2 - 1 $$T_{7}^{6} - 6T_{7}^{5} + 4T_{7}^{4} + 25T_{7}^{3} - 25T_{7}^{2} - 20T_{7} + 20$$ T7^6 - 6*T7^5 + 4*T7^4 + 25*T7^3 - 25*T7^2 - 20*T7 + 20
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{6} - 11 T^{4} + T^{3} + 29 T^{2} + \cdots - 1$$
$3$ $$T^{6}$$
$5$ $$T^{6}$$
$7$ $$T^{6} - 6 T^{5} + 4 T^{4} + 25 T^{3} + \cdots + 20$$
$11$ $$T^{6} + 3 T^{5} - 24 T^{4} - 66 T^{3} + \cdots + 244$$
$13$ $$T^{6} - 6 T^{5} - 26 T^{4} + 173 T^{3} + \cdots + 101$$
$17$ $$T^{6} + 13 T^{5} + 15 T^{4} + \cdots + 4639$$
$19$ $$T^{6} - 11 T^{5} + 11 T^{4} + \cdots - 380$$
$23$ $$T^{6} + 13 T^{5} + 6 T^{4} + \cdots - 2020$$
$29$ $$T^{6} - 3 T^{5} - 41 T^{4} + \cdots + 2105$$
$31$ $$T^{6} + 11 T^{5} - 46 T^{4} + \cdots - 2900$$
$37$ $$T^{6} - 21 T^{5} + 139 T^{4} + \cdots - 6025$$
$41$ $$T^{6} - T^{5} - 181 T^{4} + \cdots - 82655$$
$43$ $$T^{6} - 2 T^{5} - 91 T^{4} + \cdots - 6284$$
$47$ $$T^{6} + 14 T^{5} - 4 T^{4} + \cdots + 2284$$
$53$ $$T^{6} + 23 T^{5} + 61 T^{4} + \cdots - 34495$$
$59$ $$T^{6} + 9 T^{5} - 89 T^{4} + \cdots + 3920$$
$61$ $$T^{6} - 11 T^{5} - 139 T^{4} + \cdots + 168269$$
$67$ $$T^{6} - 8 T^{5} - 155 T^{4} + \cdots + 14684$$
$71$ $$T^{6} - 8 T^{5} - 150 T^{4} + \cdots - 196$$
$73$ $$T^{6} - 13 T^{5} - 74 T^{4} + \cdots + 22205$$
$79$ $$T^{6} + 5 T^{5} - 160 T^{4} + \cdots - 8000$$
$83$ $$T^{6} - 20 T^{5} + 26 T^{4} + \cdots + 41036$$
$89$ $$T^{6} - 4 T^{5} - 464 T^{4} + \cdots - 377055$$
$97$ $$T^{6} + 7 T^{5} - 215 T^{4} + \cdots + 2399$$ | 6,986 | 12,957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.562813 |
https://www.physicsforums.com/threads/lead-is-compressed-reversibly-with-t-const-find-ds.872423/ | 1,708,540,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00766.warc.gz | 971,964,415 | 17,661 | Lead is compressed reversibly with T const, find dS
• Kara386
In summary, the conversation discusses the change in entropy when lead is compressed from 1 bar to 1000 bars at a constant temperature of 300K. The Maxwell relation and other equations are used to determine the change in entropy, with the final conclusion being that the volume change is small enough to consider the integrand as constant.
Homework Statement
Lead is compressed reversibly and with temperature kept constant at 300K, and from pressure of 1bar to 1000 bar. Assuming the change in volume is small, what is the change in entropy S? Use the Maxwell relation derived from Gibbs free energy function. V = ##10^{-3}m^3##.
Homework Equations
##(\frac{\partial{V}}{\partial{T}})_p = -(\frac{\partial{S}}{\partial{p}})_T## Maxwell Relation
##\kappa = \frac{1}{V}(\frac{\partial{V}}{\partial{T}})_p = 8\times 10^{-5}K^{-1}##
##\beta = -\frac{1}{V}(\frac{\partial{V}}{\partial{p}})_T = 2.2\times 10^{-6}bar^{-1}## Where the T is meant to be a subscript denoting T constant for that derivative, but I can't make it look much like a subscript.
3. The Attempt at a Solution
From the Maxwell Relation,
##\kappa V = -(\frac{\partial{S}}{\partial{p}})_T##
And since the process is isothermal and I know dP, I'm wondering whether I can't just rearrange so that
## \kappa V \partial p = \partial S##
But that seems like wrong maths in terms of how you're allowed to treat derivatives. I suppose I'm essentially asking if that manipulation is allowed, and if not, I've tried using cyclic relationships, the central equation and other Maxwell relations but I can't seem to get anywhere. So what would I do instead?!
Thanks for any help!
Looks correct to me.
gcuf and Kara386
Chestermiller said:
Looks correct to me.
OK, thanks! But why can I do that without integrating? Because doesn't ##dp## represent a small change in ##p##, while we're looking at a change of ##999bar## in this case?
Kara386 said:
OK, thanks! But why can I do that without integrating? Because doesn't ##dp## represent a small change in ##p##, while we're looking at a change of ##999bar## in this case?
Between 1 bar and 1000 bars, the volume only changes of 0.2%. So, it is essentially constant. So, the integrand is essentially constant.
Kara386
1. What is the definition of "Lead is compressed reversibly"?
Lead is compressed reversibly means that the compression process is carried out slowly and without any friction or energy loss, allowing the system to reach equilibrium at each step.
2. How is the compression of lead reversible with constant temperature (T)?
The compression of lead is reversible with constant temperature because the system is compressed slowly and smoothly, allowing the temperature to remain constant throughout the process.
3. How is the entropy (S) affected by the reversible compression of lead?
The reversible compression of lead does not affect the entropy of the system, as the process is carried out at a constant temperature and there is no change in the number of microstates.
4. What is the mathematical equation for finding the change in entropy (dS) during the reversible compression of lead?
The equation for finding the change in entropy during reversible compression of lead is dS = nRln(V2/V1), where n is the number of moles, R is the gas constant, and V2 and V1 are the final and initial volumes, respectively.
5. What is the significance of finding dS during the reversible compression of lead?
Finding dS during the reversible compression of lead allows us to calculate the change in entropy of the system, which is a measure of the disorder or randomness of the particles. This information can be useful in understanding the thermodynamic properties and behavior of lead under different conditions. | 895 | 3,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-10 | latest | en | 0.917974 |
https://rdrr.io/cran/EcoSimR/src/R/algorithms.R | 1,537,534,183,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157070.28/warc/CC-MAIN-20180921112207-20180921132607-00025.warc.gz | 607,458,259 | 21,608 | # R/algorithms.R In EcoSimR: Null Model Analysis for Ecological Data
#### Documented in ra1ra2ra3ra4sim1sim10sim2sim3sim4sim5sim6sim7sim8size_gammasize_source_poolsize_uniformsize_uniform_uservector_sample
```#' Vector Sample Function
#' @description Takes an input binary vector and a weight vector.
#' Reassigns 1s randomly in proportion to vector weights.
#' @details This function takes an input vector of binary presence-absence
#' values and a vector of non-negative probability weights. Both vectors
#' must be of identical length.
#' @param speciesData binary vector representing species presences and
#' absences.
#' @param weights a vector of non-negative read numbers representing
#' probabilistic weights for species occurrencs of the same length
#' as speciesData.
#' @return Returns a re-ordered binary vector in which the occurrences
#' are placed in cells with probabilities proportional to values given
#' in weights.
#' @references Gotelli, N.J., G.R. Graves, and C. Rahbek. 2010. Macroecological
#' signals of species interactions in the Danish avifauna. Proceedings of the
#' National Academy of Sciences, U.S.A. 107: 530-535.
#' @note Several of the randomization algorithms use this function
#' to assign species occurrences with probabilities that reflect
#' species or site differences. It is an effective method for
#' conditioning the marginal probabilities of a null matrix on
#' independent measurements of site or species characteristics.
#' @examples
#' myColonizer <- vector_sample(speciesData=rbinom(10,1,0.5),weights=runif(10))
#' @export
vector_sample <- function(speciesData,weights)
{
x <- mat.or.vec(length(speciesData),1)
x[sample(1:length(speciesData),size=sum(speciesData),prob=weights)] <- 1
return(x)
}
#' Sim1 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling
#' all of its elements equiprobably.
#' @details This algorithm assumes species and sites are equiprobable.
#' It preserves the total matrix fill, but places no other constraints on
#' row or column totals.
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and percent fill as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This is the simplest of all randomization algorithms for a presence-
#' absence matrix. However, it assumes that both species and sites are
#' equiprobable, and has poor Type I error frequencies when tested with
#' purely random matrices. If the input matrix is sparse, it will often
#' generate null matrices with empty rows or columns. Not recommended for
#' co-occurrence analysis.
#' @examples
#' randomMatrix <- sim1(speciesData=matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim1 <- function(speciesData)
{
matrix(sample(speciesData), ncol=ncol(speciesData))
}
#' Sim2 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling elements
#' within each row equiprobably.
#' @details This algorithm assumes sites are equiprobable, but preserves
#' differences among species (= row sums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and rowsums as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. ecology 81: 2606-2621.
#' @note This algorithm preserves differences in the commonness and rarity of
#' species (= rowsums), but assumes that all sites are equiprobable. It would
#' not be appropriate for islands that vary greatly in area, but it would be
#' appropriate for quadrat censuses in a relatively homogeneous environment.
#' sim2 can sometimes generate matrices with empty columns, but this is
#' unlikely unless the matrix is very sparse. sim2 has good Type I error
#' frequenceis when tested against random matrices. However, if sites do vary
#' in their suitability or habitat quality, it will often identify aggregated
#' patterns of species co-occurrence. sim2 and sim9 have the best overall
#' performance for species co-occurrence analyses. However, because they differ
#' in their assumptions about site quality, they often differ in their results,
#' with sim9 often detecting random or segregated patterns for matrices in
#' which sim2 detects aggregated patterns.
#' @examples
#' randomMatrix <- sim2(speciesData=matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim2 <- function(speciesData)
{
t(apply(speciesData,1,sample))
}
#' Sim3 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling elements
#' within each column equiprobably.
#' @details This algorithm assumes species are equiprobable, but preserves
#' differences among sites (= column sums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and colsums as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This algorithm preserves differences in species richness among sites
#' (= colsums), but assumes that all species are equiprobable. This assumption
#' is usually unrealistic, and sim3 has a high frequency of Type I errors with
#' random matrices, so it is not recommended for co-occurrence analysis.
#' @examples
#' randomMatrix <- sim3(speciesData=matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim3 <- function(speciesData)
{
apply(speciesData,2,sample)
}
#' Sim4 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling elements
#' within each row. Sampling weights for each column are proportional to
#' column sums. Makes a call to the vector_sample function.
#' @details This algorithm preserves differences among species in occurrence
#' frequencies, but assumes differences among sites in suitability are
#' proportional to observed species richness (= column sums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and rowsums as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This algorithm preserves differences in the commonness and rarity of
#' species (= rowsums), but assumes differences among sites in suitability are
#' proportional to observed species richness (= column sums). sim4 has a
#' somewhat high frequency of Type I errors with random matrices, so it is not
#' recommended for co-occurrence analysis.
#' @examples
#' randomMatrix <- sim4(speciesData = matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim4 <- function(speciesData)
{
t(apply(speciesData,1,vector_sample,weights=colSums(speciesData)))
}
#' Sim5 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling elements
#' within each column. Sampling weights for each row are proportional to
#' row sums. Makes a call to the vector_sample function.
#' @details This algorithm preserves differences among sites in species
#' richness, but assumes differences among species in commonness and rarity
#' are proportional to observed species occurrences (= row sums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and colsums as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This algorithm preserves differences among sites in species richness
#' (= colsums), but assumes differences among species in commonness and rarity
#' are proportional to observed species occurrences (= rowsums). sim5 has a
#' high frequency of Type I errors with random matrices, so it is not
#' recommended for co-occurrence analysis.
#' @examples
#' randomMatrix <- sim5(speciesData = matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim5 <- function(speciesData)
{
apply(speciesData,2,vector_sample,weights=rowSums(speciesData))
}
#' Sim6 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling all
#' elements. Rows are equiprobable, and columns are proportional to column sums.
#' Makes a call to the vector_sample function.
#' @details This algorithm assumes that species are equiprobable, but that
#' differences in suitability among sites are proportional to observed species
#' richness (=colsums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and fill as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This algorithm assumes that species are equiprobable, and that
#' differences among sites are proportional to observed species richness
#' (=colsums). sim6 has a high frequency of Type I errors with random matrices,
#' so it is not recommended for co-occurrence analysis.
#' @examples
#' randomMatrix <- sim6(speciesData = matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim6 <- function(speciesData)
{
matrixWeights <- outer(rep(1,nrow(speciesData)),colSums(speciesData))
out <-matrix(vector_sample(speciesData, weights=matrixWeights),ncol=ncol(speciesData))
}
#' Sim7 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling all
#' elements. Columns are equiprobable, and rows are proportional to row sums.
#' Makes a call to the vector_sample function.
#' @details This algorithm assumes that sites are equiprobable, but that
#' differences in frequency of occurrence among species are proportional to
#' observed species richness (=colsums).
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and fill as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#' @note This algorithm assumes that species are equiprobable, and that
#' differences among sites are proportional to observed species richness
#' (=colsums). sim7 has a high frequency of Type I errors with random matrices,
#' so it is not recommended for co-occurrence analysis.
#' @examples
#' randomMatrix <- sim7(speciesData = matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim7 <- function(speciesData)
{
matrixWeights <- outer(rowSums(speciesData),rep(1,ncol(speciesData)))
matrix(vector_sample(speciesData, weights=matrixWeights),ncol=ncol(speciesData))
}
#' Sim8 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling all
#' elements. Columns are proportional to column sums, and rows are proportional
#' to row sums. Makes a call to the vector_sample function.
#' @details This algorithm assumes that the probability that a species occurs
#' in a site is depends on the joint independent probability of randomly
#' selecting the species and randomly selecting the site, with these
#' probabilities set proportional to row and column sums of the matrix.
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and fill as the input matrix.
#' @references Gotelli, N.J. 2000. Null model analysis of species co-occurrence
#' patterns. Ecology 81: 2606-2621.
#'
#' Ulrich, W. and N.J. Gotelli. 2012. A null model algorithm for presence-
#' absence matrices based on proportional resampling. Ecological Modelling
#' 244:20-27.
#' @note This algorithm is theoretically attractive because it incorporates
#' heterogeneity in species occurrences and species richness per site in a
#' probabilistic way that does not fix row and column frquencies. However,
#' in spite of its appeal, sim8 does not generate average row and column sums
#' that match the original matrix, and it is susceptible to Type I errors when
#' tested with random matrices. It is not recommended for co-occurrence
#' analysis. See Ulrich and Gotelli (2012) for a more complicated algorithm
#' for probabilistic row and column totals that has better statistical behavior.
#' @examples
#' randomMatrix <- sim8(speciesData = matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim8 <- function(speciesData)
{
matrixWeights <- outer(rowSums(speciesData),colSums(speciesData))
matrix(vector_sample(speciesData,weights=matrixWeights),ncol=ncol(speciesData))
}
#' Sim10 Co-occurrence Randomization Algorithm
#' @description Randomizes a binary matrix speciesData by reshuffling all
#' elements. Rows and column probabilities are proportional to user-supplied
#' row and column weights, which define relative suitability probabilities for
#' species and sites. Makes a call to the vector_sample function.
#' @details This function incorporates vectors of weights for species and/or
#' sites to condition the simulation. These two vectors are used as outer
#' products to set cell probabilities for the entire matrix. Thus:
#' \deqn{p(cell_{ij})=p(row_i)p(col_j)}{p(cell_ij)=p(row_i)p(col_j)}
#' Weights must be positive real numbers. The algorithm will scale them so they
#' sum to 1.0, so they can be used in their natural units (e.g. island area,
#' species abudance), and will be scaled properly. If all species (or sites)
#' are assumed to be equally likely, the weight vector should be set to the
#' same constant for all elements.
#' @param speciesData binary presence-absence matrix
#' (rows = species, columns = sites).
#' @param rowWeights vector of positive values representing species weights.
#' @param colWeights vector of positive values representing site weights.
#' @return Returns a binary presence-absence matrix with the same
#' dimensions and fill as the input matrix.
#' @references Jenkins, D.G. 2006. In search of quorum effects in metacommunity
#' structure: species co-occurrence analyses. Ecology 87:1523-1531
#'
#' Gotelli, N.J., G.R. Graves, and C. Rahbek. 2010. Macroecological
#' signals of species interactions in the Danish avifauna. Proceedings of the
#' National Academy of Sciences, U.S.A. 107: 530-535.
#' @note sim10 allows users to incorporate independent data on species
#' occurrence probabilities and site suitabilities. This represents an
#' important conceptual advance over standard co-occurrence analyses, which
#' must infer these probabilities from the matrix itself. sim10 may generate
#' empty rows or columns, especially if weights are very small for some species
#' or sites. Also, the results may be sensitive to algebraic transformations
#' of the weights (x, x^2, log(x), etc.), and these transformations may be hard
#' to justify biologically. Nevertheless, sim10 is worth exploring for rich
#' data sets with site and species attributes.
#' @seealso \code{\link{vector_sample}} for weighted vector sampling.
#' @examples
#' randomMatrix <- sim10(speciesData=matrix(rbinom(40,1,0.5),nrow=8))
#' @export
sim10 <- function(speciesData,rowWeights = runif(dim(speciesData)[1]),colWeights = runif(dim(speciesData)[2]))
{
matrixWeights <- outer(rowWeights,colWeights)
matrix(vector_sample(speciesData, weights =matrixWeights),ncol=ncol(speciesData))
}
#' RA1 Niche Overlap Randomization Algorithm
#' @description Randomizes a numeric utilization matrix speciesData by
#' replacing all elements with a random uniform [0,1] value.
#' @details The resource utilization matrix (rows = species, columns = discrete
#' resource categories) may include zeroes, but no negative numbers or missing
#' values. Relative resource within a species is first calculated, so the rows
#' need not sum to 1.0.
#' @param speciesData a resource utilization matrix (rows = species, columns = discrete
#' resource states) filled with non-negative real numbers.
#' @return Returns a random utilization matrix with the same dimensions as the
#' input matrix.
#' @references Kobayashi, S. 1991. Interspecific relations in forest floor
#' coleopteran assemblages: niche overlap and guild structure. Researches
#' in Population Ecology 33: 345-360.
#' @note Because all matrix elements, including zeroes, are replaced with a
#' random uniform distribution, the null expectation is based on an assemblage
#' of generalist species with maximum niche breadth. This algorithm retains
#' neither the niche breadth of the individuals species nor the placement of 0
#' values (= unutilized resource states) in the matrix. These assumptions are
#' unrealistic, and a random matrix with zeroes will generate significantly low
#' niche overlap values with this metric. It is not recommended for niche
#' overlap analysis.
#' @examples
#' ranUtil <- ra1(speciesData=matrix(rpois(40,0.5),nrow=8))
#' @export
ra1 <- function(speciesData=matrix(rpois(80,1),nrow=10)){
matrix(runif(prod(dim(speciesData))),ncol=ncol(speciesData))
}
#' RA2 Niche Overlap Randomization Algorithm
#' @description Randomizes a numeric utilization matrix speciesData by
#' replacing all non-zero elements with a random uniform [0,1] value.
#' @description Randomizes a numeric utilization matrix speciesData by
#' replacing all elements with a random uniform [0,1].
#' @details The resource utilization matrix (rows = species, columns = discrete
#' resource categories) may include zeroes, but no negative numbers or missing
#' values. Relative resource within a species is first calculated, so the rows
#' need not sum to 1.0.
#' @param speciesData a resource utilization matrix (rows = species, columns = discrete
#' resource states) filled with non-negative real numbers.
#' @return Returns a random utilization matrix with the same dimensions as the
#' input matrix.
#' @references Kobayashi, S. 1991. Interspecific relations in forest floor
#' coleopteran assemblages: niche overlap and guild structure. Researches
#' in Population Ecology 33: 345-360.
#'
#' Winemiller, K.O. and E.R. Pianka. 1990. Organization in natural
#' assemblages of desert lizards and tropical fishes. Ecological Monographs
#' 60: 27-55.
#' @note This algorithm retains the number and position of zero states in the
#' original matrix. However, all non-zero values are again replaced by a random
#' [0,1] value, which tends to inflate niche breadths of the simulated
#' assemblage. Although the results are not as severe as for RA1, this
#' algorithm is still prone to Type I errors, and is not recommended for niche
#' overlap analysis.
#' @examples
#' ranUtil <- ra2(speciesData=matrix(rpois(40,0.5),nrow=8))
#' @export
ra2 <- function(speciesData=matrix(rpois(80,1),nrow=10))
{
z <- which(speciesData > 0)
RM <- speciesData
RM[z] <- runif(length(z))
return(RM)
}
#' RA3 Niche Overlap Randomization Algorithm
#' @description Randomizes a numeric utilization matrix speciesData by
#' reshuffling the elements within each row.
#' @details The resource utilization matrix (rows = species, columns = discrete
#' resource categories) may include zeroes, but no negative numbers or missing
#' values. Relative resource within a species is first calculated, so the rows
#' need not sum to 1.0.
#' @param speciesData a resource utilization matrix (rows = species, columns = discrete
#' resource states) filled with non-negative real numbers.
#' @return Returns a random utilization matrix with the same dimensions as the
#' input matrix.
#' @references Winemiller, K.O. and E.R. Pianka. 1990. Organization in natural
#' assemblages of desert lizards and tropical fishes. Ecological Monographs
#' 60: 27-55.
#' @note This algorithm retains the niche breadth and zero states for each
#' species, but randomizes the assignment of each utilization value to a
#' different niche category. It performs effectively in simulation studies and
#' is recommended for analysis of niche overlap patterns.
#' @examples
#' ranUtil <- ra3(speciesData=matrix(rpois(40,0.5),nrow=8))
#' @export
ra3 <- function(speciesData=matrix(rpois(80,1),nrow=10))
{
RM <- apply(speciesData,1,sample)
RM <- t(as.matrix(RM,nrow=nrow(speciesData),ncol=ncol(speciesData)))
return(RM)
}
#' RA4 Niche Overlap Randomization Algorithm
#' @description Randomizes a numeric utilization matrix speciesData by
#' reshuffling the non-zero elements within each row.
#' @details The resource utilization matrix (rows = species, columns = discrete
#' resource categories) may include zeroes, but no negative numbers or missing
#' values. Relative resource within a species is first calculated, so the rows
#' need not sum to 1.0.
#' @param speciesData a resource utilization matrix (rows = species, columns = discrete
#' resource states) filled with non-negative real numbers.
#' @return Returns a random utilization matrix with the same dimensions as the
#' input matrix.
#' @references Winemiller, K.O. and E.R. Pianka. 1990. Organization in natural
#' assemblages of desert lizards and tropical fishes. Ecological Monographs
#' 60: 27-55.
#' @note This algorithm is similar to RA3, but adds the additional constraint of
#' retaining the positions of all of the zero elements of the matrix, and
#' reshuffling only the non-zero elements of the matrix within each row. It is
#' more conservative than RA3, but has a low Type I error rate, and, along with
#' RA3, is recommended for null model analysis of niche overlap.
#' @examples
#' ranUtil <- ra4(speciesData=matrix(rpois(40,0.5),nrow=8))
#' @export
ra4 <- function(speciesData=matrix(rpois(80,1),nrow=10))
{
#.....................................
## Nonzero row shuffle function for RA4
NonZeroRowShuffle <- function(vec=runif(10)) {
nonzero <- which(vec > 0)
shuffledvec <- vec
shuffledvec[nonzero] <- vec[sample(nonzero)]
return(shuffledvec)
}
#....................................
RM <- t(apply(speciesData,1,NonZeroRowShuffle))
return(RM)
}
#' SizeUniform Size Overlap Randomization Algorithm
#' @description Function to randomize body sizes within a uniform distribution
#' with boundaries set by the largest and smallest species in the assemblage.
#' @details If the assemblage contains n species,
#' only the body sizes of the inner n - 2 species are randomized.
#' @param speciesData a vector of positive real values representing the body
#' sizes or trait values for each species.
#' @return Returns a vector of body sizes that have been randomly assigned. The
#' largest and smallest body sizes in the randomized assemblage match those in
#' the empirical data.
#' @references Simberloff, D. and W. Boecklen. 1981. Santa Rosalia
#' reconsidered: size ratios and competition. Evolution 35: 1206-1228.
#'
#' Tonkyn, D.W. and B.J. Cole. 1986. The statistical analysis of size ratios.
#' American Naturalist 128: 66-81.
#' @note Although the distribution of body sizes may not be truly uniform,
#' it may be approximately uniform within the range of observed values,
#' particularly for small assemblages.
#' @seealso \code{\link{size_gamma}} size distribution function.
#' @examples
#' nullSizes <-size_uniform(speciesData=runif(20))
#'@export
size_uniform <- function(speciesData=runif(20)) {
endpoints <- c(min(speciesData),max(speciesData)) # save max and min boundaries
sim <- runif(n=(length(speciesData)-2),min=endpoints[1],max=endpoints[2])
randomVec <- c(endpoints,sim)
return(randomVec)
}
#' SizeUser Size Overlap Randomization Algorithm
#' @description Observed body sizes are randomized with a uniform distribution
#' for which the user has defined the minimum and maximum possible body size.
#' @details Within the user-defined limits, body sizes of all n species are
#' randomized, whereas uniform_size randomizes only n - 2 of the body
#' sizes and uses the extreme values to set the endpoints.
#' @param speciesData a vector of observed body sizes.
#' @param userLow a user-defined lower limit.
#' @param userHigh a user-defined upper limit.
#' @return Returns a vector of randomized body sizes.
#' @note As the difference between the lower and upper boundaries is increased
#' the test will yield results that are random or aggregated, even though the
#' same data might yield a segregated pattern when the uniform_size algorithm
#' is used. For this reason, this algorithm is not recommended for size ratio
#' analyses.
#' @seealso \code{\link{size_uniform}} size distribution algorithm.
#' @examples
#' nullSizes <- size_uniform_user(speciesData=runif(20,min=10,max=20),userLow=8,userHigh=24)
#' @export
size_uniform_user <- function(speciesData=runif(n=20),userLow=0.9*min(speciesData),
userHigh=1.1*max(speciesData)){
# if(!is.null(Param.List\$Special)){User.low <- Param.List\$Special[1]
# User.high <- Param.List\$Special[2]}
randomVec <- runif(n=length(speciesData),min=userLow,max=userHigh)
return(randomVec)
}
#' SizeSourcePoolDraw Size Overlap Randomization Algorithm
#' @description Function to randomize body sizes by drawing species from a
#' user-defined source pool. Species are drawn without replacement,
#' and there is a specified probability vector for the source pool species
#' @param speciesData a vector of observed body sizes.
#' @param sourcePool a vector of body sizes of species in the user-defined
#' pool of potential colonists.
#' @param speciesProbs a vector of relative colonization weights of
#' length 'sourcePool'.
#' @return Returns a vector of body sizes of an assemblage randomly drawn
#' from a user-defined source pool.
#' @references Strong, D.R. Jr., L.A. Szyska, and D. Simberloff. 1979. Tests of
#' community-wide character displacement against null hypotheses. Evolution 33:
#' 897-913.
#' Schluter, D. and P.R. Grant. 1984. Determinants of morphological patterns in
#' communities of Darwin's finches. American Naturalist 123: 175-196.
#' @note Although delineating a source pool of species and estimating their
#' relative colonization probabilities is difficult, this is the most realistic
#' approach to constructing a null distribution.
#' @examples
#' obsOverlap <- size_source_pool(dataRodents\$Sonoran)
#' @export
size_source_pool <- function(speciesData=21:30,sourcePool=
runif(n=2*length(speciesData),min=10,max=50),
speciesProbs=rep(1,length(sourcePool))) {
speciesDraw <- sample(sourcePool,size=length(speciesData),replace=FALSE,
prob=speciesProbs)
return(speciesDraw)
}
#' SizeGamma Size Overlap Randomization Algorithm
#' @description Function to generate a random distribution of body sizes by
#' drawing from a gamma distribution. Shape and rate parameters of the gamma
#' are estimated from the empirical data.
#' @param speciesData a vector of body sizes or other trait measurements of
#' species. All values must be positive real numbers.
#' @return Returns a vector of simulated body sizes as the same length as speciesData.
#' @note The shape and rate parameters are estimated from the real data using
#' the maximum likelihood estimators generated from the fitdr function in
#' the MASS library. The flexible gamma distribution can be fit to a variety of
#' normal, log-normal, and exponential distributions that are typical for trait
#' data measured on a continuous non-negative scale.
#' @seealso \code{\link{fitdistr}} in the MASS library.
#' @examples
#' obsOverlap <- size_gamma(speciesData=rnorm(50,mean=100,sd=1))
#' @import MASS
#' @export
size_gamma <- function (speciesData=rnorm(50,mean=100,sd=1)) {
mleFit <- fitdistr(speciesData, 'gamma')
a <- mleFit\$estimate["shape"]
b <- mleFit\$estimate["rate"] | 6,665 | 27,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-39 | longest | en | 0.667141 |
https://jp.mathworks.com/matlabcentral/answers/365215-find-max-gradient-in-a-slope?s_tid=prof_contriblnk | 1,603,875,669,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107897022.61/warc/CC-MAIN-20201028073614-20201028103614-00114.warc.gz | 394,195,576 | 23,532 | # Find max gradient in a slope
111 ビュー (過去 30 日間)
Marthe Fenne Vestly 2017 年 11 月 5 日
I have computed displacement depending on slope angle and the result is the plot below. Now I want to find the slope angle where the displacement increases the most, but i can´t figure out how the gradient commands works. Does anyone have a suggestion how to get this information?
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### 回答 (2 件)
Star Strider 2017 年 11 月 5 日
The gradient (link) function assumes a constant step in each direction you want to calculate. You have to experiment with it with your data to get the result you want.
This should get you started:
t = linspace(0, 25);
x = (1:5)/125;
f = exp(bsxfun(@times, t(:), x)); % Column-Major Array (‘t’ Increases Down Columns)
[drow,dcol] = gradient(f, x, t); % ‘drow’: Across Rows; ‘dcol’: Down Column
figure(1)
semilogy(t, f)
grid
figure(2)
surf(x, t, f)
hold on
mesh(x, t, dcol)
hold off
grid on
view([120 25])
#### 0 件のコメント
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Image Analyst 2017 年 11 月 5 日
It looks like the displacement increases the most at the last slope angle of each layer. Thus, for layer1, where you have arrays displacement1 and slopeAngle1, the slope where the displacement increases the most would simply be slopeAngle(end). Same for the other 5 layers.
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Translated by | 394 | 1,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.487338 |
https://mmf.bsu.by/en/bachelors-programs/mathematics-science-industrial-activity/subjects/13130/4-semester/ | 1,680,416,103,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00263.warc.gz | 465,119,001 | 11,917 | # 4 semester
1 Course title Differential equations 2 Year of study, study programme 2, 1-31 03 01-01 Mathematics (Research Activities and Production) 3 Semester of study 4 4 Number of credits 4 5 Lecturer Gromak Valery Ivanovich 6 Course Objective Formation of students’ skills and abilities in conducting mathematical research on the basis of the theory of differential equations, learning the basic analytical, qualitative and asymptotic methods of the theory of differential equations, constructing and analyzing basic mathematical models on the basis of the theory of differential equations. As a result of the training, the student must know: – basic concepts and methods of general theories of linear differential equations, existence and uniqueness theorems; – basic methods of integrating linear differential equations and systems, Euler’s method, elimination method, matrix method; – formulation of the Cauchy problem and boundary value problems; – the basic concepts and methods of the theory of stability by Lyapunov. be able to: – apply methods of general theories of linear differential equations for solving basic types of linear differential equations and systems; – to set initial and boundary-value problems, to solve questions of existence and uniqueness of the solution of initial and boundary value problems – to build phase portraits of the simplest autonomous systems on the plane. – apply the main theorems of the second Lyapunov method for solving the problems of stability of motion. 7 Prerequisites Algebra and Number Teory. Mathematical analysis. Geometry. 8 Course content Linear differential equations of the n-th order. Properties of solutions of linear homogeneous differential equations of n-th order. The determinant of Vronsky. A fundamental system of solutions. Formula Ostrogradsky – Liouville. Method of variation of arbitrary constants for inhomogeneous linear differential equations. The Cauchy method. Linear homogeneous equations of the n-th order with constant coefficients. Euler’s method. Linear inhomogeneous equations of the n-th order with constant coefficients and the right-hand side of a special form. Second-order linear equations and oscillatory phenomena. Sturm’s theorems on the zeros of solutions. The concept of boundary value problems. Linear differential equations with holomorphic coefficients. Generalized power series. Integration of linear differential equations by means of power and generalized power series. The Airy and Bessel equation. Bessel functions. Linear differential systems. Properties of solutions Linear independence of vector functions. Formula Ostrogradsky-Liouville. A fundamental system of solutions. The fundamental matrix. Method of variation of arbitrary constants for inhomogeneous linear systems. Exponential function of the matrix argument. The Lappo-Danilevsky theorem. Matrix method of integrating a linear homogeneous system with constant coefficients. The structure of the fundamental matrix. Euler’s method. Linear systems with periodic coefficients. Reduced systems. Solution of an inhomogeneous system with a right-hand side of a special form. Autonomous systems of differential equations. Autonomous systems. Phase portraits of a linear autonomous system of two equations. Special points: node, saddle, focus, center. The concept of a limit cycle. The problem of focus and focus. Lyapunov stability of solutions of differential equations. Functions of Lyapunov. Lyapunov’s theorems on stability and asymptotic stability. A criterion for the asymptotic stability of the zero solution of linear autonomous systems and the nth order equation. Lyapunov’s theorem on stability in the first approximation. 9 Recommended Literature Основная: 1. Амелькин В.В. Дифференциальные уравнения, Минск, БГУ, 2012. 2. Бибиков Ю.Н. Курс обыкновенных дифференциальных уравнений. Москва: «Высшая школа», 1991. 3. Федорюк М.В. Обыкновенных дифференциальные уравнения. СПб.; Издательство «Лань», 2003. 4. Филиппов А.Ф. Сборник задач по дифференциальным уравнениям. Москва «Наука», 1992. Дополнительная: 1. Богданов Ю.С., Мазаник С.А., Сыроид Ю.Б. Курс дифференциальных уравнений. Минск: «Университетское», 1996. 2. Еругин Н.П. Книга для чтения по общему курсу дифференциальных уравнений. Минск: «Наук»,1972. 3. Степанов В.В. Курс дифференциальных уравнений. Москва: Физматгиз, 1959. 10 Teaching Methods Lecture with practical exercises, using elements of distance learning and electronic materials. 11 Teaching language Russian 12 Requirements, current control Test papers. Exam score consist of the current mark (40%) and the oral exam mark (60%). 13 Method of certification Credit, Exam | 1,047 | 4,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.867376 |
https://math.stackexchange.com/questions/2100955/prove-disprove-the-following | 1,561,543,000,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000266.39/warc/CC-MAIN-20190626094111-20190626120111-00549.warc.gz | 528,875,078 | 35,237 | # Prove/disprove the following:
I'm given the following problem:
Prove that if $f$ is negative and increasing between $a$ and $b$, then the left Riemann sum is an underestimate and the right Riemann sum is an overestimate.
After drawing a graph, I concluded that this statement is false but I do not know how to prove this algebraically. Any pointers?
• It is actually true; you must be confusing something! – RGS Jan 17 '17 at 0:12
• Remember that for functions that are negative, area and integral aren't quite the same thing. – Arthur Jan 17 '17 at 0:14
• image.prntscr.com/image/277120c505564af2bd5fb0301d03d325.png – Question asker Jan 17 '17 at 0:16
• It seems to me that there is an overestimate for the area of a left reimann sum in this graph, or is is that the area of a negative function is negative, meaning that it is less area in the end? – Question asker Jan 17 '17 at 0:18
• @Questionasker The left sum overestimates the area, but underestimates the integral. – Arthur Jan 17 '17 at 0:19
Consider the partition $[x_0, x_1, x_2, \cdots, x_n]$ of $[a, b]$ where $x_0 = a, x_n = b$;
Notice that $f(x_0) \leq f(x_1) \leq \cdots \leq f(x_{n-1}) \leq f(x_n)$. (the $\leq$ may be exchanged by $<$ if the problem statement means that $f$ is strictly increasing. | 390 | 1,274 | {"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} | 3.421875 | 3 | CC-MAIN-2019-26 | latest | en | 0.883521 |
https://programs.programmingoneonone.com/2021/09/leetcode-design-twitter-problem-solution.html | 1,718,682,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00780.warc.gz | 408,171,698 | 33,719 | # Leetcode Design Twitter problem solution
In this Leetcode Design Twitter problem solution Design a simplified version of Twitter where users can post tweets, follow/unfollow another user, and is able to see the 10 most recent tweets in the user's news feed.
2. void postTweet(int userId, int tweetId) Composes a new tweet with ID tweetId by the user userId. Each call to this function will be made with a unique tweetId.
3. List<Integer> getNewsFeed(int userId) Retrieves the 10 most recent tweet IDs in the user's news feed. Each item in the news feed must be posted by users who the user followed or by the user themself. Tweets must be ordered from most recent to least recent.
4. void follow(int followerId, int followeeId) The user with ID followerId started following the user with ID followeeId.
5. void unfollow(int followerId, int followeeId) The user with ID followerId started unfollowing the user with ID followeeId.
