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https://cs.stackexchange.com/questions/57546/are-coevolutionary-free-lunches-really-free-lunches/96473#96473 | 1,652,783,294,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00151.warc.gz | 233,632,802 | 69,505 | # Are coevolutionary "Free Lunches" really free lunches?
In their paper "Coevolutionary Free Lunches" David Wolpert and William Macready describe a set of exceptions to the No Free Lunch theorems they proved in an earlier paper. The exceptions involve two-player games in which a player tries to minimize expected search cost given optimal play, or at least good play, by an opponent.
Free lunches are "allowed" in this case because the exact choice of cost function to minimize changes depending on which fitness (i.e. objective) functions previous rounds of play have "ruled out." In other words, given that an opponent already knows something about the game, and chooses responses that minimize the player's expected return, the player can eliminate certain fitness functions without having to evaluate them. To illustrate how this works, W&M provide this chart:
Here, $\langle g \rangle$ represents the search algorithm that pays no attention to the opponent's moves; $\langle g \rangle_{1}$ represents the search algorithm that considers all possible opponent replies to each move; and $\langle g \rangle_{2}$ represents the search algorithm that samples just one possible opponent reply to each move.
This illustrates the character of the free lunch provided: algorithms that take into account information provided by the opponent do better than algorithms that don't, and those that collect as much information as possible from the opponent do better than those that only collect some. W&M amplify this point in their later discussion of opponent intelligence. They show that even when an opponent is not omniscient, but has partial knowledge, the player can exploit its partial knowledge. In the case of total ignorance this won't work because the opponent always replies with a random move. In that case, there appears to be no gain:
the expected performance of the agent will be the average over the antagonist’s possible responses.
I guess this looks like $\langle g \rangle$ above. But in cases where the opponent has some knowledge, algorithm performance appears to increase monotonically with opponent intelligence.
What confuses me about all this is that the argument appears to boil down to the following trivial-seeming claim: across all problem domains, algorithms that bother to learn from knowledgable opponents do better than algorithms that do not. That is, as long as you don't turn down the free lunch, you can have it.
In effect, lunch is free because the opponent is buying.
If that's true, then couldn't we imagine a whole range of games in which, say, players play cooperative games with oracles, and those that refuse to cooperate perform worse on all problems? That's a free lunch in the same sense, right? But then we haven't explained where the oracle's knowledge comes from.
Does this mean that if we had to model the source of the other player's knowledge, we'd be back in the No Free Lunch zone? Or is the claim really that this sort of competitive play yields better outcomes even when both players start out in total ignorance, as the phrase "free lunch" seems to imply?
Coevolutionary algorithms can't magically accelerate progress on any arbitrary problem class. So in that sense, the conclusion at the end of the question is correct. However, it doesn't follow that all coevolutionary free lunches are trivial, as the conclusion of the question suggests.
I can't offer an exhaustive account of all the kinds of coevolutionary free lunches, but I can offer two examples, the first of which is trivial, and the second of which I would argue is nontrivial. The second is nontrivial because it helps explain why the no free lunch theorem really must hold.
The important difference between the two examples is this: in the first, the competing algorithms are competing to achieve the same overarching goal, while in the second, the competing algorithms are trying to achieve different goals. In the second case, the mismatch between the two algorithms' goals allows interesting things to happen. I'll begin with the trivial example.
### Opponents seeking the same goal
Imagine a very simple optimization problem in which the search landscape is a 7x7 grid of cells. The primary goal is to find the cell with the maximum value. 48 of the cells have a value of 0, and one randomly chosen cell on the grid has a value of 1.
Our secondary goal is to discover a search strategy that finds the maximum value more quickly. But it follows from the initial problem that no strategy could possibly beat random search here, because nothing can be learned from one cell about another. Nonetheless, the coevolutionary free lunch theorem holds! Here's why:
Suppose you have two optimization algorithms, A and B, both searching the grid. It doesn't really matter what strategy they use, but for concreteness, we'll stipulate that they both use a random search strategy. The only difference between them is that B pays attention to A's moves, and when it sees that A has found the maximum value, it jumps to that cell too. In some sense, when that happens, B has still "lost" the contest. But if you run a lot of competitions, and then compare the average performance of A to the average performance of B, you'll see that B finds the maximum value faster on average.
The explanation is simple. The average time to first discovery -- whether by A or B -- stays the same. But whenever A beats B to the win, B doesn't bother looking anywhere else. It skips ahead to the best cell. When B beats A to the win, on the other hand, A just continues searching, plodding along until it finds the maximum value on its own.
This looks like a win if you're only counting the number of moves B makes. If you look at the total number of moves that A and B make together, they actually do worse together than either would on its own, on average. That's because of A's obliviousness. If we change A to behave the same way as B, then they do about as well together as either would on its own -- but no better.
So here, modeling both algorithms together returns us directly to the no free lunch zone, just as the question argues. In effect, A and B are just performing random search algorithms in parallel. The final number of search operations remains the same.
### Opponents seeking different goals
Now imagine a very different scenario. Suppose we have a classification problem: recognizing sheep. Here, A's job is to look at a stream of pictures and say whether there are sheep in them. Simple enough.
But B's job is very different. B has the power to inject pictures of its own into the stream! Its goal is not to identify sheep; it just wants to slow A down.
Does anyone have a picture of sheep in a really unusual place? It's for pranking a neural net.
And here's one of the first replies:
Here's the orange sheep:
Turns out, this was perfect for pranking:
You totally got it. Orange sheep are not a thing it was expecting. "a brown cow laying on top of a lush green field"
Here are a couple of other examples from Shane's blog post:
So what does this have to do with our problem? We can connect them by being more precise. Suppose A's goal is to reach greater than 99% accuracy, and B has the power to inject one picture into A's stream for every nine "natural" pictures. B looks for patterns in A's behavior and uses them to find pictures that mess with its model. This will keep A's accuracy below 99% for much longer than if A saw only "natural" pictures.
Two things follow from this. First, B will do much better if it pays attention to what A does. If B just picks images on the basis of some randomly chosen general principle like "sheep in odd places," then there's a good chance that A will be prepared for them already. If not, it will quickly learn to handle them correctly, and B will then have to adopt a new strategy. On the other hand, if B watches A's behavior, it can pick out the specific things that A is worst at, and focus on those. As soon as A improves at one of them, B can have another one ready to go. As long as B can find patterns in A's behavior, B will always present the most challenging images for A.
Second, A will do much better if it pays attention to which images B picks. After all, B is looking for patterns in A's behavior. If it finds patterns, then it will use those patterns to send fake or troublesome pictures to A. In turn, that means that there will be noticeable patterns driving B's choices. Here again, if A is paying attention to the patterns in B's behavior, it will more quickly be able to identify which pictures B is injecting.
What's important is that in this scenario, both A and B are relying on data that is guaranteed to have patterns. It is guaranteed to have patterns because if A is doing its best, A is doing something other than random search. And if B is doing its best, then B is doing something other than random search.
So initially, this looks like a really compelling free lunch situation. But what have we actually shown? We've shown this:
As long as A is doing something other than random search, B can always find out-of-band samples that A's methods cannot handle.
That's the no free lunch theorem in a nutshell!
The only way A can prevent B from finding out-of-band images is by behaving in ways that look random to B. But if A's behavior isn't really random, then in the very long run, B will always be able to find the pattern -- even if B itself is only doing random search.
The very same argument works in the other direction. The only way B can prevent A from noticing patterns in its images is by behaving in random-seeming ways. But if B's behavior isn't really random, A will eventually find the pattern, even if it's only doing random search.
In this scenario, both algorithms will try to fool each other by adopting more and more complex, random-seeming behaviors. So in the very, very long run, they will slowly converge to a truly random search -- which is the best any algorithm can do on average across all problem domains.
### An infinite pool of randomness
It might be that these coevolutionary learning strategies do still have this advantage over others: they may encourage both algorithms to explore the space of relevant non-random behaviors more quickly or extensively. I am not even sure of that. Either way, the no free lunch theorem holds in general, because the space of possible non-random behaviors is far, far smaller than the space of possible random behaviors.
How do we know? It would be off topic to go into a detailed proof, but consider the related question of how many long strings can be compressed into shorter strings. Regardless of the compression method, the majority of all strings cannot be compressed at all. This is easy to prove with a bin-counting argument. Suppose we consider binary strings, and start with the empty string. Assuming there can be no negative-length strings, that's incompressible. Now consider length-1 strings. There are two, but there's only one length-0 string, so only one of them can be compressed. Now we have two incompressible strings, and a third string that can be compressed by one bit. Moving on to length-2 strings: there are four, but there are only two length-1 strings, and the one length-0 string is already taken, so we can only compress two of the length-2 strings. The other two are incompressible. That's three compressible strings and four incompressible strings... and so on.
As the numbers get higher, one of the things you notice is that even among the compressible strings, half of them are only compressible by one bit, because they compress to strings that are themselves incompressible. A quarter are only compressible by two bits; an eighth are only compressible by three bits. No matter how you slice it, the number of substantially compressible strings is always much lower than the number of strings that are incompressible or barely compressible.
The line of reasoning for random behavior is similar. You could also connect these ideas to the proof that there are vastly more real numbers than integers. In the global scheme of things, the no free lunch theorem is true because the scope of randomness is unimaginably vast. | 2,570 | 12,248 | {"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} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.938627 |
https://www.physicsforums.com/threads/finding-b-a-c-for-real-roots-of-sqrt-2014-x-3-4029x-2-2-0.1042801/ | 1,723,414,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00145.warc.gz | 734,822,736 | 17,917 | # Finding $b(a+c)$ for Real Roots of $\sqrt{2014}x^3-4029x^2+2=0$
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• anemone
In summary, to find the value of $b(a+c)$ in the given equation, use the quadratic formula with the values of $a=\sqrt{2014}$, $b=-4029$, and $c=2$. There are various methods, such as factoring, completing the square, and using the quadratic formula, that can be used to solve this type of equation. To check if the solution for $b(a+c)$ is correct, plug the value back into the original equation. The equation can only have one solution for $b(a+c)$ and there are restrictions on the values of $a$, $b$, and $c$ in order for the equation to have real roots.
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Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$.
To avoid radicals let $\sqrt{2014}=p$
So we get $px^3-(2p^2+1)x^2 +2 = 0$
Or factoring we get $(px-1)(x^2-2px-2)$ = 0
So one root is $x= \frac{1}{p}$ and other two roots are roots of $x^2-2px-2=0$
For the equation $x^2-2px-2=0$ sum of the roots is 2p and product is -2. so one root has to be -ve and
the postiive root shall be above 2p
So $b=\frac{1}{p}\cdots(1)$
And c is the -ve root and $a> 2p$
a,c are roots of $x^2-2px-2=0$ so $a+c = 2p\cdots(2)$
Hence $b(a+c) = 2$ using (1) and (2)
## 1. What is the formula for finding $b(a+c)$?
The formula for finding $b(a+c)$ in the equation $\sqrt{2014}x^3-4029x^2+2=0$ is $b(a+c) = \frac{-b}{a}$, where $a$ and $b$ are the coefficients of the quadratic term and the linear term, respectively.
## 2. How do you know if a given equation has real roots?
A given equation has real roots if the discriminant, $b^2-4ac$, is greater than or equal to 0. If the discriminant is less than 0, the equation will have complex roots.
## 3. Can the value of $b(a+c)$ be negative?
Yes, the value of $b(a+c)$ can be negative. This would occur if the coefficient of the linear term, $b$, is negative and the coefficient of the quadratic term, $a$, is positive.
## 4. What is the significance of finding $b(a+c)$ for real roots?
Finding $b(a+c)$ for real roots is significant because it allows us to determine the sum of the roots of the equation. This can help us in solving the equation and understanding the behavior of the graph of the equation.
## 5. Can the value of $b(a+c)$ be 0?
Yes, the value of $b(a+c)$ can be 0. This would occur if the coefficient of the linear term, $b$, is 0. In this case, the equation would simplify to $ax^2+c=0$ and the value of $b(a+c)$ would be 0 regardless of the value of $a$ and $c$.
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1. Output
2.
3. Print a wrong answer of A-B. Your answer must be a positive integer containing the same number of digits as the correct answer, and exactly one digit must differ from the correct answer. Leading zeros are not allowed. If there are multiple answers satisfying the above conditions, anyone will do.
4. The correct answer of 5858-1234 is 4624. So, for instance, 2624, 4324, 4623, 4604 and 4629 will be accepted, but 0624, 624, 5858, 4624 and 04624 will be rejected.
5.
6. a,b = map(int, raw_input().split())
7. n = a-b
8. count = 0
9. while n!=0:
10. if n%10 == 0:
11. n = n/10
12. count += 1
13. else:
14. print (n-1)*(10**count)
15. n = 0
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# GMAT Study Plan - How to Start your GMAT Prep
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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12 Aug 2010, 15:02
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Question for BB – Should I read the Kaplan Premier book simultaneously with the MGMAT quant series or do I first read the Kaplan Premier book, then start the MGMAT series? I plan to start studying next week and take the GMAT in Dec/Jan. Thanks for the advice. Great forum.
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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14 Aug 2010, 00:59
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CheriInfinite wrote:
<continued from my previous post>
The Next two days
In the next two days, i plan to step up my preparation. I am planning to take off from work so that i will be able to give intensive preparation. Simply speaking i will be increasing the scale of preparation like what i said for the first five days. this will keep me confident and i will give at least 3 mocks during this period.
The Final Three days
The final three days, i will set aside for revising the concepts. i will go back to the notes i ve made during my preparation, various tips given online in the forums. This will be the relaxation phase for me. even during relaxation, i wanna feel confident that i will crack the examination. With a positive attitude i will carry out the tasks for the day. I will read lots of poems that my friend has selected for me and reflect at the beauty of the language. understanding these poems will take me a long way to understanding RCs.
The final day before the examination, i will go to the beach, chill out, drink lots and lots of water and relax to the max. i will catch up on my fav hindi movie(jaane tu ya jaane na) and english movie ( V for vendatta ) and relax. a short quick look at the notes for half hour will be the only studies i will do before the day.
This is my strategy. Please advise and give more suggestions for the same.
Please add kudos if you like my post
Nice post . I am writing my Gmat again in another 20 days or so and I can feel the tension / nervousness building in me. I will see how to unwind from work few days before my test .
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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14 Aug 2010, 08:17
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i wish i'd gone through this post before taking my GMAT. Anway, this should come in handy for my second attempt at GMAT.
thanks for the post.
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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17 Aug 2010, 07:50
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bb,
I seem to be having trouble finding any information as far as pacing advice is concerned. Have we had any topics on the issue specifically about what we should target or is this issue something that by nature will fluctuate from member to member?
Disclaimer, I have searched on the forum for this information but I can't seem to find any threads with expert advice.
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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17 Aug 2010, 07:58
Do you mean something like this?
timing-strategies-on-the-gmat-80176.html
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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17 Aug 2010, 10:13
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bb wrote:
Do you mean something like this?
timing-strategies-on-the-gmat-80176.html
Exactly, the notation got the best of me. Thanks
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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17 Aug 2010, 20:23
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Excellent post BB..
I think gmatclub is amazing place to be.. thanks to all the members for creating such an wonderful community
Really loving it
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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17 Aug 2010, 20:40
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jcurry wrote:
bb wrote:
Do you mean something like this?
timing-strategies-on-the-gmat-80176.html
Exactly, the notation got the best of me. Thanks
This is basically what I was looking for except more of a 'what do you recommend' instead of allowing myself to make it up.
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17 Aug 2010, 20:43
Or see this one: gmat calendar
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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19 Aug 2010, 04:40
I agree. You'll really loose track of time thinking that 3 months is more than enough
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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19 Aug 2010, 09:57
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Hi,
I have been reading the GMAT study plan suggested by all bb and other members.
Here's my plan, based on several combinations which I think will suit me. Would like to have your insights on this one -
1. Day Zero - Princeton Review CAT1
2. Princeton Review, after completing half of the PR book, CAT2
3. After completion of PR, CAT3
4. OG 11th edition - Here I plan to complete the book in 3 parts, after 1/3rd of the questions in all the sections are exhausted, I will take a test and gauge my progress. I will go on doing this until I am done with the OG.
5. After building on the basics, I'll take upon the MGMAT SC and such other specific material ( like the PowerScore CR, or the Verbal and/or Quant review) to iron out the flaws
6. Take the CAT7
7. Kaplan 800 - Quant & Verbal to solve some of the tougher problems - here I'll use the same strategy as with the OG.
8. GMAT Prep 1 - After exhausting the MGMAT, PR and Gmat Club tests and on completion of the Kaplan 800
9. Towards the end of my prep, I'll go back to the OG, so as to remain fresh with actual questions for my exam.
10. GMAT prep 2.
11. OG
12. Exam day!!!
The reason, I am preparing in this distributed manner, and not sticking to a single section for a month is because I believe after a month of Quant and thereafter a month of Verbal, my Quant would again become dodgy and so on.
I've booked a late december date for the exam, so I've roughly 4 months.
Is this a good enough plan or do I need any improvements?? Your suggestions would be really welcome
Thanks
Rahul
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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19 Aug 2010, 10:04
I don't see enough books that would cover basics well (arithmetic, grammar, etc). I would strongly encourage you look into other books here: top-gmat-prep-books-guides-reviews-comments-77703.html?highlight=ml306
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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21 Aug 2010, 09:50
1
KUDOS
Thanks for all your posts.
I am brand new here. I start this long journey to GMAT and I have many questions about How to plan my work.
I started my revisions a week ago. Most of my work will be on the quantative section as I don't have a profile background in engineering. In fact, all this stuff sounds relatively "new" to me, but anyway, I expect at least 680, and why not 700+.
I am 25, currently working as junior key account manager in a MNC (Cadbury, for 2 years ), graduated from a French business school.
I really want to improve my knowledge in arithmetic, algebria and geometry before registering to the test.
How long do you think is needed to be good enough to hope a 700 GMATscore (Q41;V41)???
I'll take time if 4months of work or more, I won't hesitate!!
thank you for your report!
I have not registered to the GMAT yet.
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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21 Aug 2010, 13:37
Welcome to GMAT Club!
I think you can (and should) do better than Q41 - just get the MGMAT Guides or Kaplan Math Workbook or the Veritas Prep Guides and you should be good to go.
As to how long it takes, it really depends on your starting point - have you taken a GMAT Prep or another diagnostic test?
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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23 Aug 2010, 08:15
1
KUDOS
bb wrote:
bipolarbear wrote:
excellent post bb.
One thing I would emphasize is how quickly procrastination time goes by. It is way too easy to begin studying with 3 months left, take a short break, and resume studying with only weeks left. On the one hand, it seems like one month is plenty of time to study for just a single test, but after really starting, you realize that you have a LOT to cover and begin panicking. I know that's the case with me. 2 months ago I thought I had an eternity to study, but a couple of days ago I realized I am screwed and wish I had more time.
Thank you!
Great point - the message should be clear: If you only have 3 months left and need 700+ score - you are already late!
bb just one thing..iv asked u this before also...i badly want a 700+ score..so if i follow the 3 month plan (even though uv mentioned above that its not enuf) its not possible to get the desired score?? need some clarification here..cos iv set a 5 to 6 month study plan..then i started considering the 3 month plan u have suggested..now a lil confused.. help!
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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23 Aug 2010, 09:30
Thanks for keeping me honest Jayanth2689!
1. There is no absolute. Your improvement depends on your starting point. Obviously if you are at 650, a month maybe enough to 700
2. The study plan is about 3 months, but the full cycle is closer to 4 months (2 extra weeks to get books and 2 extra weeks to get the exam scheduled and enough courage to schedule the test)
Bottom line - if you have no clue about GMAT and rusty, chances are you need 3-4 months commitment to get a solid score. If you are at a decent place and have an idea about what GMAT is, you can do it much faster. The key is the starting point!
Jayanth2689 wrote:
bb wrote:
bipolarbear wrote:
excellent post bb.
One thing I would emphasize is how quickly procrastination time goes by. It is way too easy to begin studying with 3 months left, take a short break, and resume studying with only weeks left. On the one hand, it seems like one month is plenty of time to study for just a single test, but after really starting, you realize that you have a LOT to cover and begin panicking. I know that's the case with me. 2 months ago I thought I had an eternity to study, but a couple of days ago I realized I am screwed and wish I had more time.
Thank you!
Great point - the message should be clear: If you only have 3 months left and need 700+ score - you are already late!
bb just one thing..iv asked u this before also...i badly want a 700+ score..so if i follow the 3 month plan (even though uv mentioned above that its not enuf) its not possible to get the desired score?? need some clarification here..cos iv set a 5 to 6 month study plan..then i started considering the 3 month plan u have suggested..now a lil confused.. help!
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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23 Aug 2010, 09:48
1
KUDOS
bb wrote:
Thanks for keeping me honest Jayanth2689!
1. There is no absolute. Your improvement depends on your starting point. Obviously if you are at 650, a month maybe enough to 700
2. The study plan is about 3 months, but the full cycle is closer to 4 months (2 extra weeks to get books and 2 extra weeks to get the exam scheduled and enough courage to schedule the test)
Bottom line - if you have no clue about GMAT and rusty, chances are you need 3-4 months commitment to get a solid score. If you are at a decent place and have an idea about what GMAT is, you can do it much faster. The key is the starting point!
Jayanth2689 wrote:
bipolarbear wrote:
excellent post bb.
One thing I would emphasize is how quickly procrastination time goes by. It is way too easy to begin studying with 3 months left, take a short break, and resume studying with only weeks left. On the one hand, it seems like one month is plenty of time to study for just a single test, but after really starting, you realize that you have a LOT to cover and begin panicking. I know that's the case with me. 2 months ago I thought I had an eternity to study, but a couple of days ago I realized I am screwed and wish I had more time.
Thank you!
Great point - the message should be clear: If you only have 3 months left and need 700+ score - you are already late!
bb just one thing..iv asked u this before also...i badly want a 700+ score..so if i follow the 3 month plan (even though uv mentioned above that its not enuf) its not possible to get the desired score?? need some clarification here..cos iv set a 5 to 6 month study plan..then i started considering the 3 month plan u have suggested..now a lil confused.. help!
[/quote]
lol no problem at all! i have to be honest that iam very much rusty with the concepts! like my post in a different topic..i scored oly a 460 on the GMATPrep Test but it duznt bog me down cos i havnt started an indepth prep for the test! and when u ask if i know what the GMAT is..its like how Neo asks Morpheus what the Matrix is Maybe u can tell me how deep the rabbit hole goes!? ergo what the GMAT is..so that next time somebdy asks me what the GMAT is..i can explain to them in the same way as u are going to (i hope)
Manager
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Schools: Fuqua(no int), Darden(no int), McCombs , Kenan-Flagler, McDonough, Smith(w/$$) Followers: 1 Kudos [?]: 1 [0], given: 0 Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink] ### Show Tags 23 Aug 2010, 20:16 First of all - thanks to those responsible for putting together all of the great information on this site. I've only just begun to delve into the site and am extremely grateful to have found it before getting too far into the process of studying. I just received my MGMAT books, including the Fundamentals of GMAT Math book, and OG12 yesterday, so I decided to take my first practice test with GMATPrep tonight after work. I took it cold - no review prior, just downloaded the software and started. Here's how it went: Quant: 35 (39%) Verbal: 41 (92%) Overall score: 640 Clearly my quant needs more work than my verbal, but I plan on bumping up the verbal too. My questions for the experts around here (BB and anyone else) are how accurate is the GMATPrep practice test as an indicator of future scores and is it feasible to increase my quant score enough to get me to a 700+ score in the next 2ish months? I haven't scheduled my test yet, but will need to take it in the next few months. Probably by the end of Oct. I work full time, but am able to study ~2 hours a night and will probably study (realistically) 8-10 hours total a weekend. What do you think? Founder Affiliations: AS - Gold, HH-Diamond Joined: 04 Dec 2002 Posts: 14428 Location: United States (WA) GMAT 1: 750 Q49 V42 GPA: 3.5 Followers: 3715 Kudos [?]: 22966 [0], given: 4510 Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink] ### Show Tags 23 Aug 2010, 20:28 Kilian - that's a great result! Fantastic Verbal Score. I have scored 540 on my first one See this tool: http://gmatclub.com/gmat-score-calculator/ In general MGMAT tests are pretty accurate. You should be good with 100 point increase in the next 2 months, just make sure you can actually remember something after you close the book at night. The next day review what you covered the day before and you should be good with that. Go ahead and start with the Foundations book. You may also want to get the Test Simulation Booklet if have not already (probably worth 10 points ) _________________ Founder of GMAT Club US News Rankings progression - last 10 years in a snapshot - New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books Co-author of the GMAT Club tests GMAT Club Premium Membership - big benefits and savings Manager Joined: 23 Aug 2010 Posts: 111 Schools: Fuqua(no int), Darden(no int), McCombs , Kenan-Flagler, McDonough, Smith(w/$$)
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Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink]
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23 Aug 2010, 20:43
bb wrote:
Kilian - that's a great result! Fantastic Verbal Score.
I have scored 540 on my first one
See this tool: (won't let me post a link as a new member)
In general MGMAT tests are pretty accurate.
Very cool tool - thanks for posting it.
bb wrote:
You should be good with 100 point increase in the next 2 months, just make sure you can actually remember something after you close the book at night. The next day review what you covered the day before and you should be good with that.
Go ahead and start with the Foundations book. You may also want to get the Test Simulation Booklet if have not already (probably worth 10 points )
Will do. I'll post updates as things progress. Thanks again!
Re: GMAT Study Plan - How to Start your GMAT Prep [#permalink] 23 Aug 2010, 20:43
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# GMAT Study Plan - How to Start your GMAT Prep
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 5,984 | 22,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.65625 | 3 | CC-MAIN-2017-04 | latest | en | 0.869995 |
http://list.seqfan.eu/pipermail/seqfan/2017-January/031777.html | 1,713,951,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819089.82/warc/CC-MAIN-20240424080812-20240424110812-00681.warc.gz | 24,184,797 | 2,141 | # [seqfan] A281181 - Need Formula for Terms
Paul Hanna pauldhanna.math at gmail.com
Fri Jan 20 00:01:55 CET 2017
```Seqfans,
Let C(x) be the e.g.f. described by https://oeis.org/A281181,
then we have these beautiful results:
(1) C(x)^1 = d/dx Series_Reversion( Integral sqrt(1 - x^2) dx ),
(2) C(x)^2 = d/dx Series_Reversion( Integral cos(x)^2 dx ),
(3) C(x)^3 = d/dx Series_Reversion( Integral 1/cosh(x)^3 dx ),
(4) C(x)^4 = d/dx Series_Reversion( Integral 1/(1 + x^2)^2 dx ).
I have not found a simple formula for C(x)^n for n>4, but here is a nice
surprise:
(5) C(x)^5 = d/dx Series_Reversion( Integral C(i*x)^5 dx ).
[Note that an infinite number of functions satisfies condition (5).]
A function this lovely must have a nice formula for the coefficients,
and surely there is a combinatorial interpretation yet to be divulged.
Can anyone find a formula for the terms in A281181?
(That might be asking too much, but I had to ask.)
Thanks,
Paul
```
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what is the math behind dice game?how we calculate dice value and roll value?
Posted 1-Feb-13 15:54pm
Hard to imagine what you background might be...
—SA
Marcus Kramer 1-Feb-13 23:59pm
The computer picks a pseudo-random value between 1 and however many sides the dice should have?
If that isn't what you mean, then you'll need to clarify yourself.
Rate this:
## Solution 1
It's all pretty simple: you use an instance of the Random class to simulate rolling a real dice:
```private Random rand = new Random();
public int RollDice()
{
return RollDice(6);
}
public int RollDice(int numberOfSides)
{
return rand.Next(numberOfSides) + 1;
}
```
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https://wiki-helper.com/question/if-a-and-b-are-given-co-prime-numbers-find-the-least-number-divisible-by-both-a-and-bkitu-38155720-74/ | 1,638,020,553,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00214.warc.gz | 698,650,565 | 12,991 | ## if a and b are given co prime numbers.find the least number divisible by both a and b
Question
if a and b are given co prime numbers.find the least number divisible by both a and b
in progress 0
4 months 2021-07-17T12:00:26+00:00 1 Answers 0 views 0
## Answers ( )
1. Answer:
a×b
Step-by-step explanation:
since they are co prime, the only number they divisible by a and b will be a×b. | 115 | 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} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.875639 |
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# Python program to insert an element into sorted list
Given a sorted list and an element, Write a Python program to insert the element into the given list in sorted position.
Examples:
```Input : list = [1, 2, 4], n = 3
Output : list = [1, 2, 3, 4]
Input : list = ['a', 'b', 'c', 'd'], n = 'e'
Output : list = ['a', 'b', 'c', 'd', 'e']```
Approach #1 : This approach is the brute force method. Since the list is already sorted, we begin with a loop and check if the list element is greater than the given element. If yes, the given element need to be inserted at this position.
## Python3
`# Python3 program to insert``# an element into sorted list` `# Function to insert element` `def` `insert(``list``, n):` ` ``index ``=` `len``(``list``)`` ``# Searching for the position`` ``for` `i ``in` `range``(``len``(``list``)):`` ``if` `list``[i] > n:`` ``index ``=` `i`` ``break` ` ``# Inserting n in the list`` ``if` `index ``=``=` `len``(``list``):`` ``list` `=` `list``[:index] ``+` `[n]`` ``else``:`` ``list` `=` `list``[:index] ``+` `[n] ``+` `list``[index:]`` ``return` `list` `# Driver function``list` `=` `[``1``, ``2``, ``4``]``n ``=` `3` `print``(insert(``list``, n))`
Output:
`[1, 2, 3, 4]`
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach #2 : Python comes with a bisect module whose purpose is to find a position in list where an element needs to be inserted to keep the list sorted. Thus we use this module to solve the given problem.
## Python3
`# Python3 program to insert``# an element into sorted list``import` `bisect` `def` `insert(``list``, n):`` ``bisect.insort(``list``, n)`` ``return` `list` `# Driver function``list` `=` `[``1``, ``2``, ``4``]``n ``=` `3` `print``(insert(``list``, n))`
Output:
`[1, 2, 3, 4]`
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach #3 : Another way to insert an element into a sorted list is to use the built-in sorted() function. This function takes an iterable and returns a sorted list. We can pass the original list and the element to be inserted as arguments to the sorted() function, and it will return a new sorted list with the element inserted at the correct position.
## Python3
`def` `insert(``list``, n):`` ``# Insert the element into the list`` ``list``.append(n)`` ``# Sort the list`` ``sorted_list ``=` `sorted``(``list``)`` ``return` `sorted_list` `list` `=` `[``1``, ``2``, ``4``]``n ``=` `3``print``(insert(``list``, n))``#This code is contributed by Edula Vinay Kumar Reddy`
Output
`[1, 2, 3, 4]`
Time Complexity: O(Nlog(N)) since the sorted() function uses a sorting algorithm with a time complexity of O(Nlog(N))
Auxiliary Space: O(N) since the sorted() function creates a new list with the same size as the original list
#### Approach#4: Using while loop
This approach first appends the new element to the end of the list, then iterates from the end of the list to the beginning. For each pair of adjacent elements, if the element on the right is smaller than the element on the left, the two elements are swapped. This process continues until either the beginning of the list is reached or a pair of adjacent elements is found where the element on the right is greater than or equal to the element on the left. This ensures that the list remains sorted after the new element is inserted.
#### Algorithm
1. Define a function named ‘insert_sorted’ that takes a list ‘lst’ and an element ‘n’ as input.
2. Append the element ‘n’ to the list ‘lst’.
3. Initialize a variable ‘i’ with the last index of the list.
4. Loop through the list from the last index to the first index and swap adjacent elements if they are not in sorted order until the first element is reached or the adjacent elements are in sorted order.
5. Return the updated list ‘lst’.
6. Define the input list and the element to be inserted for each test case.
7. Call the ‘insert_sorted’ function for each test case and print the updated list.
## Python3
`def` `insert_sorted(lst, n):`` ``lst.append(n)`` ``i ``=` `len``(lst) ``-` `1`` ``while` `i > ``0` `and` `lst[i] < lst[i``-``1``]:`` ``lst[i], lst[i``-``1``] ``=` `lst[i``-``1``], lst[i]`` ``i ``-``=` `1` `lst ``=` `[``'a'``, ``'b'``, ``'c'``, ``'d'``]``n ``=` `'e'``insert_sorted(lst, n)``print``(lst)`
Output
`['a', 'b', 'c', 'd', 'e']`
Time complexity: O(n), where n is the length of the list.
Auxiliary Space: O(1), as the list is modified in-place.
Approach #5: Using heapq.heappush() function
Python’s heapq module implements heap queue algorithm. The most important feature of a heap is that the element with the highest priority is always at the root of the heap, and any operations to add or remove elements take O(log n) time where n is the number of elements in the heap. This makes heapq a great candidate for this problem.
1. Define a function named ‘insert_sorted’ that takes a list ‘lst’ and an element ‘n’ as input.
2. Append the element ‘n’ to the list ‘lst’.
3. Use heapq.heapify() function to convert the list into a heap in O(n) time.
4. Use heapq.heappush() function to insert the new element ‘n’ into the heap in O(log n) time.
5. Use a list comprehension to convert the heap back into a sorted list.
6. Return the sorted list.
7. Define the input list and the element to be inserted for each test case.
8. Call the ‘insert_sorted’ function for each test case and print the updated list.
## Python3
`import` `heapq` `# Function to insert in the sorted order``def` `insert_sorted(lst, n):`` ``lst.append(n)`` ` ` ``# Convert the list to a heap`` ``heapq.heapify(lst)`` ``# Use heapq.heappop() function to`` ``# convert the heap into a sorted`` ``# list`` ``sorted_lst ``=` `[heapq.heappop(lst) ``for` `i ``in` `range``(``len``(lst))]`` ``return` `sorted_lst` `# Driver Code``lst1 ``=` `[``1``, ``2``, ``4``]``n1 ``=` `3``print``(insert_sorted(lst1, n1))` `lst2 ``=` `[``'a'``, ``'b'``, ``'c'``, ``'d'``]``n2 ``=` `'e'``print``(insert_sorted(lst2, n2))`
Output
```[1, 2, 3, 4]
['a', 'b', 'c', 'd', 'e']```
Time Complexity: O(n log n) since the heapq.heappop() function is used to convert the heap into a sorted list
Auxiliary Space: O(n) since a heap is created from the input list
METHOD 6:Using Quicksort method
APPROACH:
The code above implements the quicksort algorithm to sort a list of integers and insert a new integer element n at the correct position in the sorted list.
ALGORITHM:
1.The quicksort algorithm sorts the list in place by partitioning it into two sub-lists around a pivot element, and then recursively sorting the sub-lists until the entire list is sorted.
2.The partitioning process involves selecting a pivot element from the list and then reordering the list such that all elements less than the pivot appear before it, and all elements greater than the pivot appear after it.
3.This process is then repeated on the two sub-lists until the entire list is sorted.
## Python3
`def` `quicksort(lst):`` ``if` `len``(lst) <``=` `1``:`` ``return` `lst`` ``pivot ``=` `lst[``len``(lst) ``/``/` `2``]`` ``left ``=` `[x ``for` `x ``in` `lst ``if` `x < pivot]`` ``middle ``=` `[x ``for` `x ``in` `lst ``if` `x ``=``=` `pivot]`` ``right ``=` `[x ``for` `x ``in` `lst ``if` `x > pivot]`` ``return` `quicksort(left) ``+` `middle ``+` `quicksort(right)` `lst ``=` `[``1``, ``2``, ``4``]``n ``=` `3` `# Insert the new element at the appropriate index``sorted_lst ``=` `quicksort(lst ``+` `[n])``index ``=` `sorted_lst.index(n)``lst.insert(index, n)` `print``(lst)`
Output
`[1, 2, 3, 4]`
Time complexity:
1.The worst-case time complexity of quicksort is O(n^2), which occurs when the pivot is chosen poorly and partitions the list into sub-lists of uneven size.
2.However, on average, quicksort has a time complexity of O(n log n), which makes it a very efficient sorting algorithm.
Space complexity:
1.The space complexity of quicksort is O(log n) on average, as it requires recursive function calls and space for the call stack.
2.However, in the worst case (when the recursion depth is equal to the length of the list), the space complexity can be as high as O(n).
### Using insert() method:
Approach:
Define a function insert_element_in_sorted_list that takes a sorted list lst and an element n as input.
Iterate over the indices of the list using a for loop and the range() function.
For each index i, check if the element at that index is greater than the element n.
If the element at index i is greater than n, insert n at index i using the insert() method of the list and return the modified list.
If all elements in the list are less than or equal to n, append n to the end of the list and return the modified list.
Here are the steps applied to the example input list1 and n1:
list1 is [1, 2, 4] and n1 is 3.
The function iterates over the indices 0, 1, and 2.
At index 2, the element 4 is greater than n1 so the function inserts n1 at index 2 using the insert() method, resulting in the modified list [1, 2, 3, 4].
The modified list is returned and printed, resulting in the output [1, 2, 3, 4].
Here are the steps applied to the example input list2 and n2:
list2 is [‘a’, ‘b’, ‘c’, ‘d’] and n2 is ‘e’.
The function iterates over the indices 0, 1, 2, and 3.
At index 3, the element d is less than n2 so the function appends n2 to the end of the list, resulting in the modified list [‘a’, ‘b’, ‘c’, ‘d’, ‘e’].
The modified list is returned and printed, resulting in the output [‘a’, ‘b’, ‘c’, ‘d’, ‘e’].
## Python3
`def` `insert_element_in_sorted_list(lst, n):`` ``for` `i ``in` `range``(``len``(lst)):`` ``if` `lst[i] > n:`` ``lst.insert(i, n)`` ``return` `lst`` ``lst.append(n)`` ``return` `lst` `# Example usage:``list1 ``=` `[``1``, ``2``, ``4``]``n1 ``=` `3``print``(insert_element_in_sorted_list(list1, n1)) ``# Output: [1, 2, 3, 4]` `list2 ``=` `[``'a'``, ``'b'``, ``'c'``, ``'d'``]``n2 ``=` `'e'``print``(insert_element_in_sorted_list(list2, n2)) ``# Output: ['a', 'b', 'c', 'd', 'e']`
Output
```[1, 2, 3, 4]
['a', 'b', 'c', 'd', 'e']```
The time complexity of this approach is O(n) in the worst-case scenario, where n is the length of the list. This is because we may need to traverse the entire list to find the appropriate position to insert the element.
The auxiliary space of this approach is O(1) as we are not creating any new data structures. | 3,146 | 10,453 | {"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-2023-40 | latest | en | 0.534473 |
https://forum.arduino.cc/t/motor-recommendation-for-slider-for-inverted-pendulum/694640 | 1,638,453,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00149.warc.gz | 328,019,159 | 5,271 | # Motor recommendation for slider for inverted pendulum
Hello everyone,
I am currently building an inverted pendulum. Initially I wanted to use a stepper motor but people in the project guidance section recommended to use a brushed motor, in order to easily reach the high speeds that are required.
Here is a short clip showing the slider with a stepper motor: https://youtu.be/SuZGTLXVWGs
Could you please recommend a motor and a driver for this application?
I have very little knowledge about elektric motors and would like to avoid having to guess and ending up buying multiple motors.
Cheers,
Mark
A DC motor would be fine - brushless are more complex to drive as you need a commutating circuit,
although sometimes you find a brushless motor with built in driver.
So I'd suggest a simple DC motor, possibly with some modest reduction gearing.
A PID loop would be needed to generate the correction feedback from some measure of the pendulum
slope (an IMU mounted on the pendulum is one way - pick off the pitch angle from the orientation).
An outer control loop can correct for linear drift along the slide, for that you need some measure of slide
position.
The parameters of the motor (power, torque) depend on the mass and length of the pendulum, without
an idea of this there's no way to choose a likely motor.
MarkT:
So I'd suggest a simple DC motor, possibly with some modest reduction gearing.
The parameters of the motor (power, torque) depend on the mass and length of the pendulum, without
an idea of this there's no way to choose a likely motor.
Would you be able to make a ballpark estimate of the needed motor parameters given the pendulum specifications?
Pendulum length 20-30 cm
pandulum mass (concentrated at the end) 200-300 g
Cheers,
Mark
Mass at the top?
To recover from 45 degree angle you need > 1g acceleration, lets say 15m/s^2, and a load mass of 0.3kg,
so force = 15 x 0.3 = 4.5N.
The typical speed needed can be figured from acceleration distance with time compared to pendulum size,
so 1/2 a t^2 = 0.3m, so 7.5 t^2 = 0.3, t = 0.2s. So velocity during that acceleration might reach 0.2 x 15 = 3m/s.
force x speed = 4.5 x 3 = 13.5 watts.
However that's likely an overestimate, in practice if the accleration continues that far, you've lost control | 562 | 2,293 | {"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-2021-49 | latest | en | 0.906938 |
https://metanumbers.com/183074 | 1,627,862,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00648.warc.gz | 397,881,263 | 10,927 | ## 183074
183,074 (one hundred eighty-three thousand seventy-four) is an even six-digits composite number following 183073 and preceding 183075. In scientific notation, it is written as 1.83074 × 105. The sum of its digits is 23. It has a total of 3 prime factors and 8 positive divisors. There are 90,916 positive integers (up to 183074) that are relatively prime to 183074.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 23
• Digital Root 5
## Name
Short name 183 thousand 74 one hundred eighty-three thousand seventy-four
## Notation
Scientific notation 1.83074 × 105 183.074 × 103
## Prime Factorization of 183074
Prime Factorization 2 × 239 × 383
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 183074 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,074 is 2 × 239 × 383. Since it has a total of 3 prime factors, 183,074 is a composite number.
## Divisors of 183074
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 276480 Sum of all the positive divisors of n s(n) 93406 Sum of the proper positive divisors of n A(n) 34560 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 427.871 Returns the nth root of the product of n divisors H(n) 5.29728 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,074 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 183,074) is 276,480, the average is 34,560.
## Other Arithmetic Functions (n = 183074)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 90916 Total number of positive integers not greater than n that are coprime to n λ(n) 45458 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16549 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 90,916 positive integers (less than 183,074) that are coprime with 183,074. And there are approximately 16,549 prime numbers less than or equal to 183,074.
## Divisibility of 183074
m n mod m 2 3 4 5 6 7 8 9 0 2 2 4 2 3 2 5
The number 183,074 is divisible by 2.
## Classification of 183074
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
• Sphenic
## Base conversion (183074)
Base System Value
2 Binary 101100101100100010
3 Ternary 100022010112
4 Quaternary 230230202
5 Quinary 21324244
6 Senary 3531322
8 Octal 545442
10 Decimal 183074
12 Duodecimal 89b42
20 Vigesimal 12hde
36 Base36 3x9e
## Basic calculations (n = 183074)
### Multiplication
n×i
n×2 366148 549222 732296 915370
### Division
ni
n⁄2 91537 61024.7 45768.5 36614.8
### Exponentiation
ni
n2 33516089476 6135924564729224 1123328253763237954576 205652196729451025296046624
### Nth Root
i√n
2√n 427.871 56.7818 20.6851 11.2856
## 183074 as geometric shapes
### Circle
Diameter 366148 1.15029e+06 1.05294e+11
### Sphere
Volume 2.57021e+16 4.21176e+11 1.15029e+06
### Square
Length = n
Perimeter 732296 3.35161e+10 258906
### Cube
Length = n
Surface area 2.01097e+11 6.13592e+15 317093
### Equilateral Triangle
Length = n
Perimeter 549222 1.45129e+10 158547
### Triangular Pyramid
Length = n
Surface area 5.80516e+10 7.23126e+14 149479
## Cryptographic Hash Functions
md5 ce34eaf57f0716099707c3ee896e9daa 9922ab1c02545d4a7adba61faea2729c82167d7e ededbcaf101c5d9df6d306e69c6cf21c095b0eb23c8c492cef8497e0ff4e4388 969bfd976bae897bd5973586f0f8e6dda8d6a7cb351e5c6bd3b1859d23f475bf57282c20e478725da960e2eb8026e30bd3620133d51f48e18ef069dc7f81d2e6 f40ebc86657a784465e57413193953f104553192 | 1,450 | 4,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-31 | latest | en | 0.815961 |
https://cloud.tencent.com/developer/article/1452153 | 1,566,558,482,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318375.80/warc/CC-MAIN-20190823104239-20190823130239-00001.warc.gz | 411,124,459 | 17,745 | # 基于matplotlib和keras的神经网络结果可视化
2.1 开发环境以及要求
2.2 训练数据的产生
2.3 网络的结构
3.1 网络的定义以及实现
3.2 训练模型保存
3.3 模型的搭建和保存代码
2.1 开发环境以及要求
2.2 训练数据的产生
----------------------------------------------
x y
1 0.093 -0.81
2 0.58 -0.45
3 1.04 -0.007
4 1.55 0.48
5 2.15 0.89
6 2.62 0.997
7 2.71 0.995
8 2.73 0.993
9 3.03 0.916
10 3.14 0.86
11 3.58 0.57
12 3.66 0.504
13 3.81 0.369
14 3.83 0.35
15 4.39 -0.199
16 4.44 -0.248
17 4.6 -0.399
18 5.39 -0.932
19 5.54 -0.975
20 5.76 -0.999
----------------------------------------------
2.3 网络的结构
3.1 网络的定义以及实现
```import math;
import random;
from matplotlib import pyplot as plt
from keras.models import Sequential
from keras.layers.core import Dense
import numpy as np
from keras.callbacks import ModelCheckpoint
import os
#采样函数
def sample(low, up, num):
data = [];
for i in range(num):
#采样
tmp = random.uniform(low, up);
data.append(tmp);
data.sort();
return data;
#sin函数
def func(x):
y = [];
for i in range(len(x)):
tmp = math.sin(x[i] - math.pi/3);
y.append(tmp);
return y;
#获取模型拟合结果
def getfit(model,x):
y = [];
for i in range(len(x)):
tmp = model.predict([x[i]], 10);
y.append(tmp[0][0]);
return y;
#删除同一目录下的所有文件
def del_file(path):
ls = os.listdir(path)
for i in ls:
c_path = os.path.join(path, i)
if os.path.isdir(c_path):
del_file(c_path)
else:
os.remove(c_path)
if __name__ == '__main__':
path = "E:/Model/";
del_file(path);
low = 0;
up = 2 * math.pi;
x = np.linspace(low, up, 1000);
y = func(x);
# 数据采样
# x_sample = sample(low,up,20);
x_sample = [0.09326442022999694, 0.5812590520508311, 1.040490143783586, 1.5504427746047338, 2.1589557183817036, 2.6235357787018407, 2.712578091093361, 2.7379109336528167, 3.0339662651841186, 3.147676812083248, 3.58596337171837, 3.6621496731124314, 3.81130899864203, 3.833092859928872, 4.396611340802901, 4.4481080339256875, 4.609657879057151, 5.399731063412583, 5.54299720786794, 5.764084730699906];
y_sample = func(x_sample);
# callback
filepath="E:/Model/weights-improvement-{epoch:00d}.hdf5";
checkpoint= ModelCheckpoint(filepath, verbose=1, save_best_only=False, mode='max');
callbacks_list= [checkpoint];
# 建立顺序神经网络层次模型
model = Sequential();
model.fit(x_sample, y_sample, nb_epoch=1000, batch_size=20,callbacks=callbacks_list);
#测试数据
x_new = np.linspace(low, up, 1000);
y_new = getfit(model,x_new);
# 数据可视化
plt.plot(x,y);
plt.scatter(x_sample, y_sample);
plt.plot(x_new,y_new);
plt.show();```
3.2 训练模型保存
（1）filename： 字符串，保存模型的路径
（2）verbose： 0或1
（3）mode： ‘auto’，‘min’，‘max’
（4）monitor： 需要监视的值
（5）save_best_only： 当设置为True时，监测值有改进时才会保存当前的模型。在save_best_only=True时决定性能最佳模型的评判准则，例如，当监测值为val_acc时，模式应为max，当监测值为val_loss时，模式应为min。在auto模式下，评价准则由被监测值的名字自动推断
（6）save_weights_only： 若设置为True，则只保存模型权重，否则将保存整个模型（包括模型结构，配置信息等）
（7）period CheckPoint之间的间隔的epoch数
3.3 模型的搭建和保存代码
``` # callback
filepath="E:/Model/weights-improvement-{epoch:00d}.hdf5";
checkpoint= ModelCheckpoint(filepath, verbose=1, save_best_only=False, mode='max');
callbacks_list= [checkpoint];
# 建立顺序神经网络层次模型
model = Sequential();
model.fit(x_sample, y_sample, nb_epoch=1000, batch_size=20,callbacks=callbacks_list);```
```import math;
import random;
from matplotlib import pyplot as plt
from keras.models import Sequential
from keras.layers.core import Dense
import numpy as np
import matplotlib.animation as animation
from PIL import Image
#定义kdd99数据预处理函数
def sample(low, up, num):
data = [];
for i in range(num):
#采样
tmp = random.uniform(low, up);
data.append(tmp);
data.sort();
return data;
def func(x):
y = [];
for i in range(len(x)):
tmp = math.sin(x[i] - math.pi/3);
y.append(tmp);
return y;
def getfit(model,x):
y = [];
for i in range(len(x)):
tmp = model.predict([x[i]], 10);
y.append(tmp[0][0]);
return y;
def init():
fpath = "E:/imgs/0.jpg";
img = Image.open(fpath);
plt.axis('off') # 关掉坐标轴为 off
return plt.imshow(img);
def update(i):
fpath = "E:/imgs/" + str(i) + ".jpg";
img = Image.open(fpath);
plt.axis('off') # 关掉坐标轴为 off
return plt.imshow(img);
if __name__ == '__main__':
low = 0;
up = 2 * math.pi;
x = np.linspace(low, up, 1000);
y = func(x);
# 数据采样
# x_sample = sample(low,up,20);
x_sample = [0.09326442022999694, 0.5812590520508311, 1.040490143783586, 1.5504427746047338, 2.1589557183817036, 2.6235357787018407, 2.712578091093361, 2.7379109336528167, 3.0339662651841186, 3.147676812083248, 3.58596337171837, 3.6621496731124314, 3.81130899864203, 3.833092859928872, 4.396611340802901, 4.4481080339256875, 4.609657879057151, 5.399731063412583, 5.54299720786794, 5.764084730699906];
y_sample = func(x_sample);
# 建立顺序神经网络层次模型
model = Sequential();
plt.ion(); #开启interactive mode 成功的关键函数
fig = plt.figure(1);
for i in range(100):
filepath="E:/Model/weights-improvement-" + str(i + 1) + ".hdf5";
#测试数据
x_new = np.linspace(low, up, 1000);
y_new = getfit(model,x_new);
# 显示数据
plt.clf();
plt.plot(x,y);
plt.scatter(x_sample, y_sample);
plt.plot(x_new,y_new);
ffpath = "E:/imgs/" + str(i) + ".jpg";
plt.savefig(ffpath);
plt.pause(0.01) # 暂停0.01秒
ani = animation.FuncAnimation(plt.figure(2), update,range(100),init_func=init, interval=500);
ani.save("E:/test.gif",writer='pillow');
plt.ioff() # 关闭交互模式```
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# Answer to Question #126533 in Mechanics | Relativity for Mwansa Kunda
Question #126533
Two objects are connected by a light string passing over a frictionless pulley. m1=5.00kg, m2=3.00kg, h=4.00m
The object of mass 5kg was released from rest. Using the principle of conservation of energy,
Determine;
i. The speed of the 3.0kg object just as the 5.0kg object hits the ground.
ii. The maximum height to which the 3.0kg object rises.