## Problem solution in Python.
```class Twitter:
def __init__(self):
self.d={}
self.f={}
self.o=[]
def postTweet(self, a: int, b: int) -> None:
if a in self.d:
self.d[a]=[b]+self.d[a]
print(self.d[a])
else:
self.d[a]=[b]
self.o=[[a,b]]+self.o
def getNewsFeed(self, a: int) -> List[int]:
fol=[]
if a in self.f:
fol=self.f[a]
res=[];c=0;
for i,j in self.o:
if i==a or i in fol:
res.append(j)
c+=1
if c>=10:
break
return res
def follow(self, a: int, b: int) -> None:
if a in self.f:
self.f[a].append(b)
else:
self.f[a]=[b]
def unfollow(self, a: int, b: int) -> None:
if a in self.f:
if b in self.f[a]:
self.f[a].remove(b)
```
## Problem solution in Java.
```public class Twitter {
public class Tweet{
int id; int userId;
public Tweet (int id, int userId) { this.id = id; this.userId = userId; }
}
HashMap<Integer, HashSet<Integer>> users = new HashMap<>();
public void postTweet(int userId, int tweetId) {
if (!users.containsKey(userId)) users.put(userId, new HashSet<>());
tweetStack.push(new Tweet(tweetId, userId));
}
public List<Integer> getNewsFeed(int userId) {
if (!users.containsKey(userId)) users.put(userId, new HashSet<>());
List<Integer> result = new ArrayList<>();
for (Tweet tweet : tweetStack){
if (users.get(userId).contains(tweet.userId) || tweet.userId == userId) result.add(tweet.id);
if (result.size() == 10) break;
}
return result;
}
public void follow(int followerId, int followeeId) {
if (!users.containsKey(followerId)) users.put(followerId, new HashSet<>());
}
public void unfollow(int followerId, int followeeId) {
if (!users.containsKey(followerId)) users.put(followerId, new HashSet<>());
users.get(followerId).remove(followeeId);
}
}
```
## Problem solution in C++.
```class Twitter {
public:
unordered_set<int> a={};
vector<unordered_set<int>> followers;
vector<pair<int,int>> news;
for(int i=0;i<10000;++i)
followers.push_back({i});
}
void postTweet(int userId, int tweetId) {
news.push_back(make_pair(userId,tweetId));
}
vector<int> getNewsFeed(int userId) {
vector<int> res;
int n=news.size();
for(int i=n-1;(i>=0&&res.size()<10);--i)
if(followers[userId].find(news[i].first)!=followers[userId].end())
res.push_back(news[i].second);
return res;
}
void follow(int followerId, int followeeId) {
followers[followerId].insert(followeeId);
}
void unfollow(int followerId, int followeeId) {
if(followerId!=followeeId)
followers[followerId].erase(followeeId);
}
};
``` | 892 | 3,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-26 | latest | en | 0.504122 |
https://math.stackexchange.com/questions/1830512/every-group-of-order-111-is-cyclic-true-or-false | 1,632,194,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00108.warc.gz | 428,968,454 | 32,702 | # Every group of order 111 is cyclic. (True or False) [duplicate]
By Lagrange's theorem the possible order of the subgroups of the group will be 1 or 3 or 37 or 111
Now by Sylow's 1st theorem this group must have a subgroup of order 3 and as well as 37.
Then we can prove that every abelian group of order 111 is cyclic.
Now my question is:
Is there be any non-abelian group of order 111?
If there be a non abelian group of order 111 then the given statement is False otherwise the statement is True .
Thanks.
Jungnickel gave a nice result in On the Uniqueness of the Cyclic Group of Order $n$ that any group of order $n$ is cyclic iff $(n,\varphi(n))=1$, which addresses this problem quickly.
Since $(111,\varphi(111))=(111,72)=3$, there are noncyclic groups of order $111$. The first paragraph of the above paper shows how to construct a nonabelian group of order $n$, when $(n,\varphi(n))\neq 1$.
• @IndrajitGhosh $\mathbb{Z}_n$ is cyclic for any $n$. If $(n,\varphi(n))=1$, $\mathbb{Z}_n$ is the only group of order $n$. If $(n,\varphi(n))\neq 1$, there is still a cyclic group of order $n$, but there are others as well. For instance, the dihedral group of order $10$ is nonabelian. Jun 18 '16 at 7:27 | 361 | 1,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-39 | latest | en | 0.884972 |
http://mathhelpforum.com/calculus/17750-limits-problems-without-s.html | 1,527,103,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00064.warc.gz | 187,324,763 | 12,117 | # Thread: Limits problems without #'s
1. ## Limits problems without #'s
Does this problem make anyones else's eyes bug out?
If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0
b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
2. But Dan, we all know that if $\displaystyle f(x)=\sqrt x\implies f'(x)=\frac1{2\sqrt x}$
But he's asking for $\displaystyle f(x)=\frac1{\sqrt x}$ which is $\displaystyle f'(x)=2\sqrt x$
3. Wow! You make it seem so easy! Thanks a bunch!
[EDIT] So wait...then the answer is 2√x?
4. Originally Posted by KennyYang
Does this problem make anyones else's eyes bug out?
If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0
b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
Krizalid isn't quite right here.
Round 2!
$\displaystyle f(x) = \frac{1}{\sqrt{x}}$
$\displaystyle f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}$
Simplify a bit first:
$\displaystyle f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \cdot \frac{\sqrt{x + h} \sqrt{x} }{\sqrt{x + h} \sqrt{x} }$
$\displaystyle = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}}$
Again, rationalize the numerator:
$\displaystyle = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}} \cdot \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}}$
$\displaystyle = \lim_{h \to 0} \frac{x - (x + h)}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$
$\displaystyle = \lim_{h \to 0} \frac{-h}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$
$\displaystyle = -\lim_{h \to 0} \frac{1}{\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}$
$\displaystyle = -\frac{1}{\sqrt{x} \sqrt{x}(\sqrt{x} + \sqrt{x})}$
$\displaystyle = -\frac{1}{x(2\sqrt{x})}$
$\displaystyle = -\frac{1}{2x \sqrt{x}}$
Which is typically written as $\displaystyle -\frac{1}{2}x^{-3/2}$
-Dan
5. Originally Posted by Krizalid
But he's asking for $\displaystyle f(x)=\frac1{\sqrt x}$ which is $\displaystyle f'(x)=2\sqrt x$
Oops!, I made a mistake here , I confused the derivative for an integral
6. Originally Posted by Krizalid
Oops!, I made a mistake here , I confused the derivative for an integral
It's okay. We still love you. (In a platonic way, of course! )
-Dan
7. I gave part (b) a shot and came up with such a convoluted answer that I'm not going to bother to post it. Did anyone else get a ridiculously overcomplicated answer as well?
8. Originally Posted by KennyYang
b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
Originally Posted by topsquark
$\displaystyle f^{\prime}(x) = -\frac{1}{2}x^{-3/2}$
-Dan
Any line can be expressed in the form $\displaystyle y = mx + b$. The slope of the tangent line at a point (x, f(x)) of the graph is $\displaystyle f^{\prime}(x)$, so the slope of the tangent line is, in this case:
$\displaystyle m = f^{\prime}(4) = -\frac{1}{2 \cdot 4^{3/2}} = - \frac{1}{2 \cdot 8} = -\frac{1}{16}$
So we have the form:
$\displaystyle y = -\frac{1}{16}x + b$
Now, we know that the function f(x) and the tangent line have the same values at the point x = 4. So we know that
$\displaystyle f(4) = y(4) = -\frac{1}{16} \cdot 4 + b$
$\displaystyle f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} = -\frac{1}{16} \cdot 4 + b$
$\displaystyle b = \frac{1}{2} + \frac{1}{16} \cdot 4 = \frac{3}{4}$
So I get that the tangent line to f(x) at x = 4 is $\displaystyle y = -\frac{1}{16}x + \frac{3}{4}$.
-Dan
9. Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
10. Originally Posted by KennyYang
Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
No problem!
-Dan
11. Originally Posted by KennyYang
If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
h->0
In fact the limit is the derivative of f(x) at x. | 1,433 | 4,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.605342 |
https://inverter110.com/17kw-free-energy-generator/ | 1,721,053,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00644.warc.gz | 281,972,561 | 22,138 | Sat. Jul 13th, 2024
# 17kw free energy generator
#### Byzahid ali
Jul 16, 2021
how to make 17kw free energy generator from 3hp motor and 17kw alternator with full rpm calculation
Hi everyone welcome to my blog. today we are going to talk about an main problem which is how to calculate the rpm and what will be the size of a pulley. So the method which I am sharing with you is very useful you can use it to calculate the rpm of any type of machines rpm. We will find an accurate calculation of rpm. There will be not any type of error in this calculation. Here we are going to do calculation of free energy generators rpm. first of all we will multiply the rpm of motor with the size of motor pulley. so motor rpm is 1450 multiplied by 2.5=3625 and divided by weight pulley 9.5=381 The rpm of weight 381. we will multiply it with the size of other pulley of the weight. That pulley is of 17 inch. 381 multiplied by 17 =6486
Now we will divide the answer with generators pulley of size 4.5 inch 6486/4.5=1441.5RPM. So our answer means our alternators RPM will be 1441.5 RPM. so guys that’s the final RPM.
And by this way you can find out the RPM of any type of machine. for an sample you can watch the video giving on the top of the page.
A 17kw free energy generator can be made from inexpensive parts such as an old computer. It’s also an affordable alternative to the high-cost, highly-regulated power plants. This device can generate enough energy to supply the average home’s needs for a month. Although it’s difficult to construct, the results are very impressive. Here are the steps to follow: To build a 17kw free energy generator, first determine its power needs. Next, determine the voltage and current requirements for the device.
Before starting your project, ensure that the components you use are available and a source of electricity. In case of a voltaic system, the voltage and current must match. Alternatively, the voltage and current should be at the same level. You will also need to consider the size of the battery. For a 17-kw free-energy generator, you should choose a large-capacity battery.
To create a 17-kw free-energy generator, first gather all the materials you need. For example, you’ll need a few old computers and a good motor. You can purchase the parts from the local hardware store. As long as you have the tools to build the unit, you’re on your way to making your own source of free energy. If you’re planning to produce your own solar power system, you’ll need to purchase a battery and install a power source.
Finally, assemble the parts in a suitable location and ensure that the machine’s temperature is stable. You’ll need a battery that is sturdy and durable. You’ll need a transformer to connect the two parts together. Once you’ve finished, you’ll need a power supply that can handle the load. It’s important that your generator’s output can support the load.
#### By zahid ali
##### 5 thoughts on “17kw free energy generator”
1. […] Generate free energy generator depends on various factors, such as the size of the turbine and the strength of the wind or water source. It is difficult to provide an exact figure, but with proper design and optimization, it can generate a significant amount of electricity to power household appliances. […]
2. […] we delve into the details, let us first understand what a free energy generator is and how it operates. A energy generator is a device that generates electricity without the use […]
3. Rambabu Bikki says:
I want a 5 KV flywheel generator can you make it for me, i will buy
9949498168
Rambabu
4. […] free energy generator […]
hello
how r u, i hope you will be fine
i am interested in your project, plz send me detail of as well as cost of the free generator | 876 | 3,774 | {"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-2024-30 | latest | en | 0.920166 |
http://www.finderchem.com/what-is-a-way-that-rectangles-are-similar-to-kites.html | 1,484,681,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280065.57/warc/CC-MAIN-20170116095120-00342-ip-10-171-10-70.ec2.internal.warc.gz | 463,062,820 | 8,234 | # What is a way that rectangles are similar to kites?
If you know how to draw a man, then you know how to draw a monkey! This animal is very similar to us with some exceptions: small ears (2) (or big ones... it depends ... - Read more
Leverage ratio is a financial term used to describe the way that a company inves... Read. How to Find a Property Parcel Number. A Property Parcel Number, ... - Read more
## What is a way that rectangles are similar to kites? resources
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The average 12-foot aluminium boat weighs around 200 pounds.
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https://rhinohide.wordpress.com/2011/01/17/lines-sines-and-curve-fitting-9-girma/ | 1,501,015,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425381.3/warc/CC-MAIN-20170725202416-20170725222416-00711.warc.gz | 693,840,738 | 39,326 | Home > CRUTEMP, LSCF, models, Statistics > Lines, Sines, and Curve Fitting 9 – Girma
## Lines, Sines, and Curve Fitting 9 – Girma
2011 January 17
Dr G. Orssengo recently brought to my attention his “line+sine” model which was presented at WUWT in April 2009. In short, his model is
y = (m*(x-1880) + b) + (A * cos(((x-1880)/T)*(2*pi)))
intercept (1880) b = -0.53
slope m = 0.0059 C / yr
amplitude A = 0.3 C
period T = 60 years
Which he displays as such:
I’ve recharted Dr Orssengo’s graph as follows:
Dr Orssengo has commented here that his correlation is 0.88. However, using annualized HadCRUv3 data, the correlation I calculate from 1880-2009 is a bit lower at 0.87. However, just reducing the amplitude of the cosine to 0.25C increases the correlation to 0.89. I suspect that Girma fit his model by hand and did not seek to optimize the values used.
Dr Orssengo has put a lot of emphasis on the correlation of his model, and has allowed that if an equally simple model had a higher correlation that it would be a better model. I now present a simple model based on the same data with a higher correlation.
First a simple exponential trend.
y1 = -a/100 + b/100 * exp((k/10000)*(x-1880))
a <- 44.40
b <- 6.02
k <- 211
This model has a correlation of 0.89 for HadCRU, but 0.90 for NCDC, and 0.91 for GISTEMP all through 2009. Since HadCRU is the data set that Dr Orssengo used, we will judge on the correlation against it. And 0.89 is higher than the 0.88 that Girma has calculated for his model or the 0.87 that I calculated for it.
Based on the criteria that the best correlation makes the best model, my exponential warming model beats Dr Orssengo’s line+sine model.
Will the model hold … no, it will probably not. That will be a whole bunch of warming if it does!
To be fair, Dr Orssengo points out that the future is a better filter for selecting models than correlation. I’ll address that – with a slightly better model-of-fit – in a couple of days.
And one more note. While curve fitting is fun, divorced of a physical explanation, they aren’t much more than fun. What drives the linear (or exponential) trend? What drives the 60 year sine? Without knowing these, the mathematical model is a sterile construct.
1. 2011 January 17 at 7:01 am
Dance could be seen as curve fitting.
I wonder now in how many Girma’s drive-bys a link that post should be amended.
2. 2011 January 17 at 5:50 pm
The dance of regressions. You made me smile. 😀
As to Girma’s obfuscatory links: the principal of charity informs me that he is just being polite and keeping links short. The cynical me isn’t worried about driving up traffic at WUWT or increasing his google ranking – I’m just not that big.
3. 2011 January 18 at 2:12 am
“To be fair, Dr Orssengo points out that the future is a better filter for selecting models than correlation”
Thanks Ron.
In the next couple of years, whether we have milder or harsher winters will show us whether the exponential or the sinusoidal model represents the reality.
4. 2011 January 18 at 2:28 am
Ron
One very important point, your exponential model does not show the big freeze of the 1970s and the dust bowl of the 1940s. The sinusoidal model does.
The Big Freeze of the 1970s
http://bit.ly/g3dtPb
The Dust Bowl of the 1940s
http://bit.ly/hXNYAA
5. 2011 January 18 at 8:42 am
In the next couple of years, whether we have milder or harsher winters will show us whether the exponential or the sinusoidal model represents the reality.
Well. No. The integrated global temperature anomalies will tell us. Not the seasonal temperatures in one region or another.
One very important point, your exponential model does not show the big freeze of the 1970s and the dust bowl of the 1940s. The sinusoidal model does.
Well, how about a sinusoidal exponential model?
6. 2011 January 18 at 9:50 am
Blog comments as dances of regressions would explain a lot, perhaps a bit too much.
A sinusoidal exponential model makes my head aches.
7. 2011 January 18 at 10:11 am
Regression analysis is widely used for prediction and forecasting, where its use has substantial overlap with the field of machine learning. Regression analysis is also used to understand which among the independent variables are related to the dependent variable, and to explore the forms of these relationships. In restricted circumstances, regression analysis can be used to infer causal relationships between the independent and dependent variables.
http://en.wikipedia.org/wiki/Regression_analysis
Why do I have a vision of blog-mining bots doing social network analysis on climate blogs?
8. 2011 January 18 at 11:48 am
Ron
“Well, how about a sinusoidal exponential model?”
Much better.
However, what is its correlation coefficient? It must be higher than your previous value of 0.89.
Ron, I doubt your previous value of 0.89. How can you have such a high value when the exponential model is monotonically increasing while the data is not?
9. 2011 January 18 at 12:21 pm
Girma, the results are easy to replicate.
I would be grateful if you did so in Excel.
The script which I used is located here:
http://rhinohide.org/gw/trendtester/tt-nls.R
Note that there is a line which sets which data source is being used:
> # column idx
gis <- 3
cru <- 4
noa <- 5
uah <- 7
idx <- gis
And the script output (not including graphics) is found here:
http://rhinohide.org/gw/trendtester/tt-nls.Rout
10. 2011 January 18 at 12:47 pm
Ron
http://bit.ly/c0Jvh0
11. 2011 January 18 at 1:16 pm
It looks roughly periodic. It is also missing the data for 2010.
12. 2011 January 18 at 4:30 pm
Ron
As the above pattern was valid for the past 130 years, is it not reasonable to assume it will also be valid for the next 20 years?
As a result, cooling until 2030?
13. 2011 January 18 at 4:42 pm
Ron
I have one objection with your exponential model.
Its estimate of the local maxima of 1880 and 1940 and local minima of 1910 & 1970 are poor.
Can you create a model based ONLY on the following local minima and maxima values?
1880=>-0.2
1910=>-0.6
1940=>0.1
1970=>-0.3
2000=>0.5
14. 2011 January 20 at 3:01 pm
“As the above pattern was valid for the past 130 years, is it not reasonable to assume it will also be valid for the next 20 years?”
Because we have more information than just 130 data points? Some of it is that we have spatial patterns as well as regional, which tells us something different is going on in this warming than in the 1910-1940 one. But more importantly, we have data on the physical causes of temperature changes such as solar, volcanic, aerosol, and greenhouse gas forcings.
If I watch a kid on a swing for 3 minutes, I might ask whether the next 3 minutes will have the same oscillation pattern as the first 3. But if I hear his mom calling for him to go home, I have information which suggests that perhaps the next oscillations won’t be the same after all.
Also: have you ever heard the expression, ‘with four parameters I can fit an elephant and with five I can make him wiggle his trunk,” attributed to Enrico Fermi? (ignoring the fact that it actually takes about 30 parameters to get a good line drawing of an elephant) One good test is to use your approach on the data from 1880 to 1990, and see how well your method* captures the period from 1990 to 2010. My guess is that your method will work poorly.
*Here, I define your method as “fit the data with a combination of sine curves and linear trends, and optimize until you get the very best R^2 value”.
15. 2011 January 21 at 4:13 pm
One good test is to use your approach on the data from 1880 to 1990, and see how well your method* captures the period from 1990 to 2010
Which is pretty much what was done at the following, but with 1900-2000 as the model period.
https://rhinohide.wordpress.com/2011/01/14/lines-sines-and-curve-fitting-6-backcast-and-forecast/
16. 2011 January 24 at 7:52 am
“Which is pretty much what was done at the following, but with 1900-2000 as the model period”
Yeah, that’s a nice way of looking at things. Now, there is still the danger that if you do throw 1000 methods at a given problem, then reserving parts of the time record no longer avoids overfitting… you could do a two stage “reserve” where you throw 100 models at 80% of the time period, pick the couple models which best fit the next 10%, and then check and make sure that those models that best fit the next 10% also do reasonably at the last 10%…
-M
17. 2011 January 26 at 4:15 am
@Grima
“As the above pattern was valid for the past 130 years, is it not reasonable to assume it will also be valid for the next 20 years?
As a result, cooling until 2030?”
Err, no. This data was with a linear trend removed, and if you plot the real data you can clearly see that the linear trend is stronger than the amplitude of the presumed 60 year oscillation. So even if this oscillation would continue the coming 20 years, there would be global warming, albeit with a smaller slope. And furthermore, I cannot say it better than Ron: “While curve fitting is fun, divorced of a physical explanation, they aren’t much more than fun”
18. 2011 January 26 at 12:36 pm
milanovic,
But we do have a physical explanation. The ghg forcing curve is fit very well by an exponential function and the modulation is caused by a cyclic variation in heat transfer alternately favoring the NH and the SH. Since this cyclic variation component is not modeled by GCM’s the empirical fit has more skill for the past and is likely to have more skill over the next decade or two than current GCM’s coupled or not. Even better would be to use a rational scenario for ghg emissions rather than a simple exponential. But the IPCC emission scenarios, say, don’t deviate much from an exponential curve for the next decade or two either.
When the cyclic component is included, you don’t need all the aerosol forcing fudge factors to explain the variation of global temperature in the twentieth century so the short term climate sensitivity is at the low end of the IPCC range.
19. 2011 January 26 at 5:16 pm
Hey Ron, you gifted Girma two free parameters (actually three if you include the extra functional form), but the serious question is that if you use the same baseline, what is the quality of the fit to the combination of the three surface records? or if you prefer it that way, how do the fitting parameters vary amongt them.
20. 2011 January 26 at 8:27 pm
21. 2011 January 27 at 1:14 am
“caused by a cyclic variation in heat transfer alternately favoring the NH and the SH”
What evidence can you give for a fixed cycle with period of approx 60 years? I don’t know of any. For example, PDO or ENSO have no such cycle. (see for example http://tinyurl.com/4u3g73y). But anyway, my main point was that if such a cycle existed (which I doubt), the amplitude would clearly be too small to cause global cooling in the coming decennia, as Grima suggested. If you want to put faith in this curve-fitting procedure, then the exponential + sinusoidal curve fit does not predict any cooling. Also the link he referenced himself http://bit.ly/c0Jvh0 made that clear, because the linear trend that was subtracted to get this graph is higher than the amplitude of the resulting oscillation.
22. 2011 January 27 at 1:00 pm
milanovic,
Of course it’s not fixed. The planetary system is likely chaotic or at least shows long term persistence. That type of system frequently exhibits quasi-periodic behavior at all time scales. The best evidence of a ~60 year oscillation that is occurring now is the AMO index . It won’t continue forever. But for the relatively short term, the null hypothesis would be that the currently observed behavior will continue. And I never said or implied that global average temperature will decline. But I do think it likely that the rate of increase for the next decade or two will be significantly ( in the frequentist statistical sense) less than the current GCM runs predict.
The bi-polar seesaw effect is also well known, although research emphasis is mainly on the longer term cycles like Heinrich and Dansgaard-Oeschger events. See here for example or google polar see saw.
23. 2011 January 28 at 12:56 am
@DeWitt Payne
Thanks for the reply. I was not aware of the ~60 year oscillation in AMO. Of course I meant “fixed” as quasi-fixed, we should at expect it to continue for the next cycle.
“And I never said or implied that global average temperature will decline.”
I know you didn’t, but my post was a reply to Grima’s post (18-01, 4.30pm) who did imply that. So we are more or less talking past to each other I am afraid.
24. 2011 January 28 at 6:28 am
The best evidence of a ~60 year oscillation that is occurring now is the AMO index .
I admit that I haven’t looked at it in detail, but that doesn’t seem very convincing in and of itself. That record shows essentially one long decline to a minimum, one rise and fall to another minimum, and then another rise. Since we don’t know when the pre-1900 peak was, nor the post-1990 peak, there’s really only one complete “oscillation” in the modern record, right? That’s not much to base any conclusions on regarding the behavior of that signal over the next couple of decades, IMHO.
Again, though, I haven’t read much about the AMO. Perhaps there are other reasons (aside from the record itself) for expecting it a priori to show a ~60-year oscillation.
25. 2011 January 28 at 9:06 am
Just FYI – I am pretty neutral on the 60 year oscillation stuff. Solar? Sea-based resonance? Ice-based resonance? Or just a fluke based on a 30 year cooling due to aerosols? Although, looking at the exponential fits, it is the 1910-1940 leg that is out-of-alignment. (which might raise questions of pre1940 instruments records)
26. 2011 January 28 at 10:51 am
Ned,
What we don’t know is where or whether there was a minimum prior to 1856. We have at least two full cycles, though, which is enough for a curve fit and a testable hypothesis. If we don’t see a significant decline in the AMO index in the next decade, the hypothesis that it’s cyclical with a period of 60-70 years fails. In fact, I’ll be very suspicious if it doesn’t decline significantly in the next five years. It looked like it was declining until the recent El Nino.
27. 2011 January 29 at 4:43 am
DeWitt Payne writes: We have at least two full cycles, though, which is enough for a curve fit and a testable hypothesis.
Well, maybe. I only see one definite peak (1940s-ish). I take it you’re assuming there’s another peak some time there prior to 1900, and one some time in the past decade or so, right? Any individual year can be an outlier, but it seems a bit risky to assume that we’ve seen the most recent peak when the most recent year’s value is actually the highest in the past century. Given the noise in that signal, it might be a little hard to pin down the local maximum until many years after we’ve passed it.
Frankly, trying to discern a “period” for the AMO in the 20thC data and use that to forecast temperature trends seems pretty close to reading tea leaves, IMHO. In the literature you see it referred to as a “60-80 years” period (e.g., Poore et al. 2009, who specifically state that the instrumental record isn’t long enough to clearly establish the period of the AMO cycle). If you assume a peak in the mid-1940s and a period of 60 years, then we should have already passed the second peak. If you assume a period of 80 years, the second peak wouldn’t be until the 2020s.
Maybe you could pin down this oscillation a bit better if you could extend the record using proxies. But the only examples I’ve seen (e.g., Poore et al’s fig. 3) don’t give me much confidence in specific claims about the “period” of the AMO. It just seems to be all over the map.
Not trying to be cranky … sorry if it comes across that way.
28. 2011 January 29 at 10:45 am
Ned,
This year’s peak was the third highest:
1878 0.448
1998 0.394
2010 0.381
All were big El Nino years. 1878 was particularly bad.
If I were really gung ho, I’d do an fft power spectrum. I’m betting there’s a big peak in the spectrum near 60 years. I bet there are also smaller peaks near 11 and 22 years. Obviously it’s not proof. There is no proof in science, as opposed to math, there’s only falsification.
29. 2011 January 29 at 6:57 pm
Here’s one I did earlier – not good but gives the general impression.
The main peaks are 68 21 15 and 9 years
The lack of resoution at 68 means the peak could be anywhere between 55 an 68 years.
30. 2011 January 31 at 1:33 am
well, this post puts the AMO discussion here in a different context.
http://tamino.wordpress.com/2011/01/30/amo/#more-3425
I did not know that AMO-index was defined by temperature anomaly. Therefore, claiming that AMO is responsible for global warming has a large risk of circular reasoning
31. 2011 January 31 at 7:50 am
Is it coincidence that Tamino keeps answering questions that I’m asking? 😆
I first ran into the AMO as the detrended anomaly on wiki when I was looking harder at Hank Robertson’s (?)* David Benson’s CO2+AMO model.
* edit to correct author; here is the link
http://www.realclimate.org/index.php/archives/2010/10/unforced-variations-3-2/comment-page-5/#comment-189329
32. 2011 January 31 at 8:00 am
I don’t know, maybe he is following your blog closely 🙂
Before Tamino’s post I was not aware of the definition of AMO, but knowing that definition it is clear you have to be extremely careful when studying AMO and global temperature and causation between the two.
33. 2011 January 31 at 11:28 am
Obviously some physical phenomenon causes the AMO index to behave the way it does. The AMO index is supposed to reflect variations in the actual Atlantic Meridional Overturning circulation. That circulation is the cause of the temperature anomalies used to calculate the index, not the other way around.
The NH has more land area than the SH. That makes the average heat capacity of the NH lower than the SH. The temperature range of the annual cycle in the NH is then larger than the SH. When you add the two together, you don’t get zero, you get a cycle that looks like a smaller version of the NH temperature with a small phase shift. If you then vary the relative amount of heat transferred to the NH vs the SH by ocean circulation, then the NH will warm and cool more than the SH. The result will be a cycle in the global average temperature even though global heat content remains the same.
34. 2011 February 1 at 1:16 am
The AMO index is supposed to reflect variations in the actual Atlantic Meridional Overturning circulation.
Well, that’s the whole point is it? When you only subtract a linear function that should account for GHG increase and then define that all other variation is due to AMO then you might overlook other global effects such as aerosol cooling during the 40s. You would define those temperature variations to be part of the AMO and in that way get a 60 year oscillation. Unfortunately, due to the somewhat circular definition, there is no way of finding out wheher all these temperature variations indeed belong to the AMO
35. 2011 February 1 at 8:57 am
There is indeed no way from analysis of past data. That’s not the point. What’s going to happen over the next decade is what’s relevant. Will the AMO index decrease and even go negative? Will, as a result, global average temperature remain below the model trends? Will Arctic sea ice loss slow and Antarctic loss increase? If none of that happens, then the aerosol only explanation for mid-twentieth century cooling is strengthened and a high climate sensitivity is more likely. However, there’s still the problem of the relatively rapid warming in the early twentieth century. A solar only explanation of that, if Leif Svalgaard is correct about the lack of variation in TSI, has problems.
36. 2011 February 2 at 12:45 am
I agree for a large part with that analysis. However, even when the in the next decade AMO index and global temperature will be correlated, it will be extremely difficult to establish that AMO was the cause, because the AMO index will be influenced by global temperatures as well.
1. 2011 February 10 at 10:02 pm
2. 2012 February 16 at 6:46 pm
3. 2012 February 16 at 6:47 pm | 5,105 | 20,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-30 | longest | en | 0.948523 |
https://socratic.org/questions/how-do-you-solve-6a-5a-11 | 1,575,973,194,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00237.warc.gz | 541,854,395 | 5,755 | # How do you solve 6a+5a=-11?
Aug 3, 2016
$a = - 1$
#### Explanation:
$6 a + 5 a = - 11$
or
$11 a = - 11$
or
$a = - \frac{11}{11}$
or
$a = - 1$ | 75 | 147 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-51 | longest | en | 0.660142 |
https://learn.adafruit.com/tweet-a-watt/make-it-software | 1,721,642,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00895.warc.gz | 307,200,371 | 20,029 | ## Introduction
Now that the hardware is complete, we come to the exciting part: running the software that retrieves the data from our receiver XBee and saves it to our computer or uploads it to a database or updates our twitter feed or....whatever you'd like!
Here is how it works, the XBee inside the Kill-a-Watt is hooked up to two analog signals. One is the voltage signal which indicates the AC voltage read. In general this is a sine wave that is 120VAC. One tricky thing to remember is that 120V is the 'RMS' voltage, and the 'true voltage' is +-170VDC. (You can read more about RMS voltage at wikipedia basically it's a way to indicate how much 'average' voltage there is.) The second reading is the AC current read. This is how much current is being drawn through the Kill-a-Watt. If you multiply the current by the voltage, you'll get the power (in Watts) used!
The XBee's Analog/Digital converter is set up to take a 'snapshot' of one sine-cycle at a time. Each double-sample (voltage and current) is taken 1ms apart and it takes 17 of them. That translates to a 17ms long train of samples. One cycle of power-usage is 1/60Hz long which is 16.6ms. So it works pretty well!
Lets look at some examples of voltage and current waveforms as the XBee sees them.
For example this first graph is of a laptop plugged in. You'll see that it's a switching supply, and only pulls power during the peak of the voltage curve.
A laptop plugged in, switching power supply
Now let's try plugging in a 40W incandescent light bulb. You'll notice that unlike the switching supply, the current follows the voltage almost perfectly. That's because a lightbulb is just a resistor!
40W lightbulb
Finally, let's try sticking the meter on a dimmable switch. You'll see that the voltage is 'chopped' up, no longer sinusoidal. And although the current follows the voltage, it's still matching pretty well.
Light bulb on dimmer switch
The XBee sends the raw data to the computer which, in a python script, figures out what the (calibrated) voltage and amperage is at each sample and multiplies each point together to get the Watts used in that cycle. Since there's almost no device that changes the power-usages from cycle-to-cycle, the snapshot is a good indicator of the overall power usage that second. Then once every 2 seconds, a single snapshot is sent to the receiver XBee
## Install python & friends
The software that talks to the XBee is written in python. I used python because it's quick to develop in, has multi-OS support and is pretty popular with software and hardware hackers. The XBees talk over the serial port so literally any programming language can/could be used here. If you're a software geek and want to use perl, C, C#, tcl/tk, processing, java, etc. go for it! You'll have to read the serial data and parse out the packet but it's not particularly hard.
However, most people just want to get on with it and so for you we'll go through the process of installing python and the libraries we need.
1. Download and install python 2.5 from http://www.python.org/download/ I suggest 2.5 because that seems to be stable and well supported at this time. If you use another version there may be issues.
2. Download and install pyserial from the package repository (this will let us talk to the XBee thru the serial port).
3. If you're running windows download and install win32file for python 2.5 (this will add file support).
4. Download and install the simplejson python library (this is how the twitter api likes to be spoken to) you'll need to uncompress the tar.gz file and then run the command "python setup.py install" to install.
Now you can finally download the Wattcher script we will demonstrate here! We're going to download it into the C:\wattcher directory, for other OS's you can of course change this directory.
## Basic configure
We'll have to do a little bit of setup to start, open up the wattcher.py script with a text editor and find the line:
SERIALPORT = "COM4" # the com/serial port the XBee is connected to.
Change COM4 into whatever the serial port you will be connecting to the XBee with is called. Under windows it's some COMx port, under linux and mac it's something like /dev/cu.usbserial-xxxx check the /dev/ directory and/or dmesg.
Save the script with the new serial port name.
## Test it out
Once you have installed python and extracted the scripts to your working directory, start up a terminal (under linux this is just rxvt or xterm, under mac it's Terminal, under windows, it's a cmd window).
I'm going to assume you're running windows from now on, it shouldn't be tough to adapt the instructions to linux/mac once the terminal window is open.
Run the command cd C:\wattcher to get to the place where you uncompressed the files. By running the dir command you can see that you have the files in the directory.
Make sure your transmitter (Kill-a-Watt + Xbee) is plugged in, and blinking once every 2 seconds. Remember it takes a while for the transmitter to charge up power and start transmitting. The LCD display should be clear, not fuzzy. Make sure that there's nothing plugged into the Kill-a-Watt, too. The RSSI (red) LED on the receiver connected to the computer should be lit indicating data is being received. Don't continue until that is all good to go!
Now run python by running the command C:\python25\python.exe wattcher.py
You should get a steady print out of data. The first number is the XBee address from which it received data, following is the estimated current draw, wattage used and the Watt-hours consumed since the last data came in. Hooray! We have wireless data!
## Calibrating
Now that we have good data being received, it's time to tweak it. For example, it's very likely that even without an appliance or light plugged into the Kill-a-Watt, the script thinks that there is power being used. We need to calibrate the sensor so that we know where 'zero' is. In the Kill-a-Watt there is an autocalibration system but unfortunately the XBee is not smart enough to do it on its own. So, we do it in the python script. Quit the script by typing in Control-C and run it again this time as C:\python25\python.exe wattcher.py -d note the -d which tells the script to print out debugging information.
Now you can see the script printing out a whole mess of data. The first chunk with lots of -1's in it is the raw packet. While it's interesting we want to look at the line that starts with ampdata:
ampdata: [498, 498, 498, 498, 498, 498, 498, 498, 498, 498, 498, 498, 498, 498, 497, 498, 498, 498]
Note: If you're getting -1's instead of nice ~500 numbers, check the XBee is set up right and has the correct wires going to the correct A/D pins. If you're getting 1024's that probably means you forgot to tie VREF to VCC in the last step. Remove the Kill-a-Watt from power and go back to repair it.
Now you'll notice that the numbers are pretty much all the same. That's because there's nothing plugged into the tweetawatt and so each 1/60 Hz cycle has a flat line at 'zero'. The A/D in the XBee is 10 bits, and will return values between 0 and 1023. So, in theory, if the system is perfect the value at 'zero' should be 512. However, there are a bunch of little things that make the system imperfect and so zero is only close to 512. In this case the 'zero' calibration point is really 498. When it's off there is a 'DC offset' to the Amp readings, as this graph shows:
See how the Amp line (green) is steady but it's not at zero, it's at 0.4 amps? There is a 'DC offset' of 0.4 amps.
OK, open up the wattcher.py script in a text editor.
```vrefcalibration = [492, # Calibration for sensor #0
492, # Calibration for sensor #1
489, # Calibration for sensor #2
492, # Calibration for sensor #3
501, # Calibration for sensor #4
493] # etc... approx ((2.4v * (10Ko/14.7Ko)) / 3```
See the line that says # Calibration for sensor #1? Change that to 498.
```vrefcalibration = [492, # Calibration for sensor #0
498, # Calibration for sensor #1
489, # Calibration for sensor #2
492, # Calibration for sensor #3
501, # Calibration for sensor #4
493] # etc... approx ((2.4v * (10Ko/14.7Ko)) / 3```
Save the file and start up the script again, this time without the -d
Now you'll see that the Watt draw is 2W or less, instead of 40W (which was way off!) The reason it's not 0W is that, first off, there's a little noise that we're reading in the A/D lines, secondly there's power draw by the Kill-a-Watt itself and finally, the XBee doesn't have a lot of samples to work with. However <2W is pretty good considering that the full sensing range is 0-1500W.
Here is the graph with the calibrated sensor:
See how the Amps line is now at 0 steady, there is no DC offset.
## Logging data
It's nice to have this data but it would be even nicer if we could store it for use. Well, that's automatically done for you! You can set the name of the log file in the wattcher.py script. By default it's powerdatalog.csv. The script collects data and every 5 minutes writes a single line in the format Year Month Day, Time, Sensor#, Watts for each sensor. As you can see, this is an example of a 40W incandescent lightbulb plugged in for a few hours. Because of the low sample rate, you'll see some minor variations in the Watts recorded. This data can be easily imported directly into any spreadsheet program.
## Tweeting
Finally we get to the tweeting part of the tweet-a-watt. First open up the wattcher.py script and set: | 2,352 | 9,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-30 | latest | en | 0.934512 |
http://openstudy.com/updates/4fe562f6e4b06e92b873ab9b | 1,448,936,114,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464396.48/warc/CC-MAIN-20151124205424-00351-ip-10-71-132-137.ec2.internal.warc.gz | 175,277,136 | 17,805 | ## alfers101 3 years ago Please help: Calculate the electric field at the center of a square 52.5cm on a side if one corner is occupied by a +45uC charge and the other three occupied by -27uC charges.
1. ujjwal
Find electric field due to individual charges at the center and do their vector sum. That will give you the resultant field. P.S. you mustn't make any mistake while determining the direction of electric field due to individual charges.|dw:1340434465357:dw|
2. alfers101
uhm okay. i think thats it.
3. ujjwal
|dw:1340434775644:dw|The electric fields due to charges at B and D seems to cancel each other. So, you have got only A and C.
4. alfers101
okay then?
5. ujjwal
calculate the fields due to individual charges and do the vector sum.