1
2020-07-20T15:13:03-0400
5 kg body locates 4 meters above ground. After it released it will go down because it heavier than 3 kg body. Energy conservation principle says:
"E_{p}=E_{k}"
(1) "m\\times a\\times h = \\frac{m\\times V^2}{2}"
where m - mass of body (5 kg), a - acceleration (not equals to gravitational acceleration in our case because we have 3kg body on other side), h - height (4 meters), V - velocity of body near ground.
acceleration could be found:
"F_{result} = F_{g5} - F_{g3}= m_{5}\\times g - m_{3} \\times g;"
"F_{result} = 5\\times 9.8 - 3\\times 9.8 = 49 - 29.4 = 19.6N";
"F_{result}=m_{5}\\times a;"
"a=\\frac{F_{result}}{m_{}}=\\frac{19.6}{5}=3.92 m\/s^2"
so, returning to formula (1) we can find V of body 5kg and body 3 kg (they a the same in this moment)
"V=\\sqrt{\\frac{m_{5}\\times a \\times h \\times 2}{m_{5}}}=\\sqrt{a \\times h \\times 2}=\\sqrt{3.92\\times 4\\times 2};"
"V=5.6m\/s"
Ansver for first question speed is 5.6 m/s
After 5 kg body hit ground, body 3kg will locate on height 8 m and continue move upward for some time, because it has some speed. To find maximum height which body 3 will have, let's use the same principle:
"m_{3}\\times g\\times h=\\frac{m_{3}\\times V^2}{2};"
"h=\\frac{V^2}{2\\times g}"
this time we use g because only force that affects body 3 kg is gravity.
"h=\\frac{5.6^2}{2\\times 9.8}=1.6m"
So maximal height of body 3 kg is 8 + 1.6 = 9.6m
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for any assignment or question with DETAILED EXPLANATIONS! | 695 | 2,081 | {"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} | 4.21875 | 4 | CC-MAIN-2023-06 | latest | en | 0.782981 |
https://r-laroseraie.fr/Nov_02-19245/ | 1,620,932,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00310.warc.gz | 495,079,085 | 7,128 | motor power calculation for ball and tube mill
• how to choose motor for ball mill
Ball mill IKA ULTRA-TURRAX Tube Drive Control grinding ball type, 28W motor, for 2ml to 50ml sample volumes, maximum speed 8000rpm, grinding ball set >> Get Price; BALL MILL POWER - Page 1 of 2 - International Cement Review. BALL MILL SIZE IS 4.4*11 M. GRINDING MEDIA SIZE IS MAX 30MM . SUPPLIER HAS GIVEN A BALL MILL MOTOR INSTALLED KW OF
• GROSS POWER CALCULATOR - SMC Testing
GROSS POWER CALCULATOR General ... The model predicts the gross power, which is also called the motor input power. This is not the same as the so-called "power at pinion", which is slightly lower than the gross power and relates to the power as ... Predicted Power Draw (kW) Ball mills AG/SAG Mills. 5 Figure 2 – Overflow ...
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• Ball Mill Design/Power Calculation
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum 'chunk size', product size as P80 and maximum and finally the type of circuit open/closed ...
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• gold mill motor capacity derate calculation
sagdesign. Ball Mill Motor Type Motor Type BM Specific Energy, kWh/ton Mill Motor Type Selection Code Eff. Loss A B C GRINDING CIRCUIT CAPACITY : Mill Throughput Capacity Grinding Circuit Capacity : Assumed from Bailey et al, 2009 SAGDesign Mill Throughput Calculation Using S d BWI when T80 1,700 µm.
Bearing Load Calculation A-22 (2)Loads acting on cross shafts Gear loads acting on straight tooth bevel gears and spiral bevel gears on cross shafts are shown in Figs. 4.4 and4.5. The calculation methods for these gear loads are shown in Table 4.3. Herein, to calculate .
• live project on ball and tube mills
Development of a tube ball mill mathematical model for mill . ... The project is sponsored by BCURA and EDF Energy with the technical support from E.ON The Tube Ball mill used by EDF is a motor driven tumbling barrel. A new modelbased approach for power plant Tubeball mill ...
• tube mills for calculations
The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are; material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating % solids or pulp density, feed size as F80 and maximum 'chunk size', product size as P80 and maximum and finally the type of circuit open/closed ...
• gold mill motor capacity derate calculation
sagdesign. Ball Mill Motor Type Motor Type BM Specific Energy, kWh/ton Mill Motor Type Selection Code Eff. Loss A B C GRINDING CIRCUIT CAPACITY : Mill Throughput Capacity Grinding Circuit Capacity : Assumed from Bailey et al, 2009 SAGDesign Mill Throughput Calculation Using S d BWI when T80 1,700 µm.
• Tube mill tail bin adopts ceramic ball to increase ...
The bearing mill reduces the motor power by 400kw, and the grinding machine still maintains the level of 220t, and the output of the mill is slightly increased. 3. Generally speaking, ceramic grinding bodies are applied to tube mills with roller presses and tube mills with ball breaking machines before grinding.
• calculation of grinding media in ball mill
Calculate Top Ball Size of Grinding Media -FRED C. BOND Equation & Method Although it was developed nearly 50 years ago, Bond 's method is still useful for calculating necessary mill sizes and power consumption for ball and rod mills. ball mill calculations, grinding media filling degree
• Coal Mill Power Calculation
Coal Ball Mill Design Calculations Royalcrescentgroupin. Ball mill designpower calculation.The basic parameters used in ball mill design power calculations, rod mill or any tumbling mill sizing are; material to be ground, characteristics, bond work index, bulk density, specific density, desired mill tonnage capacity dtph, operating solids or pulp density, feed size as f80 and maximum chunk ...
• Grinding in Ball Mills: Modeling and Process Control
Grinding in ball mills is an important technological process applied to reduce the size of particles which may have different nature and a wide diversity of physical, mechanical and chemical characteristics. Typical examples are the various ores, minerals, limestone, etc. The applications of ball mills are ubiquitous in mineral
• Published in: Cybernetics and Information Technologies · 2012Authors: Vladimir V Monov · Blagoy Sokolov · Stefan StoenchevAffiliation: Bulgarian Academy of SciencesAbout: Ball mill · Process control · Computer science · Information technology
• Milling Horsepower Calculator
Milling Horsepower Calculator. ... Also required is the unit power, which is a material property describing the amount of power required to cut that material. The horsepower at both the spindle and the motor are shown, as well as the spindle torque for a given spindle speed (RPM). The motor horsepower required for the milling operation can then ...
• CALCULATION OF THE POWER DRAW OF DRY .
by Erdem (2002) for dry -component cement grinding ball mills, using the results of the Bond ball mill grindability test. For all model based methods, a reliable method to calculate mill power draw for a given mill is required for the calculation of power draw. Morrell (1996) proposed a
• Published in: Physicochemical Problems of Mineral Processing · 2004Authors: A Seyfi Erdem · Ş Levent Ergun · A Hakan BenezerAbout: Ball mill · Cement
• BALL & ACME LEAD SCREW TECHNICAL INFORMATION I
the performance of ball screws is far superior. The rolling action of the balls versus the sliding action of the ACME nut provides significant advantages. Advantages of ball screw drives are increased efficiency (typically up to 90 – 95%) which allows required motor torque to be lower, predictable service life, low wear rate and maintenance ...
• Electric Motor Power Measurement and Analysis | Yokogawa ...
Part 2: Selecting Instruments for Electric Motor Power Measurement and Analysis. ... A PWM drive should maintain a constant V/Hz ratio over the operating speed of the motor. The power analyzer can calculate V/Hz using the RMS or the fundamental voltage value. The analyzer's user-defined math function is used to develop an equation for this ...
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• Best way to determine the ball-to-powder ratio in ball ...
Best way to determine the ball-to-powder ratio in ball-milling? What is the best way to determine the ball-to-powder ratio for ball-milling a new material? ... The maximum power draw in ball mill ...
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• inching auxillary drive calculation for ball mill
Calculation Of Torque For Ball Mill . central drive ball mill torque and power calculation. ball mill torque calculation calculating motor torque in a ball millimpfarrgartenorg Central Drive Ball Mill Torque and Power Calculation The inching drive, also known as a barring or auxiliary drive, is an important component of any mill or kiln . Get Price
• Mill Power
The Art Of Sharing and...Imagination. Home; About Us; Services. Grinding Software; Consultancy; Training Courses; Calculators Online
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• Motor Calculation Of Ball Mill Design
Motor Rating Calculation In Cement Ball Mill. Ball mill designpower calculation.The ball mill motor power requirement calculated above as 1400 hp is the power that must be applied at the mill drive in order to grind the tonnage of feed from one size distribution.The following shows how the size or select the matching mill required to draw this power is.
• SAGMILLING.COM .:. Home
Sagmilling is home to a collection of both free and subscription-based calculation tools to aid metallurgical process engineers perform comminution calculations. Grinding circuit design tools, including for SAG mill-ball mill circuits, and geometallurgy energy models are available to subscribers.
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• calculation of load torque in cement balls mills
Starting Torque Of Ball Mill Calculation . Any one can help me to calculate Load Torque central cement ballThe FL ball mill is designed for The FL cement ball mill has a Mill Motor*) ball mill drive motor choicesArtec Machine. ball mills, the starting torque restrictions of some of the newer mill drive configurations, and the . reflected to the motor shaft is calculated to be,ftlbs2.
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• BALL MILL DRIVE MOTOR CHOICES
The mill used for this comparison is a 4.4-meter diameter by 13.6 meter long ball mill with a 5000 HP drive motor. It is designed for approximately 90 s.ton per hour. This type two-compartment mill is a state- of-the-art shell supported cement finish mill. The torque and time requirements of the mill with the ball
• Electric Motor Power Measurement and Analysis | Yokogawa ...
Part 2: Selecting Instruments for Electric Motor Power Measurement and Analysis. ... A PWM drive should maintain a constant V/Hz ratio over the operating speed of the motor. The power analyzer can calculate V/Hz using the RMS or the fundamental voltage value. The analyzer's user-defined math function is used to develop an equation for this ...
[PDF]
• Ball Mill Motor Sizing
ball mill motor sizing . ball mill sizing | Mining & Quarry Plant- ball mill motor sizing,The mill sizing page: Please find below two calculators for sizing mills using the Bond and Rowland methods Ball mill sizing: Calculator for ball mill(s) in a single,ball mill size with motor power - The Kuber Opticsball mill size with motor power Spring Cone Crusher Product Introduction Low power ...
• inching auxillary drive calculation for ball mill
Calculation Of Torque For Ball Mill . central drive ball mill torque and power calculation. ball mill torque calculation calculating motor torque in a ball millimpfarrgartenorg Central Drive Ball Mill Torque and Power Calculation The inching drive, also known as a barring or auxiliary drive, is an important component of any mill or kiln . Get Price
• Rod mills, how they function
Rod mills operate at lower speed than ball mills since the rods are rolled and not cascaded. For an equivalent grind, a rod mill uses less steel than a ball mill because of the lower speed and better contact between the media and ore. The rod charge must be maintained in good working condition, and broken and worn rods must be removed.
• Ball Mill Motor/Power Sizing Calculation
Ball Mill Motor/Power Sizing Calculation View Larger Image A) Total Apparent Volumetric Charge Filling – including balls and excess slurry on top of the ball charge, plus the interstitial voids in between the balls – expressed as a percentage of the net internal mill volume (inside liners).
• Force, Torque, and Power
Calculate Tangential Force, Torque, and Machining Power for Face Milling Applications. MENU {{searchText}} in ({{typeAhead.level3}} / {{typeAhead.level4}}) ... P m at the motor: hp KW . These calculations are based upon theoretical values and are only intended for .
• Ball Mill Design/Power Calculation - LinkedIn
Dec 12, 2016 · Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80% passing ¼ inch (6350 ...
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• mill production line | 2,575 | 11,831 | {"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} | 2.71875 | 3 | CC-MAIN-2021-21 | latest | en | 0.782405 |
http://www.opengl.org/discussion_boards/archive/index.php/t-166992.html | 1,406,602,658,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510264575.30/warc/CC-MAIN-20140728011744-00045-ip-10-146-231-18.ec2.internal.warc.gz | 738,302,325 | 2,424 | PDA
View Full Version : Subdivision algorithm of a tetrahedron
Leonardave
03-16-2009, 02:40 AM
Does anybody know how to subdivide a tetrahedron to be a sphere, what kind of algorithm is applicable? Thanks in advance.
Nicolai de Haan Brøgger
03-16-2009, 03:14 AM
You can look in the Red Book, chapter 2 towards the end for an idea.
Leonardave
03-16-2009, 06:59 PM
Yomboprime
11-27-2009, 01:01 PM
Hey I'm doing exactly this right now. I subdivide the four triangles by calling a recursive function that takes three points as parameters (the three vertices), and an integer meaning the recursion level. Then in the function you simply stop at a given recursion level and draw the triangle (normalizing the vertices to a sphere of the radius you want), or do four recursive calls. In this case the steps are:
1 - increment the recursion level integer to pass it further in the recursive calls.
2 - compute the midpoints. If you have vertices v0, v1 and v2 you compute v01 = 0.5 * (v0 + v1), v12 = 0.5 * (v1 + v2), and v20 = 0.5 * (v0 + v2)
3 - make the four recursive calls by using the vertices:
first call: v0, v01, v20
second call: v01, v1, v12
third call: v12, v2, v20
fourth call: v01, v12, v20
I'm not using a tetrahedron, instead I use an octoedron wich gives me no problems of texture coordinate wrapping (a tetrahedron doesn't have a line from pole to pole through triangle segments), but I think that it is giving me problems in the culling.
I use culling of triangles (if a triangle at a given recursion level does have the vertex normals and the triangle normal facing outwards the camera, I discard it, i.e. I don't draw nor make more recursion with it), but the culling doesn't work well in some cases and I think i's because I'm using an octoedron and the triangles are not normal to the sphere anymore.
I suppose I'll have to use a tetrahedron and split the triangles falling at the "meridian 0" so no texture wrapping problems appear. | 541 | 1,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2014-23 | longest | en | 0.878977 |
https://thebikeyear.com/1003029/amazing-free-worksheets-number-bonds-to-20/ | 1,638,108,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00466.warc.gz | 647,185,185 | 11,623 | HomeWorksheet Template ➟ 1 Fresh Free Worksheets Number Bonds To 20
# Fresh Free Worksheets Number Bonds To 20
Number bonds worksheets this product includes 2 files. Number bonds to 20 include 12 8 14 6 13 7 etc.
Number Bonds Rainbow Worksheet Inspirationa Number Bonds Worksheet To 10 Google Kereses Kindergarten Worksheets Number Bonds Number Bonds Worksheets
### Number bonds are the process of adding two numbers together to get an answer total for example.
Free worksheets number bonds to 20. There are 3 sheets for each number bond from 11 to 19. Number bonds are simply pairs of numbers that add up to a given number. Differentiated Year 1 worksheets for children to attempt to work out number bonds to 20 independently.
A number bond is just a maths fact that a pair of numbers when added together make a given total. Number bonds to 20 include 12 8 14 6 13 7 etc. These worksheets contain numbers from.
Worksheets numbers 11 20 b. Number Bonds are the process of adding two numbers together to get an answer total for example. 13 7 is similar to 17 3.
Number Bonds to 20 Worksheet. There are 39 pages with 5 number bonds. Addition – 2 Digit Grade – 1st Grade – Kindergarten Math Numbers – 1 to 20 Numbers – 1 to 30 Seasons – Fall Worksheets Seasons – Spring Worksheets.
While math tends to be a subject that lends itself to boring math worksheets weve come up with these really clever fun Number Bonds Worksheets to make practicing adding within 20 fun for preschoolers kindergarteners and grade 1 students. – Draw a line from each number in a green circle to join it with a yellow circle to make a total of 20. Kite Themed Number Bond Worksheets to 20.
The first sheet contains a visual number line to support kids to count along. Our customer service team will review your report and will be in touch. Filling in the missing numbers to reinforce number bonds to 20 and to encourage quick recall.
Number bond is a special concept to teach addition and subtraction. It will test kids knowledge of the bonds of 20. My kids seem to learn materials better when the activity is presented in a fun engaging way.
Addition or subtraction will help kids solve these number bonds. How many can you put into pairs that add to 20. Adding three and four digit numbers.
Twenty frames are a great resource for teaching kids number bonds to 20. To learn number bonds to 20 we can use patterns. This Kindergarten number bond worksheets that address students of different levels in your classroom from 1 to 3 grade Includes practice for addition number bonds to 10 with missing parts.
Three addends addition tree and number bond templates. These free printable activity worksheets focus on Number Bonds math subject. Explore fun printable activities for K-8 students covering math ELA science more.
The second sheet is harder than the first and contains a range of matching and number bond search activities. Numbers Bonds Worksheets to 11 to 20. These free printable activity worksheets focus on Number Bonds math subject.
The best way to learn the number bonds to 20 is to learn the number bonds to 10 and add 10 to one of the numbers in each pair. The children who are less confident are given the number bonds to workout. We just have the 1 tens digit on the other number.
Number bonds like times tables are something which a child should know instantly without needing to think – therefore lots of practice. Number Bonds are the process of adding two numbers together to get an answer total for example. Number Bonds provides Math related worksheets for activity.
Number bonds for number X are all the possible combinations of two numbers. Number bond worksheets can help students solve a variety of questions and these worksheets can be downloaded for free in PDF format. This free number bonds to 20 worksheet uses a badminton theme.
Explore fun printable activities for K-8 students covering math ELA science more. Number Bonds to 20 are pairs of numbers that add to make 20. Report this resource to let us know if it violates our terms and conditions.
This activity will keep kids engaged and they also colour the monsters and other characters in the worksheets. Number Bonds Worksheet – 2. Great supports for teaching part part whole in a kindergarten classroom.
For example the number bonds for 10 are 01010 1910. Number Bonds to 20 TCPDF Created Date. Knowing the number bonds well sets a child up for success in math.
This is a part of my number bonds worksheets series. Number bonds to 20 include 12 8 14 6 13 7 etc. Tes classic free licence.
Know how to break up number 20 into two smaller numbers which are number pairs. Write the missing numbers to make a total of 20. Learning number pairs that add together to make 20 is a fundamental math skill for first grade.
Worksheets to 10 to 20 to 40 to 60 to 80 and to 100. Full 1000×1294 thumbnail 150×150 medium 235×150 large 640×829 Free Printable Number Bond Template A great deal of individuals are browsing the web to find a free printable download. – Write the missing numbers to make a total of 20.
The third sheet is the hardest and contains a range of addition. As a fun way to review number bonds and keep them fresh in mind Ive created kite themed number bond worksheets to 20. The children who are more confident have to find their own number bonds to 20.
This is a maths worksheet designed to test pupils knowledge of the different ways that the number 20 can be formulated. This free number bonds to 20 worksheet uses a badminton theme. Free PDF Printable Maths Worksheets – Number Bonds to 20 Keywords.
Download Number Bonds Worksheet PDFs. My daughter has put in the time to really learn these well. Addition or subtraction will help kids solve these number bonds.
2810 3710 4610 5510 etc and the number bonds for 20 are 020 20 119 20 218 20 etc. Number Bonds Worksheet – 1. Number Bonds to 20 – Free PDF Worksheet Author.
My kids seem to learn materials better when the activity is presented in a fun engaging way. Learn the number bonds or the addends which make 20. Number Bonds of 20.
About This Primary Resource. Kids will need to write the missing number pair in each shuttlecock or birdie that will add up to 20. Number bonds to 20 include 12 8 14 6 13 7 etc.
2Nd Grade Math Worksheets Number Bonds To 20 2 Math Activities Free Printable Number Bond Template Downloads. Free Number Bonds Practice Worksheet Categories. For example the number bonds for 9 are 9 0 8 1 7 2 6 3 and 5 4 plus these with the two numbers switched.
Number bonds are very basic an they are an important part for understanding how numbers work. This resource consists of 1 page. Number Bonds Worksheets.
Our printable number bond worksheets for children in kindergarten through grade 3 include simple addition of two addends. The children are know their number bonds to 20 are challenged to use the commutative law. Twenty frames can be a little problematic for kids but these fun worksheets will make the process playful and easy for kids.
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Number Bonds 0 20 Poster And Blank Worksheet Ks1 Numeracy Teaching Resource 1 85 Number Bonds Number Bonds To 20 Time Worksheets | 1,727 | 8,043 | {"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} | 3.109375 | 3 | CC-MAIN-2021-49 | latest | en | 0.818569 |
https://zbmath.org/?q=rj%3A3035+se%3A1605 | 1,679,435,370,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00520.warc.gz | 1,217,330,442 | 11,468 | Moore-Penrose inverse positivity of interval matrices.(English)Zbl 1236.15012
The authors study (entrywise) nonnegativity of the Moore-Penrose inverse $$C^†$$ of a matrix $$C\in\mathbb{R}^{m\times n}$$. In particular they consider the following questions:
(1) When is $$C^†$$ nonnegative? (2) When is $$C^†$$ nonnegative if $$C\leq B$$, where $$B\in \mathbb{R}^{m\times n}$$ is a given matrix with $$B^†\geq 0$$? (3) When is $$C^†$$ nonnegative for all matrices $$C\in[A, B]$$, where $$A^†\geq 0$$ and $$B^†\geq 0$$?
Each of these three questions is answered by necessary and sufficient conditions which are similar to known ones for regular $$n\times n$$ matrices $$A$$, $$B$$, $$C$$ in the same connection. One of these known criteria comes from Ortega and Rheinboldt and uses weak regular splittings, another one goes back to Krasnosel’skij, Lifshits, and Sobolev and uses some part of the range of $$C$$, and a third one was published by Rohn and answers question (3) for regular matrices.
For the analogue of the latter one the concept of a range kernel regular interval matrix $$[A,B]$$ is introduced by $$R(A)= R(B)$$ and $$N(A)= N(B)$$, where $$R(\cdot)$$ denotes the range and $$N(\cdot)$$ the kernel of an $$m\times n$$ matrix. For such an interval matrix a particular subset $$K$$ is introduced by virtue of which various additional results could be derived.
MSC:
15A09 Theory of matrix inversion and generalized inverses 65G30 Interval and finite arithmetic
Full Text:
References:
[1] Ben-Israel, A.; Greville, T.N.E., Generalized inverses: theory and applications. pure and applied mathematics, (2003), Wiley-Interscience New York [2] Berman, A.; Plemmons, R.J., Monotonicity and generalized inverse, SIAM J. appl. math., 22, 155-161, (1972) · Zbl 0255.15005 [3] Berman, A.; Plemmons, R.J., Cones and iterative methods for best least squares solutions of linear systems, SIAM J. numer. anal., 11, 145-154, (1974) · Zbl 0244.65024 [4] Berman, A.; Plemmons, R.J., Inverses of nonnegative matrices, Linear and multilinear algebra, 2, 161-172, (1974) [5] Berman, A.; Plemmons, R.J., Nonnegative matrices in the mathematical sciences, (1994), SIAM · Zbl 0815.15016 [6] Collatz, L., Functional analysis and numerical mathematics, (1966), Academic New York · Zbl 0221.65088 [7] Debasisha Mishra, K.C. Sivakumar, On splittings of matrices and nonnegative generalized inverses, Oper. Matrices, in press. · Zbl 1247.15003 [8] Greville, T.N.E., Note on generalized inverse of a matrix product, SIAM rev., 8, 518-521, (1966) · Zbl 0143.26303 [9] Krasnosel’skij, M.A.; Lifshits, Je.A.; Sobolev, A.V., Positive linear systems: the method of positive operators, (1989), Springer Berlin, (Transl. from the Russian by Jurgen Appell) [10] Ortega, J.M.; Rheinboldt, W.C., Monotone iterations for nonlinear equations with application to gauss – seidel methods, SIAM J. numer. anal., 4, 171-190, (1967) · Zbl 0161.35401 [11] Rohn, J., Inverse-positive interval matrices, Z. angew. math. mech., 67, 492-493, (1987) · Zbl 0628.65027 [12] Rohn, J., Systems of linear interval equations, Linear algebra appl., 126, 39-78, (1989) · Zbl 0712.65029 [13] K.C. Sivakumar, M.R. Weber, On Positive Invertibility and Splittings of Operators in Ordered Banach Spaces, Preprint, Technical University Dresden, MATH-AN-02-2010. · Zbl 1326.47042 [14] Varga, R.S., Matrix iterative analysis, Springer series in computational mathematics, (2000), Springer
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,129 | 3,807 | {"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} | 2.609375 | 3 | CC-MAIN-2023-14 | longest | en | 0.820428 |
http://stackoverflow.com/questions/11417788/calculations-between-columns-in-data-frame-week-on-week-trending-sales-detect | 1,464,085,976,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049270527.3/warc/CC-MAIN-20160524002110-00218-ip-10-185-217-139.ec2.internal.warc.gz | 276,808,158 | 19,681 | Calculations between columns in data frame: week-on-week 'trending' sales detection
I have a data.frame representing frequency book sales for a set of authors over 25 weeks:
``````author week_1 week_2 week_3 week_4 ...
author1 7 4 5 2
author2 3 6 18 5
author3 1 0 2 4
author4 0 1 1 2
author5 0 1 0 0
``````
First, I want to use this data to build a new data frame, which shows the fraction of [currentWeek / previousWeek]. Something like this perhaps:
``````author week_1 week_2 week_3 week_4 ...
author1 NA 0.57 1.25 0.2
author2 NA 2 3 0.28
author3 NA 0 2 2
author4 NA 1 1 2
author5 NA 1 0 0
``````
(I would like to substitute zeros with 1s to avoid dividing by zero.)
Second, I want to run a quick iteration over all the rows, check for any triplets of adjacent weeks where sales for that authors have increased by 100% twice in two consecutive week-pairs, and report this in some kind of output table. Perhaps like this:
``````author startTrendWeek endTrendWeek
author2 1 3
author3 2 4
``````
Any ideas for how I could solve either of these in R?
-
What have you tried? Where do you get stuck? – Andrie Jul 10 '12 at 16:35
I don't know how to create a new table where each cell is a calculation from another table. I find manipulating data like this intuitive in Perl, but (as yet) R baffles me. – Harry Palmer Jul 10 '12 at 16:42
`test1<-c(1,2,3,4,5); test2<-c(test1[-1],NA);test3<-test1/test2` Read some introductionary text for R. Also, leave your data.frame in long format and don't use `cast` when creating it (see your recent question). – Roland Jul 10 '12 at 16:48
``````x <- read.table(text=
"author week_1 week_2 week_3 week_4
author1 7 4 5 2
author2 3 6 18 5
author3 1 0 2 4
author4 0 1 1 2
author5 0 1 0 0
``````
One line of code:
``````cbind(x[1], t(apply(x[, -1], 1, function(xx)xx[-1]/xx[-length(xx)])))
author week_2 week_3 week_4
1 author1 0.5714286 1.25 0.4000000
2 author2 2.0000000 3.00 0.2777778
3 author3 0.0000000 Inf 2.0000000
4 author4 Inf 1.00 2.0000000
5 author5 Inf 0.00 NaN
``````
-
This is very helpful Andrie thank you! I can't quite figure out how all the functions used by that command are working together - like why is there a transpose, a matrix multiplication, and what is the length doing? If you can explain this briefly that would be very helpful. But anyway, thanks very much and I am starting to learn R properly with tutorials and an introductory book. – Harry Palmer Jul 11 '12 at 15:20
There's no matrix multiplication here. `xx[-length(xx)]` means take `xx` and cut off the last element. `apply(x, 2, FUN)` applies a function to each row of your data. And you need to transpose the result because the results of `apply` come out as the tranpose of what you need. Good luck with your R journey. – Andrie Jul 11 '12 at 15:51
And what is xx? – Harry Palmer Jul 11 '12 at 16:22
`xx` indicates the variable in function(xx), i.e. each row of data. – Andrie Jul 11 '12 at 16:25 | 1,037 | 3,344 | {"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-2016-22 | latest | en | 0.892825 |
http://api.flix.dev/MultiMap.html | 1,718,859,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00068.warc.gz | 2,864,146 | 7,294 | # MultiMap
`enum MultiMap[k: Type, v: Type] with Sendable`Source
`case MultiMap(Map[k, Set[v]])`
The MultiMap type.
A MultiMap is a Map that allows multiple values for a key. Multiple values are stored in a Set so duplicates are not allowed.
The key and value types must implement the Eq and Order traits.
### Instances
`instance CommutativeMonoid[MultiMap[k, v]] with Order[k], Order[v]`Source
`instance CommutativeSemiGroup[MultiMap[k, v]] with Order[k], Order[v]`Source
`instance Eq[MultiMap[k, v]] with Eq[k], Eq[v]`Source
`instance Foldable[MultiMap[k]]`Source
`instance Hash[MultiMap[k, v]] with Hash[k], Hash[v]`Source
`instance Monoid[MultiMap[k, v]] with Order[k], Order[v]`Source
`instance Order[MultiMap[k, v]] with Order[k], Order[v]`Source
`instance SemiGroup[MultiMap[k, v]] with Order[k], Order[v]`Source
`instance Sendable[MultiMap[k, v]] with Sendable[k], Sendable[v]`Source
`instance ToString[MultiMap[k, v]] with ToString[k], ToString[v]`Source
`instance UnorderedFoldable[MultiMap[k]]`Source
## Definitions
`def adjust(f: v -> v \ ef, k: k, m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Updates `m` with `k => map(f, vs)` if `k => vs` is in `m`.
Otherwise, returns `m`.
`def adjustWithKey(f: k -> (v -> v \ ef), k: k, m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Updates `m` with `k => map(f(k), vs)` if `k => vs` is in `m`. Otherwise, returns `m`.
`@ParallelWhenPure `
`def count(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Int32 \ ef`
Returns the number of mappings `k => v` in `m` that satisfy the predicate `f`.
Purity reflective: Runs in parallel when given a pure function `f`.
`def difference(m1: MultiMap[k, v], m2: MultiMap[k, v]): MultiMap[k, v] with Order[k], Order[v]`
Returns the difference of `m1` and `m2` i.e `m1 - m2` (left biased).
`def empty(): MultiMap[k, v]`
Returns the empty MultiMap.
`def exists(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Bool \ ef`
Returns `true` if and only if at least one mapping in `m` satisfies the predicate `f`.
Returns `false` if `m` is the empty MultiMap.
`def filter(f: v -> Bool \ ef, m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Returns a MultiMap of all mappings `k => v` in `m` where `v` satisfies the predicate `f`.
`def filterWithKey(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Returns a MultiMap of all mappings `k => v` in `m` where `(k, v)` satisfies the predicate `f`.
`def find(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Option[(k, v)] \ ef`
Alias for `findLeft`.
`def findLeft(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Option[(k, v)] \ ef`
Optionally returns the first mapping of `m` that satisfies the predicate `f` when searching from left to right.
`def findRight(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Option[(k, v)] \ ef`
Optionally returns the first mapping of `m` that satisfies the predicate `f` when searching from right to left.
`def foldLeft(f: b -> (v -> b \ ef), s: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `s` and all values in `m` going from left to right.
That is, the result is of the form: `f(...f(f(s, v1), v2)..., vn)`.
`def foldLeftWithKey(f: b -> (k -> (v -> b \ ef)), s: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `s` and all key-value pairs in `m` going from left to right.
That is, the result is of the form: `f(...f(f(s, k1, v1), k2, v2)..., vn)`.
`def foldMap(f: v -> b \ ef, m: MultiMap[k, v]): b \ ef with Monoid[b]`
Returns the result of mapping each value and combining the results.
`def foldMapWithKey(f: k -> (v -> b \ ef), m: MultiMap[k, v]): b \ ef with Monoid[b]`
Returns the result of mapping each key-value pair and combining the results.
`def foldRight(f: v -> (b -> b \ ef), s: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `s` and all values in `m` going from right to left.
That is, the result is of the form: `f(v1, ...f(vn-1, f(vn, s)))`.
`def foldRightWithCont(f: v -> ((Unit -> b \ ef) -> b \ ef), z: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `z` and all values in `m` going from right to left.
That is, the result is of the form: `f(v1, ...f(vn-1, f(vn, z)))`. A `foldRightWithCont` allows early termination by not calling the continuation.
`def foldRightWithKey(f: k -> (v -> (b -> b \ ef)), s: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `s` and all key-value pairs in `m` going from right to left.
That is, the result is of the form: `f(k1, v1, ...f(kn-1, vn-1, f(kn, vn, s)))`.
`def foldRightWithKeyCont(f: k -> (v -> ((Unit -> b \ ef) -> b \ ef)), z: b, m: MultiMap[k, v]): b \ ef`
Applies `f` to a start value `z` and all key-value pairs in `m` going from right to left.
That is, the result is of the form: `f(k1, v1, ...f(kn-1, vn-1, f(kn, vn, z)))`. A `foldRightWithKeyCont` allows early termination by not calling the continuation.
`def foldWithKey(f: b -> (k -> (v -> b \ ef)), s: b, m: MultiMap[k, v]): b \ ef`
Alias for `foldLeftWithKey`.
`def forAll(f: k -> (v -> Bool \ ef), m: MultiMap[k, v]): Bool \ ef`
Returns `true` if and only if all mappings in `m` satisfy the predicate `f`.
Returns `true` if `m` is the empty MultiMap.
`def forEach(f: k -> (v -> Unit \ ef), m: MultiMap[k, v]): Unit \ ef`
Applies `f` to every `(key, value)` of MultiMap `m`.
`def forEachWithIndex(f: Int32 -> (k -> (v -> Unit \ ef)), m: MultiMap[k, v]): Unit \ ef`
Applies `f` to tuple `(index, key, value)` formed of the keys and values of MultiMap `m` and the index of the traversal.
`def get(k: k, m: MultiMap[k, v]): Set[v] with Order[k]`
Returns `Some(vs)` if `k => vs` is in `m`.
Otherwise returns `Nil`.
`def insert(k: k, v: v, m: MultiMap[k, v]): MultiMap[k, v] with Order[k], Order[v]`
Updates `m` with `k => v`.
`def insertAll(k: k, vs: t[v], m: MultiMap[k, v]): MultiMap[k, v] \ Aef[t] with Order[k], Order[v], Foldable[t]`
Updates `m` with `k => vs`.
Where `vs` is any foldable container. If `vs` is empty nothing is inserted.
`def intersection(m1: MultiMap[k, v], m2: MultiMap[k, v]): MultiMap[k, v] with Order[k], Order[v]`
Returns the intersection of `m1` and `m2`.
`def isEmpty(m: MultiMap[k, v]): Bool`
Returns `true` if and only if `m` is the empty MultiMap.
`def iterator(rc: Region[r], m: MultiMap[k, v]): Iterator[(k, Set[v]), r, r] \ r`
Returns an iterator over all key-value pairs in `m` i.e. `k => Set#{v_1, ..., v_n}`.
`def keysOf(m: MultiMap[k, v]): Set[k] with Order[k]`
Returns the keys of `m`.
`@ParallelWhenPure `
`def map(f: v1 -> v2 \ ef, m: MultiMap[k, v1]): MultiMap[k, v2] \ ef with Order[v2]`
Returns a MultiMap with mappings `k => f(v)` for every `k => v` in `m`.
Purity reflective: Runs in parallel when given a pure function `f`.
`@ParallelWhenPure `
`def mapWithKey(f: k -> (v1 -> v2 \ ef), m: MultiMap[k, v1]): MultiMap[k, v2] \ ef with Order[v2]`
Returns a MultiMap with mappings `k => f(k, v)` for every `k => v` in `m`.
Purity reflective: Runs in parallel when given a pure function `f`.
`def memberOf(k: k, m: MultiMap[k, v]): Bool with Order[k]`
Returns `true` if and only if `m` contains the key `k`.
`def nonEmpty(m: MultiMap[k, v]): Bool`
Returns `true` if and only if `m` is a non-empty MultiMap.
`def reduceLeft(f: v -> (v -> v \ ef), m: MultiMap[k, v]): Option[v] \ ef`
Applies `f` to all values in `m` going from left to right until a single value `v` is obtained. Returns `Some(v)`.
That is, the result is of the form: `Some(f(...f(f(v1, v2), v3)..., vn))`
Returns `None` if `m` is the empty MultiMap.
`def reduceLeftWithKey(f: k -> (v -> (k -> (v -> (k, v) \ ef))), m: MultiMap[k, v]): Option[(k, v)] \ ef`
Applies `f` to all mappings in `m` going from left to right until a single mapping `(k, v)` is obtained. Returns `Some((k, v))`.
That is, the result is of the form: `Some(f(...f(f(k1, v1, k2, v2), k3, v3)..., kn, vn))`
Returns `None` if `m` is the empty MultiMap.
`def reduceRight(f: v -> (v -> v \ ef), m: MultiMap[k, v]): Option[v] \ ef`
Applies `f` to all values in `m` going from right to left until a single value `v` is obtained. Returns `Some(v)`.
That is, the result is of the form: `Some(f(v1, ...f(vn-2, f(vn-1, vn))...))`
Returns `None` if `m` is the empty MultiMap.
`def reduceRightWithKey(f: k -> (v -> (k -> (v -> (k, v) \ ef))), m: MultiMap[k, v]): Option[(k, v)] \ ef`
Applies `f` to all mappings in `m` going from right to left until a single mapping `(k, v)` is obtained. Returns `Some((k, v))`.
That is, the result is of the form: `Some(f(k1, v1, ...f(kn-2, vn-2, f(kn-1, vn-1, kn, vn))...))`
Returns `None` if `m` is the empty MultiMap.
`def remove(k: k, m: MultiMap[k, v]): MultiMap[k, v] with Order[k]`
Removes the mapping `k` from the MultiMap `m`.
`def removeWithValue(k: k, v: v, m: MultiMap[k, v]): MultiMap[k, v] with Order[k], Order[v]`
Removes the mapping of `(k, v)` from the MultiMap `m` it it exists.
`def singleton(k: k, v: v): MultiMap[k, v] with Order[k], Order[v]`
Returns the singleton MultiMap where key `k` is mapped to value `v`.
`def sumKeys(m: MultiMap[Int32, v]): Int32`
Returns the sum of all keys in the MultiMap `m`.
`def sumValues(m: MultiMap[k, Int32]): Int32`
Returns the sum of all values in the MultiMap `m`.
`@ParallelWhenPure `
`def sumWith(f: k -> (v -> Int32 \ ef), m: MultiMap[k, v]): Int32 \ ef`
Returns the sum of all key-value pairs `k => v` in the MultiMap `m` according to the function `f`.
Purity reflective: Runs in parallel when given a pure function `f`.
`def toAscList(m: MultiMap[k, v]): List[(k, v)]`
Returns the MultiMap `m` as a list of singleton key-value pairs in ascending order.
`def toAssocList(m: MultiMap[k, v]): List[(k, Set[v])]`
Returns the MultiMap `m` as a list of `(key, Set[value])` pairs in ascending order.
`def toDescList(m: MultiMap[k, v]): List[(k, v)]`
Returns the MultiMap `m` as a list of singleton key-value pairs in descending order.
`def toList(m: MultiMap[k, v]): List[(k, v)]`
Returns the MultiMap `m` as a list of singleton key-value pairs.
`def toMap(m: MultiMap[k, v]): Map[k, Set[v]]`
Returns the MultiMap `m` as a list of singleton key-value pairs in ascending order.
`def toMutDeque(rc: Region[r], m: MultiMap[k, v]): MutDeque[(k, Set[v]), r] \ r`
Returns the MultiMap `m` as a list of singleton key-value pairs in ascending order.
`def toString(m: MultiMap[k, v]): String with ToString[k], ToString[v]`
Returns a string representation of `m`.
`def union(m1: MultiMap[k, v], m2: MultiMap[k, v]): MultiMap[k, v] with Order[k], Order[v]`
Returns the union of `m1` and `m2`.
`def update(f: v -> Option[v] \ ef, k: k, m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Updates `m` with `k => v1` if `k => v` is in `m` and `f(v) = Some(v1)`. Otherwise, returns `m`.
`def updateWithKey(f: k -> (v -> Option[v] \ ef), k: k, m: MultiMap[k, v]): MultiMap[k, v] \ ef with Order[k], Order[v]`
Updates `m` with `k => v1` if `k => v` is in `m` and `f(k, v) = Some(v1)`. Otherwise, returns `m`.
`def valuesOf(m: MultiMap[k, v]): List[v] with Order[k]`
Returns the values of `m`.
Answer may contain duplicates where values were ascribed to multiple keys. | 3,694 | 11,189 | {"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} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.468121 |
https://www.mrexcel.com/board/threads/dynamic-ranges-summing-the-last-xx-variables.225259/ | 1,723,401,824,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00379.warc.gz | 705,666,953 | 19,244 | Dynamic Ranges - summing the last xx variables
doobage
New Member
how would i sum the last xx cells in a dynamic range?
for example, if i wanted to sum the last three cells in the following range, the formula would yield -- 8.
2
3
5
2
1
and for this range, again if i wanted to sum the last three cells, the formula would yield -- 10.
3
4
5
2
1
4
5
thanks in advance for the help
Excel Facts
Move date out one month or year
Use =EDATE(A2,1) for one month later. Use EDATE(A2,12) for one year later.
Try,
=SUM(OFFSET(A2,MATCH(9.99999999999999E+307,A2:A65536)-3,,3))
or if ya wanna do it in VB, one way is as follows..
Code:
`````` Function condsum()
lastrow = Range("A" & Rows.Count).End(xlUp).Row
For x = 0 To 2
t_num = Cells(lastrow - x, 1) + t_num
Next x
condsum = t_num
End Function``````
This works fine if this is the only dynamic range on the worksheet; however, what if I need a formula at the top of multiple dynamic ranges?
For instances, there's a dynamic range in column B starting in row 2 and a dynamic range in column C starting in row 2 and so on and so forth.
Thanks for the quick response. Appreciate any help.
Regards
This isn't the most elegant solution, but it will work for columns A-Z
put this in the Cell you want the result in
=condsum("A") ' Where "A" is the column identifier A B C etc
Code:
`````` Function condsum(col)
lastrow = Range(col & Rows.Count).End(xlUp).Row
celcol = Asc(UCase(col)) - 64
For x = 0 To 2
t_num = Cells(lastrow - x, celcol) + t_num
Next x
condsum = t_num
End Function``````
Still not elegant, but will work with columns beyond Z now
Code:
``````Function condsum(col)
lastrow = Range(col & Rows.Count).End(xlUp).Row
celcol = ((Len(trim(col))-1)*26) + (Asc(ucase(right(col,1)))-64)
For x = 0 To 2
t_num = Cells(lastrow - x, celcol) + t_num
Next x
condsum = t_num
End Function``````
One more change ... to account for the possibility that there isn't 3 rows of data (Assuming the data starts on Row 2, but you can change it at will)
Code:
``````Function condsum(col)
lastrow = Range(col & Rows.Count).End(xlUp).Row
If lastrow < 4 Then
condsum = "NOT ENOUGH DATA"
Exit Function
End If
celcol = ((Len(Trim(col)) - 1) * 26) + (Asc(UCase(Right(col, 1))) - 64)
For x = 0 To 2
t_num = Cells(lastrow - x, celcol) + t_num
Next x
condsum = t_num
End Function``````
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Go back | 1,017 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.808323 |
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# Silas Henderson
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
mer än 2 år ago | 873 | 3,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-25 | latest | en | 0.73405 |
https://hackage.haskell.org/package/diagrams-core-1.3.0.5/docs/Diagrams-Core-Measure.html | 1,566,601,995,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00417.warc.gz | 484,301,799 | 3,129 | diagrams-core-1.3.0.5: Core libraries for diagrams EDSL
Diagrams.Core.Measure
Synopsis
# Documentation
newtype Measured n a Source
'Measured n a' is an object that depends on `local`, `normalized` and `global` scales. The `normalized` and `global` scales are calculated when rendering a diagram.
For attributes, the `local` scale gets multiplied by the average scale of the transform.
Constructors
Measured Fieldsunmeasure :: (n, n, n) -> a
Instances
Source Source Source Source Source Source Source Floating a => Floating (Measured n a) Source Fractional a => Fractional (Measured n a) Source Num a => Num (Measured n a) Source Monoid a => Monoid (Measured n a) Source Semigroup a => Semigroup (Measured n a) Source HasOrigin t => HasOrigin (Measured n t) Source Qualifiable a => Qualifiable (Measured n a) Source HasStyle b => HasStyle (Measured n b) Source Juxtaposable a => Juxtaposable (Measured n a) Source MonadReader (n, n, n) (Measured n) Source type Rep (Measured n) = (n, n, n) Source type N (Measured n a) = N a Source type V (Measured n a) = V a Source
type Measure n = Measured n n Source
A measure is a `Measured` number.
fromMeasured :: Num n => n -> n -> Measured n a -> a Source
`fromMeasured globalScale normalizedScale measure -> a`
output :: Num n => n -> Measure n Source
Output units don't change.
local :: Num n => n -> Measure n Source
Local units are scaled by the average scale of a transform.
global :: Num n => n -> Measure n Source
Global units are ?
normalized :: Num n => n -> Measure n Source
Normalized units get scaled so that one normalized unit is the size of the final diagram.
normalised :: Num n => n -> Measure n Source
Just like `normalized` but spelt properly.
scaleLocal :: Num n => n -> Measured n a -> Measured n a Source
Scale the local units of a `Measured` thing.
atLeast :: Ord n => Measure n -> Measure n -> Measure n Source
Calculate the smaller of two measures.
atMost :: Ord n => Measure n -> Measure n -> Measure n Source
Calculate the larger of two measures. | 513 | 2,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-35 | latest | en | 0.846157 |
http://physicscentral.com/explore/action/doppler-spin.cfm | 1,579,452,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594662.6/warc/CC-MAIN-20200119151736-20200119175736-00205.warc.gz | 133,665,287 | 11,538 | # A Spin on Doppler
August 27, 2013
We have a number of devices to measure the speed of an object. We measure moving stars, galaxies, bugs, baseballs, and perhaps most commonly, vehicles caught in speed traps on the highway. Law enforcement uses Radar and LIDAR guns to measure our speed on the highway, but these devices make use of the same principle scientists use to measure how fast celestial objects move – the linear Doppler effect.
The Doppler effect can also help predict the weather and image fetuses in the womb. Recently, some attention has been given to the less popular angular Doppler effect used to measure rotational speed.
Lots of things rotate around an axis: planets, stars, galaxies, DVDs, baseballs, wheels, tops, tornadoes, atoms, and more. Being able to measure properties of rotation can provide insight into the structure of such objects, and research published in August 2013 has revealed some new twists on the rotational Doppler shift.
## Linear Doppler Effect
The linear Doppler effect occurs whenever the source of a wave and the observer of that wave have some relative linear motion to each other. It's also important to note that there are slight differences when the source is moving toward the observer and when the observer is moving toward the source.
For example, a stationary person listening to the siren of an emergency vehicle will hear an increase in frequency as the siren approaches and a decrease in frequency as it passes by. This is because the sound waves emitted by the siren are emitted from points closer to the observer as the siren approaches, and from points further from the observer as the siren passes.
In this case, the observer detects no change in the speed of the wave. The observer detects a shorter wavelength and higher frequency on approach and longer a wavelength and smaller frequency as the siren passes.
A passenger on the moving vehicle would receive a steady wave front from the siren and would not detect any change in frequency.