6. salini
break the problem into parts: force(here electric field) always acts through the line joining the charges so it is enough if u consider forces along AC and BD imagine placing a positive test charge(with no electric field of its own at the centre. along BD , since u have two equal charges attracting a test charge(+ve) equally the effect due to them is zero same way think wat should happen along AC then use formulae q/(4 pi epsilonnt r^@) |dw:1340437859485:dw| u can see that we are taking only charge q in the eqn as electric field is nothing but force/unit charge first u have to draw a diagram like ujjwal for clarity | 347 | 1,393 | {"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} | 3.4375 | 3 | CC-MAIN-2015-48 | longest | en | 0.92309 |
http://lotsasplainin.blogspot.com/2017/08/we-by-yevgeny-zamyatin.html | 1,550,406,766,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481992.39/warc/CC-MAIN-20190217111746-20190217133746-00237.warc.gz | 165,054,710 | 20,988 | ## Sunday, August 13, 2017
### We by Yevgeny Zamyatin
Most of the books I read are recently published, but occasionally I read an older book to continue my education, as I like to put it. This week I read the 1921 novel We by Yevgeny Zamyatin, this edition translated by Natasha Randall in 2006. Some reviews call it one of the first dystopian novels, but that isn't accurate. After Edward Bellamy's successful 1888 utopian novel Looking Backward: 2000-1887, both utopias and dystopias became the fashion for many years to come. Most of these books and their authors are now forgotten, but in English, readers will still recognize 1895's The Time Machine by H.G. Wells, and will know the name of Jack London, though his 1908 book The Iron Heel about the future dystopian struggles leading to utopian socialism is not nearly as popular as his thrilling boy's adventures.
The book We is compact, but for me it was hard to keep focused. It is written in the first person, so the narrator has to describe what is everyday to him (or her) in ways that will make sense to people like his readers who have never seen this world. There are other such books, but it's a tricky proposition. Zamyatin decides to describe things in mathematical terms and colors. As a mathematician, much of his mathematics irritates the hell out of me. For example, his narrator D-503, a mathematically trained engineer in charge of building an interplanetary spaceship, has a particular distaste for the square root of -1, which he calls "irrational". We usually call this number imaginary, but technically he is right. The number we often call i is not the ratio of two integers. Mathematicians call it algebraic.
What irritates me more is that an engineer should know that this very odd idea is of immense practical value in electrical engineering. Using both real and "imaginary" numbers together creates complex numbers and this system very cleanly represents the physical fact that electric currents naturally produce counter-currents that run in a perpendicular direction. Electrical engineers find this idea so useful, then call the square root of -1 j instead of i, so any reference to imaginary is erased. The great mathematician Gauss hated that "imaginary number" was already stuck in the mathematical vocabulary even in his era a century before Zamyatin, and wanted positive to be replaced by direct, negative by inverse and the imaginary directions to be coined lateral and inverse lateral. It is completely possible Zamyatin was never taught this.
Now that I have indulged myself to two paragraphs of mathematical quibbling, let me get to my complaints as a reader of speculative fiction. It's hard to understand some never seen world when the writing relies heavily on bad mathematical descriptions and a made up language of his own personal feelings about colors. Worse still, the first person narrator has a breakdown in the middle of the story where he believes he has died, and several chapters after this point are later to be understood as dreams or hallucinations caused by fever.
The world where D-503 lives is a city made of glass where all lives are supposed to be completely visible to everyone else to make sure everyone is doing exactly what they should be, but there is an exception for when people have sex. The sex component is excessively important to the plot, and anyone who has seen through Hugh Hefner's idea of utopia can see it for the juvenile male fantasy it is. People can have sex with anyone who can agree to have sex with them, and men are completely free from the burdens of fatherhood. It also presents women who have once given consent and wish to rescind it as horrible and duplicitous creatures. It never assumes to a man he might not be an ideal lover.
In short, if you have never read We, you have my leave to never read it. The book has fans that range from Garry Kasparov, the former world chess champion who is strongly capitalist and just as strongly anti-Putin, to Noam Chomsky, the renowned linguist whose political views are sometimes described as libertarian socialist. Chomsky has said We is superior to Nineteen Eighty Four, which he considers wooden. Just to add a little more interest to reading this book I have said you shouldn't read, Orwell considers it completely superior to Brave New World.
Here's where I stand on these provisos to my bold and underlined main position above. Huxley and Orwell did not get along and I am 100% on Team Orwell. As a prose stylist, Orwell runs rings around Huxley and Zamyatin, though I will admit I cannot read Zamyatin in the original Russian, which is my problem, not his. A point on which I agree with Orwell that We is better than Brave New World is both books have characters who are considered great poets in morally empty times. What would such a poet write? Zamyatin gives examples, Huxley does not.
Point to Zamyatin.
More importantly than any political position or literary merit, Orwell understood the connection between politics of any stripe and lying. Here are his six rules of writing, from his essay Politics and the English Language.
1. Never use a metaphor, simile or other figure of speech which you are used to seeing in print. (Many of Orwell's examples are now thankfully out of date. The best modern example is the completely meaningless cliche "thoughts and prayers".)
2. Never use a long word where a short one will do.
3. If it is possible to cut a word out, always cut it out.
4. Never use the passive where you can use the active.
5. Never use a foreign phrase, a scientific word, or a jargon word if you can think of an everyday English equivalent.
6. Break any of these rules sooner than say anything outright barbarous.
To summarize, if you are intrigued by my description or the testimonials, by all means read We. If you want to take my advice instead, find some Orwell you haven't read, especially his collections of essays. In particular, Shooting an Elephant should be at least as famous as The Declaration of Independence or the preamble to The Constitution.
It's a Sunday, so I will write: Here endeth the lesson.
It's a cliche, but any other way of writing it is barbarous.
#### 1 comment:
nancy namaste said...
I think I will pass on the book but save the rules for writing. Good ones to remember! I am enjoying your revived blog. Thoughtful and well written. Nancy | 1,380 | 6,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-09 | latest | en | 0.974574 |
http://uk.ask.com/question/how-many-minutes-are-in-24-hours | 1,386,394,691,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053578/warc/CC-MAIN-20131204131733-00099-ip-10-33-133-15.ec2.internal.warc.gz | 195,309,581 | 15,539 | # How minutes in a 24hours?
1 hour = 60 minutes. therefore. 24 hours 1,440 minutes. | 28 | 84 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2013-48 | longest | en | 0.888964 |
http://www.codeproject.com/Articles/44370/Triangulation-of-Arbitrary-Polygons?fid=1552457&df=90&mpp=10&sort=Position&spc=None&tid=3280909 | 1,469,709,815,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828282.32/warc/CC-MAIN-20160723071028-00050-ip-10-185-27-174.ec2.internal.warc.gz | 377,197,141 | 23,231 | 12,402,889 members (73,734 online)
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# Triangulation of Arbitrary Polygons
, 22 Nov 2009 CPOL
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This article describes a way of triangulating polygons that is intuitively easy to understand and explain diagrammatically
## Introduction
This algorithm demonstrates a way of triangulating polygons that is intuitively easy to understand and explain diagrammatically. However, this technique produces a sub-optimal number of triangles, a condition slightly ameliorated by a greedy merge of adjacent triangles. A second disadvantage is that the resulting triangulation of a polygon is inherently tied to its rotation.
## Description and Explanation of the Algorithm
Polygon decomposition into triangles is a subject classed under the computer science branch of computational geometry, and is generally solved using:
### Convex Polygons
For objects with more than 3 vertices, the test for convex polygons is performed by taking the cross product of the two vectors at each vertex of the polygon to obtain the perpendicular vector v. A convex polygon will have all vs facing the same direction, whereas a concave polygon will have vs facing both directions. A convex polygon with n vertices can always be triangulated into n-2 triangles in O(n) time by taking any vertex as the origin and adding edges to all vertices except its two adjacent vertices without further calculations.
### Concave Polygons
Polygons detected as concave will be put through the regular triangulation algorithm. This begins with horizontal decomposition, where imaginary horizontal lines are extended from each vertex of the dynamic object, and intersections of these imaginary lines with polygon edges of the dynamic object are added as vertices. The new vertices added by horizontal decomposition facilitate the decomposition of the polygon into triangles and trapezoids. The polygon is traversed top-down, from left to right.
Figure 1: An example of horizontal decomposition at work.
Considering two horizontal rows at a time, we extract edge pairs from left to right and, based on information on whether the two edges of a pair intersect, deduce if the edge pair forms a triangle or trapezoid. Intersecting edge pairs form triangles, and trapezoids otherwise. As an example, take Figure 1’s y1 and y2 as the two rows. The first two edges from the left form the first edge pair, and since they do not intersect, we know this pair forms a trapezoid. The second edge pair formed by edges 3 and 4 intersects at y2, so we know this pair forms a triangle. Using this algorithm, we can decompose the polygon into triangles and trapezoids. Trapezoids then need merely be further decomposed into two triangles by adding a diagonal edge between two opposite corner vertices. Upon reaching the bottom of the polygon, triangulation is done. This algorithm takes O(n log n) time.
### Greedy Merge
Figure 2: Merge-able triangles.
At the next stage, the greedy merge algorithm considers, for each triangle, whether it is possible to merge with any other triangle of the polygon and result in a larger combined triangle. Merge-able triangles have two properties, depicted in Figure 2. They each have a coincident, identical edge, and a parallel non-coincident edge that coincides at a vertex. Merge-able triangles are found, merged, and the algorithm repeats until no more merge-able triangles can be found. The time complexity of this algorithm is O(n2), scaling with the number of triangles in the polygon. The maximum number of edge-edge comparisons for each polygon required is f(n) = (n-1)*9 + f(n – 1), where n is the number of triangles fed into the algorithm.
Figure 3: Three different triangulation methods for the same polygon.
Left: Base triangulation method. Middle: Triangulation with merge. Right: Handled as convex.
While horizontal decomposition into triangles does not result in an optimal number of triangles, the implemented optimizations to reduce the final number of triangles bring us closer to the optimal number. The merits of this technique are in its elegance and ease of implementation. It localizes the problem of triangulation by a systematic polygon traversal of edge pairs, then compensates for non-optimality with a greedy merge.
The triangulation algorithm flows thus:
1. if(object has 3 vertices)
2. add to triangle vertex list
3. end if
4. else if(object is convex)
5. decompose as convex object
6. add decomposed triangles’ vertices to triangle vertex list
7. end else if
8. else
9. add vertices to object via horizontal decomposition
10. triangulate and merge
11. store triangles in temporary list 1
12. re-wind trapezoid
13. triangulate and merge
14. store triangles in temporary list 2
15. add triangles’ vertices from list with lower triangle count to triangle vertex list
16. end else
## Using the Code
This algorithm was coded in C# and utilizes the Farseer Physics engine. To create a polygon for triangulation, use the left mouse button to select vertices, then press ‘a’ to finalize and triangulate the polygon. Right mouse button allows you to undo last vertex or delete a polygon. Pressing ‘z’ toggles greedy merge on and off.
## History
• 21st November, 2009: Initial version
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Last Visit: 31-Dec-99 18:00 Last Update: 27-Jul-16 22:43 Refresh 1 | 1,249 | 5,667 | {"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-2016-30 | latest | en | 0.921168 |
https://www.nagwa.com/en/explainers/643174628736/ | 1,701,216,028,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00069.warc.gz | 1,042,490,777 | 20,422 | # Lesson Explainer: Using Permutations to Find Probability Mathematics
In this explainer, we will learn how to find the probability of an event by calculating the number of outcomes using permutation.
You should already know how to count the number of ordered arrangements of items and how to count the number of ordered arrangements of items chosen from a group of . We will briefly recap each of these situations with two examples.
As an example, consider a group of 5 friends: Anna, Billy, Charlie, Danny, and Emma. They belong to a club and want to choose a captain and a vice-captain. To do this fairly, they have put each of their names into a hat and will pull out one name to be the captain and then (without replacing the first name) will select a second name to be the vice-captain. How many possible outcomes are their for the choice of captain and vice-captain?
We know there are 5 possibilities for the first name which is drawn from the hat and 4 possibilities for the second name.
To find the total number of choices for the captain and vice-captain, we multiply these numbers together. To see why this works, you could represent each of the 5 friends by the first letter of their names (so A stands for Anna and B stands for Billy) and draw a tree diagram of the outcomes.
Hence, the number of outcomes is .
Notice that each outcome can be represented by a permutation.
### Definition: Permutation
A permutation is an ordering of items in a set.
Two permutations of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are
Two 4-letter permutations of the letters A, B, C, D, E, and F are
In the example about choosing captains, each outcome is a 2-letter permutation of the letters A, B, C, D, and E. The permutation AB, which is also shown in the tree diagram, represents choosing Anna to be captain and Billy to be vice-captain. What we have shown is that the number of permutations of 2 letters chosen from a group of 5 letters is equal to .
For a second example, suppose there are 3 levels of prizes in a raffle: gold, silver, and bronze, but only one of each type of prize. The three prize winners are Mr. Sameh, Mr. Adam, and Mr. Maged and they will randomly choose which of the three prizes they will get. How many ways are there for the prizes to be assigned?
Note that each assignment of prizes can be represented as a permutation of the letters G, S, and B, where G stands for the gold prize, S for the silver, and B for the bronze. So, the permutation GSB means that Mr. Sameh gets the gold, Mr. Adam gets the silver, and Mr. Maged gets the bronze.
The total number of ways to assign the prizes is then the product of the number of choices for the first, second, and third letters, which is
In both these examples, our answer was a product of consecutive, decreasing integers. One way to record products like this is to use the factorial symbol.
### Definition: Factorial
Given a nonnegative integer , the factorial of is the product of and all the nonnegative integers less than . This can be written as .
So,
We define .
For example,
We will briefly return to the first example: choosing a captain and a vice-captain from a group of 5 people. We found that the number of ordered choices of 2 people from a group of 5 was which we can write with the factorial notation as
### Counting the Number of Ordered Arrangements of π Items Chosen from a Group of π Items
The number of ways to choose an ordered arrangement of objects from a group of objects is
In both of the above examples, we found that each outcome could be described by a permutation. This is because the order in which we selected the items mattered. In the first example, the outcome AB (Anna for captain and Billy for vice-captain) is different from the outcome BA. However, if we instead wanted to choose two vice-captains, then the outcome AB would be the same as the outcome BA because both outcomes result in Anna and Billy becoming vice-captains. Counting orderings of this type involves counting combinations, which we do not discuss further here.
Recall that when calculating the probability of an event, we need to calculate the total number of outcomes. Counting the number of outcomes often requires finding the number of permutations or combinations of a set of objects. We will look at some examples where the order of the objects matters, so we have to count permutations.
### Example 1: Counting Outcomes When Order Matters and Repetition Is Not Allowed
A student ID number consists of 7 digits where each digit is a number from 0 to 9. Given that no digit can be repeated, find the probability that a randomly generated ID number is 7βββ651βββ932.
The probability that the ID number is 7βββ651βββ932 is because each ID number is a unique arrangement of 7 distinct digits from 0 to 9.
So, we have to count how many 7-digit ID numbers can be generated from the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 with no repeated digits.
There are 10 choices for the 1st digit, and each time we use a digit we cannot use it again, so the number of choices for each subsequent digit is reduced by 1 until there are only 4 choices for the 7th digit. The total number of ways of choosing a 7-digit ID in this way is obtained by finding the product of the number of choices for each digit, which is
Alternatively, we could observe that each student ID is a permutation of 7 digits chosen from a group of 10 digits. Since the number of permutations of objects chosen from a group of objects is equal to we can substitute the values for and to get that the number of ID numbers is
Finally, obtaining the ID 7βββ651βββ932 is one of 604βββ800 outcomes, so the probability is
Often, there is more than one arrangement that belongs to the event we are trying to calculate the probability of. So, we have to count more than one type of arrangement. The next two examples demonstrate this.
### Example 2: Counting Outcomes
Eleven members of a marching band, two girls and nine boys, always line up in a row when they march. What is the probability that a girl will be at each end of the row if they line up in random order?
We need to find the probability that when the students line up in a random order the two girls are at either end of the line. Each of the possible orders that they line up in can be described as a permutation of the 11 students.
So, the probability is
First, we find the number of permutations starting and ending with a girl.
There are ways to order the girls (2 choices for the girl at the front and then 1 choice for the girl at the back). The number of ways to order the 9 boys in the middle is equal to (9 choices for the boy in the second position, 8 choices for the boy in the third position, and so on).
Hence, the total number of permutations starting and ending with a girl is
Next, we must count the total number of permutations of the 11 students. This is equal to .
Finally, we can state the probability of a random ordering having the girls at either end. This is
By expanding out the factorials and cancelling terms, we can simplify this expression as follows:
### Example 3: Counting Outcomes
The letters E, R, R, R, and O are placed in a bag. Determine the probability of randomly selecting letters from the bag in an order that spells the word βERRORβ.
Since there are three copies of the letter R, we will denote them by , , and . Then, we have to find the probability that a permutation of the letters E, , , , and O spells the word ERROR.
There is more than one permutation that would achieve this, for example,
Notice that in each of these permutations there is exactly one position in which E can appear and exactly one position in which O can appear. So, there is only one choice for the positions of E and O and, therefore, to count the number of outcomes spelling ERROR, we have to count the number of ways that the s can be chosen.
To find the total number of permutations spelling ERROR, we have to find the total number of orders in which the three s can appear. This is equal to the number of permutations of , , and which is
Therefore, the probability of a random permutation being one of these is
The total number of permutations of these 5 letters is equal to
Hence, the probability of a random permutation spelling ERROR is
Once you are familiar with recognizing the situations where the number of outcomes is a permutation, you can simply state the formula to calculate the number of outcomes.
### Example 4: Counting Outcomes
A number is formed at random using 2 distinct digits from the set . What is the probability that both the digits are odd?
The probability of forming two two-digit numbers from both odd digits is equal to
First, we calculate the total number of two-digit numbers that can be formed from the set of 4 numbers, without repeating digits.
Observe that the order in which we choose the digit matters; 29 is different from 92. Hence, each two-digit number is a permutation.
Since the number of permutations of numbers chosen from a group of numbers is equal to we can substitute the values for and to get that the number of distinct two-digit numbers is
Next, we note that there are two of these numbers which will have both digits odd: 59 and 95.
Hence, the probability that a random number chosen in this way has two odd digits is | 2,167 | 9,398 | {"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} | 5.0625 | 5 | CC-MAIN-2023-50 | latest | en | 0.932087 |
https://convertoctopus.com/131-8-milliliters-to-cubic-feet | 1,685,386,833,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00607.warc.gz | 216,540,695 | 7,396 | Conversion formula
The conversion factor from milliliters to cubic feet is 3.5314666572208E-5, which means that 1 milliliter is equal to 3.5314666572208E-5 cubic feet:
1 ml = 3.5314666572208E-5 ft3
To convert 131.8 milliliters into cubic feet we have to multiply 131.8 by the conversion factor in order to get the volume amount from milliliters to cubic feet. We can also form a simple proportion to calculate the result:
1 ml → 3.5314666572208E-5 ft3
131.8 ml → V(ft3)
Solve the above proportion to obtain the volume V in cubic feet:
V(ft3) = 131.8 ml × 3.5314666572208E-5 ft3
V(ft3) = 0.004654473054217 ft3
The final result is:
131.8 ml → 0.004654473054217 ft3
We conclude that 131.8 milliliters is equivalent to 0.004654473054217 cubic feet:
131.8 milliliters = 0.004654473054217 cubic feet
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 cubic foot is equal to 214.84709189454 × 131.8 milliliters.
Another way is saying that 131.8 milliliters is equal to 1 ÷ 214.84709189454 cubic feet.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred thirty-one point eight milliliters is approximately zero point zero zero five cubic feet:
131.8 ml ≅ 0.005 ft3
An alternative is also that one cubic foot is approximately two hundred fourteen point eight four seven times one hundred thirty-one point eight milliliters.
Conversion table
milliliters to cubic feet chart
For quick reference purposes, below is the conversion table you can use to convert from milliliters to cubic feet
milliliters (ml) cubic feet (ft3)
132.8 milliliters 0.005 cubic feet
133.8 milliliters 0.005 cubic feet
134.8 milliliters 0.005 cubic feet
135.8 milliliters 0.005 cubic feet
136.8 milliliters 0.005 cubic feet
137.8 milliliters 0.005 cubic feet
138.8 milliliters 0.005 cubic feet
139.8 milliliters 0.005 cubic feet
140.8 milliliters 0.005 cubic feet
141.8 milliliters 0.005 cubic feet | 594 | 2,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.748515 |
https://uhomework.com/%E6%95%B0%E5%AD%A6%E5%92%8C%E7%BB%9F%E8%AE%A1%E4%BB%A3%E8%80%83/ | 1,695,971,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510498.88/warc/CC-MAIN-20230929054611-20230929084611-00029.warc.gz | 621,043,901 | 24,692 | 数学和统计代考 ACSC/MATH 216代写 – 天才代写
# 数学和统计代考 ACSC/MATH 216代写
2022-06-03 12:04 星期五 所属： 统计代写 浏览：327
## Final Exam (Version 2)
Problem Total Possible Marks Earned Marks 1 20 2 20 3 20 数学和统计代考 4 20 5 20 Total 100
### Examination Directions 数学和统计代考
1. If the last digit of your student id is either 0 or 1 or 3 or 4 or 7, thisis the correct version for you.
2. This exam is being held and proctored remotely through a live Zoom meeting. Sitand face an engaged camera to enable monitoring.
3. There are 5 questions, with assigned marks as indicated, for a total of 100 points.
4. Nobooks or notes are allowed for this exam. 数学和统计代考
5. Non-programmable calculators including the following models of Texas Instru- ments calculators are permitted in the exams: BA-35, BA II Plus (Professional), TI-30Xa, TI-30X II and TI-30XS. Make sure that the memory has been cleared for TI-30X II and TI-30XS. Calculator instructions will not be permitted in the exams.
6. Uploada single pdf file with scanned pages of all your answers no later than 11:15 am.
1. Pleaseshow all of your There will be partial marks available for each step of solution. It is your responsibility to convince me that you know what you are doing. Clarity, completeness, and organization count.
2. Round your answers for prices and rates to two and three decimal places, re- spectively.
3. Goodluck!
### 1.A 25 year annual coupon bond with face and redemption amount of 2000 is sellingat an e↵ective annual yield rate equal to twice of the annual coupon rate. 数学和统计代考
The present value of the redemption amount is equal to the present value of the coupons.
(a)Determinethe purchase price of the bond at issuance.
(b)Determinethe amount for amortization of premium (or the principal repaid) in the 20th coupon.
(c)Suppose the bond was issued January 15, 2010, and is bought by a new purchaser on January 15, 2020 immediately after the coupon payment for a price of 1500. Find the internal rate of return earned by the original bondholder.
### 2.Attime 0, the term structure of e↵ective annual yield rates for zero coupon bonds is given as follows: 数学和统计代考
1- and 2-year maturity: 8%
3- and 4-year maturity: 10%
(a)Determine the price of a 4-year annual coupon bond with face amount 100 and coupon rate6%. 数学和统计代考
(b)You are given that the price of a 5-year annual coupon bond with face amount 100 and coupon rate 6% is 80. Determine the forward e↵ective annual interest rate for the period from time 4 to time 5,e., f [4, 5].
(c)A lender o↵ers to lend you 1000 for one year at rate 12% starting twoyear from now. Construct transactions that provide an arbitrage gain and give the amount of the gain.
### 3.Liability payments of 100 each are due to be paid in 2, 5 and 8 years from now.
Asset cashflow consists of A3in 3 years and A6 in 6 years. The yield for all payments is 10%. An attempt is made to have the asset cash flow immunize the liability cashflow by matching present value and duration.
(a)Determine the Macaulay duration and the Macaulay convexity of the liabil- ity cashflflow. 数学和统计代考
(b)Determine A3and A6.
(c)Determine whether or not the conditions for Redington immunization are satisfied.
### 4.Theterm structure of (annual effective) interest rates is as follows: 数学和统计代考
Suppose that the notional amount of a 4-year interest rate swap of floating in- terest rate for fixed interest rate is 1,000 for the first year and 2,000 for the second and third years, and 4,000 for the fourth year.
(a)Find the swap rate.
(b)Atthe start of the second year, the term structure is
Find the market value of the swap to the receiver at the start of the second year.
### 5.A coupon bond has a spot price of £950.
The bond will pay coupons of 50 in6 months and in one year. The risk free rates are 9% (per year continuously compounded) for 6 month maturity and 10% for one year maturity.
(a)Find the delivery price for a one year forward contract on the bond, with deliveryimmediately after the coupon payment. 数学和统计代考
(b)Suppose that immediately after the first coupon is paid, the continuously compound risk-free rate of interest is 10% for 6 month maturity, and thespot price of the bond has risen to Determine the value of the long position in the original forward contract entered at time 0. | 1,314 | 4,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-40 | latest | en | 0.692423 |
https://www.shaalaa.com/question-bank-solutions/prove-that-medians-of-a-triangle-are-concurrent-section-formula_203129 | 1,653,146,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00784.warc.gz | 1,130,912,323 | 11,018 | # Prove that medians of a triangle are concurrent - Mathematics and Statistics
Sum
Prove that medians of a triangle are concurrent
#### Solution
Consider ∆ABC.
Let P, Q, R be the midpoints of the sides BC, CA, AB respectively.
Let the medians BQ and CR intersect at G.
To prove that the third median AP also passes through G.
Let bar"a", bar"b", bar"c", bar"p", bar"q", bar"r", bar"g" be the position vectors of the points A, B, C, P, Q, R, G respectively.
Since P, Q, R are the mid-points of the sides BC, CA, AB respectively
∴ By midpoint formula, we get
bar"p" = (bar"b" + bar"c")/2 .......(i)
bar"q" = (bar"c" + bar"a")/2 .......(ii)
bar"r" = (bar"a" + bar"b")/2 .......(iii)
From (i), (ii) and (iii), we get
2bar"p" =bar"b" + bar"c" ⇒ 2bar"p" + bar"a" = bar"a" + bar"b" + bar"c"
2bar"q" = bar"c" + bar"a" ⇒ 2bar"q" + bar"b" = bar"a" + bar"b" + bar"c"
2bar"r" = bar"a" + bar"b" ⇒ 2bar"r" + bar"c" = bar"a" + bar"b" + bar"c"
∴ (2"p" + bar"a")/3 = (2bar"q" + bar"b")/3 = (2bar"r" + bar"c")/3 = (bar"a" + bar"b" + bar"c")/3
∴ (2"p" + bar"a")/(2 + 1) = (2bar"q" + bar"b")/(2 + 1) = (2bar"r" + bar"c")/(2 +1) = (bar"a" + bar"b" + bar"c")/3
= bar"g" .......(say)
This shows that the point G whose position vector is bar"g" lies on the three medians AP, BQ, CR dividing them internally in the ratio 2:1.
Hence, the three medians are concurrent.
Concept: Section Formula
Is there an error in this question or solution? | 545 | 1,446 | {"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} | 4.53125 | 5 | CC-MAIN-2022-21 | latest | en | 0.764637 |
https://www.physicsforums.com/threads/lagrange-equations.647612/ | 1,534,560,645,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00520.warc.gz | 951,952,561 | 12,428 | # Homework Help: Lagrange equations:
1. Oct 28, 2012
### Fabio010
Consider a rectilinear rod (A-B) with negligible mass that is attached, without friction, to a vertical OZ axis. The rod rotates about that axis with a constant angular velocity ω and it maintains a angle α with OZ.
A particle of mass m moves about the rod and it is attracted by the gravity.
a) Indicate, and justify the degrees of freedom of the material point.
To write the degrees of freedom i have to know the equations of x, y and z.
My problem is the z component equation.
For x and y we have the following equation:
x^2+y^2 = r^2
Now i need the equation for z..
any types?
Last edited: Oct 28, 2012
2. Oct 30, 2012
### TSny
If I'm understanding the setup, wouldn't r and z make up two legs of a right triangle with $\alpha$ as an angle in the triangle? So, z can be expressed in terms of r and $\alpha$. | 234 | 890 | {"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} | 3.015625 | 3 | CC-MAIN-2018-34 | latest | en | 0.917962 |
http://www.indiabix.com/aptitude/clock/discussion-651 | 1,371,658,647,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708882773/warc/CC-MAIN-20130516125442-00053-ip-10-60-113-184.ec2.internal.warc.gz | 508,130,788 | 10,237 | Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing
Aptitude - Clock - Discussion
@ : Home > Aptitude > Clock > General Questions - Discussion
Read more:
"Hold a true friend with both hands."
- (Proverb)
5.
How much does a watch lose per day, if its hands coincide every 64 minutes?
[A].
32 8 min. 11
[B].
36 5 min. 11
[C]. 90 min.[D]. 96 min.
Answer: Option B
Explanation:
55 min. spaces are covered in 60 min.
60 min. spaces are covered in 60 x 60 min. = 65 5 min. 55 11
Loss in 64 min. = 65 5 - 64 = 16 min. 11 11
Loss in 24 hrs = 16 x 1 x 24 x 60 min. = 32 8 min. 11 64 11
K.Kiran Kumar said: (Wed, Aug 4, 2010 12:48:55 AM) I cant get the question what it means. Please someone try to explain it.
Ravi said: (Tue, Aug 24, 2010 12:52:05 PM) I can't understand this: (60/55*60) why this step is used. Please someone explain me?
Tarak B Patel said: (Wed, Aug 25, 2010 03:21:35 PM) Solution: In 1 hour both the hands cover 55 min space. => 60/55 = 12/11 min space covered in 1min of the actual time. for the hand to coincide hands have to cover 60 min space => 12/11 * 60 = 720/11 = 65.5/11 min in actual clock. But the clock coincides every 64 min. =>65.5/11 - 64 = 1.5/11 = 16/11 min loss in 64min. =>16/11 * 1/64 = 1/44 min loss in 1min. Loss In 24 Hrs => 1/44 * (24 *60) = 360/11 = 32.8/11 min the clock looses in a day.
Soumya said: (Wed, Jan 12, 2011 01:18:58 AM) Why we multiply 1/64 ?
Priya said: (Fri, Jun 24, 2011 12:12:21 PM) How 55 spaces in 60 mins? couldnt get that.
Sharat said: (Wed, Dec 14, 2011 11:01:53 AM) What is this min. space? Does it mean the min. hand lags the hour by 5 min. space ?
Maddy said: (Fri, Jan 13, 2012 05:58:35 PM) To coincide with each other space(ticks) between them should be ZERO(no space). Consider starting from 12 noon. in 1 hr--> min hand covers spaces(ticks) =60 in 1 hr--> hour hand covers spaces(ticks) = 5 so in 1 min --> hour hand covers 5/60=1/12 space(tick) so after 1 hr=60 mins (1.00) spaces between them will be equal to 5 so have to cover 5 places for minute hand to meet hour hand total time = 60+5 ... but in 5 mins hour hand will further go ahead by 5*1/12 ticks so have to cover 5/12 ticks for minute hand to meet total time = 60+5+5/12 ... but agin in 5/12 mins hour hand will go ahead buy 5/12*1/12= 5/12^2... and so on So total time will be = 60+5+(5/12+5/12^2+5/12^3+.........) this GP total time to meet = 720/11 i.e., 65.(5/11). This time required to coincide hands in correct clock... solve accordingly.
Karthik said: (Wed, Mar 21, 2012 03:55:52 PM) How can the watch lose time if it is covering earlier than what is expected?
Himanshu Dewangan said: (Mon, Mar 26, 2012 11:18:05 AM) It is universal truth that watch does not lose any time.Every day start from 12:00 am in night, hr. and min hand positioned at 0 degree,than how lose? no lose is there.. But suppose hands coincide every 64 min. (In really it is not possible..eg if ones they coincide it take 65+5/11 min always for next) there will be lose of 65+5/11 - 64= 16/11 from actual or real time in 65+5/11 min. So calculate lose in 1 day= 24*60 min
Vallaban said: (Wed, Apr 4, 2012 11:22:18 AM) 55 min. Spaces are gained by minute hand in 60 min period. To find how many spaces it has actually gained, we need to fix a standard point first. ! With respect to it, we need to see the difference by how much is it actually varying. ! So let us assume the standard point to be the place where the minutes hand and hours hand has been coincided. ! I may be 12:00, 1.06, 2.11, 3.17. etc. from there. 60 minutes implies the minute hand must come back to the same point where it has started. ! Is it not. ? Now, 60 minute passed and so minute hand covers 60 minute spaces. And the hour hand advances by 5 minute spaces. ! So from the standard point fixed initially (we assumed the standard point is where the minute point and hours hand were coinciding. Also. 60 minutes will be passed when the minute hand comes back to the same position from where it started). Now, there is. An absolute 60 min spaces covered by minute hand in 60 min and then there is 5 min spaces advanced by hour hand in 60 min period. ! So on total. Total advancement is 60-5 = 55 minute spaces. !
Prasant said: (Fri, Apr 13, 2012 01:14:54 AM) Pleaes clarify, whether the clock looses or gain in this question?
Naresh said: (Wed, May 30, 2012 04:45:01 PM) A watch which gains uniformly, is 5 minutes slow at 8 o' clock in the morning on Sunday and it is 5 min. 48 sec fast at 8 p. M. On following Sunday. When was it correct?
Konxie said: (Thu, Jun 7, 2012 02:29:39 AM) Say its 12 o' clock: The minute hand and the hour hand points towards 12 After an hour i.e at 1 o' clock: The minute hand points at 12 and the hour hand points 1 The difference between the two hands after 60 min (1 hr) is 55 spaces but the hands did not coincide! They will coincide when 60 spaces are covered And to cover 60 spaces it takes = (60/55*60) = 65.4545 min (in a normal clock) Now given in question is that the hands coincide at 64 min (defective clock) So loss in time when the hands coincide is 65.4545 - 64 = 1.4545 min (this loss happens for every 64 min in our defective clock) So for 1 min our loss is = 1.4545/64 min For a day 1.4545/64*(24*60) = 32.7 min @Naresh : just remember this If coinciding time > 65(5/11) then our clock is going slow than normal (our watch is loosing time) and if coinciding time < 65(5/11) then its going fast than normal (or gaining time). By 65(5/11) i mean 65.4545 and not (65*5)11 And we say loosing time in questions just to state the irregularity in time. HOPE THIS CLEARS IT ALL
Dinesh said: (Sun, Jul 29, 2012 11:04:27 PM) Well said Konxie.
Priya said: (Fri, Aug 31, 2012 08:02:16 PM) I can't understand this problem 65*5/11-64
Sachin said: (Tue, Jan 1, 2013 04:32:06 PM) First consider in how many minutes, hands should coincide. 55 min. spaces are covered in 60 min. 60 min. spaces will be covered in, 65 (5/11) i.e. 65.45 minutes. But this clock is taking 64 minutes to coincide hence it is 1.45 min slow to actual time (65.45-64=1.45). In 64 min of journey clock is 1.45 min slow, then in 24*60 min it will 32.62 min slow. =(24*60*1.45*1)/64= 32.62 or 32(8/11).
Koti said: (Sun, Feb 24, 2013 12:06:59 PM) I am not able to understand how 55 min. Spaces are gained by minute hand in 60 min period. Can anybody explain?
Amit said: (Thu, May 16, 2013 03:37:48 PM) Just imagine hour hand and minute hand starting at 12 O'clock. After 1 hour, the minute hand will be at 12 whereas the hour hand will be at 1. So what is the difference between them? it's 55 "Minute Space". So, it takes 1 hour to cover 55 minute spaces, for overlapping they should cover total 60 minute space. Thus, here is the calculation: 1 hr --> 55 minute space. ? hr --> 60 minute space. (1*60) /55 hour = 60/55 * 60 minute // converting hr to mins. = 65 5/11 minute.
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Contact us: info@indiabix.com Follow us on twitter! | 2,298 | 7,297 | {"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-2013-20 | latest | en | 0.794339 |
https://keepconvert.com/360-cm-to-feet-converter.html | 1,660,204,738,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00171.warc.gz | 309,581,475 | 14,155 | # 360 cm to feet converter
## 1 Cm is Equal to How Many Feet?
Cminchesconverter is a conversion tool which assists you in converting centimeters to feet. Centimeters and feet are both units of length. We know that a cm is about equal to 0.032808399 feet. By using this connection, we can convert cm to feet.
A cm = 0.032808399 feet.
These are the issues wefrequently face.
• How many feet are in a cm?
• one cm equal to how many feet?
• What is centimeter to feet conversion?
• How to calculate 1 cm to fet?
## Centimeter
Centimeters, also known as centimetres, is the measurement unit used to measure length in metric systems. English symbols are abbreviated as cm. The meter is internationally defined as the SI unit, the centimeter is not. But a centimeter is equals one hundredth of meter. It’s also around 39.37 inches. Application:
• We typically measure height in centimeters.
• Centimeters are used to convert map scale to real world.
• A report of measured rainfall.
## Feet Definition
Feet, also known as foot (symbol: ft) is a unit of length used in the Anglo-American customary system of measuring It equals a third of a yard and 12 inches. Application:
• For measuring heights, shorter distances, field lengths.
• Human foot size.
## Formula for Calculating 360 cm to Feet
Value in ft = value in cm × 0.032808399 So, 360 cm in feet = 360 cm × 0.032808399 = 11.811023622036 feet.
## What is 360 Centimeters Converted to Feet?
You’re in the right place. We know that there is 0.032808399 feet in one cm. Let’s convert 360 cm to feet (or 360 centimeters to feet) with cm feet conversion. So what is 360 cm to ft? 360 cm to feet = 11.811023622036 feet. Calculation Process: 360 cm to feet = 360 cm × 0.032808399 = 11.811023622036 feet
cm feet 359.2 cm 11.784776902876 feet 359.3 cm 11.788057742771 feet 359.4 cm 11.791338582666 feet 359.5 cm 11.794619422561 feet 359.6 cm 11.797900262456 feet 359.7 cm 11.801181102351 feet 359.8 cm 11.804461942246 feet 359.9 cm 11.807742782141 feet 360 cm 11.811023622036 feet 360.1 cm 11.814304461931 feet 360.2 cm 11.817585301826 feet 360.3 cm 11.820866141721 feet 360.4 cm 11.824146981616 feet 360.5 cm 11.827427821511 feet 360.6 cm 11.830708661406 feet 360.7 cm 11.833989501301 feet | 667 | 2,246 | {"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} | 3.4375 | 3 | CC-MAIN-2022-33 | latest | en | 0.891449 |
https://www.airmilescalculator.com/distance/ruk-to-ppl/ | 1,653,653,770,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00501.warc.gz | 730,469,840 | 62,235 | # Distance between Rukumkot (RUK) and Phaplu (PPL)
Flight distance from Rukumkot to Phaplu (Chaurjahari Airport – Phaplu Airport) is 279 miles / 449 kilometers / 242 nautical miles. Estimated flight time is 1 hour 1 minutes.
Driving distance from Rukumkot (RUK) to Phaplu (PPL) is 528 miles / 849 kilometers and travel time by car is about 13 hours 4 minutes.
279
Miles
449
Kilometers
242
Nautical miles
1 h 1 min
## How far is Phaplu from Rukumkot?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 278.747 miles
• 448.600 kilometers
• 242.225 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 278.339 miles
• 447.943 kilometers
• 241.870 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Rukumkot to Phaplu?