Left: This graphic shows what the Doppler effect looks like to an observer outside of a passing emergency vehicle with its siren blaring. Right: This graphic shows waves emitted by a stationary source.
Image credit: Lookang via Wikimedia Commons.
When the observer is moving and the source of the sound is stationary, then the distance between wavefronts emitted does not change. Instead, the observer travels through those wavefronts, and it appears to the observer that the wave is traveling at a higher speed, and with a higher frequency.
The moving observer detects wavefronts at a greater rate on approach (higher frequency), and at a smaller rate as the observer recedes from the source (lower frequency).
Here’s the equation describing the linear Doppler effect for waves not traveling too fast (much less than the speed of light):
where the top signs, plus in the numerator and minus in the denominator, are used for moving toward, and the bottom signs for moving away. This equation does not take into account movement of the medium.
If there is relative motion between a source interacting with a reflecting object (such as a wall), then the incident and reflected wave are frequency shifted, resulting in a higher frequency detected if the reflecting object is moving toward the detector, and lower if it is moving away.
The time it takes for the signal to reach the detector assists in imaging the surface that is reflecting the signal, and the frequency shift gives information on how that surface is moving. This is used to create ultrasonic images and give information of motion, and is used to image blood flow, muscle movement, and fetuses in utero.
The frequency of light appears higher (blue shifted) when an object is coming toward an observer and lower (red shifted) when it is moving away from an observer. Astronomers look at the spectrum of a star and determine the star’s actual brightness, and then measure its apparent brightness.
From comparing the apparent brightness and actual brightness they can determine how far away the star is. For galaxies, astronomers look for stars within the galaxy to estimate how far away a galaxy is. By measuring how the frequency of light is shifted astronomers can determine if a star is moving toward or away from us, and at what speed. Hubble showed with his measurements that there is a linear relationship with the distance and the amount of frequency shift.3 This relationship is called Hubble’s Law.
By knowing the brightness of a star one can calculate it's distance. From there, one can determine distances of further galaxies by measuring how their frequency is shifted and utilizing Hubble’s Law.
## Light vs. Sound
Light waves differ from sound waves because they do not require any medium to travel through. Also, the speed of light serves as the upper speed limit for our present understanding of physics based on special relativity. For light, or any electromagnetic radiation, one cannot distinguish if the source is moving or if the observer is moving. This results in the following mathematical description:
where the plus sign is used for the source and observer approaching and the minus sign is used if they are receding. If the light is interacting with a reflective object, for example a moving mirror, then the frequency shift is doubled.
Here’s a video on the basics of the linear Doppler effect:
Note: The end of this video states that all galaxies are red shifted (moving away from each other). Most are but not all. For example the Andromeda galaxy is blue shifted (moving toward us) and is predicted to collide with our galaxy in four billion years.4
## With a Twist
First it is important to note that if the light does not have some angular part to it, no angular frequency shift occurs. The angular part one usually considers comes from circular polarization of light, or any electromagnetic wave.
Light waves are made up of electric and magnetic fields that are perpendicular to each other and travel through space. If the electric and magnetic fields are rotating around the direction they are traveling in, then the light is circularly polarized. Linearly polarized light looks like the graphic of the electromagnetic wave below.
Light reflected off of water or glass is partially linearly polarized, or fully polarized at a specific angle (called the Brewster angle). Polarized sunglasses block the passage of one polarization (or orientation) of light and hence reduce the overall light reaching the eye.
An electromagnetic wave made of an electric field E and magnetic field B, that moves in the direction k.
Image credit: SuperManu, Wikimedia Commons
The angular or rotational Doppler shift has been around for a while. In reference 2, there is a derivation of the angular Doppler frequency shift. A comparison of the linear and angular Doppler effect on light sources is provided in the table below, adapted from reference 2.
For example, if the source of light is rotating, the observed frequency is shifted by the frequency of rotation, or ±f_rotation. When circularly polarized light reflects off of a rotating mirror, its frequency is shifted by ±2f_rotation. If the light is not circularly polarized then there is no frequency shift observed, in these descriptions.
## New Results
In August of 2013, a research team in Scotland published some interesting results involving the angular Doppler effect.1 Instead of using circularly polarized light, the researchers used light that traveled forward in a spiral or helix. A helical light wave carries angular information in how the light is distributed in space.5
Helical light, phase fronts, and intensity distributions for different modes, m. Note that m = +1 spins one way, and m = -1 spins in the opposite direction, but both helical wave fronts travel in the same direction.
Image credit: E-karimi, Wikimedia Commons5
The mode m, indicates number of twists and the direction of the turn the light makes with the direction of propagation. For mode m = 0, the light is not a helix. It consists of circles that are not connected and is a form of a plane wave. For the mode m = +1, the light is a single helical surface that turns once around, counter clockwise when looking at it head on, over a single wavelength. The m = -1 mode turns once around in the clockwise direction, over a single wavelength. For m = ± 2, the light makes two helical surfaces per wavelength. To get an idea of this, follow the outer and inner helical paths in the images above for m = -1 and m = -2.
The scientists found that when the helical light interacted with a rough, rotating object, the observed frequency shift depended on the mode, m, of the helical light wave.
When the researchers used two helical beams of opposite modes (±m) shining on a rough, rotating object, an intensity modulation of frequency was observed, with the increased relative intensity (power per unit area) at a frequency of
Left: Incoming beams of opposite helical modes (green) are scattered off of a rotating surface. The beam with negative mode m is red shifted, and the positive mode beam is blue shifted.
Right: This results in an increased relative intensity at a Doppler shifted frequency dependent on the mode. The relative intensity is the relative power per unit area.
Image credit: Martin Lavery, University of Glasgow.
The rough surface means that the incoming light is reflected in many different directions and modes. The researchers used a rough surface to ensure that no matter what modes the light source contained, the scattered light would have a large range of modes.1 The scientists tested the equation for the intensity modulation of frequency with different modes, and found results supporting the equation above.
The scientists then reversed their experiment by interchanging the source and the detector, and were able to obtain experimental evidence again supporting the relationship for the frequency modulation equation.1
Based on their results, the researchers showed that they can infer the frequency of rotation of a distant object by considering the intensity modulation of the frequency detected.1
## What's Next?
Two areas of application that the scientists mention in their article are remote sensing of turbulence in backscattered light and remote detection of rotating celestial bodies, for example the rotation rate of a planet outside our solar system.1
## References and Resources
1. Lavery, M.P.J., et al., Detection of a Spinning Object Using Light’s Orbital Angular Momentum, Science 341, 537-540 (2013).
https://www.sciencemag.org/content/341/6145/537.short?related-urls=yes&legid=sci;341/6145/537
2. Garetz, B.A., Angular Doppler effect, J. Opt. Soc. Am. 71 (5), 609-611 (1981).
http://www.opticsinfobase.org/josa/abstract.cfm?uri=josa-71-5-609
3. Scientific American, Ask the Experts (Jay M. Pasachoff), 21 October 1999.
http://www.scientificamerican.com/article.cfm?id=how-do-astronomers-measur
4. Doss, H.M. , Router Vision, APS, Physics Central
http://physicscentral.com/explore/action/router-vision.cfm
5. Orbital Angular Momentum of Light, Wikipedia
http://en.wikipedia.org/wiki/Orbital_angular_momentum_of_light
Heide Doss | 2,330 | 11,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-05 | latest | en | 0.951307 |
http://mathhelpforum.com/algebra/3043-difficult-a2-logarithm.html | 1,524,299,734,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945082.84/warc/CC-MAIN-20180421071203-20180421091203-00108.warc.gz | 196,749,693 | 9,558 | 1. ## Difficult A2 logarithm
Given the simultaneous equations
2^x=3^y
x+y=1
show that x=log3/log6
2. Originally Posted by kingkaisai2
Given the simultaneous equations
2^x=3^y
x+y=1
show that x=log3/log6
2^x = 3^y ----(1)
x +y = 1 ---------(2)
Take the logs of both sides of (1),
x*log(2) = y*log(3) ----(1a)
From (2),
y = 1-x
Substitute that into (1a),
x*log(2) = (1-x)log(3)
x*log(2) = log(3) -x*log(3)
x*log(2) +x*log(3) = log(3)
x[log(2) +log(3)] = log(3)
x[log(2*3)] = log(3)
x*log(6) = log(3)
x = log(3) / log(6) ---------answer.
3. Originally Posted by kingkaisai2
Given the simultaneous equations
2^x=3^y
x+y=1
show that x=log3/log6
eliminate y
taking log on both sides of the first equation
xlog2=ylog3
substituting y from second equation
xlog2=(1-x)log3
rearranging
x(log2+log3)=log3
therefore,x=log3/log6 | 336 | 819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-17 | latest | en | 0.726453 |
http://www.efunda.com/math/ode/sturm_liouville.cfm | 1,529,674,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864482.90/warc/CC-MAIN-20180622123642-20180622143642-00466.warc.gz | 395,427,320 | 6,629 | Sturm-Liouville Boundary Value Problem
Sturm-Liouville Boundary Value Problem
Ordinary Differential Equations ODE Home General Terms First Order ODE Higher Order ODE Systems of ODE Sturm-Liouville Special Cases Special Functions Bessel Legendre Hermite Laguerre Chebyshev Transform Methods Numerical Methods Resources Bibliography
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Sturm-Liouville Boundary Value Problem A differential equation defined on the interval having the form of and the boundary conditions is called as Sturm-Liouville boundary value problem or Sturm-Liouville system, where , ; the weighting function are given functions; , , , are given constants; and the eigenvalue is an unspecified parameter.
Orthogonality and General Fourier Series The non-trivial (non-zero) solutions , , of the Sturm-Liouville boundary value problem only exist at certain , . is called eigenvalue and is the eigenfunction. The eigenvalues of a Sturm-Liouville boundary value problem are non-negative real numbers. In addition, the associated eigenfunctions are orthogonal to each other with respect to the weighting function , The complete set of the solutions forms a complete orthogonal set of functions defined on the interval . Therefore, a piecewise continuous function can be expressed in terms of , , such that where The completeness allows us to express any piecewise continuous function in terms of these eigenfunctions while the orthogonality makes the expression unique and compact (no redundant terms). In addition, it can be shown that the orthogonal series is the best series available, i.e., each additional term fine tunes but not overhauls the sum of the existing terms. These properties generalize the conventional Fourier series and to any complete orthogonal series . The series is hence called the generalized Fourier series. The method of forming solutions by the general Fourier series is called the method of eigenfunction expansion which is an important technique in solving partial differential equations. Examples of generalized Fourier series can be found in Bessel functions, Legendre polynomials, and other orthogonal polynomials such as Laguerre polynomials, Hermite polynomials, and Chebyshev polynomials. Each of these polynomials represents a complete orthogonal set in different coordinates or circumstances and can be considered as a special case of the Sturm-Liouville boundary value problem. | 478 | 2,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-26 | longest | en | 0.840251 |
https://www.coursehero.com/file/11308701/C102Set2SolnRevS15/ | 1,600,639,013,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198652.6/warc/CC-MAIN-20200920192131-20200920222131-00094.warc.gz | 816,880,844 | 179,009 | C102Set2SolnRevS15 - Page 1 SOLUTION TO PROBLEM SET 2(1a C2H5OH(l 3 O2(g-> 2 CO2(g 3 H2O(l Balanced combustion reaction for one mole of ethanol(1b
# C102Set2SolnRevS15 - Page 1 SOLUTION TO PROBLEM SET 2(1a...
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Page 1 SOLUTION TO PROBLEM SET # 2 (1a) C 2 H 5 OH(l) + 3 O 2 (g) ------> 2 CO 2 (g) + 3 H 2 O(l) - Balanced combustion reaction for one mole of ethanol. (1b) q system or total = 0 = q rxn + q cal (insulated). Also, in a bomb calorimeter , all of the q's are measured at constant volume, i.e., q v . Thus: q rxn = - q cal = - C cal T ; where T = T f - T i , it is the same for all of the contents, since they are in equilibrium. T = 26.941ºC - 25.000 o C = +1.941ºC. q rxn = - 15.259 kJ ºC (+1.941 ºC)} = - 29.618 kJ for 1.000 g of ethanol. q rxn (per mole) = - 29.618 kJ 1.000 g ethanol 46.08 g ethanol 1 mole ethanol = - 1364.8 kJ/mole ethanol. q v,rxn (per mole) = E rxn (per mole) = -1364.8 kJ/mole ethanol . Note: q v = E since E = q + w = q - P ext V = q v ; because at constant volume, V, and hence w is zero . (1c) As discussed in class, q v,rxn = E rxn measured in (b) is at the initial temperature of the system . Thus, we have: H rxn,298 E rxn,298 + (PV) gases = E rxn,298 + n g RT ; where T is equal to 298 K ( T initial for E rxn = q v,rxn ). n g = (2 - 3) mol = - 1 moles. R = 8.314 x 10 - 3 kJ/mol-K. Plugging in: H rxn,298 = - 1364.8 kJ + ( - 1 moles)(8.314 x 10 - 3 kJ/mol-K)(298 K) H rxn,298 = -1367.28 kJ/(mole ethanol) . (1d) Assuming that H rxn,298 measured in (c) is equal to H o rxn,298 (i.e., assuming standard conditions), then we can say (according to Hess): H rxn,298 = - 1367.28 kJ/mol - 1367.28 kJ/mol = (2) H o f,298 (CO 2 ,g) + (3) H o f,298 (H 2 O,l) - (1)• H o f,298 (C 2 H 5 OH,l). Solving (suppressing the 298 K subscript throughout): H o f (C 2 H 5 OH,l) = ( ) (2) H o f (CO 2 ,g) + (3) H o f (H 2 O,l) - ( - 1367.28) kJ mole ethanol . Thus, plugging in: H o f (C 2 H 5 OH,l) = ((2)( - 393.5)+ (3)( - 286) - ( - 1367.28)) kJ/mole ethanol. H o f (C 2 H 5 OH,l) = -277.72 kJ/mol ethanol.
Page 2 2. Rxn .: C 2 H 4 (g) + 3 O 2 (g) ------> 2 CO 2 (g) + 2 H 2 O(l) Given : reversible, constant pressure of 1 atm and a constant temperature of 25 o C or 298 K and H o rxn,298 = - 1410.8 kJ. Thus, from the given conditions: H rxn = H o rxn,298 = q p,rxn . So, for the reaction, as written, H rxn = - 1410.8 kJ and q rxn = -1410.8 kJ . From the definition of H, we have : E rxn = H rxn - (PV) rxn H rxn - [(PV) product,gases - (PV) reactant,gases ] ; where we have neglected the “PV” term (i.e., volume) for H 2 O(l). Using the ideal gas law, PV = n g RT, for each of the gases, realizing that temperature (T) is constant - we have: E rxn H rxn + [(n g,prod - n g,reac )RT] = H rxn - n g RT ) (where the “ ” applies only to n g not to “RT”) Since we are at standard conditions (1 atm) and also, since the temperature is 298 K, we have: E rxn = 298 and H rxn = 298 ) - by definition. Thus, H o 298 = -1410.8 kJ . Substituting, we have: 298 = 298 - n g RT = ( - 1410.8 kJ) - (2 mol - 4 mol) 8.314 x 10 - 3 kJ mol K (298 K) 298 = - 1410.8 kJ + 4.955 kJ = - 1405.8 kJ => 298 = -1405.8 kJ . We can get work (w) from the first law of thermodynamics: E = q + w . Solving for w: w = E - q = - 1405.8 kJ - ( - 1410.8 kJ) = +5 kJ => w = +5 kJ . Thus, work is done on the system by the surroundings .
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Want to read all 17 pages? | 1,385 | 3,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-40 | latest | en | 0.70607 |
https://mathoverflow.net/questions/280817/reference-request-mapping-class-group-action-on-homology-of-surface-with-bounda | 1,597,332,068,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00207.warc.gz | 407,054,668 | 27,832 | # Reference request: Mapping class group action on homology of surface with boundary
This is a request for a reference to a proof of a result. The result is not very hard, but I'd rather cite than reprove.
I'm looking for a generalization of the following result (Farb and Margalit, A Primer on Mapping Class Groups) to surfaces with boundary.
Let $Y$ be a closed surface of genus $g$, and let $\Psi$ be the representation $$\Psi: \operatorname{Mod}(Y) \rightarrow \operatorname{Sp}(H_1(Y)) \cong \operatorname{Sp}(2g, \mathbb{Z})$$ given by the action of the mapping class group on homology. Then $\Psi$ is surjective.
My question is the following.
Suppose $Y$ is a surface of genus $g$, with $r > 0$ boundary components. What is the image of $$\Psi: \operatorname{Mod}(Y) \rightarrow \operatorname{SL}(H_1(Y))?$$
Again, I'm pretty sure I know what the answer is, and I could prove it myself, but this sounds like something that is well-known and I'd rather be able to cite it.
Here's what I think the answer is: First, 'cap off' the boundary components of $Y$ to embed $Y$ in a closed genus-$g$ surface $Y'$. We get an exact sequence $$0 \rightarrow V \rightarrow H_1(Y) \rightarrow H_1(Y') \rightarrow 0,$$ where $V$ is generated by the classes of the boundary components.
The MCG action must respect this exact sequence, restrict to the identity on $V$, and induce an automorphism of $H_1(Y')$ that respects the symplectic form.
I expect that every such automorphism of $H_1(Y)$ will lie in the image of the MCG action -- and that it's not very hard to produce Dehn twists on $Y$ to prove it.
Does anyone know of a reference where this is proven?
• I don't think you need to produce explicit Dehn twists. The map from MCG(Y) to MCG(Y') is surjective (the capping homomorphism) and since the latter surjects onto Sp(2g,Z) you get all symplectic automorphisms of H_1(Y'). the rest of your question seems to follow in a straightforward way. – Dan Margalit Oct 31 '17 at 13:02
• I don't see how to finish without doing something explicit. For lack of better notation, let Sp(H_1(Y)) be the set of automorphisms which preserve the (degenerate) pairing on H_1(Y). The image of MCG(Y) in Sp(H_1(Y)) is a subgroup which surjects onto Sp(H_1(Y')), but this is not enough to specify the subgroup. (For example, choosing a splitting H_1(Y') --> H_1(Y) gives a section Sp(H_1(Y')) --> Sp(H_1(Y)), whose image is a subgroup surjecting onto Sp(H_1(Y')).) – Brian Lawrence Nov 1 '17 at 20:24 | 687 | 2,491 | {"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} | 2.546875 | 3 | CC-MAIN-2020-34 | latest | en | 0.918136 |
http://social.msdn.microsoft.com/Forums/sqlserver/en-US/439120a8-c7a0-4519-b5b5-7086a497a312/dax-to-determine-different-transaction-types-by-same-customer-on-same-day?forum=sqlkjpowerpivotforexcel | 1,386,748,589,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164033438/warc/CC-MAIN-20131204133353-00031-ip-10-33-133-15.ec2.internal.warc.gz | 173,153,128 | 10,644 | # DAX to determine different transaction types by same customer on same day
### Question
• Hi,
We effectively have two types of transactions, 'customer orders' and 'cash and carry'. To state the fairly obvious, 'Customer Orders' are those placed by the customer in advance and collected later and 'cash and carry' is where the customer walks into the store and purchases an in stock item on the spot. We recognise the 'Customer Order' sale when the customer collects the item and pays the remaining balance.
My aim is to analyse where in the process of Finalising a Customer Order, the customer also makes additional purchases from the store that would be processed as a totally separate transaction.
I either need a calculated column that flags the 'Cash&Carry' records where that customer has also conducted a 'Customer Order' transaction on the same day or a measure that will sum the sales under the same conditions.
Any help appreciated.
Jacob
Tuesday, August 07, 2012 8:05 AM
• Hi Jacob,
The following formula will create the calculated column:
=IF([Transaction Type]="Cash&Carry" &&
CALCULATE(
COUNTROWS(Transactions);
ALLEXCEPT(Transactions;Transactions[Customer Reference];Transactions[Date]);
Transactions[Transaction Type]="Customer Order") > 0;
"Yes";
"No"
)
• Proposed as answer by Tuesday, August 07, 2012 12:21 PM
• Edited by Tuesday, August 07, 2012 2:13 PM
• Marked as answer by Wednesday, August 08, 2012 8:13 AM
Tuesday, August 07, 2012 12:16 PM
### All replies
• Hi Jacob,
The following formula will create the calculated column:
=IF([Transaction Type]="Cash&Carry" &&
CALCULATE(
COUNTROWS(Transactions);
ALLEXCEPT(Transactions;Transactions[Customer Reference];Transactions[Date]);
Transactions[Transaction Type]="Customer Order") > 0;
"Yes";
"No"
)
• Proposed as answer by Tuesday, August 07, 2012 12:21 PM
• Edited by Tuesday, August 07, 2012 2:13 PM
• Marked as answer by Wednesday, August 08, 2012 8:13 AM
Tuesday, August 07, 2012 12:16 PM
• Frederik, this works like a charm! Thank You.
Can you explain how this bit works: ALLEXCEPT(Transactions;Transactions[Customer Reference];Transactions[Date])
I understand the rest its just its not intuitive to me - I guess it has to create a table that includes everything except where Customer Reference and Date are the same as the current row!
Wednesday, August 08, 2012 8:18 AM
• Exactly Jacob,
CALCULATE() turns the row context into a filter context. With the 2nd, 3rd, ... argument we can create a new filter context.
ALLEXCEPT(Transactions;Transactions[Customer Reference];Transactions[Date])
means ignore all filters but keep those on Customer Reference and Date, or other words look for rows on the same date with the same customer reference
The 3rd argument of CALCULATE (Transactions[Transaction Type]="Customer Order") means: only look for Customer Orders.
If we find more than 1 row return Yes, otherwise return No.
Wednesday, August 08, 2012 12:34 PM
• Powerful stuff! Thanks again.
Wednesday, August 08, 2012 11:30 PM | 740 | 3,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2013-48 | latest | en | 0.911063 |
https://wikipedikia.org/what-does-1-gram-of-gold-weigh-2/ | 1,638,152,831,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00356.warc.gz | 703,405,718 | 27,053 | 1 GRAM Gold Bar: Weight Comparison = Paperclip
GRAMS KILOGRAMS TROY OUNCES
1 .001
.032
although, What is scrap gold?
“Scrap gold” is an expression that refers to any gold that is no longer of value in its original marketed form—for example, broken jewelry or industrial by-products. Scrap gold is often sent to a refinery, in order to be melted down and recycled to manufacture something else.
equally important, How much is 2 Tola in grams?
Tola to Gram Conversion Table
Tola [tola] Gram [g]
1 11.66
2
23.32
3 34.98
4 46.64
otherwise How is gold price calculated?
Now, if you wish to purchase a gold chain of 9.6 grams, then price will be calculated as: Price of
1 gram of gold = Rs 27,350 divided by 10
= Rs. 2,735. Price of 9.60 grams’ gold chain = Rs 2,735 times 9.60 grams = Rs 26,256.
Gold purity in jewellery Fineness Number
22KT 22K916
18KT 18K750
14KT 14K585
Oct 10, 2017
How much is a kilo of gold? Gold Price per Kilo USD
Current Price
\$58,051.74
Week High \$58,087.43
Week Low \$57,071.66
Week Change \$+553.06 (0.96%)
22 Related Questions Answers Found
## What is scrap gold price?
Aug 17, 2021 Scrap Gold Prices
24K (pure, .999+)
\$57.44/gram
18 Karat
\$43.08/gram
14 Karat
\$33.51/gram
Basis: \$1786.71 / troy ounce
## What is a fair price for scrap gold?
However, since most scrap gold buyers will offer 60% to 80% of the market price, you’ll get between \$23.15 and \$30.86 per gram, which translates to \$134.96 to \$179.91 in total for your item (a price closer to the high end will likely only be offered if you have hundreds of grams to sell at once).
## How is gold weighed for cash?
U.S. scales will measure 28 grams per ounce, while gold is measured at 31.1 grams per Troy ounce. … A pennyweight is the equivalent of 1.555 grams. Be alert that a dealer does not weigh your gold by pennyweight but pay you by the gram, a sneaky way for the dealer to pay you less for more weight of gold.
## How many grams is 7.5 Tola?
This is the equivalent of 87.48 grams (7.5 tola) of gold and 612.36 grams (52.5 tola) of silver, respectively. This is to be noted carefully that if someone possesses only gold, then the Nisab level for gold (87.48 grams) must be used.
## How much is 5 Tola in grams?
Tola to Gram Conversion Table
Tola [tola] Gram [g]
5
58.3
6 69.96
7 81.62
8 93.28
## How much is 1 Tola in grams?
At present, one tola gold is equivalent to 10 grams of gold. Once used by Ancient Indian and South Asian, the weight of one tola today is equal to 180 troy grains (11.6638038 grams) or 3/8 troy ounce. During the rule of British India, tola was used as the base unit of mass to weight and measure grains.
## What is a 916 gold?
916 gold is nothing but 22 carat gold. 916 is basically used to denote the purity of gold in the final product, i.e. 91.6 grams of pure gold in 100 gram alloy. The figure 916 is basically 22/24 (22 carat by 24 carat).
## What’s a gold price right now?
MONEX Live Gold Spot Prices
Gold Spot Prices Today Change
Gold Prices
Per Ounce
\$1,782.00
+0.00
Gold Prices Per Gram \$57.29 +0.00
Gold Prices Per Kilo \$57,291.30 +0.00
## How much is the GST on gold?
On gold jewellery, a GST of 3% is levied. The GST is charged on the total value of gold purchased as well as the making charges.
## How much is a kilo of 24K gold?
24K Gold Price in Different Quantities
Units of 24K Gold Price in USD
1 Kilo Gold Price
\$57252 USD
5 Kilo Gold Price \$286262 USD
10 Kilo Gold Price \$572524 USD
20 Kilo Gold Price \$1145047 USD
## How big is 1kg gold?
1kg Gold Bar Cast
Specification Value
Bar Size
53.00 mm x 118.00 mm
Bar Thickness 8 mm
## How many Tola is 1kg?
1 kilogram of gold is equivalent to 85.735 tola.
## What is gold at now?
MONEX Live Gold Spot Prices
Gold Spot Prices Today Change
Gold Prices
Per Ounce
\$1,782.00
+0.00
Gold Prices Per Gram \$57.29 +0.00
Gold Prices Per Kilo \$57,291.30 +0.00
## How is scrap gold price calculated?
For each group of gold, divide the karat by 24, then multiply that number by today’s gold price per gram. For example, if you have 10K gold and the current price of gold is USD\$1,600 per ounce or USD\$51.45 per gram (\$1,600/31.1), then the price of your scrap gold is USD\$51.45 x . 4167 = USD\$21.44 per gram.
## How do I calculate gold price?
Now, if you wish to purchase a gold chain of 9.6 grams, then price will be calculated as: Price of
1 gram of gold = Rs 27,350 divided by 10
= Rs. 2,735. Price of 9.60 grams’ gold chain = Rs 2,735 times 9.60 grams = Rs 26,256.
Gold purity in jewellery Fineness Number
22KT 22K916
18KT 18K750
14KT 14K585
Oct 10, 2017
## How much do Jewelers pay for gold?
If you are near The Los Angeles area, National Gold Market is located in Pasadena, California. They pay
75% to 80% for Gold Jewelry
and 90% for Gold Coins. You can walk in and get your items tested for free to determine the Karat metal purity and the gram weight.
LIVE GOLD PRICE CHART.
Karats Grams
24 K
\$58.08
## Can I sell gold to a bank?
Many consumers consider selling gold to a bank. The bad news is that most banks do NOT accept gold due to missing evaluation possibilities. During the last 10 years many counterfeit coins and bars appeared because the gold price raised so rapidly.
## What is gold spot today?
MONEX Live Gold Spot Prices
Gold Spot Prices Today Change
Gold Prices
Per Ounce
\$1,782.00
+0.00
Gold Prices Per Gram \$57.29 +0.00
Gold Prices Per Kilo \$57,291.30 +0.00
## What is the price of 14 karat gold per gram?
Currently, the average 14K gold price per gram is around \$33–\$34. | 1,665 | 5,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.917874 |
https://engineeringdiscoveries.com/how-to-calculate-quantity-of-bricks/ | 1,716,546,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00150.warc.gz | 196,489,909 | 14,998 | It Is Quite Simple How To Work Out The Number Of Bricks You Will Require, But There Are a Few Considerations You Must Consider
1- Thickness of the wall to be built
2- Amount of openings
3- Amount of detail in the wall ie: Decorative panels, Dentilation, or Dogtooth String Courses
Knowing the thickness of the wall is the most important thing to know.
The rule of thumb for working out the number of bricks, you will have to work out the total area of your building you are going to clad with brickwork, for every square m2 you have,
you have to multiply this by 60. So if you have a wall that is 6 meters long and 2 meters high, you will have a square area of 12 m. So we will need to order 720 bricks.
Please allow about 10% wastage, this can be due to transport or just moving the bricks around your project, some bricks can be quite fragile: so to clarify:
Single Skin Wall
→ 6m x 2m = 12 square metres x 60 (number of bricks in a square metre) = 720 + 10% wastage = 792 bricks required
Double Skin Wall
→ 6m x 2m = 12metres x 120 = 1440+ 20% wastage = 1728 bricks
Note: On the double skin wall I would allow for a little less wastage. | 304 | 1,142 | {"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-2024-22 | latest | en | 0.91818 |
http://www.homebrewtalk.com/f14/so-who-good-math-324876/ | 1,369,029,321,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698411148/warc/CC-MAIN-20130516100011-00041-ip-10-60-113-184.ec2.internal.warc.gz | 503,631,138 | 20,409 | Home Brew Forums > So, who is good with math?
04-29-2012, 02:47 AM #1
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So, who is good with math?
If I add 375ml of 180 proof whiskey to a 5 gallon batch how much would this raise my ABV. I came up with about 1.8%, sound right?
04-29-2012, 02:55 AM #2
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I suck at math. Sorry! Curious what your trying to do though.
04-29-2012, 02:56 AM #3
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AHS Oaked Imperial Whiskey Stout
04-29-2012, 03:02 AM #4
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Ok. Sounds like something I would try.
04-29-2012, 03:14 AM #5
Breaking the Interwebs
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Yeah I get 1.75%
04-29-2012, 11:25 AM #6
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180 proof = 90%
90% of 375ml=337.5ml
5gallons=18.93 liters=18930ml
337.5 / 18930 = .0178 or 1.78%
bosco
04-29-2012, 12:24 PM #7
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We need to know the alcohol content of your beer before we can give you an accurate answer. The answers above are based on adding the alcohol to water, and fail to take into account any alcohol that may already be in the beer.
The differences are probably pretty minor, but to illustrate - if you were to pour the whiskey into a beer that was already 99% alcohol, you would actually be lowering the alcohol content of your mix. A silly example, but it proves the point.
So, if you want an accurate answer, we need to know the aldohol content of your beer before the addition. It probably won't make a 1/10% difference from what has already been submitted, but it is required to do the math correctly.
DJG
04-29-2012, 02:46 PM #8
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The calc above doesn't take into account the increase in total volume:
whiskey bump = 337.5 / (18930 + 375) =0.01745 or 1.745%
Teeny difference Assuming that the beer is lower, the current abv isn't necessary to calculate the bump, but would of course be necessary to calculate the final abv.
04-29-2012, 03:32 PM #9
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Not to digress, but what exactly is 180-proof whiskey and how is it different than everclear?
04-29-2012, 04:02 PM #10
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Quote:
Originally Posted by blakelyc The calc above doesn't take into account the increase in total volume: whiskey bump = 337.5 / (18930 + 375) =0.01745 or 1.745% Teeny difference Assuming that the beer is lower, the current abv isn't necessary to calculate the bump, but would of course be necessary to calculate the final abv.
bosco | 1,063 | 3,426 | {"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-2013-20 | longest | en | 0.896437 |
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# Please rate my first AWA and point out areas of improvement.
Author Message
Intern
Joined: 14 Feb 2018
Posts: 5
Please rate my first AWA and point out areas of improvement. [#permalink]
### Show Tags
24 Feb 2018, 02:55
The following appeared as part of an annual report sent to stockholders by Olympic Foods, a processor of frozen
foods:
"Over time, the costs of processing go down because as organizations learn how to do things better, they
become more efficient. In color film processing, for example, the cost of a 3-by-5-inch print fell from 50 cents for
five-day service in 1970 to 20 cents for one-day service in 1984. The same principle applies to the processing of
food. And since Olympic Foods will soon celebrate its 25th birthday, we can expect that our
long experience will enable us to minimize costs and thus maximize profits."
Discuss how well reasoned....
Explanation
The argument that since the color film processing cost fell over 14 years, the cost of processing food will also fall leaves out some important concerns that must be addressed to substantiate the argument.
Firstly, it obliviates any other cause to fall down the price of the color film processing. Secondly, it assumes that even if the learning curve will reduce cost then any other factor will not reduce the profit. And lastly, it compares two different industries without outlining similarities. The author assumes that there is only one factor which caused the fall in the cost of color film processing over 14 years i.e. the learning curve. Failing to recognize that in 14 years economy undergoes many changes, author negates the existence of any other cause and erroneously credits the fall of cost to the learning curve. For e.g. Over 14 years the government might have introduced subsidy on processing materials which led to the fall in the cost. Taking into account this assumption, the author could not substantiate the time period required to learn to do things better in food industry industries and without sufficient statistical proof makes a judgement that 25 years are sufficient for food processing industries. For e.g. in last few decades, the eating habits and food preference of people have changed rapidly. With ever-increasing competition, it is inevitable to sustain in long run without updating and introducing new techniques in the processing channels. Introduction of new techniques makes the learning curve steeper and difficult to achieve. Lastly, the author has compared two different industries without providing any comparable data. For e.g. the number of distinct machines required is much less in the color film processing than the food processing making the learning curve steeper in the later thus allowing cost to fall only after a very long time. The author did not address the issue of labour turnover in both the industries, if the labour turnover is higher in the food processing industry then it will make learning to do things better a gradual process or at times not feasible to achieve.
Having discussed above mentioned points, we can conclude that if author addressed above mentioned points, it would make the argument more persuasive and convincing.
Senior Manager
Joined: 13 Oct 2016
Posts: 274
GMAT 1: 600 Q44 V28
Re: Please rate my first AWA and point out areas of improvement. [#permalink]
### Show Tags
27 Feb 2018, 09:40
rishirajnarooka wrote:
The following appeared as part of an annual report sent to stockholders by Olympic Foods, a processor of frozen
foods:
"Over time, the costs of processing go down because as organizations learn how to do things better, they
become more efficient. In color film processing, for example, the cost of a 3-by-5-inch print fell from 50 cents for
five-day service in 1970 to 20 cents for one-day service in 1984. The same principle applies to the processing of
food. And since Olympic Foods will soon celebrate its 25th birthday, we can expect that our
long experience will enable us to minimize costs and thus maximize profits."
Discuss how well reasoned....
Explanation
The argument that since the color film processing cost fell over 14 years, the cost of processing food will also fall leaves out some important concerns that must be addressed to substantiate the argument.
Firstly, it obliviates any other cause to fall down the price of the color film processing. Secondly, it assumes that even if the learning curve will reduce cost then any other factor will not reduce the profit. And lastly, it compares two different industries without outlining similarities. The author assumes that there is only one factor which caused the fall in the cost of color film processing over 14 years i.e. the learning curve. Failing to recognize that in 14 years economy undergoes many changes, author negates the existence of any other cause and erroneously credits the fall of cost to the learning curve. For e.g. Over 14 years the government might have introduced subsidy on processing materials which led to the fall in the cost. Taking into account this assumption, the author could not substantiate the time period required to learn to do things better in food industry industries and without sufficient statistical proof makes a judgement that 25 years are sufficient for food processing industries. For e.g. in last few decades, the eating habits and food preference of people have changed rapidly. With ever-increasing competition, it is inevitable to sustain in long run without updating and introducing new techniques in the processing channels. Introduction of new techniques makes the learning curve steeper and difficult to achieve. Lastly, the author has compared two different industries without providing any comparable data. For e.g. the number of distinct machines required is much less in the color film processing than the food processing making the learning curve steeper in the later thus allowing cost to fall only after a very long time. The author did not address the issue of labour turnover in both the industries, if the labour turnover is higher in the food processing industry then it will make learning to do things better a gradual process or at times not feasible to achieve.
Having discussed above mentioned points, we can conclude that if author addressed above mentioned points, it would make the argument more persuasive and convincing.
Hi rishirajnarooka
Please find below the link which contains almost all the essays related to the GMAT exam (you might find similar essays like the one posted by you as well). These essays will guarantee you a score of 5-6 in the AWA section of the GMAT Exam.
https://gmatclub.com/forum/awa-compilations-109-analysis-of-argument-essays-86274.html
For any further queries please do get back to me. All the best for your exam preparation
_________________
_______________________________________________
If you appreciate the post then please click +1Kudos
Re: Please rate my first AWA and point out areas of improvement. [#permalink] 27 Feb 2018, 09:40
Display posts from previous: Sort by | 1,592 | 7,699 | {"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} | 2.78125 | 3 | CC-MAIN-2019-30 | latest | en | 0.927335 |
http://metamath.tirix.org/mpests/scandx | 1,721,636,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00487.warc.gz | 22,646,326 | 1,786 | # Metamath Proof Explorer
## Theorem scandx
Description: Index value of the df-sca slot. (Contributed by Mario Carneiro, 14-Aug-2015)
Ref Expression
Assertion scandx ${⊢}\mathrm{Scalar}\left(\mathrm{ndx}\right)=5$
### Proof
Step Hyp Ref Expression
1 df-sca ${⊢}\mathrm{Scalar}=\mathrm{Slot}5$
2 5nn ${⊢}5\in ℕ$
3 1 2 ndxarg ${⊢}\mathrm{Scalar}\left(\mathrm{ndx}\right)=5$ | 131 | 376 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.488367 |
https://qoto.org/@Absinthe?max_id=107418399500078831 | 1,675,710,412,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00796.warc.gz | 463,014,359 | 15,869 | I just completed "Seven Segment Search" - Day 8 - Advent of Code 2021 adventofcode.com/2021/day/8
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Okay, here is the challenge for this week.
Tennis is a cool sport to watch. Especially, if either Serena or Venus or both are playing. :) However, the scoring is really odd, because it is based on the clock face. To make it even more fun, it did come from France, so instead of Zero we refer to l'œuf or "the egg". However, as we now pronounce it , it is "love". I am certain that there can be some interesting jokes in there, but I am not going for any of them. :)
Anyway, when you begin a game both players have scored zero times so their score is said to be "love - love" or "love all". When a player has scored 1 time, they are said to have "15" as advancing around the clock face 1/4 of the way. Their second score would be called "30" since that would be 1/2 way around. Obviously, their third score should be "40" because that forces you to win by 2 :) Not sure it makes sense, but that is how it works. So as an example a game might go like this:
love - love
15 - love
30 - love
40 - love
Game - player1
Not all games work out that one player gets to run the score, so it could also go like this:
love - love
15 - love
15 - 15
15 - 30
30 - 30
30 - 40
Game - player2
However, after the 3rd point is scored if both players have 40 the score is said as
"Deuce"
Because, France. :)
At any point when a player becomes 2 points ahead after 40 they win, and we say Game - <player name>
But after the score is tied at 40 or higher, the next point is called "advantage" so we no longer care about the underlying clock face scores, and merely go through "Deuce", "Advantage - player1" "Deuce" "Advantage - player2" "Game - player2". In theory, one could continually bounce from "Deuce" to advantage one player or the other until one of them players simply falls out or dies. Though I am not aware of that ever actually happening. There have been some pretty long games.
"Back in 1975 on May 26, at the Surrey Grass Court Championships at Surbiton, Anthony Fawcett and Keith Glass racked up a record 37 deuces in a single game for a grand total of 80 points."
But I digress. Here is your challenge if you choose to accept it.
Write a method/function that takes any legal tennis game score, such as (0, 0) or (5,3) and so on, and have it output the traditional tennis score as described above, "love - love" or "Game - player1" in my example of legal scores. Keep in mind, (12, 2) is not a legal score, as once the first player achieved 4 he would have already won. You don't need to test for this, but be aware it is not a scenario you should have to deal with. You can represent any tie up to 2 simply as 'love - love' or '15 - 15' or '30 - 30' but starting at 3 you would not say '40 - 40' but rather "Deuce" at that level and any higher. If you would like to do so, as a bonus you could represent the simple ties less than 3:3 (Deuce) with the 'all' so 'love all', '15 all', or '30 all' would be a bonus representation, but '40 all' would be incorrect, as it should merely be 'Deuce'.
Use whatever language or methods you like. I am using Java. I will also have unit tests. You are free not to use them, but I am not interested in testing anyone's code for accuracy, so if you don't provide them, the best you can hope for is "Nice job. Good effort. I guess." At least from me :)
Try to focus on good quality code. Also focus on reduction of complexity, DRY, and certainly use subroutines as necessary.
Good Luck!!
I will probably be checking mine in to my GitHub or GitLab repo at some point later in the week.
@Absinthe Ha, but you saw it along with a lengthy post that starts with a paragraph about that chart being a complete lie.
Here should be a simple one. It was fun to do in Java and with chained calls it turns out to be somewhat of a one-liner. This is not a code-golf thing, but feel free to try it that way if that is your interest.
"Square Odd"
Let's process an integer. The idea is that if the number contains odd digits they need to be converted to their squared value. Even digits are passed through unchanged.
For example, given the number 232, you would output 292. Or in the case of 99, you would output 8181. But if it were 22 then the output would be 22.
If this is too simple and you would like a bit more of a challenge see if you can try a couple bonus situations.
1*. determine if a number is already in this state. For example 232 wouldn't be because 3 is not squared. However, 212 could be, as could 2814 since that could have come from 294.
and
2*. given a number possibly in this state already, decode it back to the original number.
Since 1 squared is 1 and some squares have even digits like 81 such a number could have started out as '9' or actually '81'. So in 1 and 2 you may want to come up with multiple initial beginnings. So given such '81' the answer would be both of those '81' and '9'.
Things in my life have changed. I am now in a much larger focus on Java. I am still a huge fan of TDD and Test First, but I guess that is a religious thing so I will try to contain my zeal.
Initially the challenges were intended as small problems that should be in the 1-3 hour at most. As well, I focused towards Python and the problems I was using to learn the particular concepts I was working on at the time. Since my shift to Java I am less interested in some of the finer concepts and more interested in general issues. So even though I might find a solution that uses a whole pile of chained references interesting, its solution in some other language may not be nearly as interesting to you :)
I will try to be better about posting more often. Unless I find people aren't interested, in which case I won't waste the bandwidth :)
Tried a today, was pretty fun. @namark thanks for recommending these! I expect I’m gonna spend a lot of hours on these in the next few weeks.
I just boosted the last that I had posted from over a year ago.
I still think this is an interesting problem to solve. I have yet to attack it myself, but am definitely considering it. I have recently been doing a bunch more smaller exercises, and perhaps can post a few of those. Though I haven't necessarily written them, I have found them fun to do.
I do have some ideas, but for now I am waiting to see if anyone is even still interested in the puzzles anyway.
Okay, here's one with a story :) Let's see if this is entertaining enough :D
Here is a problem that involves being jerk.
You receive a parking ticket and decide to pay in the least
convient way possible... change. This decision comes to mind
because the tickets are in strange amounts because they use
the cents portion for some kind of internal encoding.
You decide to pay all in pennies but when you start to collect
them someone informs you that although change will be accepted,
if the counts of coins exceed the quantities required for a wrapper
then you must roll them.
US Coinage count to a roll
0.01 = 50
0.05 = 40
0.10 = 50
0.25 = 40
0.50 = 20
$1.00 = 25 (small) or 20 (large) Question 1. How many rolls and free coins of each can you provide to pay your$100.37 ticket in order to use the highest count of unrolled coins?
Considering that a ticket can cost anywhere from $1.00 to$250.00.
You start telling everyone else about your plan and they decide to
play too, so you calculate how they should pay as well.
For example, 45 pennies 21 nickles for a $1.00 ticket might be pretty obnoxious :). Question 2. Which fine amount (in that range) would allow you to provide the highest number of unrolled coins? Question 3. The parking authority figures out what you are doing and decides to change things up by hiring you. Your job is to determine the best fine values to get paid in the least amount of unrolled coins. What are those amounts what are those amounts (still within that range of$1.00 to $250.00) Just to clean it up a little, and add some more entertainment. Okay, here's one with a story :) Let's see if this is entertaining enough :D Here is a problem that involves being jerk. You receive a parking ticket and decide to pay in the least convient way possible... change. This decision comes to mind because the tickets are in strange amounts because they use the cents portion for some kind of internal encoding. You decide to pay all in pennies but when you start to collect them someone informs you that although change will be accepted, if the counts of coins exceed the quantities required for a wrapper then you must roll them. You also figure out that collecting all these coins is a bit of a nuisance for you. So your goal is to reduce the overall number of coin rolls while maximizing the number of free (or unrolled individual coins). US Coinage count to a roll 0.01 = 50 0.05 = 40 0.10 = 50 0.25 = 40 0.50 = 20$1.00 = 25 (small) or 20 (large)
Question 1.
How many rolls and free coins of each can you provide to pay your
$100.37 ticket in order to use the highest count of unrolled coins while using the least number of rolled coins? Considering that a ticket can cost anywhere from$1.00 to $250.00. You start telling everyone else about your plan and they decide to play too, so you calculate how they should pay as well. To make things a little more interesting, you show up at a city council meeting and propose that because of the time involved in processing unrolled coinage that the parking authority should, rather than bother with unrolled coins, all unrolled coins they collect should just be donated to the city coucil's general fund. They pass this unanimously. For example, 45 pennies 21 nickles for a$1.00 ticket might be
pretty obnoxious :).
Question 2. Which fine amount (in that range) would allow you to
provide the highest number of unrolled coins?
Question 3. The parking authority figures out what you are doing
and decides to change things up by hiring you. Your job is to
determine the best fine values to get paid in the least amount of
unrolled coins. What are those amounts what are those amounts (still
within that range of $1.00 to$250.00)
And just in case here is the first one again. Review my Java :)
git.qoto.org/Absinthe/GameOfLi
And while I am at it here is another. Review me, I can take it. :)
git.qoto.org/Absinthe/TicTacTo
Still hoping for some Java based code reviews. Here is the next one
git.qoto.org/Absinthe/24DiceGa | 2,749 | 10,962 | {"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} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.962665 |
https://electronics2electrical.com/18322/ | 1,563,799,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528013.81/warc/CC-MAIN-20190722113215-20190722135215-00172.warc.gz | 378,235,373 | 23,841 | AC Circuits with Resistance but No Reactance
AC Circuits with Resistance but No Reactance
AC Circuits with Resistance but No Reactance
Series resistances
let two resistances R 1 and R 2 are connected in series as shows in figure and there values are 50 ohm and 50 ohm respectively.
The AC voltage source is applied.
The value of AC voltage source is 100 volts.
The current flowing in the series circuit is calculated by total voltage V T upon Total resistance R T.
So the current flowing through the series circuit is 1 ampere.
Also according to ohm's law we can find the voltage across resistance R 1 and Resistance R 2.
Voltage across R 1 will be V 1 is equal to Current I into Resistance R 1.
Voltage across R 2 will be V 2 is equal to current I into resistance R 2.
The total resistance will be R T is equal to R 1 plus R 2.
Since there is no reactance in the circuit there will be no lead or lag in the voltage or current.
In resistive circuit voltage and current are in phase.
Also voltage V 1 and V 2 are in phase with current I.
Since the current flowing through both the resistance is same the common reference will be current I.
Now parallel resistances
Let the two resistances are connected in parallel.
They have same 50 ohm and 50 ohm values and 100 volts AC voltage source.
The total resistance R T can be find by R 1 into R 2 upon R 1 plus R 2.
Both the branch are connected in parallel so they have same voltage across them.
The current I in resistance R 1 and R 2 can be find by Ohm's law.
I 1 is equal to Voltage V A upon resistance R 1.
I 2 is equal to voltage V A upon resistance R 2.
Total current is equal to current I 1 plus current I 2.
Each branch current has same voltage.
Voltage V A is common to both the branch hence voltage V A is common reference. | 437 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-30 | longest | en | 0.938037 |
https://orsha-moloko.com/qa/what-will-be-value-of-1-crore-after-20-years.html | 1,618,144,148,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00553.warc.gz | 528,377,794 | 8,326 | # What Will Be Value Of 1 Crore After 20 Years?
## What will be the value of 1 crore after 15 years?
approximately Rs.
36 lakhs1 crore in 15 years use the division factor of 2.8.
That means, Rs 1 crore today will be worth (1 crore/2.8) approximately Rs.
36 lakhs after 15 years..
## Is 10 crore enough to retire?
While some people say that 10 crore is enough to retire comfortably in India, other professionals use a much broader, 80% thumb rule. The 80% thumb rule says that after you retire, you need enough money in your bank account to live on 80% of your current income for the rest of your life. … Which adds up to 4.8 Crores!
## How can I get 5 crores in 20 years?
Assuming an annual return of 12 per cent, you should invest around Rs 43,000 every month to create a corpus of Rs 1 crore at the end of 10 years. To create Rs 5 crore at the end of 20 years, you need to invest around Rs 50,000 every month.
## Is 1 crore a lot of money?
People often take a ballpark figure as a goal and consider it adequate without going into details of its sufficiency. Most people consider Rs 1 crore to be an adequate retirement amount.