Estimated flight time from Chaurjahari Airport to Phaplu Airport is 1 hour 1 minutes.
## What is the time difference between Rukumkot and Phaplu?
There is no time difference between Rukumkot and Phaplu.
## Flight carbon footprint between Chaurjahari Airport (RUK) and Phaplu Airport (PPL)
On average flying from Rukumkot to Phaplu generates about 66 kg of CO2 per passenger, 66 kilograms is equal to 146 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Rukumkot to Phaplu
Shortest flight path between Chaurjahari Airport (RUK) and Phaplu Airport (PPL).
## Airport information
Origin Chaurjahari Airport
City: Rukumkot
Country: Nepal
IATA Code: RUK
ICAO Code: VNRK
Coordinates: 28°37′37″N, 82°11′41″E
Destination Phaplu Airport
City: Phaplu
Country: Nepal
IATA Code: PPL
ICAO Code: VNPL
Coordinates: 27°31′4″N, 86°35′4″E | 582 | 2,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | longest | en | 0.841705 |
http://ir.lib.ncu.edu.tw/handle/987654321/63557 | 1,657,003,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00546.warc.gz | 26,782,787 | 9,523 | English | 正體中文 | 简体中文 | Items with full text/Total items : 73032/73032 (100%) Visitors : 23358574 Online Users : 517
Scope All of NCUIR 理學院 數學研究所 --博碩士論文 Tips: please add "double quotation mark" for query phrases to get precise resultsplease goto advance search for comprehansive author search Adv. Search
NCU Institutional Repository > 理學院 > 數學研究所 > 博碩士論文 > Item 987654321/63557
Please use this identifier to cite or link to this item: `http://ir.lib.ncu.edu.tw/handle/987654321/63557`
Title: 伽羅瓦理論;Galois Theories Authors: 曾宇揚;Tseng,Yu-yang Contributors: 數學系 Keywords: 伽羅瓦理論;伽羅瓦範疇;亞歷山大·格羅滕迪克;Galois theory;Galois category;Grothendieck Date: 2014-01-28 Issue Date: 2014-04-02 15:50:24 (UTC+8) Publisher: 國立中央大學 Abstract: 在第二章，我重新整理了Lenstra的筆記和Grothendieck的SGA I的5.4章節。基本上可以分成兩個部分： 第一個部分根據Lenstra筆記的第三章。有一些比較省略的步驟，還有一些步驟留作習題，我把這些地方補上並想辦法寫得更流暢些。第二個部分根據Grothendieck的SGA I的5.4章節，用pro-objects的技術，首次出現是在Seminaire Bourbaki一篇Grothendieck的文章裡。我從原來文章裡簡短的描述中，給了定理4.1一個詳細的證明。; In chapter 2, I reorganize some part of \cite{Le08} and \cite{SGA1}. Basically it can be divided into two parts: the first part follows \cite{Le08} Chapter 3. Some steps are written a bit roughly and some steps are exercises in original texts, I just make them more fluent, and write down the exercises; The second part follows \cite{SGA1} Section 5.4, using the technique so called pro-objects, first introduced by Grothendieck in his article in Seminaire Bourbaki \cite{Gr59}. I give a proof of \cite{SGA1} Expos\'{e} V. Theorem 4.1, following the brief sketch in the original article. Appears in Collections: [數學研究所] 博碩士論文
Files in This Item:
File Description SizeFormat
index.html0KbHTML431View/Open | 650 | 1,698 | {"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} | 3.140625 | 3 | CC-MAIN-2022-27 | latest | en | 0.502017 |
https://www.jiskha.com/display.cgi?id=1284952943 | 1,503,257,317,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106984.52/warc/CC-MAIN-20170820185216-20170820205216-00678.warc.gz | 912,222,584 | 3,811 | # statistics
posted by .
In a box of thirty AAA batteries, there are three defective batteries. Two batteries are randomly selected and tested. What is the probability that both are defective, if the first one is not replaced after being tested?
• statistics -
3/30 * 2/29 = ?
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More Similar Questions | 587 | 2,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-34 | latest | en | 0.958038 |
https://repl.it/repls/TrueHonestPython | 1,553,338,467,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202781.83/warc/CC-MAIN-20190323101107-20190323123107-00057.warc.gz | 579,555,519 | 80,444 | @anonymous/
# TrueHonestPython
## No description
Files
• main.cpp
main.cpp
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```
```/*
Problem: Calculate and print the monthly paycheck for an employee given a list of deductions.
Algorithm
1. Get Employee Name
2. Get Gross Income
3. Calculate all deductions
4. Output all deductions to two decimal places
a. Left align name of all outputs
b. Right align all numerical values with their respective name
*/
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
int main()
{
//Variable Declaration and Initialization
string employeeFullName;
double grossIncome;
double fedTax = .15;
double fedDeduction;
double stateTax = .035;
double stateDeduction;
double ssTax = .0575;
double ssDeduction;
double medTax = .0275;
double medDeduction;
double pensionTax = .05;
double pensionDeduction;
double healthInsurance = 75.00;
double netPay;
//Input
cout << "==================================" << endl;
cout << "Description: Employee Paycheck" << endl;
cout << "School: Long Beach City College" << endl;
cout << "Author: Christian Araya" << endl;
cout << "Date: March 1st 2018" << endl;
cout << "Program: CArayaProject3" << endl;
cout << "==================================\n" << endl;
cout << "Please Enter Employee's Full Name:" << endl;
getline(cin, employeeFullName);
cout << "Please Enter Employee's Gross Income (ex. 75876.34):" << endl;
cin >> grossIncome;
//Process
fedDeduction = grossIncome * fedTax;
stateDeduction = grossIncome * stateTax;
ssDeduction = grossIncome * ssTax;
medDeduction = grossIncome * medTax;
pensionDeduction = grossIncome * pensionTax;
netPay = grossIncome - fedDeduction - stateDeduction - ssDeduction - medDeduction - pensionDeduction;
//Output
cout << "***********************************************************" << endl;
cout << "Net Pay is calculated after taking the following deductions" << endl;
cout << "Federal Income Tax: 15%" << endl;
cout << "State Tax: 3.5%" << endl;
cout << "Social Security Tax: 5.75%" << endl;
cout << "Medicare/Medicaid Tax: 2.75%" << endl;
cout << "Pension Plan: 5%" << endl;
cout << "Health Insurance: \$75.00" << endl;
cout << "***********************************************************\n\n\n" << endl;
cout << "TOTAL DEDUCTIONS for " << employeeFullName << endl;
cout << "----------------------------------------------------" << endl;
cout << fixed << setprecision(2);
cout << setfill('.') << setw(40) << left << "Gross Amount: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << grossIncome << endl;
cout << setfill('.') << setw(40) << left << "Federal Tax Deduction: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << fedDeduction << endl;
cout << setfill('.') << setw(40) << left << "State Tax Deduction: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << stateDeduction << endl;
cout << setfill('.') << setw(40) << left << "Social Security Tax Deduction: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << ssDeduction << endl;
cout << setfill('.') << setw(40) << left << "Medicare/Medicaid Tax Deduction: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << medDeduction << endl;
cout << setfill('.') << setw(40) << left << "Pension Plan Deductions: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << pensionDeduction << endl;
cout << setfill('.') << setw(40) << left << "Health Insurance Deduction: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << healthInsurance << endl;
cout << setfill('.') << setw(40) << left << "Net Pay: "
<< setw(2) << " \$"
<< setfill(' ') << setw(8) << right << netPay << endl;
return 0;
}``` | 1,213 | 3,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-13 | latest | en | 0.609506 |
https://discuss.leetcode.com/topic/87/swap-nibbles | 1,521,496,332,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647153.50/warc/CC-MAIN-20180319214457-20180319234457-00652.warc.gz | 566,920,370 | 11,459 | # Swap Nibbles
• Given a 64-Bit Unsigned Integer, write a function to accept that integer as an argument and return an integer with all the nibbles swapped.
Example:
Input: 0x0123456789ABCDEF
• Welcome @pkhatri to the new Discuss! Thanks for posting this question, it looks interesting. Is this a bit-manipulation kind of problem?
• Yes. It has to be done using Bit Manipulation.
• Interesting! I have never heard of the term nibble before, until I look up on Wikipedia, which means a four-bit aggregation a.k.a. half byte.
mask1 = `0F0F0F0F0F0F0F0F`
mask2 = `F0F0F0F0F0F0F0F0`
The answer should be: `(n & mask1) << 4) | (n & mask2) >> 4)`. Does this sound right to you?
• This solution looks great!
• @1337c0d3r I came up with another way of doing it during my interview. I'll post that as soon as I get back home.
You solution is the easiest way to accomplish this. Good job :)
I believe this could be added to the Bit Manipulation section.
• @pkhatri I was wondering other ways to do it and I am looking forward to your method. Yeah will add this question to our library soon, thanks!
• ``````uint64_t p = 0x0123456789ABCDEF;
uint64_t swapped = 0x0000000000000000;
int i=0;
for(i=0;i<64;i+=8)
{
uint64_t nib1 = p & mask1>>i;
uint64_t nib2 = p & mask2>>i;
swapped |= nib1>>4;
swapped |= nib2<<4;
}
printf("%lu\n",swapped);.
``````
I put this through the phone, please edit it for code formatting if possible. Luckily I had it in my email :)
• Awesome! Thanks for sharing your solution with us @pkhatri !
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 472 | 1,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | latest | en | 0.887656 |
https://documen.tv/question/in-a-class-of-50-students-24-like-football-21-basketball-and-18-cricket-6-like-football-and-bask-19695477-31/ | 1,675,609,908,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00047.warc.gz | 238,367,608 | 19,631 | # In a class of 50 students 24 like football, 21 basketball and 18 Cricket, 6 like football and basketball, only 3 like basketball only, 5 lik
Question
In a class of 50 students 24 like football, 21 basketball and 18 Cricket, 6 like football and basketball, only 3 like basketball only, 5 like any of the three games. Illustrate the information in a Venn diagram. Find the number of students who like football and crickets only and exactly one of the game.
in progress 0
1 year 2021-08-31T02:04:56+00:00 1 Answers 39 views 0 | 146 | 527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.931819 |
https://edulissy.com/new-projects/assignment-4-solution-158/ | 1,660,131,710,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00184.warc.gz | 226,889,512 | 23,795 | # Assignment 4 Solution
\$35.00 \$30.80
Category:
## Description
In this assignment you will sort trajectories of Whales based on the lengths of the trajectories, using GPU.
The Whale trajectories will be provided using a text file. The format of the file is following. The file line will have two numbers N and M. N represents the number of trajectories, and M represents the number of stops each trajectory has. For simplicity, you can consider all Whales will take the same number of stops.
Then, there will be N lines in the file, each line will contain M Cartesian coordinates (M pair of numbers in each line) where each pair indicates a stop. All Whale starts from the location (0,0).
When completed, you GPU based program will read two file names from command line argument list. The first file will be the whale trajectory file. The program, then read the trajectory file, upload the data to GPU, and perform calculation as follows. For each data, your GPU will calculate the length of the data. The length will be calculated by summing the Euclidian distances between consecutive stops, starting from (0,0). Then the GPU will sort the trajectory based on the length of the trajectories. Once the sort complete, the host program will download the sorted the data from GPU, and save it to file, as name given in the second argument.
You need to develop the program using computing programming model either CUDA or OpenCL.
Sample input file (input.dat)
4 2
051010
2246
102034
104 410
Sample output file (output.dat)
2246
051010
104 410
102034
Explanation
0 5 10 10 (distance 16.18)
2 2 4 6 (distance 7.30)
10 20 3 4 (distance 39.83)
10 4 4 10 (distance 19.26)
Individual Distance Calculation:
051010
(0,0) -> (0,5) => 5
(0,5) -> (10,10) => 11.18034
Total Trajectory (5+11.18034) = 16.18
Page 2 of 2 | 467 | 1,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | latest | en | 0.7901 |
https://9lib.co/document/oz110ez9-%E6%8C%87%E6%95%B8%E8%88%87%E5%B0%8D%E6%95%B8%E5%87%BD%E6%95%B8.html | 1,657,149,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104678225.97/warc/CC-MAIN-20220706212428-20220707002428-00030.warc.gz | 119,711,712 | 29,996 | # 指數與對數函數
## 全文
(1)
### 指數對數函數
(2)
https://sites.google.com/site/hysh4math · (d) y = 3x+1 例題4 解指數不等式 (3x − 9)(3x − 27) ≤ 0 [Ans:2 ≤ x ≤ 3] 例題5 解指數方程式 22x − 7 · 2x − 23 = 0 [Ans:x = 3 ] 例題6 解下列指數不等式 (a). 2x > 4 (b). (12)x+2 > 1 (c). (14)x + (12)x − 2 < 0 [Ans:(a)x > 2 (b) x < −2 (c) x > 0] 例題7 比較 (0.3)1.3,(0.3)0.3,(0.3)−0.3,0.3 與 1 這五個數的大小關係? [Ans:(0.3)−0.3 > 1 > (0.3)0.3 >0.3 > (0.3)1.3 對數與對數定律 例題1 求下列對數值:
(a) 2 log102 + log1015 − log106 = [Ans:1]
(b) log 59 − log 37 + log 2735 = [Ans:0]
(c) (log23 + log49)(log34 + log92) = [Ans:5]
a+ab ]
[Ans:−√2] 例題5 已知 log102 ≈ 0.3010, log103 ≈ 0.4771, log107 ≈ 0.8451 , 試求下列各數的近 似值: a = log102120, b = log102400, c = log100.375 [Ans:a = 36.12, b =
3.3801, c = −0.4259]
10 ]
x + 3zy [Ans:6] 例題4 解 x1+log5x = 25x2 [Ans: x = 25,1
5]
− 2x + 10) 在 0 ≤ x ≤ 4 範圍內的最小值? [Ans:2]
(3)
https://sites.google.com/site/hysh4math · 2 4 6 8 10 -6 -4 -2 2 4 6 y=logax y=log2x y=logbx y=logcx
0.6, n ≥ 5
Updating...
Updating... | 588 | 972 | {"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-2022-27 | latest | en | 0.290783 |
http://mathhelpforum.com/math-topics/17334-understanding-sat-problem-print.html | 1,529,523,475,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863834.46/warc/CC-MAIN-20180620182802-20180620202802-00388.warc.gz | 204,458,523 | 3,134 | # understanding SAT problem
• Jul 30th 2007, 01:41 PM
Dergyll
understanding SAT problem
Hey guys
Just need to know, on the SAT/PSAT, a phrase called "Units digit" keeps popping up and I have no idea what it means...Here is an example problem
"The unit digit of the sum of three consecutive integers is 8. What is the units digit of the least of these integers?"
A) 4
B) 5
C) 6
D) 7
E) 8
I know this is noobie stuff but just need to know what it means real quick
Thanks alot
Derg
• Jul 30th 2007, 02:00 PM
topsquark
Quote:
Originally Posted by Dergyll
Hey guys
Just need to know, on the SAT/PSAT, a phrase called "Units digit" keeps popping up and I have no idea what it means...Here is an example problem
"The unit digit of the sum of three consecutive integers is 8. What is the units digit of the least of these integers?"
A) 4
B) 5
C) 6
D) 7
E) 8
I know this is noobie stuff but just need to know what it means real quick
Thanks alot
Derg
The "units digit" of a number is the same as the "ones digit": it is the digit of a number that corresponds to how many times 1 you have.
Probably not a good explanation. Here are some examples:
1234 The "units digit" is 4
5678 The "units digit" is 8
-Dan
• Jul 30th 2007, 02:02 PM
Dergyll
Wow dan thanks a million, you really saved my hide with this...:D
Now this clears everything apart
Thanks again
Derg
• Jul 30th 2007, 02:04 PM
topsquark
Quote:
Originally Posted by Dergyll
Hey guys
Just need to know, on the SAT/PSAT, a phrase called "Units digit" keeps popping up and I have no idea what it means...Here is an example problem
"The unit digit of the sum of three consecutive integers is 8. What is the units digit of the least of these integers?"
A) 4
B) 5
C) 6
D) 7
E) 8
I know this is noobie stuff but just need to know what it means real quick
Thanks alot
Derg
Let the first number be x. Then the three numbers are x, x + 1, and x + 2. The sum of these is 3x + 3. Call the units digit of x n. So the units digit of the sum will be
3n + 3 = 8 (If 3n + 3 is not greater than 9)
or
3n + 3 - 10 = 3n - 7 = 8 (else)
In the first case we have:
3n + 3 = 8
3n = 5 <-- No such n exists.
In the second case we have:
3n - 7 = 8
3n = 15
n = 5.
Thus the units digit of x (the least of the three numbers) is 5.
-Dan | 728 | 2,283 | {"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} | 3.703125 | 4 | CC-MAIN-2018-26 | latest | en | 0.917216 |
https://www.geeksforgeeks.org/check-if-characters-of-a-string-can-be-made-non-decreasing-by-replacing-s/?ref=rp | 1,713,828,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00143.warc.gz | 702,117,773 | 48,102 | # Check if characters of a string can be made non-decreasing by replacing ‘?’s
Last Updated : 18 Jun, 2021
Given a string S of length N consisting of lowercase English alphabets and ‘?’, the task is to check if it is possible to make the string non-decreasing by replacing all the ‘?’ characters with English alphabets. If found to be true, print “Yes”. Otherwise, print “No”.
Examples:
Input: S = “abb?xy?”
Output: Yes
Explanation:
Replacing the ‘?’s at index 3 and 6 with ‘b’ and ‘z’, modifies the string S to “abbbxyz”, which is non-decreasing.
Input: S = “??z?a?”
Output : No
Naive Approach: The simplest approach to solve the problem is to generate all possible strings using English alphabets and check if the resulting string is non-decreasing or not. If found to be true, print “Yes”, Otherwise, print “No”.
Time Complexity: O(N*26N)
Auxiliary Space: O(N)
Efficient Approach: The given problem can be solved based on the following observations:
• It can be observed that the only position that matters is the position with S[i]! = ‘?’, as one can replace the ‘?’ with the character at the nearest index that is not ‘?’
• Therefore, the task is reduced to checking if characters of the strings that are not ‘?’, are in non-decreasing order or not.
Follow the steps below to solve the problem:
• Initialize a variable, say last, to store the last character which is not a ‘?’.
• Traverse the string and for every ith character, check if S[i]! = ‘?’ and S[i]<last, then print “No”
• Otherwise, update last as last = S[i] if S[i]!=’?’.
• Print “Yes” if none of the above cases are satisfied.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `// Function to check if it's possible` `// to make the string s non decreasing` `int` `nonDecreasing(string s)` `{` ` ``// Stores length of string` ` ``int` `size = s.length();` ` ``// Stores the last character` ` ``// that is not '?'` ` ``char` `c = ``'a'``;` ` ``// Traverse the string S` ` ``for` `(``int` `i = 0; i < size; i++) {` ` ``// If current character` ` ``// of the string is '?'` ` ``if` `(s[i] == ``'?'``) {` ` ``continue``;` ` ``}` ` ``// If s[i] is not '?'` ` ``// and is less than C` ` ``else` `if` `((s[i] != ``'?'``)` ` ``&& (s[i] < c)) {` ` ``return` `0;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ``// Update C` ` ``c = s[i];` ` ``}` ` ``}` ` ``// Return 1` ` ``return` `1;` `}` `// Driver Code` `int` `main()` `{` ` ``string S = ``"abb?xy?"``;` ` ``if` `(nonDecreasing(S))` ` ``cout << ``"Yes"` `<< endl;` ` ``else` ` ``cout << ``"No"` `<< endl;` ` ``return` `0;` `}`
## Java
`// Java program for the above approach` `import` `java.util.*;` `class` `GFG` `{` `// Function to check if it's possible` `// to make the String s non decreasing` `static` `int` `nonDecreasing(``char` `[]s)` `{` ` ` ` ``// Stores length of String` ` ``int` `size = s.length;` ` ``// Stores the last character` ` ``// that is not '?'` ` ``char` `c = ``'a'``;` ` ``// Traverse the String S` ` ``for` `(``int` `i = ``0``; i < size; i++)` ` ``{` ` ``// If current character` ` ``// of the String is '?'` ` ``if` `(s[i] == ``'?'``) ` ` ``{` ` ``continue``;` ` ``}` ` ``// If s[i] is not '?'` ` ``// and is less than C` ` ``else` `if` `((s[i] != ``'?'``)` ` ``&& (s[i] < c))` ` ``{` ` ``return` `0``;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ``// Update C` ` ``c = s[i];` ` ``}` ` ``}` ` ``// Return 1` ` ``return` `1``;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ``String S = ``"abb?xy?"``;` ` ``if` `(nonDecreasing(S.toCharArray())==``1``)` ` ``System.out.print(``"Yes"` `+``"\n"``);` ` ``else` ` ``System.out.print(``"No"` `+``"\n"``);` `}` `}` `// This code is contributed by 29AjayKumar`
## Python3
`# python 3 program for the above approach` `# Function to check if it's possible` `# to make the string s non decreasing` `def` `nonDecreasing(s):` ` ` ` ``# Stores length of string` ` ``size ``=` `len``(s)` ` ``# Stores the last character` ` ``# that is not '?'` ` ``c ``=` `'a'` ` ``# Traverse the string S` ` ``for` `i ``in` `range``(size):` ` ` ` ``# If current character` ` ``# of the string is '?'` ` ``if` `(s[i] ``=``=` `'?'``):` ` ``continue` ` ``# If s[i] is not '?'` ` ``# and is less than C` ` ``elif``((s[i] !``=` `'?'``) ``and` `(s[i] < c)):` ` ``return` `0` ` ``# Otherwise` ` ``else``:` ` ` ` ``# Update C` ` ``c ``=` `s[i]` ` ``# Return 1` ` ``return` `1` `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` ` ``S ``=` `"abb?xy?"` ` ``if``(nonDecreasing(S)):` ` ``print``(``"Yes"``)` ` ``else``:` ` ``print``(``"No"``)` ` ` ` ``# This code is contributed by SURENDRA_GANGWAR.`
## C#
`// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{` ` ``// Function to check if it's possible` ` ``// to make the String s non decreasing` ` ``static` `int` `nonDecreasing(``char` `[]s)` ` ``{` ` ``// Stores length of String` ` ``int` `size = s.Length;` ` ``// Stores the last character` ` ``// that is not '?'` ` ``char` `c = ``'a'``;` ` ``// Traverse the String S` ` ``for` `(``int` `i = 0; i < size; i++)` ` ``{` ` ``// If current character` ` ``// of the String is '?'` ` ``if` `(s[i] == ``'?'``) ` ` ``{` ` ``continue``;` ` ``}` ` ``// If s[i] is not '?'` ` ``// and is less than C` ` ``else` `if` `((s[i] != ``'?'``)` ` ``&& (s[i] < c))` ` ``{` ` ``return` `0;` ` ``}` ` ``// Otherwise` ` ``else` `{` ` ``// Update C` ` ``c = s[i];` ` ``}` ` ``}` ` ``// Return 1` ` ``return` `1;` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `Main(String[] args)` ` ``{` ` ``String S = ``"abb?xy?"``;` ` ``if` `(nonDecreasing(S.ToCharArray())==1)` ` ``Console.Write(``"Yes"` `+``"\n"``);` ` ``else` ` ``Console.Write(``"No"` `+``"\n"``);` ` ``}` `}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output:
`Yes`
Time Complexity: O(N)
Auxiliary Space: O(1)
Previous
Next | 2,347 | 6,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.837307 |
https://en.m.wikipedia.org/wiki/Rotational_symmetry | 1,709,438,314,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00527.warc.gz | 237,575,614 | 25,748 | Rotational symmetry
Rotational symmetry, also known as radial symmetry in geometry, is the property a shape has when it looks the same after some rotation by a partial turn. An object's degree of rotational symmetry is the number of distinct orientations in which it looks exactly the same for each rotation.
Certain geometric objects are partially symmetrical when rotated at certain angles such as squares rotated 90°, however the only geometric objects that are fully rotationally symmetric at any angle are spheres, circles and other spheroids.[1][2]
Formal treatment
Formally the rotational symmetry is symmetry with respect to some or all rotations in m-dimensional Euclidean space. Rotations are direct isometries, i.e., isometries preserving orientation. Therefore, a symmetry group of rotational symmetry is a subgroup of E +(m) (see Euclidean group).
Symmetry with respect to all rotations about all points implies translational symmetry with respect to all translations, so space is homogeneous, and the symmetry group is the whole E(m). With the modified notion of symmetry for vector fields the symmetry group can also be E +(m).
For symmetry with respect to rotations about a point we can take that point as origin. These rotations form the special orthogonal group SO(m), the group of m × m orthogonal matrices with determinant 1. For m = 3 this is the rotation group SO(3).
In another definition of the word, the rotation group of an object is the symmetry group within E +(n), the group of direct isometries; in other words, the intersection of the full symmetry group and the group of direct isometries. For chiral objects it is the same as the full symmetry group.
Laws of physics are SO(3)-invariant if they do not distinguish different directions in space. Because of Noether's theorem, the rotational symmetry of a physical system is equivalent to the angular momentum conservation law.
Discrete rotational symmetry
Rotational symmetry of order n, also called n-fold rotational symmetry, or discrete rotational symmetry of the nth order, with respect to a particular point (in 2D) or axis (in 3D) means that rotation by an angle of ${\displaystyle {\tfrac {360^{\circ }}{n}}}$ (180°, 120°, 90°, 72°, 60°, 51 37°, etc.) does not change the object. A "1-fold" symmetry is no symmetry (all objects look alike after a rotation of 360°).
The notation for n-fold symmetry is Cn or simply n. The actual symmetry group is specified by the point or axis of symmetry, together with the n. For each point or axis of symmetry, the abstract group type is cyclic group of order n, Zn. Although for the latter also the notation Cn is used, the geometric and abstract Cn should be distinguished: there are other symmetry groups of the same abstract group type which are geometrically different, see cyclic symmetry groups in 3D.
The fundamental domain is a sector of ${\displaystyle {\tfrac {360^{\circ }}{n}}.}$
Examples without additional reflection symmetry:
• n = 2, 180°: the dyad; letters Z, N, S; the outlines, albeit not the colors, of the yin and yang symbol; the Union Flag (as divided along the flag's diagonal and rotated about the flag's center point)
• n = 3, 120°: triad, triskelion, Borromean rings; sometimes the term trilateral symmetry is used;
• n = 4, 90°: tetrad, swastika
• n = 6, 60°: hexad, Star of David (this one has additional reflection symmetry)
• n = 8, 45°: octad, Octagonal muqarnas, computer-generated (CG), ceiling
Cn is the rotation group of a regular n-sided polygon in 2D and of a regular n-sided pyramid in 3D.
If there is e.g. rotational symmetry with respect to an angle of 100°, then also with respect to one of 20°, the greatest common divisor of 100° and 360°.
A typical 3D object with rotational symmetry (possibly also with perpendicular axes) but no mirror symmetry is a propeller.
Examples
C2 (more) C3 (more) C4 (more) C5 (more) C6 (more)
Double Pendulum fractal
Roundabout traffic sign
US Bicentennial Star
The starting position in shogi
Snoldelev Stone's interlocked drinking horns design
Multiple symmetry axes through the same point
For discrete symmetry with multiple symmetry axes through the same point, there are the following possibilities:
• In addition to an n-fold axis, n perpendicular 2-fold axes: the dihedral groups Dn of order 2n (n ≥ 2). This is the rotation group of a regular prism, or regular bipyramid. Although the same notation is used, the geometric and abstract Dn should be distinguished: there are other symmetry groups of the same abstract group type which are geometrically different, see dihedral symmetry groups in 3D.
• 4×3-fold and 3×2-fold axes: the rotation group T of order 12 of a regular tetrahedron. The group is isomorphic to alternating group A4.
• 3×4-fold, 4×3-fold, and 6×2-fold axes: the rotation group O of order 24 of a cube and a regular octahedron. The group is isomorphic to symmetric group S4.
• 6×5-fold, 10×3-fold, and 15×2-fold axes: the rotation group I of order 60 of a dodecahedron and an icosahedron. The group is isomorphic to alternating group A5. The group contains 10 versions of D3 and 6 versions of D5 (rotational symmetries like prisms and antiprisms).
In the case of the Platonic solids, the 2-fold axes are through the midpoints of opposite edges, and the number of them is half the number of edges. The other axes are through opposite vertices and through centers of opposite faces, except in the case of the tetrahedron, where the 3-fold axes are each through one vertex and the center of one face.
Rotational symmetry with respect to any angle
Rotational symmetry with respect to any angle is, in two dimensions, circular symmetry. The fundamental domain is a half-line.
In three dimensions we can distinguish cylindrical symmetry and spherical symmetry (no change when rotating about one axis, or for any rotation). That is, no dependence on the angle using cylindrical coordinates and no dependence on either angle using spherical coordinates. The fundamental domain is a half-plane through the axis, and a radial half-line, respectively. Axisymmetric and axisymmetrical are adjectives which refer to an object having cylindrical symmetry, or axisymmetry (i.e. rotational symmetry with respect to a central axis) like a doughnut (torus). An example of approximate spherical symmetry is the Earth (with respect to density and other physical and chemical properties).
In 4D, continuous or discrete rotational symmetry about a plane corresponds to corresponding 2D rotational symmetry in every perpendicular plane, about the point of intersection. An object can also have rotational symmetry about two perpendicular planes, e.g. if it is the Cartesian product of two rotationally symmetry 2D figures, as in the case of e.g. the duocylinder and various regular duoprisms.
Rotational symmetry with translational symmetry
Arrangement within a primitive cell of 2- and 4-fold rotocenters. A fundamental domain is indicated in yellow. Arrangement within a primitive cell of 2-, 3-, and 6-fold rotocenters, alone or in combination (consider the 6-fold symbol as a combination of a 2- and a 3-fold symbol); in the case of 2-fold symmetry only, the shape of the parallelogram can be different. For the case p6, a fundamental domain is indicated in yellow.
2-fold rotational symmetry together with single translational symmetry is one of the Frieze groups. There are two rotocenters[definition needed] per primitive cell.
Together with double translational symmetry the rotation groups are the following wallpaper groups, with axes per primitive cell:
• p2 (2222): 4×2-fold; rotation group of a parallelogrammic, rectangular, and rhombic lattice.
• p3 (333): 3×3-fold; not the rotation group of any lattice (every lattice is upside-down the same, but that does not apply for this symmetry); it is e.g. the rotation group of the regular triangular tiling with the equilateral triangles alternatingly colored.
• p4 (442): 2×4-fold, 2×2-fold; rotation group of a square lattice.
• p6 (632): 1×6-fold, 2×3-fold, 3×2-fold; rotation group of a hexagonal lattice.
• 2-fold rotocenters (including possible 4-fold and 6-fold), if present at all, form the translate of a lattice equal to the translational lattice, scaled by a factor 1/2. In the case translational symmetry in one dimension, a similar property applies, though the term "lattice" does not apply.
• 3-fold rotocenters (including possible 6-fold), if present at all, form a regular hexagonal lattice equal to the translational lattice, rotated by 30° (or equivalently 90°), and scaled by a factor ${\displaystyle {\tfrac {1}{3}}{\sqrt {3}}}$
• 4-fold rotocenters, if present at all, form a regular square lattice equal to the translational lattice, rotated by 45°, and scaled by a factor ${\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}$
• 6-fold rotocenters, if present at all, form a regular hexagonal lattice which is the translate of the translational lattice.
Scaling of a lattice divides the number of points per unit area by the square of the scale factor. Therefore, the number of 2-, 3-, 4-, and 6-fold rotocenters per primitive cell is 4, 3, 2, and 1, respectively, again including 4-fold as a special case of 2-fold, etc.
3-fold rotational symmetry at one point and 2-fold at another one (or ditto in 3D with respect to parallel axes) implies rotation group p6, i.e. double translational symmetry and 6-fold rotational symmetry at some point (or, in 3D, parallel axis). The translation distance for the symmetry generated by one such pair of rotocenters is ${\displaystyle 2{\sqrt {3}}}$ times their distance.
Euclidean plane Hyperbolic plane
Hexakis triangular tiling, an example of p6, [6,3]+, (632) (with colors) and p6m, [6,3], (*632) (without colors); the lines are reflection axes if colors are ignored, and a special kind of symmetry axis if colors are not ignored: reflection reverts the colors. Rectangular line grids in three orientations can be distinguished.
Order 3-7 kisrhombille, an example of [7,3]+ (732) symmetry and [7,3], (*732) (without colors)
References
• Weyl, Hermann (1982) [1952]. Symmetry. Princeton: Princeton University Press. ISBN 0-691-02374-3. | 2,472 | 10,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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-2024-10 | latest | en | 0.908372 |
http://indigo.ie/~peter/holqua6.html | 1,503,276,574,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107065.72/warc/CC-MAIN-20170821003037-20170821023037-00132.warc.gz | 195,704,928 | 16,264 | Holistic Mathematics 3 - Part B
Q. We have now reached the radial level. Can you explain what this entails in holistic mathematical terms?
PC Yes, this is perhaps the key question of all. How does (finite) phenomenal reality emerge from the (transfinite) fundamental void?
Remarkably, our simple mathematical approach can give a compelling explanation.
We have already seen that the diagonal or null lines, which in mathematical terms have a magnitude of zero, precisely represent - in holistic terms - the (transfinite) void.
However - as these diagonal lines lie in the two dimensional complex plane - their intersection with the circumference of the circle can be represented as the sum of two coordinates, representing the distance of the points from both horizontal (real) and vertical (imaginary) axes. These complex coordinates i.e. 1+ i, 1- i, -1 + i, -1 - i (in each case multiplied by the reciprocal of the square root of 2) in fact represent the four complex solutions (in addition to the two real and two imaginary) obtained from the eight roots of unity.
Thus our transfinite diagonal lines (initially involving a two dimensional transformation using the Pythagorean Theorem), alternatively can be explained in one dimensional finite terms as the sum of equal real and imaginary components.
In psychological terms the transfinite void (representing the purely intuitive two dimensional unconscious), can be alternatively explained in reduced one dimensional conscious rational terms as the equal combination of real (quantitative) and imaginary (qualitative) experience. Thus this void - representing the supersymmetry of spiritual "forces" as pure potential, is continually actualised - in reduced asymmetrical terms - as real (cognitive) and imaginary (affective) experience.
In complementary physical terms, the transfinite void (representing the empty implicit ground of reality) can be expressed in reduced finite terms as the equal combination of real (quantitative) and imaginary (qualitative) aspects. Thus again this void - representing the supersymmetry of physical forces as pure potential, is continually actualised - in a world of broken symmetry - as (real) matter and (imaginary) dimensions.
Thus both physically and psychologically the transfinite empty void (representing pure potential) is continually actualised in reduced finite terms (quantitative and qualitative).
This reduced actuality is likewise continually transformed in some measure (through both the processes of evolution and involution) back to that empty potential void.
Q. So one can explain both physical and spiritual "forces" alternatively in simple or (reduced) "complex" terms. Is this the case?
PC Yes, this is precisely the case. We saw in our earlier discussion of (physical) light
its utterly mysterious true nature. Though providing the standard by which space and time measurements take place, in terms of itself it has no finite meaning in space and time (In other words its true nature is transfinite).
However light (as with all of the forces) has an alternative (reduced) finite interpretation (through its interaction with phenomena). Now this reduced finite translation involves
equal "real" and "imaginary" components. This at once provides a key to the understanding of the wave-particle duality of light.
The "real" component of light is most easily understood in terms of its particle aspect
(where light can be identified at a specific location in space as photons).
The "imaginary" component of light is by contrast represented by its wave aspect (whereby light travels in a generalised fashion through time).
Put alternatively in reduced finite terms, light exhibits both quantitative (particle) and qualitative (wave) aspects. In terms of my holistic mathematics, this is exactly what we would expect to find.
Now I treat all four forces in the same fashion. In holistic mathematical terms the electromagnetic and weak forces serve as (horizontal) opposites (positive and negative with respect to each other). The gravitational and strong forces likewise serve as horizontal opposites. These related pairings in turn serve as (vertical) opposites (with respect to each other) and can be converted into each other through a "real" to "imaginary" (or "imaginary" to "real" transformation) Thus in the supersymmetry of the void all the forces are diagonally complementary. Each of the four forces can be explained in finite terms by one of the complex solutions representing the coordinates of the diagonal lines in each quadrant. Each therefore will manifest both "real" (particle) and "imaginary" (wave) aspects. (The fact that it has not been yet experimentally possible to validate this contention for all four forces, in no way weakens my assertion).
Of course this thinking equally applies to the "spiritual" forces.
Once again these are represented in mystical terms as the contemplation of immanence and transcendence both of which can be given - in horizontal terms - an external (physical) or internal (psychological) interpretation.
Now these spiritual "forces" in their own terms are transfinite and absolute (providing the standard through which relative phenomena are measured). However they again have a reduced finite interpretation with both "real" and "imaginary" components.
Thus spiritual light has a "real" (quantitative) immanent aspect and an "imaginary" (qualitative) transcendent aspect. The same logically applies to spiritual "gravity".
As we have seen in relation to the dark night there is a very close relationship as between spiritual "gravity" and spiritual light. Thus what is internalised as gravity is released in "imaginary" form as radiation (light).
Thus in holistic mathematical terms if x(1 - i) (where x is the reciprocal of the square root of 2) represents the spiritual electromagnetic force then to obtain gravity we transform "real" to "imaginary" ("imaginary" to "real") and obtain negative signs. This gives us x(-1 + i). Thus "spiritual" gravity can be represented as diagonally opposite to the electromagnetic force. Likewise the other two forces (strong and weak) are diagonally opposite.
Of course if this applies to the "spiritual" forces it equally applies - in complementary fashion - to "physical" forces. In holistic mathematical terms therefore the electromagnetic and the gravitational force can be represented as physically symmetric in diagonal "complex" terms. Likewise the weak and strong forces are diagonally symmetric.
The electromagnetic and weak, and gravitational and strong are alternatively both horizontally and vertically symmetric with respect to each other.
To conclude this section the "forces" in their ultimate symmetrical state - physical and spiritual - can be represented as both absolutely simple (i.e. as transfinite null lines) and absolutely "complex" (with equal finite "real" and "imaginary" components).
So here in holistic mathematical terms simplicity and complexity are understood quite clearly as two expressions of the same reality. This interpenetration of simplicity and complexity characterises the radial level.
Q. Can you now outline the overall structure of the radial level in holistic mathematical terms?
PC As we have seen earlier, each of the transitions between levels involves a major development which has a precise holistic mathematical interpretation.
The linear level involves the specialisation of the "real" positive (conscious) direction of experience.