## Does 401k double every 7 years?
If you want to double your money, the rule of 72 shows you how to do so in about seven years without taking on too much risk. … If you invest at an 8% return, you will double your money every 9 years. (72/8 = 9) If you invest at a 7% return, you will double your money every 10.2 years.
## What will \$1 be worth in 40 years?
Value of \$1 from 1940 to 2020 The U.S. dollar experienced an average inflation rate of 3.72% per year during this period, causing the real value of a dollar to decrease. In other words, \$1 in 1940 is equivalent in purchasing power to about \$18.59 in 2020, a difference of \$17.59 over 80 years.
## Is Rs 1 crore enough to retire?
On the face of it, a nest egg of Rs 1 crore appears big enough to sustain a retiree’s expenses for life. If put into an annuity plan when the individual is 60 years old, the corpus can yield a monthly pension of about Rs 70,000 for life.
## How can I become Crorepati in 5 years?
To get to Rs 1 crore in five years, you need to invest at least Rs 1.2 lakh, assuming an annual return of 12 per cent per year. You might get around Rs 45 lakhs if you invest Rs 50,000 for five years. Ideally, you should invest for a longer term in equities.
## How can I get 1 crore?
Here are a few ideas which made more than 1 crore.Start a blog and build your audience.Affiliate marketing.Sell a Course and leverage it by Live training, workshop, mastermind and personal mentorship.Write several Books.Dropshipping.Software as a service.Build an App which helps 10 lakh people.Freelance Expert.More items…
## What will be the value of Rs 1 crore after 30 years?
With inflation growing at the rate it does, today’s princely sum of Rs 1 crore is worth a lot less in 20 years. Based on an inflation rate of 5% per year, in 20 years time, the estimated value of your Rs 1 crore money in today’s terms might be Rs 38 lakh only.
## What will 50000 be worth in 20 years?
How much will an investment of \$50,000 be worth in the future? At the end of 20 years, your savings will have grown to \$160,357.
## What will 30k be worth in 20 years?
How much will an investment of \$30,000 be worth in the future? At the end of 20 years, your savings will have grown to \$96,214. You will have earned in \$66,214 in interest.
## What will a dollar be worth in 20 years?
Suppose that for the next 20 years inflation only averaged 2% (the green line). In that case, twenty years from now your \$10,000 would be equivalent to \$6,730 in today’s dollars.
## Can we deposit 1 crore bank?
If FD interest rate is 6.5%, then you get Rs 6.5 lakh on a fixed deposit of Rs 1 crore in a year. This means you get a monthly interest of Rs 54,167. … If FD interest rate is 7.5%, then you get Rs 7.5 lakh on a fixed deposit of Rs 1 crore in a year. This means you get a monthly interest of Rs 62,500.
## How can I make 50 lakhs in 5 years?
How Can You make Rs. 50 lakh in 5 years?1.HDFC Small Cap Fund.2.L&T Emerging Businesses Fund.Parag Parikh Long Term Equity Fund.Mirae Asset India Equity Fund.The Moderate Investor.1.Invesco India Contra Fund.Axis Focused 25 Fund.3.Principal Focused Multicap Growth Fund.More items…• | 1,156 | 4,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-17 | longest | en | 0.943535 |
https://blog.myrank.co.in/superficial-expansion/ | 1,718,347,573,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00476.warc.gz | 119,126,282 | 14,404 | Superficial Expansion
Superficial Expansion
When the temperature of a two – dimensional object is changed, its area changes, then the expansion is called superficial expansion. A two – dimensional object, such as a thin metallic plate, changes area when its temperature is raised or lowered.
For an isotope material the area expansion may be expressed in terms of the linear expansion coefficient α.Consider a rectangular area of dimension l₁ and l₂.
The area (A₀) = l₁ x l₂.
Differentiating both sides with respect to T, we get:
$$\frac{dA}{dT}={{l}_{1}}\frac{d{{l}_{2}}}{dT}+{{l}_{2}}\frac{d{{l}_{1}}}{dT}$$.
dA/ dT = l₁ (αl₂) + l₂(αl₁) = αl₁l₂ + αl₁l₂
dA/ dT = 2αl₁l₂
dA = 2αl₁l₂dT … (1)
If α is constant over the temperature range of under consideration, then equation (1) can be integrated.
$$\int\limits_{{{A}_{i}}}^{{{A}_{f}}}{dA}=2\alpha {{l}_{1}}{{l}_{2}}\int\limits_{{{T}_{i}}}^{{{T}_{f}}}{dT}$$.
ΔA = 2αA₀ΔT.
Where,
A₀ = Original Area,
ΔT = Temperature Change,
ΔA = βA₀ΔT
Where,
β = Coefficient of superficial expansion, it is twice the coefficient of linear expansion α.
Final area of the plate, A = A₀ (1 + βΔT)
The average coefficient of linear expansion, β = ΔA/ A₀ΔT.
The unit of β is °C⁻¹ (or) K⁻¹. Its dimension is θ⁻¹. | 413 | 1,258 | {"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} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.771762 |
https://www.coursehero.com/file/32041167/EET110-Unit1-Review-Assignment-Template-1docx/ | 1,561,523,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00061.warc.gz | 705,026,671 | 156,452 | EET110_Unit1_Review_Assignment_Template (1).docx
# EET110_Unit1_Review_Assignment_Template (1).docx - EET110...
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EET110 Unit 1 Review Assignment Name: DAVID HUNT Assigned Questions Chapter 2 3. How many coulombs of charge do 50 X 10 31 electrons possess? ANSWER:8.00000000e+49 JUSTIFICATION: Given: Required: 50x10^31/6.25x10^18 Solution: 8.00000000e+49 4. How many electrons does it take to make 80 µC (microcoulombs) of charge? ANSWER: 5e14 electrons JUSTIFICATION: Given: ( 80 X 10^ -6 ) / (1.6 x 10^ -19) Required: Solution: 5e14 electrons 5. Determine the voltage in each of the following cases: (a) 10 J/C ANSWER:10v JUSTIFICATION: Given: v = 10 j / 1 c Required: Solution:10v (b) 5 J/2 C ANSWER:2.5v JUSTIFICATION: Given:v=5j/2c Required: Page 1 of 5
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EET110 Unit 1 Review Assignment Solution:2.5v (c) 100 J/25 C ANSWER: 4v JUSTIFICATION: Given: Required: V=100j/25c Solution:4v 6. Five hundred joules of energy are used to move 100 C of charge through a resistor. What is the voltage across the resistor? ANSWER:5v JUSTIFICATION: Given: v=500j/100c Required: Solution:5v 8. How much energy does a 12 V battery use to move 2.5 C through a circuit?
• Fall '16
• Resistor, 5J, 5 j, 10 J
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https://im.kendallhunt.com/k5_es/teachers/grade-1/unit-3/lesson-21/lesson.html | 1,721,672,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00381.warc.gz | 259,333,478 | 27,181 | # Lesson 21
Día 3 de centros
## Warm-up: ¿Qué sabes sobre 20? (10 minutes)
### Narrative
The purpose of this What Do You Know About _____? is to invite students to share what they know and how they can represent the number 20.
### Launch
• Display the number 20.
• “¿Qué saben sobre 20?” // “What do you know about 20?”
• 1 minute: quiet think time
### Activity
• Record responses.
### Student Facing
¿Qué sabes sobre 20?
### Activity Synthesis
• “¿Qué conexiones ven entre las diferentes respuestas?” // “What connections do you see between different answers?”
• Pick two addition expressions that use different addends and ask, “¿Cómo se relacionan estas dos expresiones?” // “How are these two expressions related?”
## Activity 1: Conozcamos “¿Qué tan cerca?: Suma para obtener 20” (20 minutes)
### Narrative
The purpose of this lesson is for students to learn stage 1 of the center, How Close? Students pick a given number of digit cards and then choose a subset of those to make an expression that yields a number as close as possible to 20.
### Required Materials
Materials to Gather
Materials to Copy
• How Close? Stage 1 Recording Sheet, Spanish
### Launch
• Groups of 2
• Give each group a set of number cards, two recording sheets, and access to double 10-frames and connecting cubes or two-color counters.
• “Vamos a aprender un juego llamado ‘¿Qué tan cerca?: Suma para obtener 20’. Juguemos la primera ronda juntos” // “We are going to learn a game called How Close? Add to 20. Let's play the first round together.”
• “Primero saquemos todas las tarjetas que tengan el número 10. No vamos a usar esas tarjetas en este juego” // “First we take out any card that has the number 10. We will not use those cards for the game.”
• Display 5 cards.
• “Puedo escoger dos o tres de estas tarjetas para sumar y acercarme a 20 lo que más pueda. ¿Qué tarjetas debería escoger?” // “I can choose two or three of these cards to add to get as close to 20 as I can. What cards should I choose?”
• 1 minute: quiet think time
• 2 minutes: partner discussion
• Share responses.
• “Escribo una ecuación que tenga los números que escogí y la suma de los números” // “I write an equation with the numbers I chose and the sum of the numbers.”
• Demonstrate writing the equation on the recording sheet.
• “La persona que obtenga la suma más cercana a 20 gana un punto en esa ronda. Después, toman más tarjetas. Cada uno debe tener siempre cinco tarjetas para escoger. Jueguen otra vez. Gana la persona que obtenga más puntos” // “The person who gets a sum closer to 20 gets a point for the round. Then you each get more cards so you always have five cards to choose from. Play again. The person who gets more points wins.”
### Activity
• 10 minutes: partner work time
### Activity Synthesis
• Invite students to share their methods for making a sum close to 20.
## Activity 2: Conozcamos “Cinco en línea: Suma 7, 8 o 9” (20 minutes)
### Narrative
The purpose of this activity is for students to learn stage 3 of the Five in a Row center. Students choose to add 7, 8, or 9 to the number on their card and then place their counter on the sum on the gameboard. The first partner to have five counters in a row wins.
MLR8 Discussion Supports. Synthesis: For each method that is shared, invite students to turn to a partner and restate what they heard using precise mathematical language.
Action and Expression: Develop Expression and Communication. Give students access to manipulatives such as connecting cubes, counters, or 10-frames.
Supports accessibility for: Conceptual Processing, Memory
### Required Materials
Materials to Gather
Materials to Copy
• Five in a Row Addition and Subtraction Stage 3 Gameboard, Spanish
### Launch
• Groups of 2
• “Vamos a aprender una nueva forma de jugar ‘Cinco en línea’” // “We are going to learn a new way to play Five in a Row.”
• Display the gameboard and pick a number card.
• “Puedo escoger si le sumo 7, 8 o 9 al número que está en esta tarjeta. ¿Cuál debería sumar?” // “I can decide to add 7, 8, or 9 to the number on this card. What should I add?”
• 30 seconds: quiet think time
• 1 minute: partner discussion
• Share responses.
• Choose 7, 8, or 9 to add to the number and find the sum.
• “Ahora pongo una ficha donde está la suma en el tablero. Después, es el turno de mi compañero. Sigan jugando hasta que alguien complete cinco fichas en línea” // Now I put a counter on the sum on the gameboard. Then it is my partner's turn. Continue playing until someone gets five counters in a row.
### Activity
• 10 minutes: partner work time
### Activity Synthesis
• “¿Qué método vieron que su compañero usó para decidir si sumar 7, 8 o 9?” // “What method did you see your partner use to decide whether to add 7, 8, or 9?”
## Lesson Synthesis
### Lesson Synthesis
“Hoy aprendimos dos juegos nuevos que podemos jugar en los centros” // “Today we learned two new games we can play during center time.”
“¿Cómo les fue trabajando con su pareja en los centros? ¿Qué salió bien? ¿Qué podemos mejorar?” // “How did you and your partner work together during centers? What went well? What can we continue to work on?” | 1,366 | 5,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-30 | latest | en | 0.556795 |
http://web2.0calc.com/questions/geo_7 | 1,513,184,384,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00433.warc.gz | 297,000,730 | 5,315 | +0
# geo
0
164
3
what squar root of 25
Guest May 15, 2017
Sort:
#1
+49
+1
the square root of 25 is 5
JSaint113 May 15, 2017
#3
+6900
+1
First, think about what number times itself equals 25.
Obviously 5 x 5 = 25
Therefore square root of 25 is 5.
~The most stupid smartest cookie in the world
MaxWong May 15, 2017
### 25 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 184 | 581 | {"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} | 2.828125 | 3 | CC-MAIN-2017-51 | latest | en | 0.864277 |
https://tutors.com/lesson/diagonal-formula | 1,708,673,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00282.warc.gz | 598,483,747 | 44,306 | # Diagonal Formula
Written by
Malcolm McKinsey
Fact-checked by
Paul Mazzola
## The diagonal formula for squares, rectangles, cubes & polygons
Polygons are the shapes of our world. Computer and television screens, doors, and sheets of paper are all polygons. Diagonals of polygons are also useful. Learn how to instantly know how many diagonals any polygon can have by using this formula:
Get free estimates from geometry tutors near you.
## What's a simple polygon?
simple polygon is any two-dimensional (flat) shape made only with straight sides that close in a space, and with sides that do not cross each other (if they do, it is a complex polygon). A triangle is a polygon. A dart, kite, quadrilateral, and star are all polygons. Simple polygons can be concave or convex. The formula we will use works for all simple polygons.
## What's a diagonal?
diagonal of a polygon is a line from a vertex to a non-adjacent vertex. So a triangle, the simplest polygon, has no diagonals. You cannot draw a line from one interior angle to any other interior angle that is not also a side of the triangle. A quadrilateral, the next-simplest, has two diagonals. A pentagon, whether regular or irregular, has five diagonals.
In convex, simple polygons, diagonals will always be within the interior. Consider a rectangular door. You can run a line from the top hinge corner to the bottom, opposite corner. You can also run a line from the bottom hinge corner up to the top, opposite corner. Those are the only two diagonals possible.
In concave, simple polygons, the diagonals may go outside the polygon, crossing sides and partly lying in the shape's exterior. They are still diagonals. Darts and stars are typical examples of concave polygons with diagonals outside their shapes.
### Diagonals in real life
Diagonals in squares and rectangles add strength to construction, whether for a house wall, bridge, or tall building. You may see diagonal wires used to keep bridges steady. When houses are being built, look for diagonal braces that hold the walls straight and true.
Bookshelves and scaffolding are braced with diagonals. For a catcher in softball or baseball to throw out a runner at second base, the catcher throws along a diagonal from home plate to second.
The phone or computer screen you are viewing this lesson on is measured along its diagonal. A 21" screen never tells you the width and height; it is 21" from one corner to an opposite corner.
## Diagonal of a polygon formula
To find all possible diagonals of a simple polygon with just a few sides, you can easily count them. When the polygon gets a bit complicated, counting them can be very hard.
Fortunately, an easy formula exists to tell you exactly how many diagonals a polygon has. Remember that any vertex (corner) is connected by sides to two other vertices, so those connections cannot count as diagonals.
That vertex cannot connect to itself, either. So for n sides, we will immediately reduce the possible number of diagonals by three.
We also do not want to count the same diagonal twice. Our door, for example, only has two diagonals; you do not count going from the top hinge to bottom opposite and back again. You will have to divide any answer by two.
### Diagonal formula
This leaves us with an elegant formula, where n is the number of sides (or vertices):
Get free estimates from geometry tutors near you.
## How to find the diagonal of a rectangle
Test this formula with something we know: diagonals of a rectangle. A rectangle has four sides and four vertices.
Be skeptical! Try it for a pentagon (five sides):
A pentagon has only five diagonals; our formula works.
Be really skeptical! Try it for a tetracontakaiheptagon, which is a ridiculously long (but correct) name for a 47-gon:
Trust the formula. A 47-gon has 1,034 diagonals. This formula works every time to tell you exactly how many diagonals can be constructed inside (or outside) of any simple polygon, whether the shape is convex or concave.
### Diagonal of a rectangle formula
For rectangles, l is the length of the rectangle, and b is the height of the rectangle.
### Diagonal of a square formula
Now let's look at a few different diagonal formulas to find the length of a diagonal.
Where a is the side of the square.
### Diagonal of a cube formula
For a cube, we find the diagonal by using a three-dimensional version of the Pythagorean Theorem/distance formula:
## Lesson summary
You have learned a lot about particularly important parts of polygons, their diagonals. You now know how to identify the diagonals of any polygon, what some real-life examples of diagonals are, and how to use the formula, $\# of Diagonals=\frac{n(n-3)}{2}$,where n is the number of sides (or vertices) of the polygon.
Also, we briefly covered diagonal formulas to find the length of a diagonal in cubes squares, and rectangles.
Related articles | 1,099 | 4,909 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "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} | 4.65625 | 5 | CC-MAIN-2024-10 | latest | en | 0.909349 |
https://mathematicalcrap.com/category/acara-crash/ | 1,718,309,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00010.warc.gz | 364,406,660 | 23,710 | ## ACARA Crash 18: Errors in the Draft Mathematics Curriculum
The following is a list of errors – and possible/arguable errors – in the draft mathematics curriculum. Commenters are invited and encouraged to suggest additions, and deletions.
By “error” we mean a statement or instruction that is factually wrong or that makes no logical/mathematical/everyday sense. Some of the listed “errors” are clear-cut, while others are less so. Of course the fact that a statement/instruction made no sense to us does not prove that it makes no sense; we’ve attempted to be fair, being tough on the improper use of technical terms while giving weird phrasings a good-faith pondering in context. Nonetheless, there may well be reasonable interpretations that we have missed. (Of course phrasing that is difficult to interpret has no place in a curriculum document, but that is a separate category of sin.) As well, it is not always clear whether to characterise a statement as an error or simply a really dumb idea, but we’ve tried to stick pretty closely to “error”, leaving the noting of really dumb ideas to our other ninety-eight posts.
The list follows. The majority are elaborations. There are a few content descriptors, for which associated elaborations are indicated by a further indentation. Again, commenters are encouraged both to suggest additions to the list, and to argue for deletions from the list.
## ACARA CRASH 17: Algebraic Fractures
The following are Year 10 Number-Algebra content-elaborations in the current curriculum:
CONTENT
Apply the four operations to simple algebraic fractions with numerical denominators
ELABORATIONS
expressing the sum and difference of algebraic fractions with a common denominator
using the index laws to simplify products and quotients of algebraic fractions
CONTENT
Solve linear equations involving simple algebraic fractions
ELABORATIONS
solving a wide range of linear equations, including those involving one or two simple algebraic fractions, and checking solutions by substitution
representing word problems, including those involving fractions, as equations and solving them to answer the question
And what does the draft curriculum do with these?
Removed
And, why?
Not essential for all students to learn in Year 10.
God only knows how one develops fluency with expressions that cease to exist.
## ACARA CRASH 16: Unlevel Playing Field
Each Year’s content in the draft curriculum begins with a Level description, and each of the thirteen Level description begins with the exact same sentences:
The Australian Curriculum: Mathematics focuses on the development of a deep knowledge and conceptual understanding of mathematical structures and fluency with procedures. Students learn through the approaches for working mathematically, including modelling, investigation, experimentation and problem solving, all underpinned by the different forms of mathematical reasoning. [emphasis added]
Yep, “an equal focus on building fluency“, no doubt about it.
## ACARA CRASH 15: Digital Insertion
Continuing to try to rid ourselves of ACARA irritants, the following are the “calculator” elaborations from Year 1 – Year 6 Number and Algebra (sic):
YEAR 1
using the constant function on a calculator to add ten to single digit numbers, recording the numbers to make, show and explore the patterns in a 0 – 100 chart
with the use of a calculator, exploring skip-counting sequences that start from different numbers, discussing patterns
modeling skip counting sequences using the constant function on a calculator, while saying, reading and recording the numbers as they go
## ACARA CRASH 14: Backward Thinking
This one we really don’t get. It concerns Foundation and Year 1 Number, and was pointed out to us by Mr. Big.
We began the Crash series by critiquing the draft curriculum’s approach to counting in Foundation. Our main concern was the painful verbosity and the real-world awfulness, but we also provided a cryptic hint of one specifically puzzling aspect. The draft curriculum’s content descriptor on counting is as follows:
“establish understanding of the language and processes of counting to quantify, compare, order and make correspondences between collections, initially to 20, and explain reasoning” (draft curriculum)
“explain reasoning”. Foundation kids.
OK, let’s not get distracted; we’ve already bashed this nonsense. Here, we’re interested in the accompanying elaborations. There are ten of them, which one would imagine incorporates any conceivable manner in which one might wish to elaborate on counting. One would be wrong.
The corresponding content descriptor in the current Mathematics Curriculum is as follows:
“Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point” (current curriculum)
Notice how much more “cluttered” is the current descriptor… OK, OK stay focussed.
The current descriptor on counting has just (?) four elaborations, including the following two:
“identifying the number words in sequence, backwards and forwards, and reasoning with the number sequences, establishing the language on which subsequent counting experiences can be built” (current curriculum, emphasis added)
“developing fluency with forwards and backwards counting in meaningful contexts, including stories and rhymes” (current curriculum, emphasis added)
The point is, these elaborations also emphasise counting backwards, which seems an obvious idea to introduce and an obvious skill to master. And which is not even hinted at in any of the ten elaborations of the draft counting descriptor.
Why would the writers of the draft curriculum do that? Why would they consciously eliminate backward counting from Foundation? We’re genuinely perplexed. It is undoubtedly a stupid idea, but we cannot imagine the thought process that would lead to this stupid idea.
OK, we know what you’re thinking: it’s part of their dumbing down – maybe “dumbing forward” is a more apt expression – and they’ve thrown backward counting into Year 1. Well, no. In Year 1, students are introduce to the idea of skip-counting. And, yep, you know where this is going. So we’ll, um, skip to the end.
The current Curriculum has two elaborations of the skip-counting descriptor, one of which emphasises the straight, pure ability to count numbers backwards. And the draft curriculum? There are four elaborations on skip-counting, suggesting in turn the counting of counters in a jar, pencils, images of birds, and coins. Counting unadorned numbers? Forget it. And counting backwards? What, are you nuts?
OK, so eventually the draft curriculum seems, somehow, to get around to kids counting backwards, to look at “additive pattern sequences” and possibly to solve “subtraction problems”. The content descriptors are so unstructured and boneless, and the elaborations so vague and cluttered, it is difficult to tell. But how are the kids supposed to get there? Where is the necessary content description or elaboration:
Teach the little monsters to count backwards.
If it is there, somewhere in the draft curriculum, we honestly can’t see it. And if it is not there, that it is simply insane.
## ACARA Crash 12: Let X = X
(With apologies to the brilliant Laurie Anderson. Sane people should skip straight to today’s fish, below.)
I met this guy – and he looked like he might have been a math trick jerk at the hell brink.
Which, in fact, he turned out to be.
And I said: Oh boy.
Right again.
Let X=X.
You know, that it’s for you.
It’s a blue sky curriculum.
Parasites are out tonight.
Let X=X.
You know, I could write a book.
And this book would be thick enough to stun an ox.
Cause I can see the future and it’s a place – about a thousand miles from here.
Where it’s brighter.
Linger on over here.
Got the time?
Let X=X.
I got this postcard.
And it read, it said: Dear Amigo – Dear Partner.
Listen, uh – I just want to say thanks.
So…thanks.
Thanks for introducing me to the chaff.
Thanks for showing me the feedbag.
Thanks for going all out.
Thanks for showing me your amiss, barmy life and uh
Thanks for letting me be part of your caste.
Hug and kisses.
XXXXOOOO.
Oh yeah, P.S. I – feel – feel like – I am – in a burning building – and I gotta go.
Cause I – I feel – feel like – I am – in a burning building – and I gotta go.
OK, yes, we’re a little punch drunk. And drunk drunk. Deal with it.
Today’s fish is Year 7 Algebra. We have restricted ourselves to the content-elaboration combo dealing with abstract algebraic expressions. We have also included an omission from the current curriculum, together with the offical justification for that omission.
LEVEL DESCRIPTION
As students engage in learning mathematics in Year 7 they … explore the use of algebraic expressions and formulas using conventions, notations, symbols and pronumerals as well as natural language.
CONTENT
create algebraic expressions using constants, variables, operations and brackets. Interpret and factorise these expressions, applying the associative, commutative, identity and distributive laws as applicable
ELABORATIONS
generalising arithmetic expressions to algebraic expressions involving constants, variables, operations and brackets, for example, 7 + 7+ 7 = 3 × 7 and 𝑥 + 𝑥 + 𝑥 = 3 × 𝑥 and this is also written concisely as 3𝑥 with implied multiplication
applying the associative, commutative and distributive laws to algebraic expressions involving positive and negative constants, variables, operations and brackets to solve equations from situations involving linear relationships
exploring how cultural expressions of Aboriginal and Torres Strait Islander Peoples such as storytelling communicate mathematical relationships which can be represented as mathematical expressions
exploring the concept of variable as something that can change in value the relationships between variables, and investigating its application to processes on-Country/Place including changes in the seasons
OMISSION
Solving simple linear equations
JUSTIFICATION
Focus in Year 7 is familiarity with variables and relationships. Solving linear equations is covered in Year 8 when students are better prepared to deal with the connections between numerical, graphical and symbolic forms of relationships.