The transition from linear to circular levels involves the specialised development of the complementary negative (unconscious) direction.
The circular level is then defined in terms of the development of structures with both positive and negative aspects.
The transition from circular to point levels involves the specialised development of the "imaginary" (projected) conscious direction of experience.
The point level then involves the (additional) development of structures with both real and imaginary aspects.
The transition from point to radial levels involves the specialised development of the (simple) transfinite aspect of experience. Sometimes misleadingly - esp. in Eastern spirituality - the arrival at this simple non-dual experience of reality is represented as the completion of the spiritual journey. True reality is identified here solely with the hidden underlying symmetry implicit in all (manifest) phenomena.
However a complete expression of reality - more typical of Western spirituality - involves the journey into radial level with the (additional) development of structures having both finite (actual) and transfinite (potential) aspects. This return to finite reality arises from the realisation that what is absolutely simple (in intuitive transfinite terms) is equally absolutely "complex" (in rational finite terms). Thus the recognition of the complementary rational pole of experience entails the rebirth of finite reality and the world of broken symmetry.
Overemphasis on the merely rational expression of experience, confines one unduly to the finite asymmetrical world (consequently blotting out the hidden transfinite domain of perfect symmetry). This is the position which characterises Western science.
However overemphasis on the merely intuitive expression of reality, leads to over absorption in this hidden transfinite ground in an unduly passive experience (blotting out the world of finite phenomena). This is the position which too often characterises the emphasis of Eastern spirituality.
Clearly truly comprehensive experience of reality implies equal attention to both rational and intuitive poles at the radial level. Remarkably holistic mathematics can precisely demonstrate the simultaneous simple and "complex" nature of this level.
So in many ways the radial level - far from representing the final goal of spiritual experience represents a brand new beginning. In the light of spiritual realisation one is now able to return - in varying capacities - to the world of broken symmetry offering a healing presence.
Thus the radial level is properly characterised by the increasing dynamic interaction as between the transfinite and finite, emptiness and form, the simple and the "complex", symmetry and asymmetry.
Q. Can you now outline the stages of the radial level.
PC I divide the radial level into two sub-levels.
The first sub-level involves the slow emergence from the transfinite void and the gradual rebirth of "transformed" finite reality. However for some considerable time experience may suffer from what I call "process split" (where finite and transfinite remain to some degree separated). In other words involvement in actual phenomenal affairs does not yet serve as an adequate expression of one’s spiritual vision.
The second sub-level is more dynamic and active where finally one finds a satisfactory
worldly expression of deep spiritual convictions.
Though secondary personality characteristics may vary greatly, people living at the radial level are essentially centroverts (i.e. primarily centred on God).
Of course the best known exponents of this final phase are the great spiritual leaders who with superhuman energy and endurance transform the face of history.
Others - though less known - exercise a profound influence for good on those whom they encounter. Finally - as at all levels - there are many who are largely anonymous, with their intense commitment remaining a secret to others (and perhaps even to themselves).
However it is important to avoid any elitist notions in terms of the radial level. All ego comparisons with others in terms of "better" and "worse", "higher" and "lower" are only true in a relative sense. The predominant attitude of the radial level - which reflects a cosmic based personality - is one of radical equality (in horizontal, vertical and diagonal terms).
In holistic mathematical terminology the radial level is eight directional. Thus we have the successful differentiation (with consequent integration) of positive (external) and negative (internal) experience; of real (quantitative) and imaginary (qualitative) experience, of finite (actual) and transfinite (potential) experience.,
We can deal with both radial sub-levels in terms of spiritual, intellectual and emotional experience which are very closely interrelated.
Spiritual involvement involves both transcendent (masculine) and immanent (feminine) poles. On the one hand there is the discovery that divine identity (inherent in every person). Also and equally important one is at last free to be human without masks or pretensions. So human and divine identity now coincide.
Cognitive intellectual development is of course closely related with the spirit.
We have now the unfolding of a dynamic "complex" paradigm in understanding. A diagonal form of complementarity is at work. At a local immediate level the "real" analytical aspect of understanding is restored; at the global level the "imaginary" holistic aspect predominates. The task of the radial level is to progressively integrate both aspects
until they are no longer distinguishable.
Affective emotional development initially remains polarised to some degree. Again at an immediate personal level one feels intimately loved - perhaps for the first time - in an unconditional fashion; at a global level a more detached platonic awareness initially predominates.
It may take many years of ongoing development before these diagonal poles are properly harmonised.
Thus the first sub-level of the radial level involves increasing immersion in phenomenal reality (which is now reborn in transformed fashion). At both spiritual, cognitive and affective levels, however, initial difficulty will be experienced in properly harmonising finite (partial) with transfinite (holistic) experience.
The second sub-level involves is highly dynamic representing the supreme manifestation
of integrated experience. Here transfinite and finite aspects approach equal balance (the marriage of contemplation with activity). Though the manner of expression is unique to each individual, a satisfactory worldly expression of one’s deepest spiritual desires is at last obtained. Though this brings joy and fulfilment, equally it entails much suffering through growing compassionate involvement with the problems of the world.
Q. What is the relevance of the radial level for mathematics.
PC Once again each level is characterised by a distinctive mathematical approach.
The linear level is characterised by the rational quantitative approach (Standard Mathematics) and is analytical in nature.
The circular level is characterised by the intuitive qualitative approach (Holistic Mathematics 1) and is synthetic in nature.
The point level is characterised more subtly by the relationship as between the rational quantitative and intuitive holistic approaches (Holistic Mathematics 2).
Thus each level involves the specialisation of a particular mathematical system.
The radial level now involves the gradual integration of these mathematical approaches
(already successfully differentiated at the other levels). It therefore involves the mutual interaction of all three systems with the relative independence of each system.
I refer to this as Holistic Mathematics 3. Now this terminology might seem misleading as increasing attention is likewise paid now to analytical mathematics. However full integration is not obtained immediately and the qualitative holistic element is likely to remain prominent for some time.
I do recognise a final idealised dynamic state of mathematical understanding where analytical and holistic aspects are highly integrated. This can be referred to as Comprehensive Mathematics.
One of the earlier mathematical tasks of the radial level is the formulation of "A Theory of Everything" which represents an extreme in terms of a purely holistic understanding of the fundamental ground of reality (non-dual reality). (I will deal with this in detail in my next posting!)
Subsequent mathematical development involves the study of the relationship as between the underlying fundamental ground and the now (reborn) world of finite phenomena. In other words it involves the mathematical study of the relationship between symmetry and asymmetry.
The proper investigation of this symmetrical-asymmetrical reality involves therefore a "complex" rational paradigm. This comprises a "real" analytical and an "imaginary" holistic aspect. Thus Holistic Mathematics is now seen in its appropriate context as the "imaginary" component of the "complex" rational paradigm.
Q. Can you say a little more about the nature of this "complex" paradigm.
PC One of the great weaknesses of the standard "real" approach is that because it lacks an appropriate holistic framework, it can lead to much naivety where fundamental deep questions regarding the nature of reality are involved. Indeed most of the key issues in mathematics and physics are really philosophical. Holistic Mathematics provides the ideal vehicle for dealing with such philosophical issues. It also provides the appropriate intuition for interpreting mathematical hypotheses. Superstring Theory is an interesting example of where Holistic Mathematics can make a valuable contribution by giving it a meaningful intuitive explanation. (As this illustrates so well Holistic Mathematics in action, I may return to it in a future posting).
One of the interesting philosophical findings of Holistic Mathematics is that a consistent Theory of Everything (TOE) must be an expression of the empty transfinite ground of reality, and can only be adequately expressed in the language of Holistic as opposed to Standard Mathematics.
Also, when one deals with phenomenal reality, a comprehensive approach requires a "complex" paradigm comprising both "real" analytical and "imaginary" holistic aspects.
This approach is subject to the uncertainty principle. Any attempt to maximise the "real" analytical approach leads to increased fuzziness in terms of the complementary "imaginary" holistic aspect. In other words we run into paradoxes and inconsistencies which cannot be explained in "real" analytical terms. (This very problem lies at the root of the continuing difficulty in terms of reconciling Quantum Mechanics with the Theory of Relativity).
By contrast any attempt to maximise the "imaginary" holistic approach leads to a purely circular qualitative approach (thus eliminating the scope for quantitative analysis).
Thus we have two extremes. The "real" analytical component is suited to the study of the quantitative "parts" (in isolation from the qualitative whole). The "imaginary" holistic component is geared to the study of the qualitative whole (in isolation from the quantitative "parts").
Thus any truly overall understanding i.e. (a TOE) must be couched in holistic mathematical terms (and expressed in purely circular paradoxical rational fashion).
Any truly partial limited understanding i.e. (a TOS or Theory of Something) must be couched in analytical mathematical terms (in standard linear non-paradoxical fashion).
Thus the use of standard mathematics by definition will always lead to partial explanations of reality. Therefore when we get into the difficult area of mathematically interpreting the dynamic interpenetration of "parts" and "wholes" we require a "complex" paradigm.
Again the "real" analytical component will be used to formulate quantitative theories. The "imaginary" holistic component will provide a background intuitive framework to interpret such theories and establish their scope and limitations. Thus standard mathematical and enlightened philosophical understanding of reality now go hand in hand.
Q. Can you go into this relationship between the "real" and "imaginary" aspects of mathematics in more detail.
PC Remember that the linear level (which provides the paradigm for "real" mathematics) is strictly one directional.
Now as we have seen reality is more subtle. At the circular level it becomes dynamically two- directional with the complementarity of positive (external) and negative (internal) directions. At the point level reality now becomes four directional with the additional complementarity of "real" (quantitative) and "imaginary" (qualitative) aspects. At the radial level it becomes eight directional with additional transfinite as well as finite directions.
For the purposes of exposition we will restrict ourselves here to four rather than eight directions.
Thus the one directional approach of the linear level always requires the positing of one of these directions absolutely (to the exclusion of three other related aspects).
Now this in fact is all very relevant to understanding Ken Wilber’s four quadrant approach. As Ken indicates we can have theories which emphasise the external and internal aspects of experience (separately). Also we can have theories which emphasise the individual and collective aspects of experience (separately).
By the very nature of the linear level there will be a tendency to emphasise one partial explanation as the whole truth (thus excluding equally valid explanations).
Now Ken of course quite rightly emphasises the need to concentrate on all four facets to achieve a more balanced understanding. However by definition - as it is not suited to the synthesis of opposing directions of experience - this can only be done in a very limited fashion at the linear level. At best one will form a differentiated understanding of four (rather than one) alternative facets. But these will not be properly integrated in experience.
At the circular level one forms the understanding that positive (external) and negative (internal) aspects of experience are dynamically complementary (in horizontal terms). In other words they are ultimately integrated as the same unified experience. This leads to heightened intuitive awareness. One is now enabled to form one directional understanding (separately) of each facet at the linear (rational) level while dynamically unifying both aspects at the circular (intuitive) level.
At the point level one forms the additional understanding that "real" (individual) and "imaginary" (collective) understanding are dynamically complementary (in vertical terms).
Again they are ultimately unified in dynamic intuitive fashion at this level. One is now enabled to form one directional understanding (separately) of all four facets at the linear (rational) level while dynamically integrating all aspects at the ("higher" intuitive) point level.
Thus the appropriate rational differentiation of each facet of experience is properly the task of the linear level. However the appropriate integration of these facets is the task of the "higher" intuitive levels. That is why I maintain that my holistic mathematical approach (which is designed to integrate reason with intuition) provides the appropriate context to interpret Ken’s four quadrant approach. Properly understood - in such holistic mathematical terms - the four quadrants relate to the complex number axes (interpreted in qualitative terms).
(Of course a full interpretation of holons involves the relationship as between finite (phenomenal) and transfinite (formless) aspects and is eight directional requiring the specialised understanding of the radial level).
Q. Have you anything more to say here before we leave this topic?
PC Using a four directional approach, the standard scientific paradigm is built on the asymmetrical interpretation of dimensions (3 and 1) i.e. three dimensions of space and one of time.
The holistic scientific paradigm is based on the - more mathematically correct - symmetric interpretation of dimensions (2 and 2) i.e. two dimensions of space and two of time each having positive and negative directions. Indeed we can alternatively define reality here in terms of 4 space dimensions (two real, two imaginary) or alternatively 4 time dimensions (two imaginary, two real).
Now experience can be seen as a continual correction process by which the asymmetrical nature of conventional space and time ceaselessly switches to its underlying symmetrical nature (and the symmetrical in turn switches to the asymmetrical). Of course when these dynamics are limited - as at the linear level - the correction process in turn is very restricted so that the finite phenomena of broken symmetry assume an absolute rigid identity.
Now conventional science - even in linear terms - greatly misrepresents the nature of experience. We can represent actual phenomena from four different aspects.
• We can experience (perceptual) objects - in relation to the self. (This gives the common sense observation of objects being three dimensional in space. Now in this interaction the time dimension directly results from the mental contribution. So in this formulation we have 3 dimensions of space (positive) and one of time (positive).
• We can experience the self - in relation to (perceptual) objects. However whereas before, we experienced the object externally in space and time, now it has an internal direction. However conventional science makes no distinction as between these two aspects.
• We can experience (conceptual) objects - in relation to the self. Now this experience is abstract yet still misleadingly identified as three dimensional (in spatial terms). For example the new bulky printer I have purchased - as (quantitative) perception - is certainly three dimensional (in spatial terms). However it is highly misleading to refer to the (qualitative) holistic concept of "printer" as three dimensional in space. Thus there is an immediate confusion in conventional science as between the partial quantitative and holistic qualitative interpretations of objects.
• We can experience (conceptual) objects - in relation to the self. Again though these holistic qualitative objects have internal as well as external directions, this distinction is not made in conventional science.
So we can have here four distinctive interpretations of the same observation. Thus any equation - even at the linear level - has (at a minimum) four interpretations.
Thus if a represents the hypotenuse of a right angled triangle and b and c the other two sides, then a2 = b2 + c2.
Now this can be given (a) a partial quantitative (external) interpretation as objective truth (i.e. that the result will apply for an actual triangle), (b) a partial quantitative (internal) interpretation as subjective truth )i.e. that the result has a psychological interpretation), (c) a holistic qualitative (external) interpretation as objective truth (i.e. that the result applies abstractly to "all" triangles) and (d) and finally a holistic qualitative (internal) interpretation as subjective truth (i.e. again that the result has a general interpretation in psychological terms).
However though we can identify these four possible interpretations of the same equation, there is no way of reconciling them at the linear level. This is why the linear level is asymmetrical.
Now considerable progress is made at the circular level.
When we take the same four aspects of the experience of an object, the external and internal directions are clearly differentiated. If we take the objective direction as positive, then - in relative terms - the subjective direction is negative. Also the perceptual and conceptual experiences are differentiated in "real" terms. Quite simply if the perception is identified in spatial terms, then - relatively speaking - the corresponding concept is identified in temporal terms (and vice versa).
So lets now look again at the interpretation of the four aspects.
• When we experience the external perception of the object (in relation to the self), it is given a positive interpretation. Thus we identify the object as three dimensional in (positive) space and one dimensional in (positive) time.
• When we experience the internal perception (in relation to the world), it is given - in relative terms - a negative interpretation. Thus we now identify the object as three dimensional in (negative) space and one dimensional in (negative) time.
• When we experience the external concept (in relation to the self), space and time dimensions are exchanged. Thus we now identify the object as three dimensional in (positive) time and one dimensional in (positive) space.
• Finally when we experience the internal concept (in relation to the world) the positive direction turns to negative. So we now experience the object as three dimensional in (negative) time and one dimensional in (negative) space.
Though each aspect of phenomenal reality is experienced in asymmetrical terms, overall in terms of the differing four aspects there is symmetry. We have the same number of negative as well as positive directions (8 in all). Also we have the same number of temporal as well as spatial dimensions.
Thus the very ability to switch as between the differing asymmetrical aspects, implicitly depends on an underlying ground where space and time are symmetrical (two dimensions each with both positive and negative directions).
If scientists are to understand that the "same" equation can have several different interpretations (as complementary directions), then rational analytical understanding must be underlined by intuitive holistic understanding.
When we move into the point level, there is now the further appreciation that space and time in dynamic terms are "real" and "imaginary" with respect to each other. Thus we can describe an object in "complex" terms using solely spatial or temporal dimensions.
Thus using this "complex" perspective we can describe an observation of an object in the following four ways.
• The external perception of the object (in relation to mind and concept) is in "real" (positive) space. "Real" time is now "imaginary" space. Thus the observation is four dimensional (three of "real" space and one of "imaginary" space).
• The internal perception (in relation to object and concept) is in "real" (negative) space (three dimensions) and "imaginary" (negative) space (one dimension).
• The external concept of the object (in relation to mind and perception) is in "imaginary" (positive) space (three dimensions) and "real" (positive) space (one dimension).
• Finally the internal concept (in relation to object and perception) is in "imaginary" (negative) space (three dimensional) and "real" (negative) space (one dimensional).
Thus we have here described the four aspects of the object in terms of "complex" space. (We could equally describe them in "complex" time).
(An eight directional interpretation would of course be even more subtle).
Again the ability to switch between these four asymmetrical interpretations of the observation involves implicitly an underlying "complex" space symmetry (i.e. where space has two "real" and two "imaginary" dimensions (with positive and negative directions).
Thus when combined with holistic mathematics, standard mathematical equations require a far more refined multifaceted interpretation.
Q. Can you demonstrate how the "arrow of time" is eight directional at the radial level?
PC In Holistic Mathematics 1, I dealt with the two-directional arrow of time.
As all events in experience involve the dynamic relationship of (internal) observer and (external) event, relativity is always involved with the event both past and future in time.
Now the four directional interpretation involves a distinction as between the quantitative and qualitative interpretations of phenomena. The quantitative (partial) interpretation is given by the perception of "the event"; the qualitative (holistic) interpretation is given by the corresponding concept of "event". In holistic mathematical terms if the (partial) quantitative definition of a phenomenon is "real", then - in relative terms - the (holistic) qualitative definition is "imaginary".
Experience of time is now polarised in both horizontal and vertical fashion.
Thus in terms of the relationship of (internal) observer and (external) event as defined in quantitative (partial) terms, time has two directions - positive and negative (relatively past and future) - in "real" time.
In terms of the relationship of (internal) observer and (external) event as defined in qualitative (holistic) terms, time has again two directions - positive and negative (relatively past and future) - in "imaginary" time.
Now these four directions of time are measured in finite terms.
However each of these four finite directions can be given a transfinite grounding. Whereas the polarised finite approach relates to rational understanding of the experience, immediate transfinite understanding relates to corresponding intuitive understanding.
This intuitive understanding is absolute (not relative) and best described simply as the present moment.
Thus each of the four finite directions in time are grounded in a transfinite experience of the present moment. In this sense we have eight directions to the experience of time.
What is fascinating is that the transfinite (null) diagonal directions can be given a reduced finite explanation as the combination of equal "real" and "imaginary" components. And as both "real" and "imaginary" components can have positive or negative signs, this allows for four "complex" interpretations.
Thus an eight directional experience of an event - which characterises the radial level - involves two horizontal opposites (the "partial" quantitative aspect) that are - relatively - positive and negative in "real" time.
It also involves two vertical opposites (the "holistic" qualitative aspect) that are - relatively - positive and negative in "imaginary" time.
It finally involves four diagonal opposites (the essential aspect) - that are transfinite representing experience of the present moment (which underlines each of the four finite relative directions).
These four diagonal opposites can be given a reduced finite explanation (as the intersection of "partial" quantitative and "holistic" qualitative aspects) that are relative in "complex" time with equal "real" and "imaginary" components. (These are "real" + and
"imaginary" +, "real" - and "imaginary" +, "real" + and "imaginary -, "real" - and "imaginary" -).
These distinctions are far from academic. For example a properly integrated explanation of evolution would have to maintain a balance as between horizontal, vertical and diagonal complementarity.
This complementarity is often missing in practice. Indeed I would characterise my biggest criticism of Ken Wilber’s work as a lack of sufficient vertical complementarity. This leads - among other things - to an elitist view of evolution with a continued tendency to elevate "higher" above "lower" life forms. Thus the statement that the noosphere represents a "higher" holarchy than the physiosphere is only true in a relative limited sense. (What Ken seems to often forget is that our very interpretations of this "lower" physioshere are derived from the "higher" noosphere). Thus when not balanced by a recognition of the opposite polarity (i.e. that in a certain equally valid sense that the physiospere is "higher" than the noosphere) we can obtain a distorted perspective and fail to realise the ultimate radical equality of all creation.
When one considers that conventional science is still based on just a single direction of time, one realises how limited is its perspective.
Q. What do you precisely mean when you say that the radial level is multidimensional?
PC Firstly I want to clear up a possible confusion in relation to terms.
In holistic mathematics directions relate directly to dimensions. Thus to say that the radial level is eight directional, implies that it is eight dimensional.
However the correct way to interpret these "dimensions" - as I have already stated - is in terms of three binary systems (horizontal, vertical and diagonal) relating to information, transformation and essential states respectively. As the final diagonal binary system can be represented as the equal intersection of both information and transformation states, at its simplest we can represent reality dynamically in terms of a double binary system involving the processing of information and transformation respectively.
So just as the application of the horizontal binary system leads to the generation of information as multiple quantities (objects), the application of the vertical binary system leads to the generation of transformations as multiple qualities (dimensions).
Thus properly understood, corresponding to each object as quantity is a corresponding dimension as quality. Thus there are as many dimensions as objects. So just as reality consists of multiple objects, it is also multidimensional.
I already showed in Holistic Mathematics 2, how each "real" object can be viewed alternatively as an "imaginary" dimension. Likewise each "real" dimension can be viewed alternatively as an "imaginary" object.
When we identify objects solely as quantities in conventional scientific terms (and thereby in abstraction from dimensions) there is much confusion.
Properly understand the "object" represents a dynamic interaction pattern that is both quantitative and qualitative. (In psychological terms the perception of the object represents the "partial" quantitative aspect; the concept of the object represents the "holistic" qualitative aspect). Thus in dynamic terms - at all levels of reality - quantitative and qualitative aspects interpenetrate. Objects and dimensions are truly complementary (in vertical terms).
Now for a static analysis of reality we have to keep one of these poles fixed. Of course the usual convention is to view multiple objects (quantitative) against a fixed dimensional background (qualitative).
However it is equally valid to view multiple dimensions (qualities) against a fixed object background (quantitative). This latter approach leads - in very much complementary fashion to psychology to the prediction of a "shadow" world of matter. Interestingly this result is predicted by Superstring Theory (which is not surprising given the highly dynamic interaction as between "objects" and "dimensions" at this level of reality).
Of course the radial level involves the additional realisation that all "objects" and "dimensions" - which are "real" and "imaginary" with respect to each other have both finite and transfinite aspects. In other words the uniqueness of all quantitative "objects" and qualitative "dimensions" (reflected psychologically through perceptions and concepts respectively) are continually mediated through a radiant spiritual light.
Though of necessity we must make polar distinctions in communication, these have only a relative limited use with no essential value. Thus reality which appears polarised in terms of objective-subjective aspects (matter and mind) and quantitative-qualitative aspects (wholes and parts) ultimately comprises a seamless invisible web (without distinctions).
Q. Can you now deal with the important area of mathematical proof. Does this have any meaning in the context of holistic mathematics?
PC Yes, this is a very important and interesting area. I have already dealt with this in part in Holistic Mathematics 1 where I strongly questioned the accepted meaning of mathematical proof. Standard mathematics expresses "proofs" formally in explicit rational terms. However intuition is always essential to literally "see" what is implied by the logical connections established through reason. Thus in dynamic terms two elements are involved in standard analytical "proofs"; a rational explicit element (which is solely recognised in formal presentation) and an intuitive implicit element (which enables one to "see" what is implied by starting axioms). All "proofs" in fact are thus subject to the uncertainty principle and in dynamic terms strictly relative. Indeed they represent no more than an especially important form of social consensus. There are thus two types of understanding.
The rational type responds readily to logical linear connections. However when carried to extremes this leads to a lack of creative insight and a considerable narrowing of vision.
The intuitive type responds readily to holistic circular connections often enabling instant insight (without the need for formal rational connections) into what is implied by initial axioms. However once again when carried to extremes this can lead to undue vagueness and a lack of sufficient rigour.
However a significant imbalance remains. Whereas most important mathematical "proofs" initially depend essentially on creative intuition, formally they are presented in merely rational terms.
Surprisingly Holistic Mathematics has its own type of distinctive "proof". However the verification of such "proof" - in contrast to Standard Mathematics - is more personal and intuitive.
We can identify three distinctive levels of proof corresponding to each of the three levels of Holistic Mathematics.
Holistic Mathematics 1 is concerned with establishing complementary connections of the horizontal type (i.e. positive external and negative internal aspects).
Such complementary connections are obtained directly by intuition. Now the role of reason here is in providing an appropriate indirect rational translation of such relationships. Holistic Mathematics itself has been designed to provide such reduced translations.
Thus for example when I say that numbers are strictly dynamic entities with positive and negative polarities, this requires - in direct terms - the appropriate intuition to verify this insight, and - indirectly - the appropriate reason to appreciate the mathematical translation employed.
Whereas standard mathematical "proofs" require linear understanding,
proofs relating to Holistic Mathematics 1 require the understanding of the "higher" circular level. Thus there is a great danger of reductionism in relation to Holistic Mathematics. If people are unable to understand in terms of the "higher" levels, then inevitably they will try and reduce it to the "lower" linear level and thereby misrepresent its findings.
Thus "proof" in relation to Holistic Mathematics 1 is essentially circular whereby one successfully establishes horizontal complementarity (at the required level of investigation), indirectly expressed in appropriate reduced rational fashion. In the precise qualitative language of Holistic Mathematics this entails (algebraic) irrational - rather than rational - "proof".
Holistic Mathematics 2 involves vertical as well as horizontal complementarity.
This implies a far closer relationship being maintained as between rational and intuitive modes of expression.
A good example of this type of "proof" is my assertion that the entire Spectrum of Consciousness constitutes the qualitative number system.
In this approach there is both horizontal and vertical complementarity. The qualitative definition of each number type referring to a particular level or stage of development requires horizontal complementarity in the manner I have described. Vertical complementarity is then provided through establishing direct correspondence as between
pre-linear and trans-linear stages of development.
Acceptance of my number spectrum then depends on the validity of the complementary number relationships established (both horizontal and vertical).
However once again this requires the appropriate understanding of the point level.
Holistic Mathematics 3 is especially interesting as it involves the three types of "proof" thus established.
One of the practical implications of this that properly understood a comprehensive "proof" requires all three types.
Firstly we need to provide a quantitative analytical proof; then we need a qualitative holistic proof; finally we need to establish a satisfactory correspondence as between both quantitative and qualitative proofs.
One example of where I believe this has been done is in relation to the irrational nature of the square root of 2.
A quantitative analytical "proof" has been long established that the square root of 2 is irrational.
A qualitative holistic "proof" (i.e. as to why the square root of 2 is irrational) was provided in my posting on Holistic Mathematics 1).
Finally I was able to establish a direct dynamic correspondence in this posting as between both quantitative and qualitative behaviour.
The implications of this comprehensive approach are extremely interesting.
No proposition is fully proven until quantitative and qualitative "proofs" exist with a satisfactory correspondence between both demonstrated. For example a quantitative proof for Fermat’s Last Theorem now exists. However a comprehensive solution requires that a holistic qualitative proof be also found and that a satisfactory correspondence as between both quantitative and qualitative "proofs" established. My own view is that the qualitative "proof" of Fermat’s Last Theorem can be expressed very simply. Direct correspondence as between quantitative and qualitative aspects would then require the establishment of an equally simple quantitative "proof".
It could well be that since Fermat some essential key insight has been missing preventing such a simple approach. Thus I believe that it is entirely possible that Fermat did prove his theorem in an ingeniously simple (quantitative) fashion.
One highly creative approach to many of the remaining unsolved problems in quantitative mathematics (many of which can be simply stated) would be to approach them from a holistic perspective attempting to "solve" them in qualitative terms. If such a qualitative proof for a proposition could be successfully found, then this would suggest that a quantitative "proof" also exists. A proposition could therefore be deemed provisionally true pending the establishment of a satisfactory rational "proof".
Indeed there is a very important conjecture in relation to prime numbers for which I have established a qualitative "proof". This would therefore suggest that - though still lacking a rational "proof" - that the conjecture is in fact true. (As this has played an important role in much of my thinking I may return to this later).
To sum up, mathematics has an analytical quantitative aspect (which is rational) and a holistic qualitative aspect (which is intuitive). A creative approach requires combining both aspects.
Q. Can you now sum up your main findings in relation to Holistic Mathematics 3.
PC Once again each of the major levels has a distinctive mathematical approach associated with it.
The linear level is based on a rational logic and involves the specialisation of the analytical (quantitative) aspect. (Standard Mathematics).
The circular level is based on an (algebraic) irrational logic and involves the specialisation of the synthetic (qualitative) aspect (Holistic mathematics 1).
The point level is based on a transcendental logic and involves the specialisation of the relationship as between analytical and synthetic approaches. (Holistic Mathematics 2).
The radial level is based on a "complex" logic and involves the progressive integration of all three approaches.
Thus we have a "real" component which represents the rational analytical aspect.
We have an "imaginary" component which represents the intuitive holistic aspect.
Finally these are combined as a "complex" component representing the integration of analytical and holistic aspects.
A truly comprehensive paradigm must necessarily be "complex" (in a holistic mathematical sense) recognising both quantitative and qualitative approaches to understanding.
Holistic Mathematics 1 involves horizontal complementarity; Holistic Mathematics 2 involves in addition vertical complementarity; Holistic Mathematics 3 involves diagonal complementarity. This level relates directly to the radical equality of the transfinite void; in reduced terms it involves a "complex" relative hierarchy (involving "real" quantitative and "imaginary" qualitative aspects).
Ultimately both the analytical and holistic aspects of mathematics are fully integrated. This is what I term Comprehensive Mathematics. My hope is that one day that this will be understood as simply mathematics. | 9,037 | 48,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-34 | latest | en | 0.912629 |
https://brainmass.com/statistics/hypothesis-testing/filet-mignon-and-weight-hypothesis-testing-96784 | 1,601,390,044,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00695.warc.gz | 292,447,213 | 10,828 | Explore BrainMass
Filet Mignon and Weight Hypothesis Testing
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
I love filet mignon. The full cut is supposed to weigh 16 ounces when fully cooked. My restaurant serves 200 servings of filet mignon per day. The restaurant claim is they serve filets that are 16 ounces with a standard deviation of .4 ounces due to shrinking while cooking. I ate filet mignon for 16 days straight. The average weight of all the servings was 15.4 ounces. Using a = .10, can I make a public statement and back it up with hypothesis testing that the average filet is actually less that 16 ounces? | 160 | 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} | 2.796875 | 3 | CC-MAIN-2020-40 | latest | en | 0.94745 |
https://www.jiskha.com/questions/303122/What-are-the-odds-in-favor-of-getting-at-most-two-heads-in-three-successive-flips | 1,571,392,132,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00042.warc.gz | 942,007,762 | 4,778 | # Math
What are the odds in favor of getting at most two heads in three successive flips of a coin?
This problem has to be done in ratio not fraction.
Can anyone help?
1. 👍 0
2. 👎 0
3. 👁 44
1. The possibilities from three coin tosses:
HHH
HHT
HTH
THH
THT
TTH
HTT
TTT
"At most two heads" fits every alternative except HHH.
I hope that helps.
1. 👍 0
2. 👎 0
posted by PsyDAG
2. So would it be 7:1
1. 👍 0
2. 👎 0
3. Looks good!
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2. 👎 0
posted by PsyDAG
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Problem 1702. Maximum value in a matrix
Solution 276822
Submitted on 9 Jul 2013 by Richard Zapor
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 2 3; 4 5 6; 7 8 9]; y_correct = 9; assert(isequal(your_fcn_name(x),y_correct))
ans = 9
2 Pass
%% x = -10:0; y_correct = 0; assert(isequal(your_fcn_name(x),y_correct))
ans = 0
3 Pass
%% x = 17; y_correct = 17; assert(isequal(your_fcn_name(x),y_correct))
ans = 17
4 Pass
%% x = magic(6); y_correct = 36; assert(isequal(your_fcn_name(x),y_correct))
ans = 36
5 Pass
%% x = [5 23 6 2 9 0 -1]'; y_correct = 23; assert(isequal(your_fcn_name(x),y_correct))
ans = 23 | 274 | 753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-16 | latest | en | 0.565604 |
http://thescienceforum.org/topic2330.html?sid=a054e3629ca96beda7146843d053cb1a | 1,508,647,202,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061356-00014.warc.gz | 343,786,856 | 10,779 | It is currently Sun Oct 22, 2017 4:40 am
22 posts • Page 1 of 1
Author Message
M_Gabriela
Post subject: chemistry problem | Posted: Thu Mar 02, 2017 4:35 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
Hello, I have this problemH30+ + 2 NO2- + 2 I- --> I2 + 2 NO + 6 H2OThe problem asks for the order of reaction with respect to I2, the product. How do I do that? Then it asks the reaction rate if formation rate of NO is xx M/s...I have more data... For example for the first question I need to find the concentration of H30+ and I have the reaction rate, the k, and the concentration of NO2 and I-. And the expression of the reaction rate so I only have to clear H30+.But then it asks me to find a new reaction rate if NO formation rate is xx M/s... I'm lost....Thankss in advance
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 5:36 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
Could you post the whole question with the data (maybe as an image)? It might be clearer to me then how you go about solving it.Asking for a reaction order WRT to a product seems a little odd...Also your equation as written doesn't look correct, the charges don't balance... _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 5:57 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
It's in spanish .Yes, the equation is wrong.4 H30+ + 2 NO2- + 2 I- --> I2 + 2 NO + 6 H2OThe reaction rate isv= k x [NO2 -] x [I -] x [H3O+]^21) Value of H3O+ if... and they give me all the data to clear H30+ from the equation above.2) order of reaction with respect to I2.3)Value of reaction rate when formation rate of NO is xxxxx M/s.
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:05 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
OK, I guess you've done part one (the easy bit :p)...As for part two I2 does not appear in the rate equation and has no effect on the rate so the order with respect to it must be zero (although I still think it strange asking about a product in this way, I've never seen that before unless it's an equilibrium).Part 3, "v" in your equation is equal to $\displaystyle \frac{1}{2} \frac{d[NO_2]}{dt}$ and this is the reaction rate, so the rate of reaction is half of your xxxx M/s _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:25 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
PhDemon wrote:OK, I guess you've done part one (the easy bit :p)...As for part two I2 does not appear in the rate equation and has no effect on the rate so the order with respect to it must be zero (although I still think it strange asking about a product in this way, I've never seen that before unless it's an equilibrium).Part 3, "v" in your equation is equal to $\displaystyle \frac{1}{2} \frac{d[NO_2]}{dt}$ and this is the reaction rate, so the rate of reaction is half of your xxxx M/sThanksss. A few questionsAbout part 3, so you're saying that the difference between reaction rate and formation rate is only the 1/2???Cause I thought that formation rate was $\displaystyle \frac{1}{2} \frac{d[NO_2]}{dt}$. So I would have to multiply by 2 if I want to clear the $\displaystyle {d[NO_2]}/{dt}$I feel like I'm writing something really stupid....About part 2. I only read about order of reaction respect to reactants, not products, that's why I was lost. again thank you!
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:29 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
The rate of reaction (v in your equation) can be expressed in terms of any of the reactants or products as: $\displaystyle \frac{1}{n} \frac{d[X]}{dt}$ where n is the stoichiometric coefficient in front of X... This should give the same value for the rate regardless of which reactant or product you use. _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan Last edited by PhDemon on Thu Mar 02, 2017 6:34 pm, edited 3 times in total.
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:31 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
And what is this:v= k x [NO2 -] x [I -] x [H3O+]^2???
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:33 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
That is the rate law. It tells you how the rate depends on the concentration of each reactant. _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:44 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
Ok... I don't understand why I have to divide by 2. I understand where the 1/2 comes from but I don't understand why, to calculate the reaction rate, I have to do that.$\displaystyle \frac{1}{2} \frac{d[NO_2]}{dt}$ = vif v= xxxx M/s, then$\displaystyle \frac{1}{2} \frac{d[NO_2]}{dt}$ = xxxx M/sso why divide it?
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 6:58 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
You are given the rate of formation of NO, not the "rate of the reaction". The reaction rate is defined in my above post, you need to use that 1/n bit. If you didn't you would get a different answer for the rate depending on which reactant ir product you chose to measure the concentration change of...Edit: the NO2 in my equations above should be NO, does that clear up your confusion? (It's been a long day! ) _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan Last edited by PhDemon on Thu Mar 02, 2017 7:06 pm, edited 2 times in total.
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 7:05 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
I'm obviously not explaining it very well (sorry about that). Does the "Formal definition" section of this link do a better job?https://en.m.wikipedia.org/wiki/Reaction_rate _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 7:38 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
Ok. I think I understood. Thank you ph!!!
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 7:42 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
No problem, despite using it a lot myself (it became second nature) I last taught this in 2003 to a first year undergrad class (probably why my explanation wasn't great ) _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 7:45 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
The data given is the rate without dividing it by 2. Thats what you are saying. It's confusing...
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 7:48 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
You seem to be confusing rate of formation of NO with the rate of reaction, they are not the same (they would be the same if there was not a 2 in front of NO in your equation). Read how rate of reaction is defined in the link I gave, I think it mentions where this 1/n term comes from. _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 8:13 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
Yesssss. I finally understoodddd
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 8:28 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
Good! Another happy customer _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 9:54 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
I had to explain this to a student Supposedly i should know this,hehehehheheh
PhDemon
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 10:24 pm
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
If you have a chemistry degree, yes you should! _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Thu Mar 02, 2017 11:13 pm
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
I don't. I have a degree in biology. But it has been a while since i practiced it....
PhDemon
Post subject: Re: chemistry problem | Posted: Fri Mar 03, 2017 12:42 am
Joined: Wed Jun 26, 2013 10:44 am
Posts: 515
Location: Newcastle-upon-Tyne
That explains a lot! (I'm joking... at work the chemists and physicists like to tease the biologists We get together to take the piss out of the sociology and psychology departments though ) _________________"The big trouble with dumb bastards is that they are too dumb to believe there is such a thing as being smart"- Kurt Vonnegut, The Sirens of Titan
M_Gabriela
Post subject: Re: chemistry problem | Posted: Fri Mar 03, 2017 9:21 am
Joined: Tue May 03, 2016 2:25 pm
Posts: 217
Location: Bs As, Argentina
We do make fun of psychologists....
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# Central Meridian
Central Meridian
central meridian - [coordinate systems] The line of longitude that defines the center and often the x-origin of a projected coordinate system.
the longitude of the horizontal center of a coordinate system (this longitude value is often the longitude origin of the coordinate system); in the case of the transverse Mercator projection, ...
Central Meridian
The meridian that passes through the poles and origin of an ellipsoid/sphere representing the earth and is used in calculations of a specific projection.
Central Processing Unit (CPU) ...
A is a meridian that passes through the center of a projection. The is often a straight line that is an axis of symmetry of the projection.
...
Central meridian and lines perpendicular to it
Central Clindrical
Classification: ...
- A line running North and South, at the center of a graticule, along which all points have the same longitude. See also Meridian.
...
central meridian
The line of longitude that defines the center and often the x-origin of a projected coordinate system.
the is straight, half as long as the Equator, and a standard line in odd-numbered projections
poles are flat, half as long as the Equator
even-numbered projections are equal-area ...
the central meridian is the meridian where the cylinder touches the sphere
theoretically, the central meridian is the line of zero distortion ...
Distances are only correct along parallels and . Shapes become more distorted away from the and close to the poles. Slaking See wetting and drying.
Select the Projection Lambert_Azimuthal_Equal_Area. Alter the Central Meridian and Latitude of Origin.
Map makers have technical terms to describe the line of latitude or longitude where this imaginary 'piece of paper' touches the Earth. These are:
for a line of latitude - standard parallel
for a line of longitude - ...
Other projection mathods are based on more complicated flattenable projection surfaces, and instead of points of tangency, spacial cases of these projections can be made by adjusting their Standard Paraslells or Central Meridians ...
Just as the normal aspect Mercator projection has low distortion of scale near the equator, so does a transverse Mercator projection have low scale distortion near its .
The origin of each UTM zone is the intersection of its central meridian and the equator, and the parameters are applied to this origin to make it convenient to work with making all x and y values positive, or reducing their range.
The UTM easting coordinate (the X coordinate) for a feature is the distance in meters east or west from the of the UTM zone.
and is the longitude from the central meridian, and is the latitude.
The Mollweide is a pseudocylindrical projection in which the equator is represented as a straight horizontal line perpendicular to a central meridian one-half its length.
There is a false origin defined at 400,000, -100,000 such that the is the Zero easting of the National Grid - which is also aligned to the 2 degree West meridian of Longitude. Additionally, the scale along the is 0.
To eliminate the necessity for using negative numbers to describe a location, the east-west origin is placed 500,000 meters west of the central meridian. This is referred to as the zone's ‘false origin'.
Within each zone the is given an Easting value of 500,000 metres. The equator is designated as having a Northing value of 0 for northern hemisphere coordinates.
Put another way: UTM projection is used to define horizontal positions world-wide by dividing the surface of the Earth into 6 degree zones, each mapped by the Transverse Mercator projection with a central meridian in the center of the zone.
Enter Central Parallel: 0 if you want the Equator as the central parallel
Enter : 0 if you want the Greenwich meridian as
Enter Scale Factor at the ...
Transverse Mercator projections result from projecting the sphere onto a cylinder tangent to a central meridian. Transverse Mercator maps are often used to portray areas with larger north-south than east-west extent.
UTM zones have an origin on the equator at the point where the of the zone intersects. Coordinates are measured in meters from the false origin followed by the zone number and the hemisphere.
The projection is true to scale along the central meridian and along each parallel. It is neither conformal nor equal-area, and it is only free of distortion along the central meridian.
UTM easting coordinates are referenced to the center line of the zone known as the . The is assigned an easting value of 500,000 meters East.
The easting coordinates are measured from an artificial reference line drawn parallel and 500,000 meters to the west of the zone's central meridian.
Using Spatial Analyst this tile was clipped, warped with an order-three polynomial to approximate an Equidistant Conic projection (Clarke 1866 spheroid, 71 degrees west, reference latitude 19 degrees north, ...
A coordinate system is usually defined by a map projection, a spheroid of reference, a datum, one or more standard parallels, a central meridian, and possible shifts in the x- and y-directions to locate x,y positions of point, line, ...
[ESRI software] In Survey Analyst for field measurements, any meridian that is parallel to the , used when computing points in planar rectangular coordinate systems of limited extent.
A class of map projections in which the parallels are represented by a system of non-concentric circular arcs with centers lying on the straight line representing the central meridian (Lee 1944).
Each zone extends 3 degrees east and west from its and are numbered consecutively west to east from the 180-degree meridian. Transverse Mercator projections may then be applied to each zone.
There are two geoprocessing tools available from the Data Driven Pages toolset in the Cartography toolbox to help you populate a spatial reference field: Calculate Central Meridian and Parallels and Calculate UTM Zone.
Eastings are in meters with respect to a drawn through the center of each grid zone (and given an arbitrary easting of 500,000 meters). In the northern hemisphere, northings are read in meters from the equator (0 meters).
A transverse cylindric projection uses a meridian of longitude as its central meridian. travelling salesman problem p. | 1,335 | 6,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-35 | latest | en | 0.861086 |
https://www.cfd-online.com/Forums/cfx/24415-symmetry-problem-pie-section-print.html | 1,513,486,096,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592972.60/warc/CC-MAIN-20171217035328-20171217061328-00252.warc.gz | 708,016,117 | 4,519 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- CFX (https://www.cfd-online.com/Forums/cfx/)
- - symmetry problem with a pie section (https://www.cfd-online.com/Forums/cfx/24415-symmetry-problem-pie-section.html)
Jonathan Lemay August 16, 2007 13:58
symmetry problem with a pie section
Dear CFD user
I produced a 15degree pie to model flow inside a 170mm diameter pie. I did this meshing with ICEM. I have a problem with the BC (boundary condition) when I define my CFX case.
To avoid simulating the whole 360 degree of the pipe I modeled only a small 15 degrees angle pie. I defined each of the vertical plane of my cylinder as symmetry plane when I defined the BC. Unfortunalty when I try to run this case the CFX solver give me an error that my two plane are not parellel and the solver cannot solve the problem...
Is there something I defined wrong in the BC? Should I change something to my ICEM mesh?
Robin August 16, 2007 14:08
Re: symmetry problem with a pie section
Hi Jonathan,
Check your mesh on these faces. ICEM CFD is notorious for producing errors like this because it assigns faces to regions (parts) by finding the surface it projects closest to. You will probably find that at the edges you have a few faces which are non-planar.
If it is a simple model, consider meshing it in the ANSYS Meshing app instead. You can define a sweep method on the body and assign the two sides as the start and end of the sweep. Also, if it is perfectly axisymmetric you can reduce you wedge to ~3 degrees and specify only one element in the sweep direction.
Regards, Robin
Jonathan Lemay August 16, 2007 16:27
Re: symmetry problem with a pie section
Hi Robin,
Thanks a lot for your help. I chose ICEM CFD because I wanted to mesh with hexa element (I mean full cubic types, with 6 faces volumes). So is there a way to mesh with CFX MESH (workbench) and get this kind of element? I know that CFX mesh is much more convenient for easy application.
By the way I am simulating the mixing cold flow of a bluff body burner.
Regards, Jonathan
Robin August 16, 2007 16:37
Re: symmetry problem with a pie section
You cannot generate a pure hex mesh with the CFX-Mesh method, but you can with the sweep method. However, the topology is much more limited than with Hexa. That said, I wouldn't worry too much about getting a pure hex mesh. A revolved mesh with inflation will be very accurate too (it will result in hexes where you have inflation and prisms through the rest). The main thing is to make sure you have sufficient mesh where you need it.
I hadn't realized you were using Hexa. In that case I would stick with ICEM CFD. The problem in that case is most likely that you have not properly projected your edges to all of the curves around the plane, projected your vertices to the points at the corners and/or projected the faces to surfaces. Small misalignments in these projections will lead to a non-planar face.
Regards, Robin
Anantha August 17, 2007 01:55
Re: symmetry problem with a pie section
Hello,
I am bit skeptical here. Did you apply "symmetry" BC or "rotational periodicity"... I hope the later one best suits your problem...
Regards Anantha
Jonathan Lemay August 17, 2007 11:36
Re: symmetry problem with a pie section
Hi Anantha
I simply applied a Symmetry boundary condition as it is shown in the CFX tutorial of Reacting flows in a mixing tube. Would you inform me about this "rotational periodicity" BC? Never heard about that in CFX.
Regards. Jonathan
Jonathan Lemay August 17, 2007 11:38
Re: symmetry problem with a pie section
Hi Robin
I would ask your help for something is ICEM. Thanks to your advice I associated the faces of my blocks to the different surfaces. I either associate point with vertex. Do you know if it possible to associate more then one vertex to a point?
Another useful information I would need. How do I manage my surfaces in ICEM so they could be recognize by CFX when it is coming the time to define the region of my BC?
I am getting confuse with two other things. Sometimes when I choose my region for BC in CFX, their is two available regions similar to each other. One has the mesh of the face only and the other region has the mesh plus the contour. What is the difference for those regions?
I am a beginner in ICEM so thanks a lot for your precious help.
Robin August 17, 2007 13:16
Re: symmetry problem with a pie section
Hi Jonathan,
I'll do my best to answer your questions, but I suggest you go through some of the tutorials as well.
I would ask your help for something is ICEM. Thanks to your advice I associated the faces of my blocks to the different surfaces. I either associate point with vertex. Do you know if it possible to associate more then one vertex to a point? Yes. You can assign multiple vertices to a point, but be careful that you don't collapse a block by doing so.
Another useful information I would need. How do I manage my surfaces in ICEM so they could be recognize by CFX when it is coming the time to define the region of my BC? Before you generate your mesh, group surfaces into "Parts" by right clicking on the Part object and creating new parts. These parts will then appear in Pre as separate regions.
I am getting confuse with two other things. Sometimes when I choose my region for BC in CFX, their is two available regions similar to each other. One has the mesh of the face only and the other region has the mesh plus the contour. What is the difference for those regions? I'm not sure what the answer is. I haven't seen such a problem so this may be a result of what you have done specifically in ICEM. What do you see in terms of regions under your mesh object in Pre?
Regards, Robin
Jonathan Lemay August 17, 2007 19:13
Re: symmetry problem with a pie section
Hi Robin
Thanks for the tips. I did several tutorial on ICEM but it seems I missed something. Anyway your help is really appreciated
About producing part in ICEM a already did that including surfaces in my parts. The thing is when I open CFX-PRE, all those part defined previously in ICEM are not there, some ones are missing. So should I change something when exporting? I thought ICEM was either exporting all the primitive 2D surfaces to CFX-Pre. Unfortunatly it is not my case. I currently use as solver ANSYS CFX an it produce a .cfx5 mesh file.
About those region in CFX that you have not understand, I will work on that to resolve that issue.
Hehehe Ok Thank you professor Regards
Jonathan Lemay August 21, 2007 09:52
Thanks you
Hi Robin
Thanks you so much for your help. It has been great help!! My meshing is now running in CFX..
Felix August 30, 2007 11:12
Re: Thanks you
Hi Jonathan,
I think I know why you don't have all your faces in PRE. In ICEM, start with creating/importing your 2D geometry. Then, as Robin suggested, name every part you want to use in PRE (inlet line, outlet line, symmetry plane, etc..). Next, create your 2D Planar blocks.
The following step was probably missing and is very important. **Assign every block's face, edges and vertex to their corresponding geometry**. This will allow the mesh elements to take the name of the corresponding part. If you then convert the Pre-Mesh to an Unstructured mesh, you should see Line elements in the Inlet part, for example.
Finally, when you extrude the mesh, leave the "new side name" as "inherited", name the "new top name" symmetry2 and give a cool name to the volume part.
Save your mesh, export it and everything should be fine in PRE !
Regards, Felix
All times are GMT -4. The time now is 00:48. | 1,822 | 7,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.902054 |
https://electronics.stackexchange.com/questions/130580/what-is-a-safe-max-discharge-rate-for-a-12v-lead-acid-battery/430463 | 1,561,015,781,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00206.warc.gz | 419,489,425 | 37,026 | # What is a safe max. discharge rate for a 12V lead acid battery?
I've got a 12V 2.4Ah lead acid battery which I plan to connect a water pump to. I've looked at various pumps, but the one I'm most interested in draws 2.2A.
I'm not so interested in how long the pump can run, as it only will need to run for about 5 - 10 minutes/day. So, I'm assuming the battery is plenty for that. The battery will be charged via solar cell panels.
However, I'm more concerned about the discharge rate. I've read that lead acid battery not should be discharged too quickly, as this might result in overheating the battery (and cause damage to it).
How do I figure out what a safe maximum discharge rate is for a 12V lead acid battery?
• I agree with explanation about the charge/discharge rates. Bur I think the problem is that the battery is too small to suply that pump. Remember that start current may be 3 times higher than the nominal. – Bruno Y Oct 21 '15 at 17:38
• I have a 12 volt 9 amp hour battery pack and I use it mostly for charging my phones and a light and a radio but I have used it to run my 2.7 amp water pump from time to time. I noticed it doesn't go down but maybe halfway. After a 15 min shower the battery bank go down maybe from 13.6v to 12.8v I have been living on batteries for the past 5 years. In most times you want to get something that's at least double or triple the amount of amps you going to be pulling from it. As per personal experience I always try for at least double the amps needed to the amp hours. Say I need it 2 amps I would lo – user102136 Mar 2 '16 at 0:41
An easy rule-of-thumb for determining the slow/intermediate/fast rates for charging/discharging a rechargeable chemical battery, mostly independent of the actual manufacturing technology: lead acid, NiCd, NiMH, Li...
• We will call C (unitless) to the numerical value of the capacity of our battery, measured in Ah (Ampere-hour).
• In your question, the capacity of the battery is 2.4 Ah, hence, C=2.4 (unitless).
• The vast majority of the batteries in the market will safely charge/discharge at a rate of less than 1C Amperes.
• In an ideal world (without losses), this would translate into a 1 hour charge/discharge process. In practice, the charging/discharging operation may require up to twice/half the time.
• Without further information (datasheet), I would not charge/discharge any battery at a rate higher than 1C, for safety and endurance reasons.
• In your question, less than 2.4 A would be a nice charge/discharge rate, as the manufacturer datasheet confirms.
• By applying a charge/discharge rate much less than 1C, you usually extend considerably the life of a chemical battery.
• Rates << 1C are commonly known as "SLOW" rates: 0.5C, 0.2C, 0.1C...
• Charge/discharge rates higher than 1C are best avoided unless working with a properly known battery.
• Rates >> 1C rates are commonly known as "FAST" rates: 2C, 3C...
• In the past, batteries designed for rates >1C were usually marketed as "high current" batteries, because not all batteries were capable of sustaining such rates safely or without compromising its endurance.
• Nowadays, most batteries can safely be used at rates >1C, up to the rating specified by the manufacturer. However, a reduction in the battery life is to be expected.
• Forcing a battery to rates >5-10C involves serious risks.
Disclaimer: this is a rule-of-thumb, useful as an starting point when the datasheet is not available or when dealing with a no-brand/unknown battery.
• Thanks! Not that I doubt your knowledge (I've accepted your answer), but how do you know this? Is this a well known rule-of-thumb which you can pick up from any "teaching book", or is it something you've learned through experience? – sbrattla Sep 28 '14 at 18:44
• This is just a rule-of-thumb so please take it with a grain of salt! However it proved very useful to me through the years. It is based on my experience and knowledge of the market, though I have observed some batteries and professional chargers manufacturers using this "C" terminology and setting similar limits when defining a slow/standard/fast rate. – jose.angel.jimenez Sep 29 '14 at 7:14
• Most deep cycle lead-acid batteries charge at 0.2 to 0.3 C . – Nick Alexeev Mar 2 '16 at 5:53
• This rule of thumb is problematic as a 12V lead-acid battery is actually 6x2V cells in series. If a 2V cell of a particular size was able to be charged at, say 0.5A, six of them in series (six times the capacity) should also be charged at 0.5A. Voltage and power will need to be higher but the current should be identical. Rules of thumb can be handy but it pays to understand why they apply and when they don't. – Richard Thomas Jun 26 '16 at 16:02
• @richard-thomas You are right. As any rule of thumb, you are entirely responsible for knowing the underlying physics involved. However, the much less than 1C rule for charging 12V lead-acid batteries is perfectly adequate and according to the recommendation of most manufacturers. Should to want to stay on the safe side, you can limit the charge rate to 0.1C or 0.2C. – jose.angel.jimenez Jun 27 '16 at 16:50
Ideally the manufacturer supplies the discharge rates on the battery datasheet.
A quick point: You mention you have a 12 V 2.4 A SLA (sealed lead acid) battery, but batteries are rated in amp-hours not amperes. Therefore I suspect you have a 12 V 2.4 Ah battery.
Now that we have that out of the way, a 12 V 2.5 Ah SLA battery from Power Sonic, as an example (a company that has datasheets for their batteries) shows several discharge rates that may be of interest:
• Nominal Capacities:
• 125 mA discharge rate = 20 hours (2.5 Ah)
• 220 mA discharge rate = 10 hours (2.2 Ah)
• 400 mA discharge rate = 5 hours (2 Ah)
• 1.5 A discharge rate = 1 hour (1.5 Ah)
• 4.5 A discharge rate = 15 minutes (1.13 Ah)
• Max Discharge Current (7 Min.) = 7.5 A
• Max Short-Duration Discharge Current (10 Sec.) = 25.0 A
This means you should expect, at a discharge rate of 2.2 A, that the battery would have a nominal capacity (down to 9 V) between 1.13 Ah and 1.5 Ah, giving you between 15 minutes and 1 hour runtime.
• Your'e absolutely right about the Ah vs. A. My mistake - thanks for correcting! And also thank's for the examples above! – sbrattla Sep 24 '14 at 5:35
• He said that the pump he is interested in draws 2.2A, not the battery! – Leon Heller Sep 24 '14 at 8:02
• @Leon Oops you're right. I misread. I'll amend that. – JYelton Sep 24 '14 at 8:07
Jose's answer states that the discharge rate isn't related to chemistry. However, this is not correct. It can vary up to a factor of 1000 depending on chemistry.
Different batteries chemistries have different properties.
Beyond the chemistry, this is also related to the battery design itself. Size of the electrodes, the thickness of electrode coatings, electrolyte so it can also vastly vary upon this.
Some are designed for a lower self discharge rate, some for higher energy density or higher instant power output.
Larger electrode with thinner coating will have a higher discharge rate, while the opposite will lead to higher energy density.
The best is to check at the manufacturer datasheet if it is available.
Here is also a table with common values.
Concerning specifically on lead-acid, there are also several types, but two are most common, the car starter battery and the stationary battery.
Because of its construction, a starter battery is only suitable for short loads with high current, which most commonly take place when starting an engine of a car, truck … The main characteristic of a starter battery is that they have big, thin, flat plates. Starter batteries are not suitable for cyclic use (continuous charging & discharging) A starter battery is relatively cheap. Source
With your pump, make sure to use a stationary battery, since you are below 1C do that is totally fine.
## protected by Dave Tweed♦Apr 9 '16 at 13:40
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2,046 | 8,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-26 | latest | en | 0.95268 |
https://i.fluther.com/223935/can-you-calculate-what-percentage-of-the-remaining-ballots-trump-would/ | 1,638,377,992,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00466.warc.gz | 383,754,470 | 4,390 | ## Social Question
#### Can you calculate what percentage of the remaining ballots Trump would need to win in Arizona?
Asked by LostInParadise (29141) November 11th, 2020
2 responses
According to this site , as of yesterday Biden has a lead of about 12,800 and there are about 61,500 ballots to be counted. What percentage of those votes would Trump need to win? There is a simple way of solving this without any algebra. Do you see it?
Observing members: 0
Composing members: 0
## Answers
Here is the answer. Imagine the 61,500 ballots are in two stacks, one for Trump and one for Biden. We will start by imagining that they are the same size, with both getting 50% of the vote. For the total number of votes to remain the same, any additional Trump votes must be matched by on equal decrease in Biden’s votes. For Trump to get 12,800 more votes than Biden, Trump would need an additional 6,400 votes beyond 50% and Biden would need to compensate by having 6,400 fewer votes than 50%. Trump’s share of the vote would then be 50% + 6400/61500 = 60.4%, which is very unlikely.
LostInParadise (29141)
he needs 37,151 votes out of 61,500 to take the lead. That would be over 60% of the remaining ballots. Not impossible but highly unlikely, given that ballots left to be counted would lean towards Democratic ballots.
zenvelo (36883)
or
to answer.
` | 343 | 1,358 | {"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-2021-49 | latest | en | 0.964384 |
https://blogs.nvcc.edu/elanthier/psy-213/chapter-02/ | 1,726,409,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00512.warc.gz | 118,540,530 | 10,816 | # Chapter 02
This is a chapter that requires you to perform operations with your data. You need to understand the concepts, but you also need to be able to perform the operations.
The concepts to understand:
• raw score
• outlier
• N
• simple frequency distribution
• percentage distribution
• relative frequency and proportion of area under the normal curve
• cumulative frequency
• percentile
• normal distribution—what is it? can you graph one? what are the features of a normal distribution? Can you identify the raw score (the location on the X axis) that would be the 50th percentile? Where are the scores that are the 20th and 80th percentiles?
• skew—define and draw graphs of negative skew and positive skew
• be able to recognize, give, and/or interpret examples of skewed distributions
• bimodal distribution
• histograms
• floor effect and ceiling effect
Now, perform some operations to practice. Use the following data set to create a simple frequency distribution, a relative frequency and percentage distribution, cumulative frequency distribution, and percentiles. (Put all of these in the same table.)
Then draw a histogram of the simple frequency distribution. | 246 | 1,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.900668 |
https://planetmath.org/AnotherDefinitionOfCofinality | 1,618,765,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038507477.62/warc/CC-MAIN-20210418163541-20210418193541-00488.warc.gz | 568,649,788 | 3,267 | # another definition of cofinality
Let $\kappa$ be a limit ordinal (e.g. a cardinal). The cofinality of $\kappa$ $\operatorname{cf}(\kappa)$ could also be defined as:
$\operatorname{cf}(\kappa)=\inf\{|U|:U\subseteq\kappa\text{ such that }\sup\;U=\kappa\}$
($\sup\;U$ is calculated using the natural order of the ordinals). The cofinality of a cardinal is always a regular cardinal and hence $\operatorname{cf}(\kappa)=\operatorname{cf}(\operatorname{cf}(\kappa))$.
This definition is equivalent to the parent definition.
Title another definition of cofinality AnotherDefinitionOfCofinality 2013-03-22 13:52:59 2013-03-22 13:52:59 x_bas (2940) x_bas (2940) 9 x_bas (2940) Definition msc 03E04 | 222 | 697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 6, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | longest | en | 0.606865 |
https://fr.mathworks.com/matlabcentral/cody/problems/12-fibonacci-sequence/solutions/611200 | 1,591,204,889,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00348.warc.gz | 348,082,798 | 15,623 | Cody
# Problem 12. Fibonacci sequence
Solution 611200
Submitted on 3 Apr 2015 by Przyczajony Tygrys, Ukryty Smok
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
%% n = 1; f = 1; assert(isequal(fib(n),f))
2 Pass
%% n = 6; f = 8; assert(isequal(fib(n),f))
3 Pass
%% n = 10; f = 55; assert(isequal(fib(n),f))
4 Pass
%% n = 20; f = 6765; assert(isequal(fib(n),f)) | 176 | 491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-24 | latest | en | 0.651415 |
https://www.maplesoft.com/support/help/Maple/view.aspx?path=DEtools%2Fconstcoeffsols | 1,652,895,395,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00144.warc.gz | 998,884,667 | 31,761 | constcoeffsols - Maple Help
DEtools
constcoeffsols
find solutions of a linear constant coefficient ODE
Calling Sequence constcoeffsols(lode, v) constcoeffsols(coeff_list, x)
Parameters
lode - homogeneous linear differential equation v - dependent variable of the lode coeff_list - list of coefficients of a linear ode x - independent variable of the lode
Description
• The constcoeffsols routine returns a basis of the space of solutions of a homogeneous linear differential equation having constant coefficients.
• There are two input forms. The first has as the first argument a linear differential equation in diff or D form and as the second argument the variable in the differential equation.
• A second input sequence accepts for the first argument a list of coefficients of the linear ode, and for the second argument the independent variable of the lode. This input sequence is useful for programming with the constcoeffsols routine.
• In the second calling sequence, the list of coefficients is given in order from low differential order to high differential order and does not include the nonhomogeneous term.
• This function is part of the DEtools package, and so it can be used in the form constcoeffsols(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[constcoeffsols](..).
Examples
> $\mathrm{with}\left(\mathrm{DEtools}\right):$
> $\mathrm{ode}≔3\mathrm{diff}\left(z\left(t\right),\mathrm{}\left(t,3\right)\right)+\mathrm{diff}\left(z\left(t\right),\mathrm{}\left(t,2\right)\right)-\mathrm{diff}\left(z\left(t\right),t\right)-3z\left(t\right)=0:$
> $\mathrm{constcoeffsols}\left(\mathrm{ode},z\left(t\right)\right)$
$\left[{{ⅇ}}^{{t}}{,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{sin}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right){,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{cos}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right)\right]$ (1)
> $\mathrm{ode}≔3{\mathrm{D}}^{\left(3\right)}\left(z\right)\left(t\right)+{\mathrm{D}}^{\left(2\right)}\left(z\right)\left(t\right)-\mathrm{D}\left(z\right)\left(t\right)-3z\left(t\right)=0$
${\mathrm{ode}}{≔}{3}{}{{\mathrm{D}}}^{\left({3}\right)}{}\left({z}\right){}\left({t}\right){+}{{\mathrm{D}}}^{\left({2}\right)}{}\left({z}\right){}\left({t}\right){-}{\mathrm{D}}{}\left({z}\right){}\left({t}\right){-}{3}{}{z}{}\left({t}\right){=}{0}$ (2)
> $\mathrm{constcoeffsols}\left(\mathrm{ode},z\left(t\right)\right)$
$\left[{{ⅇ}}^{{t}}{,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{sin}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right){,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{cos}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right)\right]$ (3)
> $\mathrm{ode}≔\left[-3,-1,1,3\right]:$
> $\mathrm{constcoeffsols}\left(\mathrm{ode},t\right)$
$\left[{{ⅇ}}^{{t}}{,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{sin}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right){,}{{ⅇ}}^{{-}\frac{{2}{}{t}}{{3}}}{}{\mathrm{cos}}{}\left(\frac{\sqrt{{5}}{}{t}}{{3}}\right)\right]$ (4)
This command also outputs the answer in RootOf form in some cases:
> $\mathrm{ode}≔\mathrm{diff}\left(z\left(t\right),\mathrm{}\left(t,5\right)\right)+\mathrm{diff}\left(z\left(t\right),\mathrm{}\left(t,2\right)\right)-\mathrm{diff}\left(z\left(t\right),t\right)-3z\left(t\right):$
> $\mathrm{constcoeffsols}\left(\mathrm{ode},z\left(t\right)\right)$
$\left[{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{+}{{\mathrm{_Z}}}^{{2}}{-}{\mathrm{_Z}}{-}{3}{,}{\mathrm{index}}{=}{1}\right){}{t}}{,}{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{+}{{\mathrm{_Z}}}^{{2}}{-}{\mathrm{_Z}}{-}{3}{,}{\mathrm{index}}{=}{2}\right){}{t}}{,}{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{+}{{\mathrm{_Z}}}^{{2}}{-}{\mathrm{_Z}}{-}{3}{,}{\mathrm{index}}{=}{3}\right){}{t}}{,}{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{+}{{\mathrm{_Z}}}^{{2}}{-}{\mathrm{_Z}}{-}{3}{,}{\mathrm{index}}{=}{4}\right){}{t}}{,}{{ⅇ}}^{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{5}}{+}{{\mathrm{_Z}}}^{{2}}{-}{\mathrm{_Z}}{-}{3}{,}{\mathrm{index}}{=}{5}\right){}{t}}\right]$ (5) | 1,469 | 4,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "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} | 3.140625 | 3 | CC-MAIN-2022-21 | latest | en | 0.705258 |
https://docs.aimsun.com/next/22.0.1/UsersManual/StaticScenarios.html | 1,696,393,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00325.warc.gz | 241,356,347 | 17,052 | Static Scenarios¶
In a Static Assignment Scenario, flows are assigned to the network using a deterministic algorithm. A Static Assignment does not use individual vehicles, it is oriented about trip volumes and about speed and flows on road sections. It is typically used in wide area models with time periods, used to define the demand in an OD matrix measured in hours, i.e peak, interpeak time periods.
A Static Assignment uses one of five methods to distribute traffic in the network:
The All or Nothing Assignment calculates a single shortest path for each OD pair in free flow conditions (empty network) and assigns the whole OD pair demand to its shortest path. Costs are not updated after assignment, so results will show free-flow costs.
The Incremental Assignment adds iterations to the process. The user will specify the load/unload percentages for each iteration and costs will be updated always after the load and before the unload step. A shortest path with updated costs will be calculated per OD pair at each step.
The MSA Assignment is a method based on Frank and Wolfe but simpler and faster because it skips one calculation, the search for the optimal step lambda, substituting this value by 1/n (being n the iteration number). In general, with a sufficient number of iterations this method will behave well enough and tend toward an equilibrium situation, though this is not assured.
The Equilibrium Traffic Assignment is based on Wardrop's user optimal principle: No user can improve his travel time by changing routes. In Aimsun Next, the Frank and Wolfe algorithm is used to calculate the flows according to this principle. The algorithm is based in a Shortest Paths Algorithm and an ad hoc implementation of a Linear Approximation Algorithm. When using Junction Delay Functions, uniqueness and convergence of solution are compromised. For a wider theoretical explanation about the Assignment, and algorithms used to solve it, refer the section on Static Traffic Assignment: User Equilibrium Models.
The Stochastic Assignment calculates the k-shortest paths for each OD pair and splits the OD pair demand among them according to a Discrete Choice function defined by the user.
The first four are deterministic methods and they are listed in order of complexity. All five are based on the calculation of shortest paths and path percentage usage. These calculations use the cost of the different network elements.
Multi-User Traffic Assignment¶
In Aimsun Next , matrices are defined for each User Class. A User Class consists of a paired Vehicle Type and Purpose. For example, the trips can be made by the user class: Car – Work. Note the Vehicle Type is mandatory, the Purpose is optional. A Traffic Demand will contain matrices for the different user classes and can contain multiple matrices for each class.
A Multi-User Traffic Assignment consists of a traffic assignment where different types of users are taken simultaneously into account, so that the cost for a user class considers the congestion caused by the volumes produced by the rest of the users. Each user class can perceive a different cost for every section and turn, but the calculation is based on the total volume.
The results correspond to each user class (in vehs), as well as a global assignment result computed by adding the assigned volumes for each vehicle type, additional volumes and the Transit, all of them being considered either in PCU’s (Passenger Car Units) (under the label All (pcu)) or in number of vehicles (when choosing All (veh)). Paths results, however, are not aggregated for All; they are only available per user class.
Static Assignment Scenario Editor¶
To create a new static assignment scenario, select New > Scenarios > Static Assignment Scenario in the Project Menu. If you are working on a subnetwork, the new scenario can be created from the subnetwork's context menu. The minimum requirement for a static assignment scenario is a base transport network and a traffic demand.
The scenario editor is divided into several tabs which describe what is to be simulated, the outputs to be collected, the variables used to modify the scenario, and some parameters to describe the scenario.
Main Tab¶
The parameters of the scenario, which will be the defaults for the experiments in this scenario, are:
• The name and external ID of the scenario.
• The simulation time and duration. Note the date is for information and not used by the simulation.
• The Traffic Demand: An OD based Traffic Demand. which can contain one or more OD matrices for one or more user classes.
• A Transit Plan with a set of Transit routes and schedules. The PCU’s corresponding to the volume represented by Transit vehicles along their transit lines will be automatically taken into account in the total volume for all the travel time calculations.
• A Path Assignment Plan with a set of routes derived from a prior static assignment experiment can be selected in MSA and Frank and Wolfe Assignments, so that the Assignment calculations start from this and not from an empty network. Setting a Path Assignment Plan and choosing 1 iteration for the experiment would mean applying the APA file to the demand, with no extra calculation. At least 2 iterations would be needed to calculate new paths and redistribute the demand.
• A Master Control Plan with the signals data used in this scenario. The average cycle times (Calculating Aggregate Control Times) and phases of each turn will then be available for use in Turn Penalty Functions to determine the costs of traversing a signalized junction.
• A Validation Data Set: This is data used to compare simulation outputs with observed data.
• A set of Geometry Configurations: These are the optional variations in the network applied to this scenario.
Outputs to Generate Tab ¶
The possible outputs for a static assignment are as follows:
• Flow, assigned volume, and cost data for road sections, centroid connections, and detectors
• Flow, assigned volume, and cost data for junctions and supernode trajectories
• Data for groupings
• Path assignments
• Flow, assigned volume, and cost data for subpaths
• Skim cost matrices.
These can be stored in the project database or in an external database as described in Storing Simulation Results.
For a complete list of output data for static scenarios, see the Output Database Definition topic.
Ticking the Groupings box will recalculate and refresh the groupings statistics (available in the Time Series tab of each group’s dialog) when the static assignment is executed.
The calculation of shortest paths (Path Assignment) is optional: if the Keep in Memory option is ticked, the paths will be explicitly calculated and path outputs will be available after the process, together with other assignment results. Otherwise, no data about the paths used in the assignment will be available, which makes the execution of the scenario faster.
When you tick Path Assignment, you can choose to store the path assignment results in an APA file by selecting a Path Assignment object in the associated experiment's dialog. Statistics for subpaths can also be generated and stored in the database if you tick the options Generate Time Series and Store in Database.
Finally, skim matrices for the generalized cost and distance, as well as for all the Function Components will be calculated and made available in the OD Matrices folder if the Generate checkbox is ticked.
An alternative method of obtaining output matrices is available in the Path Assignment folder in the experiment editor. However, output matrices generated from the experiment will only have values for cells corresponding to non-zero trips, while skim matrices generated from the scenario's Outputs to Generate tab will be full matrices, containing values of the shortest path with 0 trips but taking into account the current costs on the network.
Variables Tab¶
The variables that will be used in the static assignment are initialized here. Note that these variables can also be defined at the experiment level and in that case, the value will be taken from the experiment. When a simulation starts, the variables are assigned their values by looking first at the experiment and, if no value is defined at that level, then at the scenario.
A variable is an arbitrary string that starts with the dollar sign ($speed and$demand are examples of valid variables).
Static Assignment Experiment Editor¶
The Static Assignment Experiment is used to run the assignment specified in the scenario with some additional parameters or overridden values for variables. Experiments are created using the "New" option of the scenario context menu.
The experiment context menus is used to run the experiment, it can also be used to:
• Retrieve Static Traffic Assignment Results: This feature reads the assignment results (sections, turns, and convergence data) which were stored in a database when the Static Assignment Experiment was assigned. The database to be read is the one stated in the Static Assignment Scenario editor, Outputs to Generate folder.
• Retrieve Path Assignment Results: This option can be used to read the shortest path information stored in the Path Assignment results file when the Static Assignment Experiment was assigned. The path assignment results file read will be the one stated in the Static Assignment Scenario editor, Outputs to Generate folder.
Depending on the type of static experiment you are running, the wording of the Retrieve [Experiment Name] Results option will differ.
As a complement to the retrieval options just described, to save memory and enhance performance, there are two options that enable you to unload replication data and path assignment results:
• Unload Static Traffic Assignment Results: Unloads the assignment results that were stored in the database, to save memory and enhance performance.
• Unload Path Assignment Results: Unloads the path statistics and path assignment information from the replication to save memory and enhance performance.
You might find it helpful to select one or both of these options before retrieving data and results from another replication.
Main Tab¶
The Static Experiment Editor has different options in the Main Tab depending on the assignment method chosen as the experiment is created.
The parameters common to all methods are:
• Attribute Overrides: The Attribute Overrides for this experiment are selected and the order in which they are applied can be modified.
• Scripts: Scripts can be defined to be run before the execution of an experiment and after the execution of an experiment. These are set in the Pre-Run and Post-Run selectors.
There is a mandatory requirement that both these scripts must contain the following function to check any prerequisite the script might have. This function will return True if the experiment can be run or False if something is missing or invalid; The run will then be cancelled. The signature of the function is:
def filter( experiment )
...
return True
All or Nothing Assignment¶
The All or Nothing Assignment model has no additional parameters.
Incremental Assignment¶
The Incremental Assignment model loads the assignment incrementally. In the example above the first iteration loads 40% of the demand, calculates the travel times on links and unloads 20%. It then loads 50% ( 70% total) and recalculates the link times before unloading 20% and finally assigning the last 50%. Note the total loaded minus the total unloaded must equal 100%.
MSA Assignment¶
The MSA Assignment process will stop when either the Maximum number of Iterations or the desired Relative Gap is reached.
Frank and Wolfe method¶
The Frank and Wolfe Assignment model, like the MSA model will stop when either the Maximum number of Iterations or the desired Relative Gap is reached. There is also an option to activate the Conjugate Frank and Wolfe method, a variant to improve convergence (Daneva 2002).
Stochastic Assignment¶
The Stochastic Assignment parameters are:
• The number of k-shortest paths to be calculated.
• The discrete choice function to be used for probabilities calculation, defined as a Stochastic Discrete Choice Function (check the Function Editor for more information).
• The pre-existing loads in the network(optional). Some models might assume there is a capacity drop on links due to traffic that has been separately modeled, these volumes can be pre-loaded here.
Cost Functions¶
There are three types of Cost Functions which model the macroscopic costs on the network:
• Volume delay functions (VDF) model the generalized cost of the sections and centroid connections.
• Turn penalty functions (TPF) model the primary generalized cost of crossing a turn.
• Junction delay functions (JDF) model the delay caused in a turn, due to the turn volume, the conflicting turn volumes, and the volume on the turn origin section.
The definition of functions in Aimsun Next is done with Python code, so they admit a wide range of possibilities. For example, the user can choose different values for different user classes (depending on the vehicle type or the purpose, etc.) and also define Function Components for specific subfunctions (for example a subfunction calculating the distance in km.) inside the VDF, TPF or JDFs, so that the corresponding new assignment outputs are available. Please check the Functions section for more information on cost functions and function components.
The default template offers a set of VDFs associated with each Road Type in it. The units of the VDFs in the template are minutes, they model the travel time needed to go through a section depending on the volume it has.
The user can add or change the Road Types as well as the VDFs; VDFs can be set by Road Type or defined on a per section basis.
If no VDF is assigned by the user to the centroid connections, they have a default VDF. Users can edit their own functions and assign them to centroid connections.
• Connections to sections or nodes (that is, to the private network) have a default function proportional to its length, with a free flow speed of 1000 km/h (0.06*connection.length3D()/1000). That is quite a low value. Turns between centroid connections and sections are always assigned a zero cost, as they are not editable.
• Connections to Transit Stops or Stations (that is, to the Transit network) have a default function equivalent to walking at 5km/h (12.0*connection.length3D()/1000.0).
The turns in a node also have an associated a default cost (in minutes) for the TPF, which depends on the length and speed of the turn, corresponding to the travel time needed to cross the node in free flow (0.06*turn.length3D()/turn.getSpeed()). Again, users can edit their own functions and assign them to turns.
No Junction Delay Function is assigned by default to turns, so by default, only the travel time needed to cross the junction in free flow is assumed as the turn cost.
Not only the cost functions but also all the parameters they depend on have to be taken care of to get the appropriate results. It is important to check in the definition of each Vehicle Type the value set for the Passenger Car Units (PCUs). Each vehicle has its equivalent value in PCUs in terms of capacity, that is, for example, if the effect of a truck on the network is equivalent to the effect of 2 cars, then trucks should be accounted as 2 PCUs in the calculations.
Cost Function Errors and Constraints¶
The functions TPF, JDF, and VDF are subject to certain constraints which will trigger errors and the cancellation of dynamic simulations, static assignments, and static adjustments. For more information, see Cost Function Errors and Constraints.
Experiment Outputs to Generate Tab¶
The outputs to generate tab specifies the Path Assignment be to used to store the paths. If only a subset of the paths is to be stored, this can be stored in a separate object. There is an option to store only the first n most used paths, for later use in a dynamic model. The Paths object can be used either to restore the information without recalculating the assignment or with vehicle-based simulators as user defined shortest path trees to simulate dynamically the static equilibrium situation.
Variables Tab¶
In this tab, the value of the variables that will be used in the simulation can be defined. If they are defined, the value defined here will be used. If they are not defined, the value defined by the Scenario editor under the Scenario Variables tab, will be used.
Running a Static Assignment¶
A Static Assignment is run from the context menu of the Experiment. The progress is indicated in the Progress Bar. It is always advisable to run the Check and Fix Experiment option before starting a static traffic assignment. This tool is available from the Experiment context menu.
Static Assignment Results¶
Data Retrieval¶
The options for Data Retrieval – moving the results from the experiment to the project results database – are located in the context menu of the Static Experiment.The data retrieval options are in the experiment editor. These are:
• Retrieve Static Traffic Assignment results: The data outputs requested for the scenario are loaded to the database specified in the Outputs Tab.
• Retrieve Path Assignment Results: Loads the path statistics and path assignment information from the assignment. In order to load this information an output file must be defined and the related Store option activated.
Depending on the type of static experiment you are running, the wording of the Retrieve [Experiment Name] Results option will differ.
Note: You can also unload this data to save memory and enhance performance; see Unloading Data and Results.
Outputs¶
The experiment editor for a static assignment will be extended after the experiment has run with an Outputs tab of which contains seven subtabs.
Where appropriate, the data in each subtab can be broken down by user class and if the data in the table is linked to a simulation object, selecting the line in the table moves the main 2D view to that object. Clicking on the header for each column sorts the rows on the contents of that column, then reverses the order. Hence it would be possible for example to identify the turns with the highest flows or the sections with the highest costs.
The Copy button takes a copy of the selected area of the results table and places it on the system clipboard. The data can then be pasted into other applications, e.g. MS Excel, for further processing. The Create Traffic State Button creates new traffic states for the vehicle types in the assignment and inserts them in the Project : Demand Data: Traffic States* list. These can be used to provide demand data for later traffic simulations or exported to signal optimization software such as Synchro.
Summary Tab¶
Summarizes the overall network throughput in mean network occupancy, distance traveled, and cost. These measures do not include centroid connection values. If Function Components have been defined, a global value will be calculated for each one of them (so the previous measures can be obtained including centroid connections by the use of function components).
Sections Tab¶
Gives the assigned volume, flow, cost, V/C Ratio (an occupancy measure, volume divided by capacity, in percentage) and any Function Components for each section. If the Show Only Entrances box is checked, only entrance sections will be listed; that is sections which are connected to a centroid with a centroid-to-section connection.
The ‘Copy Data’ button allows copying all the data in the table into a file, as in the example given below.
Turns Tab¶
Gives the assigned volume, cost, and turn percentage for the turns in each node. The Show all Turns button includes the nodes where there was no turn choice( i.e the sole turn has 100% allocated).
The ‘Copy’ button allows copying the turns results into a file, as in the example given below:
Connections Tab¶
Gives the assigned volume, flow, and cost for each centroid connection.
Supernode Trajectories Tab¶
Gives the assigned volume, cost, and turn percentage for the turns in each supernode.
Convergence Tab¶
This tab summarizes the assignment convergence in tabular and graphical form with the option in the graph to restrict which iterations are shown. For each iteration, the achieved Relative Gap and the Lambda (step length) (when applicable), as well as execution times are listed.
The Relative Gap is a measure of how close the current state of the assignment is to equilibrium (when, for each OD pair, all paths have the same cost). It compares the current costs with the costs on the current shortest paths:
Where:
• h~ijp~ are the trips from origin i to destination j that use path p.
• d~ij~ is the demand (trips) for origin i and destination j.
• s~ijp~ is the cost of path p.
• s~ij~* is the minimum cost of all used paths with origin i and destination j.
Validation Tab¶
The Validation Tab compares the observed detection data with the flow or volume calculated in the assignment. This uses a Real Data Set for the comparison, that must be loaded and set in the scenario editor Main tab as the Real Data Set for validation.
The options in selecting data used in validation are first, to select one or all time series in the real data set (i.e. compare all counts, or just the car counts and truck counts, if the Real Data Set contained those 2 series of counts). The validation calculations can also be configured to omit groups of sections, detectors or turns where the section is congested, i.e. the demand exceeds capacity. Sections can be grouped and each grouping optionally removed from the validation calculations. The effect of omitting measurements from over-congested sections is shown below, helping in the task of focusing on discrepancies on non-congested sites, due to other reasons rather than demand being higher than physical throughput of the network in congested locations.
The validation can be shown as a Graph, a Regression chart or a Table. The options to display the validation data in a static assignment are similar to those used in a dynamic replication.
Validation¶
Once the Time Series to compare with the Assignment volumes is selected in the Validation folder, there are three different possible representations available: a Graph, a Regression chart or a Table shown in the three figures below.
In the Regression graphic, the blue line is the regression line, the black lines are the confidence intervals at 95% and the red line represents the fixed line y = x.
In the Table representation, the observed and assigned volumes or flows are listed, as well as their Absolute Difference and their Relative Difference computed as:
100 * ( (Assigned-Observed)/Max(Assigned, Observed))
Data can be copied into a text file, in the table format with the Copy Date option and the Graph and the Regression Graph can be copied as a picture with the Snapshot option. The limits of the Graph can be set in the Adjust Limits option.
Graphical assignment results¶
The graphical representation used to show the assignment results in the Outputs folder is defined in the "Assigned Volume" View Mode.
Aimsun Next has a default View Mode for Assignment results, but any of the view modes in the model can be set as the default for showing assignment results by setting the view mode Automatic Activation field to Macro in the view mode’s editor.
The figure below shows an example of graphical assignment results.
The View Styles included in the Assigned Volume View Mode are:
• Styles for the width of each section, proportional to its volume.
• Styles for the labels of the assigned volumes.
• Styles to hide section objects and nodes.
• Styles for coloring objects based on the volume/capacity percentage (occupancy). By default, a color ramp with six different intervals from green to red is used for sections, and a color ramp in blue tones is used for centroid connections.
Creation of a Traffic State from Assignment results¶
The section volumes and turn percentages from the assignment can be used to generate a Traffic State with the “Create Traffic State” button in the Outputs folder in the Static Assignment experiment editor .Traffic States for each of the user classes in the demand will be created in the Project window, Traffic States folder.
Path Assignment Tab¶
The Path Assignment Tab is only shown if the Path Assignment calculation is Activated in the Outputs to Generate folder of the Static Assignment Scenario, and it is available for all user classes except for the aggregated All. It is described in the Path Analysis section and its usage is the same for static and dynamic scenarios. | 4,867 | 24,812 | {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.906839 |
http://mrburkemath.blogspot.com/2016/02/january-2016-new-york-geometry-common.html | 1,601,271,858,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00691.warc.gz | 91,242,517 | 31,143 | ## Monday, February 08, 2016
### January 2016 New York Geometry (Common Core) Part 1
Sorry for the delays. These things happen.
Below are the questions with answers and explanations for Part 1 the Geometry (Common Core) Regents exam for January 2016. Part II questions appeared in a another post.
### Part I
1. William is drawing pictures of cross sections of the right circular cone below. (image omitted)
Which drawing can not be a cross section of a cone?
(1) the square. You can make a slice through that cone and get an oval (or circle), a semicircle with a diameter on the bottom or even triangle if you split it vertically. You can't get a square.
2. An equation of a line perpendicular to the line represented by the equation y = -(1/2)x - 5 and passing through (6, -4) is
(4) y = 2x - 16. A line perpendicular to a line with a slope of -1/2 would have a slope of 2, so (1) and (2) are out. If you plug in 6 for x, -4 = 2(6) + b, b = -16.
3. In parallelogram QRST shown below, diagonal TR is drawn, U and V are points on TS and QR, respectively, and UV intersects TR at W.
If m<S = 60°, m<SRT = 83°, and m<TWU = 35°, what is m< WVQ?
(3) 72°. Look at quadrilateral QTWV, which has 360°. Angle Q = S = 60. Angle QTV = SRT = 83. Angle TWV = 180 - TWU = 180 - 35 = 145. 60 + 83 + 145 = 288. 360 = 288 = 72°.
4. A fish tank in the shape of a rectangular prism has dimensions of 14 inches, 16 inches, and 10 inches. The tank contains 1680 cubic inches of water. What percent of the fish tank is empty?
(2) 25. The volume of the tank is 14 * 16 * 10 = 2240. 2240 - 1680 = 560 gallons empty. 560 / 2240 = .25 = 25%
5. Which transformation would result in the perimeter of a triangle being different from the perimeter of its image?
(3) (x,y)--> (4x,4y). A dilation would change the distance between the vertices, and therefore the perimeter. Reflections and translations do not affect the size, so the image would have the same perimeter.
6. In the diagram below, FE bisects AC at B, and GE bisects BD at C.
Which statement is always true?
(1) AB = DC. Because of bisecting AB = BC and BC = CD. Therefore, AB = CD. We have no information about where points F or G are, so you cannot make any assumptions about those lines being bisected or the segments being equal.
7. As shown in the diagram below, a regular pyramid has a square base whose side measures 6 inches. (image omitted)
If the altitude of the pyramid measures 12 inches, its volume, in cubic inches, is
(2) 144. The Volume is (1/3)(Area of the Base)(height) = (1/3)(6 * 6)(12) = (12)(12) = 144.
8. Triangle ABC and triangle DEF are graphed on the set of axes below.
Which sequence of transformations maps triangle ABC onto triangle DEF?
(1) a reflection over the x-axis followed by a reflection over the y-axis. It is also a 180 degree rotation about the origin, but it would NOT be followed by a reflection in y = x.
9. In triangle ABC, the complement of <B is <A. Which statement is always true?
(4) sin <A =cos <B. The sine of one complementary angle is the cosine of the other in a right triangle. The side opposite angle A will be adjacent to angle B.
10. A line that passes through the points whose coordinates are (1,1) and (5,7) is dilated by a scale factor of 3 and centered at the origin. The image of the line
(2) is parallel to the original line. As long as the line does not go through the origin, its dilation will be a parallel line (having the same slope). If the line went through the origin, it would be the same line. The slope of the line is 6/4 = 3/2. The line y = 3/2x does not go through (1, 1), so the line being dilated does not go through the origin.
11. Quadrilateral ABCD is graphed on the set of axes below.
When ABCD is rotated 90° in a counterclockwise direction about the origin, its image is quadrilateral A' B 'C 'D'. Is distance preserved under this rotation, and which coordinates are correct for the given vertex?
(4) yes and B'(-3,4) . Yes, distance is preserved, (1) and (2) are eliminated. A(-2, 6) will move to A'(-6, -2). Choice (3) would have been correct for a clockwise rotation.
12. In the diagram below of circle 0, the area of the shaded sector LOM is 2(pi) cm2.
If the length of NL is 6 cm, what is m<N?
(3) 40°. The diameter is 6 cm, so the radius is 3 cm. That makes the Area of the entire circle = (pi)(3)2 = 9pi.
if the shaded sector is 2pi, then that sector and that arc is 2/9 of the circle, which is 2/9(360) = 80 degrees. Angle N is an inscribed angle that intercepts arc LM, so it is half of 80 degrees, or 40 degrees.
13. In the diagram below, triangle ABC ~ triangle DEF.
If AB = 6 and AC = 8, which statement will justify similarity by SAS?
(1) DE = 9, DF = 12, and <A = <D. Angles A and D are the included angles, so (3) and (4) are eliminated. 6/8 = .75 and 9/12 = .75, so they are proportional.
14. The diameter of a basketball is approximately 9.5 inches and the diameter of a tennis ball is approximately 2.5 inches. The volume of the basketball is about how many times greater than the volume of the tennis ball?
(3) 55. The radius of the basketball is 4.75. The radius of the tennis ball is 1.25. Volume requires cubing the radii. Divide (4.753 / 1.253) = 54.872, which is approximately 55.
15. The endpoints of one side of a regular pentagon are (-1,4) and (2,3). What is the perimeter of the pentagon?
(2) 5*SQRT(10). Using the Distance formula (or the Pythagorean Theorem), we can find the length of the segment joining those two endpoints to be SQRT(32 + 12), which is SQRT(10). Since there are five sides to a pentagon, the perimeter is 5*Sqrt(10). [5 radical 10]
16. In the diagram of right triangle ABC shown below, AB = 14 and AC= 9. (image omitted)
What is the measure of <A, to the nearest degree?
(3) 50. AC is the adjacent side and AB is the hypotenuse, so cos A = 9/14. That means <A = cos-1(9/14) = 49.99... degrees.
17. What are the coordinates of the center and length of the radius of the circle whose equation is x2 + 6x + y2 - 4y = 23?
(4) (-3,2) and 6 You need to complete the squares to find the equation for the circle. (I could make a comic out of that sentence.)
Half of +6 is +3, and half of -4 is -2, but you have to flip the signs to find (h, k). Remember: (x - h)2 + (y - k)2.
So the center is at (-3, 2) and you eliminate (1) and (2).
To complete the square you need to add (3)2 add (-2)2 to both sides of the equations. That adds 9 + 4 to 23, giving 36, which is r2. So the radius is 6 (not 36).
18. The coordinates of the vertices of triangle RST are R(-2, -3), S(8,2), and T(4,5). Which type of triangle is triangle RST?
(1) right. Eliminate (4) equiangular, because equiangular/equilateral triangles are always acute, and you can't have two correct answers.
If the triangle is right, then two of the sides will have perpendicular slopes -- that is, they will be negative reciprocals. If that is not true, then you have to find the lengths of the three sides to determine if the triangle is acute or obtuse.
Slope of RS = (2 - -3)/(8 - -2) = 5/10 = 1/2. Slope of ST = (5 - 2)/(4 - 8) = 3/(-4) = -(3/4). Slope of TR = (-3 - 5)/(-2 - 4) = -8/-6 = 4/3. ST is perpendicular to TR because (-3/4)(4/3) = -1. It is a right triangle.
19. Molly wishes to make a lawn ornament in the form of a solid sphere. The clay being used to make the sphere weighs .075 pound per cubic inch. If the sphere's radius is 4 inches, what is the weight of the sphere, to the nearest pound?
(2) 20. Multiply the Volume of the sphere by .075, so .075(4/3)(pi)(4)3 = 20.1..., which is about 20.
20. The ratio of similarity of triangle BOY to triangle GRL is 1:2. If BO = x + 3 and GR = 3x - 1, then the length of GR is
(4) 20. Because GR is twice as big as BO, start with 2(x + 3) = 3x - 1,
So 2x + 6 = 3x - 1
and 7 = x. (note that this is choice (2).)
The length of GR is 3(7) - 1 = 20.
21. In the diagram below, DC, AC, DOB, CB, and AB are chords of circle O, FDE is tangent at point D, and radius AO is drawn. Sam decides to apply this theorem to the diagram: "An angle inscribed in a semi-circle is a right angle."
Which angle is Sam referring to?
(3) <DCB. DOB is a diameter of the triangle and arc DAB is a semi-circle. Angle DCB is an inscribed angle that intercepts the semi-circle. Since it is half the measure of the 180 degrees, it must be 90 degrees, making it a right angle.
22. In the diagram below, CD is the altitude drawn to the hypotenuse AB of right triangle ABC. (image omitted)
Which lengths would not produce an altitude that measures 6*SQRT(2)?
(2) AD = 3 and AB = 24 Square (6*SQRT(2)) and you get 36 * 2 = 72 as the altitude. The product of AD and DB must be 72 for the altitude to be 6*SQRT(2). Read the choices carefully. Two of the choices give you AB, not DB. You need to subtract AD from AB to get DB. While 3 * 24 equals 72, you are supposed to be multiplying 3 * 21, which is only 63.
23. A designer needs to create perfectly circular necklaces. The necklaces each need to have a radius of 10 cm. What is the largest number of necklaces that can be made from 1000 cm of wire?
(1) 15. If the radius is 10 cm, then the length of the wire to produce one necklace, (2)(10)(pi), is approximately 62.83 cm. Divide 1000/62.83 and you get 15.9. However, you cannot round up because you don't have enough wire to finish the 16th necklace.
24. In triangle SCU shown below, points T and O are on SU and CU, respectively. Segment OT is drawn so that <C = <OTU.
If TU = 4, OU = 5, and OC = 7, what is the length of ST?
(3) 11. Triangles OUT and CUS are similar, and the sides are proportional, but make sure you set up the proportion correctly. OT is NOT parallel to CS. The correct proportion is 4/(7 + 5) = 5/(4 + x), so 16 + 4x = 60.
4x = 44, x = 11.
END OF PART 1.
Anonymous said...
I don't get #22
Anonymous said...
I don't understand 22
Anonymous said...
In 22 where do you get 21 from???
(x, why?) said...
AD X DB = (CD)^2
Choices 2 and 4 do NOT say DB, they say AB. AB is the length of AD + DB.
To find the length of DB you need to subtract DB = AB - AD.
So if AB = 24 and AD = 3, then DB = 21
Hope this helps
(x, why?) said...
I wrote more about question 22 here:
http://mrburkemath.blogspot.com/2016/06/daily-regents-right-triangle-altitude.html
Anonymous said...
Thank you so much!!!
Unknown said...
This comment has been removed by the author.
Unknown said...
I dont under stand the proportion you set up for question #24, wouldent it be 4/5 = x + 4/12?
(x, why?) said...
When you set up the proportion with similar triangles, you need to compare the corresponding sides.
If you notice from the picture, the bases of the two triangles are NOT parallel. This is because the smaller triangle is basically a flipped over version of the bigger one.
You can't trust the image (because it doesn't have to be drawn to scale), but if you read the problem, you can see which angles are congruent. The opposite sides are the corresponding sides. | 3,225 | 10,982 | {"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} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.913332 |
http://www.jiskha.com/display.cgi?id=1349061811 | 1,495,887,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608953.88/warc/CC-MAIN-20170527113807-20170527133807-00172.warc.gz | 681,171,860 | 3,967 | # chemistry
posted by on .
a 250.0 ml sample of an aqueous solution at 25 degrees celsius contains 35.8 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 11.25 mmHg, what is the molar mass of the unknown compound?
I know you need to have the answer in g/mol. I have grams (0.358 g), but am having a hard time figuring out how to find mols.
• chemistry - ,
pi = MRT. Substitute and solve for M = molarity. Then n = grams/molar mass. You know grams and mols, solve for molar mass. Note that pressure should be in atm for M to be molarity.
• chemistry - ,
So you invert the equation to M=pi/RTi to solve for Molarity. I plug in the numbers ((0.1480 atm)/(0.08206 L atm/K mol)(298K) (1))
That equals 6.1554*10^-4 mols.
Dont you divide 0.358g/6.1554*10^-4?? | 240 | 795 | {"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} | 3.40625 | 3 | CC-MAIN-2017-22 | latest | en | 0.900971 |
http://sites.math.washington.edu/~mesiket/Teaching/R17%20M308E/index.html | 1,558,723,790,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257731.70/warc/CC-MAIN-20190524184553-20190524210553-00223.warc.gz | 187,542,516 | 5,385 | ## Math 308E Summer 2017 A term
Welcome to our M308 course page. Here you can find announcements, the course calendar, and the syllabus.
### Announcements
• (S 7-22) Your final scores are now available on Catalyst. I gave the overall median percentage of 75.5% a 3.1.
Thanks for a great A term! Have a good rest of your summer, and never forget that homogeneous systems are awesome.
• (S 7-15) I will hold extra office hours Tuesday 7-18 from 11:00-12:00.
• (Updated S 7-15) Wednesday is our final.
• Cheat sheet: this time you can use both sides of one 8.5x11 page of hand-written notes. You may not have any written proofs on the notes. Anything else (theorem statements, definitions, motivational slogans, etc) is fine.
• As before, you may only use a scientific calculator.
• The test will be comprehensive, but will emphasize the material since the midterm, namely, chapters 4 and beyond.
• There will not be anything from the "Additional topics" segment last Friday except that I expect you to know the statement of Theorem 5.20, p.209, describing how determinants give the area distortion factor.
• Expect several proofs. These will fall into two categories:
1. "Baby proofs": proving simple statments like: "If c is an eigenvalue of A, then c^2 is an eigenvalue of A^2," or "The only vector in S and S^perp is 0."
2. Non-baby proofs: I will ask you to prove at least one of the following theorems (final list):
• Theorem 6.3, p.221 (your argument would need to include the proof of Theorem 6.2)
• Theorem 8.11, p.309
• Theorem 8.l2, p.310
• Theorem 8.14, (c) and/or (d), p.315
• (M 7-10) WebAssign will not work from 3-9pm on Saturday. Your 8.2 HW is still due Saturday at 11:59pm. Please plan accordingly.
• (F 7-7)
• The midterm solutions are posted on the calendar below.
• Here is a page from a fellow instructor with a lot of practice final exams. The sample finals on our other page are also good practice. Since we have skipped a couple of sections the content on our final may be slightly less than everything these cover. But it will be very similar.
• The final will have at least one proof. Any proofs I ask will be from the following list: (which will be updated more next week)
• Theorem 6.3, p.221 (your argument would need to include the proof of Theorem 6.2)
• (M 7-3) I said that I would post solutions for the midterm today. Unfortunately this is going to have to wait until I clear up some details about some tests.
• (F 6-30) I will be in the office Monday morning (July 3) from 8:30-9:30 if you have last-minute questions for the test.
• (W 6-28) Thursday's office hours, starting tomorrow, will be from 3-4.
• (M 6-26)
• Wednesday's office hours will now be 2:30-3:30.
• Our midterm next Monday will cover all the sections we have done through 3.3 (so no chapter 4). You may use a scientific (=non-graphing) calculator and one side of an 8.5x11" page of notes that you have hand written. You may not have any proofs on your note sheet. Anything else is fine.
• (F 6-23) Some have asked about practice midterms. A Google search gave this page, and I like the flavor of questions on these midterms. The "Midterm 1" practice tests are the relevant ones. I don't guarantee that mine will be identical, but I think if you can answer these you have a good conceptual understanding that will serve you well. The material covered on our test may also not be identical to that in these, because this review page was not for an A-term class. But it is still good practice.
• (W 6-21) Our midterm is re-scheduled to be on Monday, July 3rd instead of Wednesday, July 5th, as initially stated.
• (T 6-20) You may need to reload this page from time to time to get the below calendar to update.
### Course calendar
M 6-19 1.1-1.2: Systems of linear equations and how to solve them
W 6-21 2.1-2.2: Vectors and their span 1.1-1.2 HW due R 6-22 Warm-up problems
F 6-23 2.3: Linear independence Warm-up problems
M 6-26 3.1-3.2: Linear transformations and matrix algebra 2.1-2.2 HW due T 6-27 Warm-up, review and Big Theorem slides
W 6-28 3.2, 3.3, 4.1: Matrices, inverses, subspaces 2.3 HW due; 3.1 HW due R Slides
F 6-30 4.1, 4.2: Subspaces, basis and dimension; review for MT1 3.2, 3.3 HW due Sat Slides
M 7-3 4.2: Finish basis and dimension; Midterm 1; solutions. The median was 40/50. Slides
W 7-5 4.3, 5.1: Row and column space; Determinants 4.1, 4.2 HW due Slides
F 7-7 6.1, 6.3: Eigenvalues and vectors; Change of basis 4.3, 5.1 HW due Sat 7-8 Slides
M 7-10 Finish 6.3; 8.1: Dot products, orthogonal sets 6.1, 6.3 HW due T 7-11 Slides
W 7-12 8.2: Projections, Gramm-Schmidt algorithm (course evals open) 8.1 HW due R, 7-13 Slides
F 7-14 8.5: Least-squares regression; Additional topics (more on determinants; Cramer's rule) 8.2 due S, 7-15
M 7-17 Review for final (course evals close T, 7-18) 8.5 HW due T 7-18 Slides
W 7-19 Final exam
### Syllabus
• When and where: MWF 9:40-11:50 in SMI 102
• Instructor: Tim Mesikepp
• Office: PDL C-34
• Office hours: M 1-2, W 2:30-3:30, R 3-4, F 1-2, or by appointment
• Email: mesiket (at) math.washington.edu
• Warning: This course will be extremely compressed. We cover all the material from the entire quarter in 4.5 weeks. This will be a frenetic, drinking-from-the-firehose pace. You will need a lot of effort and time outside of class to digest the material. If you cannot or are not willing to invest so much effort, then you should switch to a regular-length section.
• Course content and text: See the math department's 308 syllabus.
• Homework: Homework is online through WebAssign. The first problem set (1.1-1.2) is due Thursday, 6-22, and then homeworks are due every subsequent Tuesday and/or Thursday. You need to get access to WebAssign as soon as possible and login to see your assignments.
• Exams: There will be one midterm and a final exam. There are no make-up exams. In order to pass the course you must take both the midterm and the final. Schedule your travel and obligations around these dates, or otherwise do not take this class.
• Midterm: Monday, July 3
• Final: Wednesday, July 19 (the final day of A-term)
• Homework: 15%
• Midterm: 35%
• Final: 50%
• Some things you are responsible for:
• Knowing the policies in this syllabus.
• Knowing when homework is due. This will always be visible on WebAssign and above on our course calendar.
• Announcements I make in class and post above in "Announcements." So you should come to class and periodically check this site.
• Resources for help:
• You! Your drive, perseverance, and study. Your lecture notes, your creativity and your ideas. Most of your learning for this class will have to take place outside of class (!) as you reflect on and digest what we discuss in lecture. | 1,951 | 6,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-22 | latest | en | 0.909961 |
https://www.coursehero.com/file/6629033/51-ch-03-Mechanical-Design-budynas-SM-ch03/ | 1,481,395,231,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543434.57/warc/CC-MAIN-20161202170903-00308-ip-10-31-129-80.ec2.internal.warc.gz | 929,616,019 | 73,650 | 51_ch 03 Mechanical Design budynas_SM_ch03
# 51_ch 03 Mechanical Design budynas_SM_ch03 - = tan 1 F y F...
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64 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-75 r i = 0 . 125 in, r o = 0 . 125 + 0 . 1094 = 0 . 2344 in From Table 3-4 for h = 0 . 1094 r c = 0 . 125 + 0 . 1094 / 2 = 0 . 1797 in r n = 0 . 1094 / ln(0 . 2344 / 0 . 125) = 0 . 174 006 in e = r c r n = 0 . 1797 0 . 174 006 = 0 . 005 694 in c i = r n r i = 0 . 174 006 0 . 125 = 0 . 049 006 in c o = r o r n = 0 . 2344 0 . 174 006 = 0 . 060 394 in A = 0 . 75(0 . 1094) = 0 . 082 050 in 2 M = F (4 + h / 2) = 3(4 + 0 . 1094 / 2) = 12 . 16 lbf · in σ i =− 3 0 . 082 05 12 . 16(0 . 0490) 0 . 082 05(0 . 005 694)(0 . 125) =− 10 240 psi Ans. σ o =− 3 0 . 082 05 + 12 . 16(0 . 0604) 0 . 082 05(0 . 005 694)(0 . 2344) = 6670 psi Ans. 3-76 Find the resultant of F 1 and F 2 . F x = F 1 x + F 2 x = 250 cos 60 + 333 cos 0 = 458 lbf F y = F 1 y + F 2 y = 250 sin 60 + 333 sin 0 = 216 . 5 lbf F = (458 2 + 216 . 5 2 ) 1 / 2 = 506 . 6 lbf This is the pin force on the lever which acts in a direction
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Unformatted text preview: = tan 1 F y F x = tan 1 216 . 5 458 = 25 . 3 On the 25 . 3 surface from F 1 F t = 250 cos ( 60 25 . 3 ) = 206 lbf F n = 250 sin(60 25 . 3 ) = 142 lbf r c = 1 + 3 . 5 / 2 = 2 . 75 in A = 2[0 . 8125(0 . 375) + 1 . 25(0 . 375)] = 1 . 546 875 in 2 The denominator of Eq. (3-63), given below, has four additive parts. r n = A ( d A / r ) 25.3 206 507 142 2000 lbf in...
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Ask a homework question - tutors are online | 770 | 1,689 | {"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} | 3.59375 | 4 | CC-MAIN-2016-50 | longest | en | 0.550779 |
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A241214 Solutions of the equation (n+2)' = (n+1)' + n', where n' is the arithmetic derivative of n. 0
1, 4, 20257, 43910, 563101, 18895033, 119847146, 305478634708, 7461770367940, 29820549118362 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS a(9) > 5*10^11. - Giovanni Resta, Apr 18 2014 a(11) > 5*10^13. - Hiroaki Yamanouchi, Aug 27 2015 LINKS EXAMPLE The arithmetic derivative of 43910+2 is 69948, of 43910+1 is 39201, of 43910 is 30747 and 69948 = 39201 + 30747. MAPLE with(numtheory); P:= proc(q) local a, b, c, n, p; for n from 1 to q do a:=(n+2)*add(op(2, p)/op(1, p), p=ifactors(n+2)[2]); b:=(n+1)*add(op(2, p)/op(1, p), p=ifactors(n+1)[2]); c:=n*add(op(2, p)/op(1, p), p=ifactors(n)[2]); if a=b+c then print(n); fi; od; end: P(10^9); CROSSREFS Cf. A003415. Sequence in context: A053015 A089210 A203037 * A034014 A217600 A275684 Adjacent sequences: A241211 A241212 A241213 * A241215 A241216 A241217 KEYWORD nonn,more AUTHOR Paolo P. Lava, Apr 17 2014 EXTENSIONS a(7)-a(8) from Giovanni Resta, Apr 18 2014 a(9)-a(10) from Hiroaki Yamanouchi, Aug 27 2015 STATUS approved
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Last modified June 16 04:54 EDT 2019. Contains 324145 sequences. (Running on oeis4.) | 573 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-26 | latest | en | 0.538452 |
http://www.haskell.org/haskellwiki/index.php?title=Euler_problems/151_to_160&oldid=18683 | 1,410,844,025,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657113000.87/warc/CC-MAIN-20140914011153-00162-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 562,548,093 | 11,815 | # Euler problems/151 to 160
## 1 Problem 151
Paper sheets of standard sizes: an expected-value problem.
Solution:
`problem_151 = undefined`
## 2 Problem 152
Writing 1/2 as a sum of inverse squares
Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.
Solution:
```import Data.Ratio
import Data.List
invSq n = 1 % (n * n)
sumInvSq = sum . map invSq
subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
subsets _ = [[]]
primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
all (\q -> p `mod` q /= 0) [3, 5, 7]]
-- All subsets whose sum of inverse squares,
-- when added to x, does not contain a factor of p
pfree s x p = [(y, t) | t <- subsets s, let y = x + sumInvSq t,
denominator y `mod` p /= 0]
-- Verify that we need not consider terms divisible by 11, or by any
-- prime greater than 13. Nor need we consider any term divisible
-- by 25, 27, 32, or 49.
verify = all (\p -> null \$ tail \$ pfree [p, 2*p..85] 0 p) \$
11 : dropWhile (< 17) primes ++ [25, 27, 32, 49]
-- All pairs (x, s) where x is a rational number whose reduced
-- denominator is not divisible by any prime greater than 3;
-- and s is all sets of numbers up to 80 divisible
-- by a prime greater than 3, whose sum of inverse squares is x.
only23 = foldl f [(0, [[]])] [13, 7, 5]
where
f a p = collect \$ [(y, u ++ v) | (x, s) <- a,
(y, v) <- pfree (terms p) x p,
u <- s]
terms p = [n * p | n <- [1..80`div`p],
all (\q -> n `mod` q /= 0) \$
11 : takeWhile (>= p) [13, 7, 5]
]
collect = map (\z -> (fst \$ head z, map snd z)) .
groupBy fstEq . sortBy cmpFst
fstEq (x, _) (y, _) = x == y
cmpFst (x, _) (y, _) = compare x y
-- All subsets (of an ordered set) whose sum of inverse squares is x
findInvSq x y = f x \$ zip3 y (map invSq y) (map sumInvSq \$ init \$ tails y)
where
f 0 _ = [[]]
f x ((n, r, s):ns)
| r > x = f x ns
| s < x = []
| otherwise = map (n :) (f (x - r) ns) ++ f x ns
f _ _ = []
-- All numbers up to 80 that are divisible only by the primes
-- 2 and 3 and are not divisible by 32 or 27.
all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]
solutions = if verify
then [sort \$ u ++ v | (x, s) <- only23,
u <- findInvSq (1%2 - x) all23,
v <- s]
else undefined
problem_152 = length solutions```
## 3 Problem 153
Investigating Gaussian Integers
Solution:
`problem_153 = undefined`
## 4 Problem 154
Exploring Pascal's pyramid.
Solution:
```#include <stdio.h>
int main(){
int bound = 200000;
long long sum = 0;
int val2[bound+1], val5[bound+1]; // number of factors 2/5 in i!
int v2 = 0, v5 = 0;
int i;
int n;
for(n=0;n<=bound;n++)
{val5[n]=n/5+n/25+n/125+n/625+n/3125+n/15625+n/78125;
val2[n]=n/2+n/4+n/8+n/16+n/32+n/64+n/128+n/256+n/512+n/1024
+n/2048+n/4096+n/8192+n/16384+n/32768+n/65536+n/131072;}
v2 =val2[bound]- 11;
v5 = val5[bound]-11;
int j,k,vi2,vi5;
for(i = 2; i < 65625; i++){
if (!(i&1023)){
// look how many we got so far
printf("%d:\t%lld\n",i,sum);
}
vi5 = val5[i];
vi2 = val2[i];
int jb = ((bound - i) >> 1)+1;
// I want i <= j <= k
// by carry analysis, I know that if i < 4*5^5+2, then
// j must be at least 2*5^6+2
for(j = (i < 12502) ? 31252 : i; j < jb; j++){
k = bound - i - j;
if (vi5 + val5[j] + val5[k] < v5
&& vi2 + val2[j] + val2[k] < v2){
if (j == k || i == j){
sum += 3;
} else {
sum += 6;
}
}
}
}
printf("Total:\t%lld\n",sum);
return 0;
}
problem_154 = main```
## 5 Problem 155
Counting Capacitor Circuits.
Solution:
`problem_155 = undefined`
## 6 Problem 156
Counting Digits
Solution:
```digits =reverse.digits'
where
digits' n
|n<10=[n]
|otherwise= y:digits' x
where
(x,y)=divMod n 10
digitsToNum n=foldl dmm 0 n
where
dmm=(\x y->x*10+y)
countA :: Int -> Integer
countA 0 = 0
countA k = fromIntegral k * (10^(k-1))
countFun :: Integer -> Integer -> Integer
countFun _ 0 = 0
countFun d n = countL ds k
where
ds = digits n
k = length ds - 1
countL [a] _
| a < d = 0
| otherwise = 1
countL (a:tl) m
| a < d = a*countA m + countL tl (m-1)
| a == d = a*countA m + digitsToNum tl + 1 + countL tl (m-1)
| otherwise = a*countA m + 10^m + countL tl (m-1)
fixedPoints :: Integer -> [Integer]
fixedPoints d
= [a*10^10+b | a <- [0 .. d-1], b <- findFrom 0 (10^10-1)]
where
fun = countFun d
good r = r == fun r
findFrom lo hi
| hi < lo = []
| good lo = lgs ++ findFrom (last lgs + 2) hi
| good hi = findFrom lo (last hgs - 2) ++ reverse hgs
| h1 < l1 = []
| l1 == h1 = if good l1 then [l1] else []
| m0 == m1 = findFrom l1 (head mgs - 2) ++ mgs
++ findFrom (last mgs + 2) h1
| m0 < m1 = findFrom l1 (m0-1) ++ findFrom (goUp h1 m1) h1
| otherwise = findFrom l1 (goDown l1 m1) ++ findFrom (m0+1) h1
where
l1 = goUp hi lo
h1 = goDown l1 hi
goUp bd k
| k < k1 && k < bd = goUp bd k1
| otherwise = k
where
k1 = fun k
goDown bd k
| k1 < k && bd < k = goDown bd k1
| otherwise = k
where
k1 = fun k
m0 = (l1 + h1) `div` 2
m1 = fun m0
lgs = takeWhile good [lo .. hi]
hgs = takeWhile good [hi,hi-1 .. lo]
mgs = reverse (takeWhile good [m0,m0-1 .. l1])
++ takeWhile good [m0+1 .. h1]
problem_156=sum[sum \$fixedPoints a|a<-[1..9]]```
## 7 Problem 157
Solving the diophantine equation 1/a+1/b= p/10n
Solution:
`problem_157 = undefined`
## 8 Problem 158
Exploring strings for which only one character comes lexicographically after its neighbour to the left.
Solution:
`problem_158 = undefined`
## 9 Problem 159
Digital root sums of factorisations.
Solution:
```import Control.Monad
import Data.Array.ST
import qualified Data.Array.Unboxed as U
spfArray :: U.UArray Int Int
spfArray = runSTUArray (do
arr <- newArray (0,m-1) 0
loop arr 2
forM_ [2 .. m - 1] \$ \ x ->
loop2 arr x 2
return arr
)
where
m=10^6
loop arr n
|n>=m=return ()
|otherwise=do writeArray arr n (n-9*(div (n-1) 9))
loop arr (n+1)
loop2 arr x n
|n*x>=m=return ()
|otherwise=do incArray arr x n
loop2 arr x (n+1)
incArray arr x n = do
when(ab<a+b) (writeArray arr (x*n) (a + b))
writ x=appendFile "p159.log"\$foldl (++) "" [show x,"\n"]
main=do
mapM_ writ \$U.elems spfArray
problem_159 = main
--at first ,make main to get file "p159.log"
--then ,add all num in the file```
## 10 Problem 160
Factorial trailing digits
We use the following two facts:
Fact 1:
(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0
Fact 2:
product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1
We really only need these two facts for the special case of
d == 5
, and we can verify that directly by
evaluating the above two Haskell expressions.
More generally:
Fact 1 follows from the fact that the group of invertible elements
of the ring of integers modulo
5^d
has
4*5^(d-1)
elements.
Fact 2 follows from the fact that the group of invertible elements
of the ring of integers modulo
10^d
is isomorphic to the product
of a cyclic group of order 2 and another cyclic group.
Solution:
```problem_160 = trailingFactorialDigits 5 (10^12)
trailingFactorialDigits d n = twos `times` odds
where
base = 10 ^ d
x `times` y = (x * y) `mod` base
multiply = foldl' times 1
x `toPower` k = multiply \$ genericReplicate n x
e = facFactors 2 n - facFactors 5 n
twos
| e <= d = 2 `toPower` e
| otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1)))
odds = multiply [odd | a <- takeWhile (<= n) \$ iterate (* 2) 1,
b <- takeWhile (<= n) \$ iterate (* 5) a,
odd <- [3, 5 .. n `div` b `mod` base],
odd `mod` 5 /= 0]
-- The number of factors of the prime p in n!
facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) .
-- The digits of n in base b representation
radix p = map snd . takeWhile (/= (0, 0)) .
iterate ((`divMod` p) . fst) . (`divMod` p)```
it have another fast way to do this .
Solution:
```import Data.List
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c= (b * c) `rem` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
where
f x n y
| n == 1 = x `mul` y
| r == 0 = f x2 q y
| otherwise = f x2 q (x `mul` y)
where
(q,r) = quotRem n 2
x2 = sq x
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)
productMod =foldl (mulMod (10^5)) 1
hFacial 0=1
hFacial a
|gcd a 5==1=mod (a*hFacial(a-1)) (5^5)
|otherwise=hFacial(a-1)
fastFacial a= hFacial \$mod a 6250
numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
p160 x=mulMod t5 a b
where
t5=10^5
lst=numPrime x 5
a=powMod t5 1563 \$mod c 2500
b=productMod c6
c=sum lst
c6=map fastFacial \$x:lst
problem_160 = p160 (10^12)``` | 3,273 | 8,525 | {"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-2014-41 | latest | en | 0.720508 |
https://www.cheresources.com/invision/topic/14392-ash-run-off/ | 1,591,396,523,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00007.warc.gz | 672,032,802 | 33,762 | ## Featured Articles
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# Ash Run Off
This topic has been archived. This means that you cannot reply to this topic.
|
### #1 Rey Aryana
Rey Aryana
Brand New Member
• Members
• 1 posts
Posted 08 February 2012 - 05:51 AM
Hello sir, i want to ask several question about ash run off. (Power Plant Industry)
1. How to design ash run off, there are any standard calculation for it???
2. How many time that bottom ash need to settle in the ash run off??
3. Please tell me the simplest way, How to seperated bettwen water and ash in ash run off??
Thank you.
Regards,
Rey
### #2 kkala
kkala
Gold Member
• Banned
• 1,939 posts
Posted 11 February 2012 - 06:44 AM
I do not have design experience on the subject, but probably following notes are useful and hopefully answering the query to some extent.
1. Please explain the reported ash runoff. Runoff is probably rain water collected from an area (water shed), so it may be the rainwater of the plant mixed with ashes laying on the ground. A trench is supposed around the plant to collect external run off and guide it to a river or to sea.
2. Assumed ash run off can go to a settling pond for the ash particles to settle down. Overhead liquid may or may not need treatment (depending on its chemical analysis) before disposal.
3. Volume of settling pond should cover max rainfall (that is rain water falling on the site), so that the pond shall not overflow during a storm. Suppose a site of 10000 m2 and a maximum rainfall of 250 mm. The water to be collected into the pond is 10000*0.25*φ = 2500*φ m3, where φ= percentage of rainwater absorbed by soil. If the soil is covered with asphalt / concrete φ=1.0, if there are trees it can be down to (say) 0.6, according to books on hydraulics.
4. Suppose φ=0.8, then water during max rainfall shall be 2000 m3, so volume above minimum level of settling pond is 2000 m3. That max rainfall total quantity is usually estimated to be higher (ie pond to overflow) once every 15 or 20 years. Meteorological data are needed to estimate max rainfall. If I remember well, (Porter?) theory of extreme values is useful to estimate the max runoff out of 15 - 20 years.
5. Granulometry and (not bulk) density of ash is needed to estimate settling velocity of ash.
Above is what I can remember from a settling pond of an alumina plant project (1987) to separate water run off from entrained bauxite dust (height of the pond about 3 - 4 m). An hydraulics engineer (branch of civil engineering here) is the pertinent person for more precise information on the above matters. Look for such a professional in your organization.
For other options beside the settling pond, look at Perry, Chemical Engineer's Handbook, McGraw-Hill, Gravity Sedimentation Operations.
Edited by kkala, 11 February 2012 - 06:48 AM. | 736 | 2,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-24 | latest | en | 0.932894 |
http://www.instructables.com/id/Cutting-Odd-Shaped-Bottles/ | 1,521,334,893,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00226.warc.gz | 425,606,703 | 10,028 | # Cut Square and Odd Shaped Glass Bottles
4,396
57
7
Posted
## Introduction: Cut Square and Odd Shaped Glass Bottles
In this instructable I'm going to show you a method for cutting bottles that's universal, and it can be used to cut any shape of bottle. You can still get flaws with this method, but I'll explain how to minimize them. For example, if you're working with a regular round bottle, I recommend using the method I showed in my previous instructable. The link is here: Bottle-Cutting-With-a-Soldering-Iron
I'm going to demonstrate this method with an oval bottle and a square bottle. If you've seen my previous instructable on cutting bottles, you'll notice this technique is similar. If you haven't seen that one, I recommend checking it out since it goes into more detail.
If you prefer to view this instructable in video form, you can watch it here: https://youtu.be/_jRNuVxLsfE
What you will need:
## Step 1: Mark the Bottle
The first thing we need to do is draw a line with a marker on the bottle where we want it cut. This line will be a visual guide for you to follow.
Fill the bottle with water to the level you want to cut it, then draw a line at that level. But if you prefer, you can free-hand the line or use any method you want to mark it. If the line you're drawing isn't very easy to see, pour the water out and then darken the line before you start the next step.
I did this same thing with the square bottle, but then decided I wanted to cut this one to be cut a little higher. That's why there are two lines on the square one.
## Step 2: Score the Bottle
We need to score the bottle a little bit. Since these are odd shaped bottles, I don't have a jig to help with that, so I need to score them by hand. The score line needs to be about an inch or so long. Keep in mind that it may not end up straight, so just try the best you can. And be careful not to use too much pressure when scoring. You don't want to break the bottle in your hands. This score line is a bit hard to see on the camera.
## Step 3: Start the Crack
With a hot soldering iron, heat the score line in one spot for about 8-10 seconds, then move the soldering iron a bit and heat it some more. Do this across the entire score line, and past the end of it by about half an inch. Then look and see if the crack has started. Turn the bottle around and start again going in the other direction. It will probably take a while for the crack to start, so repeat this cycle, going in opposite directions each time, until a crack starts.
If a crack doesn't start, no matter how many cycles you do, you may need more heat on that bottle. Instead of using the tip to heat a small part of the score line, use the thicker part of the soldering iron and heat up as much of the score line at once that you can. But keep in mind that if you do this, the crack will form suddenly, and it will most likely stray off the path.
## Step 4: Guide the Crack
Once the crack starts, heat the line in front of the crack and the crack should follow. If it stops following the soldering iron, move to the other end of the crack and work on that side for a while.
Go back and forth as necessary until the crack goes all the way around. Remember that when both ends of the crack get close to meeting, a lot of the time they will stop when the two ends are about a quarter inch apart. Usually this means that you're done and the bottle should separate with little force.
This square bottle needed extra heat to get the crack started, and now it has a stray crack in the top portion of the bottle. I've already dragged the crack most of the way around the bottle, so for the last little bit I'm going to pull under the stray crack the best that I can.
## Step 5: Separate the Bottle
Now all I need to do is separate the bottles. The oval shaped one cut cleanly and fairly flat. The square one separated easily, but there is a small jagged piece I'll need to spend extra time to sand down. And you can see this piece that is a result of the stray crack.
## Step 6: And That's It!
This technique works with any shape or size of glass bottle. Just remember to sand down and polish the sharp edges so you don't get cut. You can go here to see my instructable for that: Polishing-Bottle-After-Cutting
If you want to see the video version of this instructable, you can watch it here: https://youtu.be/_jRNuVxLsfE
## Recommendations
• ### WOODEN SLIDE
3,789 Enrolled
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We have a be nice policy. | 1,066 | 4,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-13 | latest | en | 0.946676 |
http://matthiasbook.de/papers/parallelfractals/performance.html | 1,701,602,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00856.warc.gz | 27,634,466 | 3,895 | Home > Papers > Parallel Fractals Introduction | Mandelbrot Set | Image Characteristics | Parallel Algorithm Design | Partitioning | Agglomeration | Output Synchronization | Token-Passing | Polling | Performance Analysis | Conclusion | Bibliography | Slides
Parallel Fractal Image Generation
Performance Analysis
Our next step is analyze to analyze the performance of the algorithm we just developed. We want to see how much time we actually save by computing the parts of the image in parallel, and we need to check if the synchronization mechanism introduces any overhead.
In order to see how the execution time changes with the problem size, we run the program with a maximum iteration count of N = 5,000, N = 50,000 and N = 500,000.
To get good measurements without too much influence of fluctuating system performance (e.g. background processes, other users' activity), we calculate the image of the whole Mandelbrot set with 160 x 120 pixels and run the program three times for each choice of N, averaging the results. After completing all the test runs, we end up with the following (averaged) timings:
average TP,N P = 1 (serial) P = 2 P = 4 P = 6 N = 5,000 1.43199333 s 0.48863167 s 0.25034200 s 0.17215833 s N = 50,000 14.18023330 s 4.67098667 s 2.36133667 s 1.58265667 s N = 500,000 141.93433300 s 46.43553330 s 23.48320000 s 15.71240000 s
Table 4: Average execution walltime
Figure 12: Execution walltime chart
Table 4 shows that the execution time grows almost proportionally with the maximum number of iterations. This makes sense since the number of points computed remains the same, but the time to compute the points inside the set grows with the number of iterations. Since the time required for most points outside the set stays the same no matter how high the iteration maximum, the relation can not be exactly proportional.
We are mainly interested in the performance gain that can be achieved when using a parallel algorithm on a certain number of processors as opposed to a serial algorithm on one processor. From Figure 12, we can deduce that with an increasing number of processors, the execution time decreases overproportionally – for example, we would expect that by doubling the number of processors, we can cut the execution time in half, but the actual execution time is even less than that. To examine this effect, we compute the speedups, defined as
where TP,N is the overall execution time when running on P processors with an iteration maximum of N.
SP,N P = 2 P = 4 P = 6 N = 5,000 2,93061915 5,72014816 8,31788581 N = 50,000 3,03581113 6,00517219 8,95976592 N = 500,000 3,05658884 6,04407973 9,03326882
Table 5: Speedups
Figure 13: Speedup chart
Surprisingly, the achieved speedups are even higher than the "ideal" linear speedup, although the parallel program does even more work than the serial program, since it has to communicate in order to synchronize the output. We can rule out an erroneous measurement since all test runs were performed three times, and the timings were within remarkably close range. One might imagine some external influence (like a background process or another user's job) slowing the serial program down, but this would only have yielded a deviation in one measurement, but not in all nine test runs of the serial program. We can therefore assume that the timings must be correct.
The overproportional speedup can be explained when we look more closely at the differences between the serial and the parallel algorithm: In the serial algorithm, we compute a point, output it, compute the next point, output it, and so on - the whole algorithm is completely serial (Figure 14):
Node 0:
compute output compute output compute output compute output compute ...
Figure 14: Operations in serial execution
In the parallel algorithm, we obviously have several nodes computing points in parallel. But in addition to this, we have a second layer of parallelism introduced by the non-blocking communication because the output of a node (sending a line to the master) is performed in the background while the computation continues:
Node 0:
compute compute compute compute compute compute compute compute compute ... output output
Node 1:
compute compute compute compute compute compute compute compute compute ... output output
Node 2:
compute compute compute compute compute compute compute compute compute ... output output
Figure 15: Operations in two-level parallel execution
It might be argued that the parallel execution of computation and output is an illusion created by the operating system's multi-tasking, since the output process running in the background takes CPU time slices away from the computation process. However, this is not really the case: The transmission of data is an I/O operation that does not involve the CPU anymore after being initiated.
The parallel execution of computation and output in every node is therefore real and a subtle, but deciding factor in the overproportional speedup that we observe. | 1,137 | 5,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.824638 |
https://math.stackexchange.com/questions/2638309/dedekind-cuts-as-a-pedagogical-tool | 1,563,462,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00016.warc.gz | 458,331,585 | 35,102 | # Dedekind Cuts as a pedagogical tool
It's sort of a meaningless and subjective question to ask how valuable a proof or construction is among others if they are all equivalent and they all lead to the same theory. I think, however, it is worth asking whether some proofs and constructions have more pedagogical value than others. (For example, the topological definition of continuity, while true in the limited domain of AB calculus, has almost no pedagogical value until the student is ready to tackle college level analysis.)
My main question is: What are the pedagogical advantages of using Dedekind cuts to construct the real numbers as opposed to the classic approach of metric space completion of the rationals through Cauchy Sequences, or Rudin's construction of the real field by assuming the least upper bound property? I have noticed that some people have strong opinions on the value of Dedekind cuts ("...the stupidest thing ever..."), so I guess what I'm asking for is a defense of the approach at least in analysis pedagogy, or even what new perspectives/theories Dedekind cuts raise that Cauchy sequences or L.U.B. do not elucidate. (This is also for my own education, as I have been taught using Cauchy sequences and L.U.B.)
I apologize if this question is not suited for this site.
• I like the Dedekind cuts, because they make it intuitively obvious that the real line has the LUB property. After all, that's what the cuts are: LUBs of bounded sets. – Arthur Feb 6 '18 at 9:31
• I like that way of thinking about it. To play devil’s advocate, why is that more intuitively obvious than the classic example of a set in an ambient space that does not have an LUB, $\{x \in \mathbb{Q} | x^2 <2 \}$? – Marcus Aurelius Feb 6 '18 at 9:47
• The construction of reals via Dedekind cuts is easiest to understand because it requires only the knowledge of arithmetic / order relations of rationals (available to 12 year old 7th graders). It's a pity that this wonderful thing is purposely hidden from students until they have done a course in calculus and are starting with a course in real analysis. – Paramanand Singh Feb 6 '18 at 10:28
• You might get better answers over at matheducators.stackexchange.com – Stig Hemmer Feb 6 '18 at 10:52
• @MarcusAurelius : the definitions for $<, >, +, -$ using Dedekind cuts are a cakewalk, but some care is needed while defining $\times, /$. You can find nice explanations for these in Dedekind's original paper Stetigkeit und irrationale zahlen (English translation available). – Paramanand Singh Feb 7 '18 at 0:52 | 624 | 2,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-30 | latest | en | 0.944491 |
http://machinedesign.com/batteriespower-supplies/difference-between-ac-electrical-power-and-ac-electrical-energy | 1,481,137,180,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542244.4/warc/CC-MAIN-20161202170902-00344-ip-10-31-129-80.ec2.internal.warc.gz | 170,274,330 | 23,062 | It's essential to understand how power and energy calculations relate, especially with the increased importance of energy efficiency.
Calculations of power and energy are relatively straightforward when the subject is strictly dc circuits.
The situation is more complex in the ac world, however.
### Here's how calculations of electrical power and energy relate to each other ... the basics of watts and watt-hours.
In a dc circuit, the power, P, developed in any component of the circuit is:
P = × I
where P is in watts, W
V is the potential difference in volts across the component, V and
I is the current through the component in amperes, A.
Power is the rate of doing work or, equivalently, the rate at which the circuit transmits electrical energy. Energy can exist whether or not there is any work done. The amount of energy converted into work, or consumed, is equal the product of power and time.
Consider the dc case where power is steady over time. Then the energy consumed J is:
J = P × t
where P in watts, t is in seconds, and J is in joules or watt-seconds. The more typical unit of energy for ac power circuits is watt-hours, Wh. In this case, t is in hours. If P is in kilowatts, kW, and t is in hours, J has units of kilowatt-hours, kWh.
In most real-world circuits, power consumption is not steady. In this case, the power must be integrated over the specified period of time to determine the energy consumed:
J = ∫T2P dt
where the period between T1 and T2 is the interval of interest.
In addition, integration takes place by summing up the power consumed over specific intervals.
### Case in point: electromechanical power meters
Consider electromechanical power meters normally found in residential housing, which use a metal disc that rotates at a speed proportional to the power passing through the meter. The number of disc revolutions is proportional to the energy use. The disc is acted upon by two sets of coils.
• One coil produces a magnetic flux in proportion to the line voltage.
• The other produces a magnetic flux in proportion to the current.
The magnetic fields of the two coils are oriented in such a way as to produce a force on the disc that is proportional to the product of the instantaneous current, voltage and phase angle (power factor) between them. A permanent magnet exerts an opposing force proportional to the speed of disc rotation. The equilibrium between these two opposing forces makes the disc rotate at a speed proportional to the rate of energy use. A register mechanism counts the revolutions to register the total energy used.
Of course, in dc circuits, power is just equal to the current multiplied by the voltage. The situation in ac circuits is more complicated because current typically lags the voltage. This relationship enters into power calculations and also can lead to some misunderstandings about differences between power and energy.
To calculate ac power, the usual approach starts with the equations for instantaneous voltage and current:
i = Im sin ωt
e = Esin (ωtθ)
where Iand Em are the maximum current and voltage respectively, θ is the angle of lag of the current behind the voltage, and ω is 2πf — where f is the ac freqency.
### Multiplying these two relationships together yields the ac power.
With some mathematical manipulation of e and i, it can be shown that the average ac power — EI — is:
### E I = Em Im / 2
After some further mathematical manipulations, the average ac power P is:
### P = E I cos θ
Here cos θ is actually the power factor. Solving the above equation for cos θ yields:
P EI
True power volt-amperes
The quantity E I — volt-amperes— is sometimes called apparent power. The name comes from the fact that E I is simply what's obtained from reading a voltmeter and ammeter and multiplying the two readings together.
To show power relationships in ac circuits, it is often convenient to start with the well-known phasor diagram representation of an ac circuit:
Here Z, R, and Xare impedance, resistance, and inductive reactance respectively.
θ is the angle of lag of the current behind the voltage, as before. To obtain what's called the power triangle shown here, each side of the triangle is multiplied by the average current I.
× (I Z) = I2 Z = E I
× ( I R) = IR = E I cos θ
× (I XL) = I2 XL = E I sin θ
Here sin θ is called the reactive factor.
The term E I sin θ is called reactive volt-amperes and is given in units of var.
One source of confusion associated with reactive volt-amps is that they are sometimes referred to as a measure of reactive power. In reality, reactive volt-amps are a measure of energy, not power. Reactive volt-amps represent the energy supplied to the circuit during part of the cycle, then returned to the system during the part of the cycle when the system inductance is discharging. | 1,082 | 4,861 | {"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} | 3.875 | 4 | CC-MAIN-2016-50 | longest | en | 0.939052 |
http://www.chegg.com/homework-help/two-dt-signals-sk-n-sl-n-said-orthogonal-interval-n1-n2-ak-1-chapter-4-problem-8p-solution-9788120330306-exc | 1,474,863,997,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660602.38/warc/CC-MAIN-20160924173740-00251-ip-10-143-35-109.ec2.internal.warc.gz | 382,621,016 | 16,324 | Digital Signal Processing 4th Edition (0th Edition) View more editions Solutions for Chapter 4 Problem 8PProblem 8P: Two DT signals, sk(n) and sl(n), are said to be orthogonal o...
• 524 step-by-step solutions
• Solved by publishers, professors & experts
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Chapter: Problem:
Two DT signals, sk(n) and sl(n), are said to be orthogonal over an interval [N1, N2] if
If Ak — 1, the signals are called orthonormal.
(a) Prove the relation
(b) Illustrate the validity of the relation in part (a) by plotting for every value of k = 1, 2, ... , 6, the signals [Note: For a given k, n the signal sk(n) can be represented as a vector in the complex plane.]
(c) Show that the harmonically related signals
are orthogonal over any interval of length N.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
(a)
Consider the following condition for two discrete time signals that are said to be orthogonal over an interval :
Here, if , the signals are said to be orthogonal.
For , always results in 1.
For
Therefore,
Hence, the expression is proved.
• Chapter , Problem is solved.
Corresponding Textbook
Digital Signal Processing 4th Edition | 0th Edition
9788120330306ISBN-13: 8120330307ISBN: Authors: | 348 | 1,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-40 | latest | en | 0.856188 |
http://www.thenakedscientists.com/forum/index.php?topic=19231.100 | 1,481,100,755,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00057-ip-10-31-129-80.ec2.internal.warc.gz | 735,906,013 | 16,464 | # The Naked Scientists Forum
### Author Topic: Lessons In Physics (Read 41729 times)
#### DoctorBeaver
• Naked Science Forum GOD!
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• A stitch in time would have confused Einstein.
##### Lessons In Physics
« Reply #100 on: 08/01/2009 18:34:17 »
I note that the last 20 or posts have been devoted to humour, while a little lightening of heavy subjects is not out of place I cannot but feel a little miffed that this goes on while my quite serious discussion for the need for better spelling is vanquished to the outer wastes of 'Geek Speak'.
Sorry [:I]
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #101 on: 08/01/2009 19:32:45 »
"Take a dice. A dice has about 10^80 grams of energy!"
Everything is wrong with it! the statement is completely obscure and meanigless in scientific terms
Firstly energy is not measured in grams
secondly 10^80 is a very big number indeed
Finally precisely how are they related to a dice (strictly the term for one is a die dice is a plural) but how big and heavy is it? it is not something that has a standard size,
As you claim to have written this statement you must be able to explain it in rather more extended and unambiguous terms.
Energy can be measured in grams: Its not unheard of. I have within the last couple weeks read something similar, where Dr Wolf examines how much energy is stored in small systems, and he also measured it in grams. It is related to the amount of matter that is stored in comparrison to the energy that can be extracted in question.
#### lyner
• Guest
##### Lessons In Physics
« Reply #102 on: 08/01/2009 19:50:50 »
You'd need to define you terms.
If the dice were 80g, then the energy equivalent would be mc2.
Which would be about 0.08 X 9e16, or 7.2e15. Where does e80 come from?
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #103 on: 08/01/2009 19:54:35 »
No no, these grams measure the energy.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #104 on: 08/01/2009 20:01:42 »
In fact, i am probably just confusing everyone this way. I apologize for this, because i promised i would not intentionally be confusing, so, instead, i will carefully measure the matter if a dice, and then convert it into energy mathematically with the usual units.
#### lyner
• Guest
##### Lessons In Physics
« Reply #105 on: 08/01/2009 21:25:01 »
The use of grams is confusing in itself because we normally use SI units.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #106 on: 08/01/2009 22:09:35 »
I know. I am yet to get my scientific scales out and then mathematically transpose this into the energy. Give me time, and i will release a new configuration, with units you will recognize, as well as everyone who use them notationally everyday. I do realize, i made a mistake believing this would have been appropriate.
#### lyner
• Guest
##### Lessons In Physics
« Reply #107 on: 08/01/2009 22:16:37 »
It might be a good idea to rehearse your ideas in private and get them right before you expose them to the rest of the world.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #108 on: 08/01/2009 22:19:58 »
That could be a wiseful idea.
Are you a moderator here?
#### lyner
• Guest
##### Lessons In Physics
« Reply #109 on: 08/01/2009 22:34:00 »
Yep.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #110 on: 08/01/2009 22:44:55 »
Then i will post you my work on Cartesian Coordinates, as that will be my next chapter, expalaining time as invariant under space.
Is that ok?
#### lyner
• Guest
##### Lessons In Physics
« Reply #111 on: 08/01/2009 22:49:26 »
And you expect me to wade through yards and yards of stuff which I find boring? Most technical authors PAY someone to do that sort of thing.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #112 on: 08/01/2009 22:51:13 »
Well then, don't let me bore you.
#### Mr. Scientist
• Neilep Level Member
• Posts: 1451
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##### Lessons In Physics
« Reply #113 on: 08/01/2009 22:52:13 »
Secondly, don't suggest something, unless you are willing to do your bit. As in getting someone here would be a moderator, to moderate.
In fact, don't do your job at all, since that is what you are implying.
#### Chemistry4me
• Neilep Level Member
• Posts: 7709
##### Lessons In Physics
« Reply #114 on: 08/01/2009 22:55:04 »
Thought this was a lesson in physics not geography?
Indeed. So, let's talk about relecting light through 90o - a right-Angola.
O-man! You're losing it Doc . Maybe you should look in the atlas, there might be some good ideas In-dia !
#### lyner
• Guest
##### Lessons In Physics
« Reply #115 on: 08/01/2009 22:58:08 »
Moderators are not there to do other people's research for them. We are there to deal with spam and other sources of wasted time. Also to remind people about how the Forum expects them to behave.
#### lyner
• Guest
##### Lessons In Physics
« Reply #116 on: 08/01/2009 22:59:37 »
C4me
That's Africkin good joke.
#### Mr. Scientist
• Neilep Level Member
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##### Lessons In Physics
« Reply #117 on: 08/01/2009 23:06:52 »
lol
#### Chemistry4me
• Neilep Level Member
• Posts: 7709
##### Lessons In Physics
« Reply #118 on: 08/01/2009 23:55:19 »
Thank you sophiecentaur, I must say that my brain has never Ben-in such good shape!
#### Bikerman
• Sr. Member
• Posts: 119
##### Lessons In Physics
« Reply #119 on: 09/01/2009 00:23:58 »
Without wishing to distract this playful interlude I wish to make an observation:
Mr Scientist - I appreciate all offers to teach, and as a teacher myself I understand the commitment and effort needed in that role. I think your offer was well meant, but I would ask you to reconsider, given what you have posted here. It is apparent to me that you really haven't got a scheme of work, let alone detailed lesson plans. These are essential. Nor do I think you actually have a proper grasp of the subject - a major drawback. Now, I completely understand that a teacher is often his/her own best pupil and teaching something is the surest way to test that you actually understand it yourself. That said, I think it is unwise to offer to teach quite complex physics unless you have a real understanding of the basics. I'm not at all sure that you do.
Don't take that as an insult...take it as a constructive suggestion.
#### Soul Surfer
• Neilep Level Member
• Posts: 3345
• keep banging the rocks together
##### Lessons In Physics
« Reply #120 on: 09/01/2009 09:44:47 »
Far worse than that Bikerman The stuff he is purporting to "teach" is totally riddled with inconsistencies and errors that are more likely to confuse people and give them the wrong idea.
I believe that he has malevolent intents even though he whinges victimisation. These are the typical actions of an internet chat page troll. I have seen enough of them to know one when I see it.
#### yor_on
• Naked Science Forum GOD!
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• (Ah, yes:) *a table is always good to hide under*
##### Lessons In Physics
« Reply #121 on: 10/01/2009 13:16:57 »
Soulsurfer, even though I think (as you:) that Mr S. might have had an agenda of his own when he was presenting his material.
I do not think he was having 'malevolent intentions', rather that he was building up to a 'idea' of his own.
Using what he himself have learnt and trying to give us what he saw as the 'proper' references for it.
But physics and especcially mathematics is a very stringent subject, so using it one has to be prepared for 'flack'.
He might come across as a somewhat 'straightlaced' person?
And as having taken on somewhat more than he could 'chew' here:)
But then again:)
"Yes, there were times, I'm sure you knew
When I bit off more than I could chew.
But through it all, when there was doubt,
I ate it up and spit it out.
I faced it all and I stood tall;
And did it my way."
As we all have to do at times.
« Last Edit: 10/01/2009 13:22:59 by yor_on »
#### Soul Surfer
• Neilep Level Member
• Posts: 3345
• keep banging the rocks together
##### Lessons In Physics
« Reply #122 on: 11/01/2009 12:32:13 »
As far as I can see his only intent is to gain some sort of fame and notoeriety by arousing interest and argument from his activities irrespective of their scientific accuracy. This has nothing to do with most of the contributors here who wish to learn from and help others to learn some genuine and accurate science. OK we also allow people to expound some innovative and wild ideas in the right areas but Mr Scientist is not trying to do this. He appears to be concentrating on main stream science and it has been clearly demonstrated several times he does not understand the stuff he is posting and it is also often copied from banned contributors on other science chat pages.
#### Bikerman
• Sr. Member
• Posts: 119
##### Lessons In Physics
« Reply #123 on: 13/01/2009 00:33:40 »
Well, I try to take the most charitable line. As a moderator on other forums, I find it very easy to quickly categorise people as trolls (I see plenty of such behaviour). To avoid becoming over-cynical, therefore, I try to imagine the best, given that people often have problems posting articulately (in the case of Mr.S I am guessing that English might be a second language).
I have acted as 'mentor' to enough trainee teachers over the years to know that some people with the best of intentions should really not be let loose in a classroom situation. I was, perhaps over-charitably, assuming that Mr.S falls into this category rather than trying to deliberately mislead.....
#### Chemistry4me
• Neilep Level Member
• Posts: 7709
##### Lessons In Physics
« Reply #124 on: 13/01/2009 00:39:52 »
I think Mr.S has already decided To-go! *I can't believe nobody has thought of that*
#### The Naked Scientists Forum
##### Lessons In Physics
« Reply #124 on: 13/01/2009 00:39:52 » | 2,757 | 10,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-50 | latest | en | 0.904599 |
http://physics.stackexchange.com/tags/order-of-magnitude/hot | 1,467,289,012,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398628.62/warc/CC-MAIN-20160624154958-00094-ip-10-164-35-72.ec2.internal.warc.gz | 242,949,547 | 19,667 | # Tag Info
## Hot answers tagged order-of-magnitude
72
The energy of a bullet is around 735 joules (see bullet details here). This is about the same energy that I have when I'm running at about 4.6 m/s. Would you rather be hit by me or the bullet? The bullet kills you because it concentrates all the energy onto a small impact area while my impact area is rather larger (and sadly getting even larger as ...
68
7 TeV is not that much kinetic energy, that has been covered by your question and previous answers. However, in the context of a proton, with a rest mass of $1.672×10^{−27}~\mathrm {kg}$ (very, very little mass), when a single proton has 7 TeV then it is travelling at a specific speed: $$E= mc^2$$ \begin{align}E& = E_0 + \text{KE}\\ \text{KE}&=E- ...
39
Neither of those statements are true. It's an easy approximation to make: a neutron star has all of that 'space' removed from between nucleons --- so we just need to know how big a neutron star of mass equal to the solar system would be. Well, the only significant mass is the sun (jupiter is about 1% the mass of the sun---negligible). If the sun were ...
35
There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough that subleading terms in the series expansion for $f\bigl(\frac{Y}{X}\bigr) - f(0)$ can be neglected, where $f$ is some relevant function involved in the ...
11
$\gg$ doesn't mean anything about a specific order of magnitude. If you are told to assume $Y \gg x$, what that means is if you have a series expansion of the form $$\sum C_n \biggl(\frac{x}{Y}\biggr)^n\tag{1}$$ you can neglect all subleading terms (i.e. all except the first one with a nonzero coefficient). For comparison, a simple statement $Y > x$ does ...
10
This is perhaps similar to what mbq meant, but I will elaborate. The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance. If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane. I am not changing the pressure of the triple ...
10
If you could compress the mass into that small a space, it would collapse into a black hole, at which point the notion of "size" becomes harder to define, with space-time being so warped. The "event horizon" radius would be about 3 km, if I get the formula correctly. The idea of "there's a lot of space in atoms" comes from computations which state that the "...
10
Generally taking note of these relationships is a precursor to either (a) applying an approximation or (b) using a purturbative or series solution. In case (a) what qualifies is completely a matter of your sensitivity to error. If you are going to throw out terms $\mathcal{O}(C^{\pm 2})$ and require a 1% approximation then $C$ had better differ from 1 by a ...
9
Live on earth is protected from solar wind by the earth's magnetic field. Charged particles from the sun (mostly) penetrate the earth's atmosphere with great velocity. These particles can be trapped by a magnetic field to follow circular path's around the magnetic field lines, thereby losing their energy due to collisions or bremstrahlung. From first ...
9
This is a complement to dmckee's answer by way of an example. Suppose that I want to determine the acceleration due to gravity of an object near the surface of the earth. Let $h$ be the height of the object above the surface, then according to Newton's Law of Gravitation, I get \begin{align} a = \frac{GM}{(R+h)^2} = \frac{GM}{R^2}\left(1-2\frac{h}{R}+O\...
9
For forces that can be expressed in terms of a quantum field theory, you can compare the size of the coupling constants (which are dimensionless). In short this means that perturbative expansions of weaker forces are well represented by a small number of leading terms because the series converges quickly, while those of stronger forces require more terms (or ...
8
a great topic. First, ten gigatesla is only the magnetic field near a magnetar - a special type of neutron star. They were discussed e.g. in this Scientific American article in 2003: http://solomon.as.utexas.edu/~duncan/sciam.pdf Ordinary neutron stars have magnetic fields that are 1000 times weaker than that. It is true that in the magnetar stars, atoms ...
8
There is a great answer (with references) to this at http://answers.google.com/answers/threadview/id/539329.html which I'll summarize as follows: From http://www.hawaii.edu/suremath/jsand.html the estimate for the grains of sand is 7.5 x 10^18 or 7.5 billion billion. From http://www.faqs.org/faqs/astronomy/faq/part8/section-3.html the estimate for the ...
7
Hmmm...some back-of-the-envelope calculations: The depth of the air column at sea level is $14\text{ lbs/in}^2 = 2 \times 10^5\text{ g/cm}^2$, so neglecting space-charge effects and assuming minimum ionization the whole way we get about $4 \times 10^5\text{ MeV} = 0.4\text{ TeV}$ energy loss. We are actually above minimum ionization, so we can multiply that ...
7
The "removing the space" and "atoms are mostly empty" memes for atomic nuclei are interesting, but I do grit my teeth every time I hear this. A description that fits better with me might be "remove the electromagnetic force". Concepts of size and space of particles are based on how they interact using forces. There is no evidence that fundamental particles ...
7
The most common case where this comes up is when you're dealing with a problem where it's helpful to linearize using a Taylor expansion. For example, a decaying exponential $$e^{-t/t_0} = 1 - \frac{t}{t_0} + \frac12\left(\frac{t}{t_0}\right)^2 - \frac1{3!}\left(\frac{t}{t_0}\right)^3 + \cdots$$ or a smallish logarithm $$\ln(1+x) = x - \frac{x^2}{2} + \... 6 One notable class of exceptions are what are called "hierarchy problems" in particle physics. For example, if you identify the Planck mass as a fundamental energy scale, you end up with huge dimensionless numbers which don't have an obvious explanation (i.e. ratio of Planck to electroweak scale, etc.). Explaining these large (or small depending on how you ... 6 Let's make some assumptions. First, assume the fish is rigid. Second, let's assume he's not flapping. Third, I guess let's assume it's a male fish since I said "he." We'll also assume this is 2D because we're looking for an approximation. I would approximate the fish as an airfoil. NACA airfoils are a pretty good choice because they are analytically defined ... 6$$ 110 \text{ hp} = 82 \text{ kW} $$This is 1000x what the laptop draws. You won't notice. To put that change in perspective, you would see a similar increase in the power (85 W) used by the car if your speed changed from 65.00 mph to 65.02 mph, since P \propto v^3 (at high speed, the power goes as the velocity cubed) as per this answer, so \frac{dP}{... 5 According to http://ga.water.usgs.gov/edu/earthhowmuch.html the total volume of water on the earth is 1.386\times 10^9km^3, which is about 1.4 \times 10^{21}kg (I'm rounding because I don't know the average temperature and therefore density of the water). According to http://en.wikipedia.org/wiki/Biomass_(ecology)#Global_rate_of_production the annual ... 5 Consider a spherical drop of water, initial temp 40C, radius 3mm, mass 0.1g To get it down to 0C, you need to remove 4.18 (J/gK) * 0.1 g * 40 K = 17 J then, to freeze it solid, you need to remove latent heat of fusion 333 (J/g) * 0.1 g = 33 J for a total of 50 J. The heat conductivity equation is H=\frac{\Delta Q}{\Delta t} = k A\frac{\Delta T}{x} ... 5 Molecules vibrate with frequencies in the range 10^{12} to 10^{14}Hz. Although I don't know of any strict definition, I would take the view that a molecule must hold together for a few vibrations otherwise what you have is a collision not a molecule. That means the lifetime must be greater than 10^{-14} to 10^{-12} seconds, depending on the molecule. ... 5 I think the answer is no. It generally precedes some approximation method with a bounded error, but there are so many approximations methods in physics -- some rigorous, some nonrigorous -- that it's way too presumptuous to give it a rigorous definition. Generally, it means one of several things: If a\ll b, expanding in powers of \frac{a}{b} is ... 4 Well, of course you have to pick the quantities in your dimensional analysis right. Example: Use dimensional analysis to estimate the potential energy of a star, hold together only by gravitation. Solution: Newtons gravitational constant G better show up somewhere. This requires us to include something with units kg^2 / m. We can get this by inserting ... 4 Yes, I just did it. Arced a small spheriod of spit (no phlem) and it hit the ground and rolled instead of splattered. -15 degrees F, light wind in parking garage (not sure if that matters) attempted with numerous quantaties of spit and trajectories but small quantity with upward trajectory is what got the tiny frozen droplet. 4 I suppose the OP is looking for some general rule to be used when you want to say "A is N orders of magnitude bigger (smaller) than B". In that case, consider$$N = || \log_b(A/B)) || (where $|| \dots ||$ is taken to mean round to the nearest integer, and negative values just mean chose "A is smaller than B", but the magnitude retains the same ...
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