I – feel – feel like – I am – in a burning building
## ACARA Crash 11: Pulped Fractions
We’re still crazy-nuts with work, so, for today, it’s just another fish. This one is from Year 7 Number. and appears to be the sum of fraction arithmetic in Year 7.
LEVEL DESCRIPTION
As students engage in learning mathematics in Year 7 they … develop their understanding of integer and rational number systems and their fluency with mental calculation, written algorithms, and digital tools and routinely consider the reasonableness of results in context
ACHIEVEMENT STANDARD
By the end of Year 7, students use all four operations in calculations involving positive fractions and decimals, using the properties of number systems and choosing the computational approach. … They determine equivalent representations of rational numbers and choose from fraction, decimal and percentage forms to assist in computations. They solve problems involving rational numbers, percentages and ratios and explain their choice of representation of rational numbers and results when they model situations, including those in financial contexts.
CONTENT
determine equivalent fraction, decimal and percentage representations of rational numbers. Locate and represent positive and negative fractions, decimals and mixed numbers on a number line
ELABORATIONS
investigating equivalence of fractions using common multiples and a fraction wall, diagrams or a number line to show that a fraction such as is equivalent to and and therefore
expressing a fraction in simplest form using common divisors
applying and explaining the equivalence between fraction, decimal and percentage representations of rational numbers, for example, and , using manipulatives, number lines or diagrams
representing positive and negative fractions and mixed numbers on various intervals of the real number line, for example, from -1 to 1, -10 to 10 and number lines that are not symmetrical about zero or without graduations marked
investigating equivalence in fractions, decimals and percentage forms in the patterns used in the weaving designs of Aboriginal and Torres Strait Islander Peoples
CONTENT
carry out the four operations with fractions and decimals and solve problems involving rational numbers and percentages, choosing representations that are suited to the context and enable efficient computational strategies
ELABORATIONS
exploring addition and subtraction problems involving fractions and decimals, for example, using rectangular arrays with dimensions equal to the denominators, algebra tiles, digital tools or informal jottings
choosing an appropriate numerical representation for a problem so that efficient computations can be made, such as or
developing efficient strategies with appropriate use of the commutative and associative properties, place value, patterning, multiplication facts to solve multiplication and division problems involving fractions and decimals, for example, using the commutative property to calculate of giving of
exploring multiplicative (multiplication and division) problems involving fractions and decimals such as fraction walls, rectangular arrays, algebra tiles, calculators or informal jottings
developing efficient strategies with appropriate use of the commutative and associative properties, regrouping or partitioning to solve additive (addition and subtraction) problems involving fractions and decimals
calculating solutions to problems using the representation that makes computations efficient such as 12.5% of 96 is more efficiently calculated as of 96, including contexts such as, comparing land-use by calculating the total local municipal area set aside for parkland or manufacturing and retail, the amount of protein in daily food intake across several days, or increases/decreases in energy accounts each account cycle
using the digits 0 to 9 as many times as you want to find a value that is 50% of one number and 75% of another using two-digit numbers
CONTENT
model situations (including financial contexts) and solve problems using rational numbers and percentages and digital tools as appropriate. Interpret results in terms of the situation
ELABORATIONS
calculating mentally or with calculator using rational numbers and percentages to find a proportion of a given quantity, for example, 0.2 of total pocket money is spent on bus fares, 55% of Year 7 students attended the end of term function, 23% of the school population voted yes to a change of school uniform
calculating mentally or with calculator using rational numbers and percentages to find a proportion of a given quantity, for example, 0.2 of total pocket money is spent on bus fares, of Year 7 students attended the end of term function, of the school population voted yes to a change of school uniform
interpreting tax tables to determine income tax at various levels of income, including overall percentage of income allocated to tax
using modelling contexts to investigate proportion such as proportion of canteen total sales happening on Monday and Friday, proportion of bottle cost to recycling refund, proportion of school site that is green space; interpreting and communicating answers in terms of the context of the situation
expressing profit and loss as a percentage of cost or selling price, comparing the difference
investigating the methods used in retail stores to express discounts, for example, investigating advertising brochures to explore the ways discounts are expressed
investigating the proportion of land mass/area of Aboriginal Peoples’ traditional grain belt compared with Australia’s current grain belt
investigating the nutritional value of grains traditionally cultivated by Aboriginal Peoples in proportion to the grains currently cultivated by Australia’s farmers
## ACARA Crash 10: Dividing is Conquered
This Crash is a companion to, and overlaps with, the previous Crash, on multiplication. It is from Year 5 and Year 6 Number. and is, as near as we can tell, the sum of the instruction on techniques of division for F-6.
ACHIEVEMENT STANDARD (YEAR 5)
They apply knowledge of multiplication facts and efficient strategies to … divide by single-digit numbers, interpreting any remainder in the context of the problem.
CONTENT (YEAR 5)
choose efficient strategies to represent and solve division problems, using basic facts, place value, the inverse relationship between multiplication and division and digital tools where appropriate. Interpret any remainder according to the context and express results as a mixed fraction or decimal
ELABORATIONS
developing and choosing efficient strategies and using appropriate digital technologies to solve multiplicative problems involving multiplication of large numbers by one- and two-digit numbers
solving multiplication problems such as 253 x 4 using a doubling strategy, for example, 253 + 253 = 506, 506 + 506 = 1012
solving multiplication problems like 15 x 16 by thinking of factors of both numbers, 15 = 3 x 5, 16 = 2 x 8; rearranging the factors to make the calculation easier, 5 x 2 = 10, 3 x 8 = 24, 10 x 24 = 240
using an array model to show place value partitioning to solve multiplication, such as 324 x 8, thinking 300 x 8 = 2400, 20 x 8 = 160, 4 x 8 = 32 then adding the parts, 2400 + 160 + 32 = 2592; connecting the parts of the array to a standard written algorithm
investigating the use of digital tools to solve multiplicative situations managed by First Nations Ranger Groups and other groups to care for Country/Place including population growth of native and feral animals such as comparing rabbits or cane toads with platypus or koalas, or the monitoring of water volume usage in communities
LEVEL DESCRIPTION (YEAR 6)
use all four arithmetic operations with natural numbers of any size
ACHIEVEMENT STANDARD (YEAR 6)
Students apply knowledge of place value, multiplication and addition facts to operate with decimals.
CONTENT (YEAR 6)
apply knowledge of place value and multiplication facts to multiply and divide decimals by natural numbers using efficient strategies and appropriate digital tools. Use estimation and rounding to check the reasonableness of answers
ELABORATIONS
applying place value knowledge such as the value of numbers is 10 times smaller each time a place is moved to the right, and known multiplication facts, to multiply and divide a natural number by a decimal of at least tenths
applying and explaining estimation strategies to multiplicative (multiplication and division) situations involving a natural number that is multiplied or divided by a decimal to at least tenths before calculating answers or when the situation requires just an estimation
deciding to use a calculator in situations that explore multiplication and division of natural numbers being multiplied or divided by a decimal including beyond hundredths
explaining the effect of multiplying or dividing a decimal by 10, 100, 1000… in terms of place value and not the decimal point shifting
## ACARA Crash 9: Their Sorrows Shall Be Multiplied
We still have no time for the deep analysis of this shallow nonsense. So, we’ll just continue with the fish.
Below are two content-elaborations combos, from Year 5 and Year 6 Number. As near as we can tell, that’s about the sum of the instruction on techniques of multiplication for F-6.
ACHIEVEMENT STANDARD (YEAR 5)
They apply knowledge of multiplication facts and efficient strategies to multiply large numbers by one-digit and two-digit numbers
CONTENT (YEAR 5)
choose efficient strategies to represent and solve problems involving multiplication of large numbers by one-digit or two-digit numbers using basic facts, place value, properties of operations and digital tools where appropriate, explaining the reasonableness of the answer
ELABORATIONS
interpreting and solving everyday division problems such as, ‘How many buses are needed if there are 436 passengers, and each bus carries 50 people?’, deciding whether to round up or down in order to accommodate the remainder
solving division problems mentally like 72 divided by 9, 72 ÷ 9, by thinking, ‘how many 9 makes 72’, ? x 9 = 72 or ‘share 72 equally 9 ways’
investigating the use of digital technologies to solve multiplicative situations managed by First Nations Ranger Groups and other groups to care for Country/Place including population growth of native and feral animals such as comparing rabbits or cane toads with platypus or koalas, or the monitoring of water volume usage in communities
LEVEL DESCRIPTION (YEAR 6)
use all four arithmetic operations with natural numbers of any size
ACHIEVEMENT STANDARD (YEAR 6)
Students apply knowledge of place value, multiplication and addition facts to operate with decimals.
CONTENT (YEAR 6)
apply knowledge of place value and multiplication facts to multiply and divide decimals by natural numbers using efficient strategies and appropriate digital tools. Use estimation and rounding to check the reasonableness of answers
ELABORATIONS
applying place value knowledge such as the value of numbers is 10 times smaller each time a place is moved to the right, and known multiplication facts, to multiply and divide a natural number by a decimal of at least tenths
applying and explaining estimation strategies to multiplicative (multiplication and division) situations involving a natural number that is multiplied or divided by a decimal to at least tenths before calculating answers or when the situation requires just an estimation
deciding to use a calculator in situations that explore multiplication and division of natural numbers being multiplied or divided by a decimal including beyond hundredths
explaining the effect of multiplying or dividing a decimal by 10, 100, 1000… in terms of place value and not the decimal point shifting
## UPDATE (29/50/21)
we’ve just discovered some multiplication techniques tucked inside some division elaborations, as indicated in this companion Crash. The two Crashes should be considered together (and should have been just one Crash, dammit.)
## ACARA Crash 8: Multiple Contusions
OK, roll out the barrel, grab the gun: it’s time for the fish. Somehow we thought this one would take work but, really, there’s nothing to say.
It has obviously occurred to ACARA that the benefits of their Glorious Revolution may not be readily apparent to us mathematical peasants. And, one of the things we peasants tend to worry about are the multiplication tables. It is therefore no great surprise that ACARA has addressed this issue in their FAQ:
When and where are the single-digit multiplication facts (timetables) covered in the proposed F–10 Australian Curriculum: Mathematics?
These are explicitly covered at Year 4 in both the achievement standard and content descriptions for the number strand. Work on developing knowledge of addition and multiplication facts and related subtraction and division facts, and fluency with these, takes place throughout the primary years through explicit reference to using number facts when operating, modelling and solving related problems.
Nothing spells sincerity like getting the name wrong.* It’s also very reassuring to hear the kids will be “developing knowledge of … multiplication facts”. It’d of course be plain foolish to grab something huge like 6 x 3 all at once. In Year 4. And, how again will the kids “develop” this knowledge? Oh yeah, “when operating, modelling and solving related problems”. It should work a treat.
That’s the sales pitch. That’s ACARA’s conscious attempt to reassure us peasants that everything’s fine with the “timetables”. How’s it working? Feeling good? Wanna feel worse?
What follows is the relevant part of the Year Achievement Standards, and the Content-Elaboration for “multiplication facts” in Year 4 Algebra.
ACHIEVEMENT STANDARD
By the end of Year 4, students … model situations, including financial contexts, and use … multiplication facts to … multiply and divide numbers efficiently. … They identify patterns in the multiplication facts and use their knowledge of these patterns in efficient strategies for mental calculations.
CONTENT
recognise, recall and explain patterns in basic multiplication facts up to 10 x 10 and related division facts. Extend and apply these patterns to develop increasingly efficient mental strategies for computation with larger numbers
ELABORATIONS
using arrays on grid paper or created with blocks/counters to develop and explain patterns in the basic multiplication facts; using the arrays to explain the related division facts
using materials or diagrams to develop and record multiplication strategies such as skip counting, doubling, commutativity, and adding one more group to a known fact
using known multiplication facts for 2, 3, 5 and 10 to establish multiplication facts for 4, 6 ,7 ,8 and 9 in different ways, for example, using multiples of ten to establish the multiples 9 as ‘to multiply a number by 9 you multiply by 10 then take the number away’; 9 x 4 = 10 x 4 – 4 , 40 – 4 = 36 or using multiple of three as ‘to multiply a number by 9 you multiply by 3, and then multiply the result by 3 again’
using the materials or diagrams to develop and explain division strategies, such as halving, using the inverse relationship to turn division into a multiplication
using known multiplication facts up to 10 x 10 to establish related division facts
Alternatively, the kids could just learn the damn things. Starting in, oh, maybe Year 1? But what would we peasants know.
*) It has since been semi-corrected to “times-tables”. | 5,290 | 25,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-26 | latest | en | 0.926779 |
https://nigerianscholars.com/tutorials/functions-iii/inverse-of-the-function-y-ax2/ | 1,558,897,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259452.84/warc/CC-MAIN-20190526185417-20190526211417-00284.warc.gz | 575,002,997 | 17,377 | Mathematics » Functions III » Quadratic Functions
# Inverse of the Function y = ax2
## Example
### Question
Determine the inverse of the quadratic function $$h(x) = 3x^{2}$$ and sketch both graphs on the same system of axes.
### Determine the inverse of the given function $$h(x)$$
• Interchange $$x$$ and $$y$$ in the equation.
• Make $$y$$ the subject of the new equation.
\begin{align*} \text{Let } y & = 3x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 3y^{2} \\ \cfrac{x}{3} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{3}} \qquad (x \geq 0) \end{align*}
### Sketch the graphs on the same system of axes
Notice that the inverse does not pass the vertical line test and therefore is not a function.
### To determine the inverse function of $$y=ax^{2}$$:
$\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay^{2} \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & \cfrac{x}{a} = y^{2} \\ &&\therefore y = ±\sqrt{\cfrac{x}{a}} \qquad (x \geq 0) \end{array}$
The vertical line test shows that the inverse of a parabola is not a function. However, we can limit the domain of the parabola so that the inverse of the parabola is a function.
Domain and range
Consider the previous worked example $$h(x) = 3x^{2}$$ and its inverse $$y = ±\sqrt{\cfrac{x}{3}}$$:
• If we restrict the domain of $$h$$ so that $$x\ge 0$$, then $$h^{-1}(x) = \sqrt{\cfrac{x}{3}}$$ passes the vertical line test and is a function.
• If the restriction on the domain of $$h$$ is $$x\le 0$$, then $$h^{-1}(x) = -\sqrt{\cfrac{x}{3}}$$ would also be a function.
The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.
Similarly, a restriction on the domain of the function results in a restriction on the range of the inverse and vice versa.
## Example
### Question
Determine the inverse of $$q(x) = 7x^{2}$$ and sketch both graphs on the same system of axes. Restrict the domain of $$q$$ so that the inverse is a function.
### Examine the function and determine the inverse
Determine the inverse of the function:
\begin{align*} \text{Let } y & = 7x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = 7y^{2} \\ \cfrac{x}{7} & = y^{2} \\ \therefore y & = ±\sqrt{\cfrac{x}{7}} \qquad (x \geq 0) \end{align*}
### Determine the restriction on the domain
Option $$\text{1}$$: Restrict the domain of $$q$$ to $$x \ge 0$$ so that the inverse will also be a function $$( q^{-1} )$$. The restriction $$x \ge 0$$ on the domain of $$q$$ will restrict the range of $$q^{-1}$$ such that $$y \ge 0$$.
\begin{align*} q: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \ge 0 \end{align*}
or
Option $$\text{2}$$: Restrict the domain of $$q$$ to $$x \le 0$$ so that the inverse will also be a function $$( q^{-1} )$$. The restriction $$x \le 0$$ on the domain of $$q$$ will restrict the range of $$q^{-1}$$ such that $$y \le 0$$.
\begin{align*} q: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \\ q^{-1}: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \end{align*}
## Example
### Question
1. Determine the inverse of $$f(x) = -x^{2}$$.
2. Sketch both graphs on the same system of axes.
3. Restrict the domain of $$f$$ so that its inverse is a function.
### Determine the inverse of the function
\begin{align*} \text{Let } y & = -x^{2} \\ \text{Interchange } x \text{ and } y: \quad x & = -y^{2} \\ -x & = y^{2} \\ y & = ±\sqrt{-x} \qquad (x \leq 0) \end{align*}
Note: $$\sqrt{-x}$$ is only defined if $$x \le 0$$.
### Sketch both graphs on the same system of axes
The inverse does not pass the vertical line test and is not a function.
### Determine the restriction on the domain
• If $$f(x) = -x^{2}, \text{ for } x \le 0$$: \begin{align*} f: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \le 0 \end{align*}
• If $$f(x) = -x^{2}, \text{ for } x \ge 0$$: \begin{align*} f: \qquad & \text{domain } x \ge 0 \quad \text{range } y \le 0 \\ f^{-1}: \qquad & \text{domain } x \le 0 \quad \text{range } y \ge 0 \end{align*}
Determine $$f$$.
\begin{align*} \text{Let } y & = – \cfrac{1}{2} \sqrt{x} \\ \text{Interchange } x \text{ and } y: \quad x & = – \cfrac{1}{2} \sqrt{y} \qquad (y \geq 0) \\ -2x & = \sqrt{y} \\ \therefore y & = 4x^{2} \qquad (y \geq 0) \end{align*}
Determine the coordinates of the point(s) of intersection of $$g$$ and $$f$$ intersect.
To determine the coordinates of the point(s) of intersection, we equate $$g$$ and $$f$$:
\begin{align*} – \cfrac{1}{4} x & = 4x^{2} \\ 0 & = 4x^{2} + \cfrac{1}{4} x \\ 0 & = x (4x + \cfrac{1}{4} ) \\ \therefore x = 0 & \text{ or } 4x + \cfrac{1}{4} = 0 \\ \text{If } x = 0: \enspace y &= 0 \\ \text{If } 4x + \cfrac{1}{4} &= 0: \\ 4x &= – \cfrac{1}{4} \\ x &= – \cfrac{1}{16} \\ \text{If } x = – \cfrac{1}{16}: \enspace y &= – \cfrac{1}{4} ( – \cfrac{1}{16} ) \\ &= \cfrac{1}{64} \end{align*}
Therefore, the two graphs intersect at $$( 0;0 )$$ and $$( – \cfrac{1}{16}; \cfrac{1}{64} )$$.
## Example
### Question
Given: $$h(x) = 2x^{2}, \quad x \ge 0$$
1. Determine the inverse, $$h^{-1}$$.
2. Find the point where $$h$$ and $$h^{-1}$$ intersect.
3. Sketch $$h$$ and $$h^{-1}$$ on the same set of axes.
4. Use the sketch to determine if $$h$$ and $$h^{-1}$$ are increasing or decreasing functions.
5. Calculate the average gradient of $$h$$ between the two points of intersection.
### Determine the inverse of the function
\begin{align*} \text{Let } y & = 2x^{2} \qquad (x \ge 0) \\ \text{Interchange } x \text{ and } y: \quad x & = 2y^{2} \qquad (y \ge 0) \\ \cfrac{x}{2} & = y^{2} \\ y & = \sqrt{\cfrac{x}{2}} \qquad (x \ge 0, y \geq 0) \\ & \\ \therefore h^{-1}(x) & = \sqrt{\cfrac{x}{2}} \qquad (x \geq 0) \end{align*}
### Determine the point of intersection
\begin{align*} 2x^{2} & = \sqrt{\cfrac{x}{2}} \\ ( 2x^{2} )^{2} & = ( \sqrt{\cfrac{x}{2}} )^{2} \\ 4x^{4} & = \cfrac{x}{2} \\ 8x^{4} & = x \\ 8x^{4} – x & = 0 \\ x(8x^{3} – 1) & = 0 \\ \therefore x = 0 &\text{ or } \enspace 8x^{3} – 1 = 0 \\ \text{If } x=0, \quad y & = 0 \\ \text{If } 8x^{3} – 1 & = 0 \\ 8x^{3} & = 1 \\ x^{3} & = \cfrac{1}{8} \\ \therefore x & = \cfrac{1}{2} \\ \text{If } x= \cfrac{1}{2}, \quad y & = \cfrac{1}{2} \end{align*}
Therefore, this gives the points A$$(0;0)$$ and $$B( \cfrac{1}{2} ; \cfrac{1}{2})$$.
### Examine the graphs
From the graphs, we see that both $$h$$ and $$h^{-1}$$ pass the vertical line test and therefore are functions.
\begin{align*} h: \quad & \text{as } x \text{ increases. } y \text{ also increases. } \text{ therefore } h \text{ is an increasing function. } \\ h^{-1}: \quad & \text{as } x \text{ increases. } y \text{ also increases. } \text{ therefore } h^{-1} \text{ is an increasing function. } \end{align*}
Calculate the average gradient of $$h$$ between the points A$$(0;0)$$ and $$B( \cfrac{1}{2} ; \cfrac{1}{2})$$.
\begin{align*} \text{Average gradient: } &= \cfrac{y_{B} – y_{A}}{x_{B} – x_{A}} \\ &= \cfrac{ \cfrac{1}{2} – 0 }{\cfrac{1}{2} – 0 } \\ &= 1 \end{align*}
Note: this is also the average gradient of $$h^{-1}$$ between the points $$A$$ and $$B$$.
Do you want to suggest a correction or an addition to this content? Leave Contribution | 2,709 | 7,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2019-22 | latest | en | 0.723654 |
http://www.neverendingbooks.org/quiver-representations | 1,716,192,946,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00769.warc.gz | 47,582,415 | 10,897 | In what
way is a formally smooth algebra a _machine_ producing families of
manifolds? Consider the special case of the path algebra $\mathbb{C} Q$ of a
quiver and recall that an $n$-dimensional representation is an algebra
map $\mathbb{C} Q \rightarrow^{\phi} M_n(\mathbb{C})$ or, equivalently, an
$n$-dimensional left $\mathbb{C} Q$-module $\mathbb{C}^n_{\phi}$ with the action
determined by the rule $a.v = \phi(a) v~\forall v \in \mathbb{C}^n_{\phi}, \forall a \in \mathbb{C} Q$ If the $e_i~1 \leq i \leq k$ are the idempotents
in $\mathbb{C} Q$ corresponding to the vertices (see this [post][1]) then we get
a direct sum decomposition $\mathbb{C}^n_{\phi} = \phi(e_1)\mathbb{C}^n_{\phi} \oplus \ldots \oplus \phi(e_k)\mathbb{C}^n_{\phi}$ and so every $n$-dimensional
representation does determine a _dimension vector_ $\alpha = (a_1,\ldots,a_k)~\text{with}~a_i = dim_{\mathbb{C}} V_i = dim_{\mathbb{C}} \phi(e_i)\mathbb{C}^n_{\phi}$ with $| \alpha | = \sum_i a_i = n$. Further,
for every arrow $\xymatrix{\vtx{e_i} \ar[rr]^a & & \vtx{e_j}}$ we have (because $e_j.a.e_i = a$ that $\phi(a)$
defines a linear map $\phi(a)~:~V_i \rightarrow V_j$ (that was the
whole point of writing paths in the quiver from right to left so that a
representation is determined by its _vertex spaces_ $V_i$ and as many
linear maps between them as there are arrows). Fixing vectorspace bases
in the vertex-spaces one observes that the space of all
$\alpha$-dimensional representations of the quiver is just an affine
space $\mathbf{rep}_{\alpha}~Q = \oplus_a~M_{a_j \times a_i}(\mathbb{C})$ and
base-change in the vertex-spaces does determine the action of the
_base-change group_ $GL(\alpha) = GL_{a_1} \times \ldots \times GL_{a_k}$ on this space. Finally, as all this started out with fixing
a bases in $\mathbb{C}^n_{\phi}$ we get the affine variety of all
$n$-dimensional representations by bringing in the base-change
$GL_n$-action (by conjugation) and have $\mathbf{rep}_n~\mathbb{C} Q = \bigsqcup_{| \alpha | = n} GL_n \times^{GL(\alpha)} \mathbf{rep}_{\alpha}~Q$ and in this decomposition the connected
components are no longer just affine spaces with a groupaction but
essentially equal to them as there is a natural one-to-one
correspondence between $GL_n$-orbits in the fiber-bundle $GL_n \times^{GL(\alpha)} \mathbf{rep}_{\alpha}~Q$ and $GL(\alpha)$-orbits in the
affine space $\mathbf{rep}_{\alpha}~Q$. In our main example
$\xymatrix{\vtx{e} \ar@/^/[rr]^a & & \vtx{f} \ar@(u,ur)^x \ar@(d,dr)_y \ar@/^/[ll]^b}$ an $n$-dimensional representation
determines vertex-spaces $V = \phi(e) \mathbb{C}^n_{\phi}$ and $W = \phi(f) \mathbb{C}^n_{\phi}$ of dimensions $p,q~\text{with}~p+q = n$. The arrows
determine linear maps between these spaces $\xymatrix{V \ar@/^/[rr]^{\phi(a)} & & W \ar@(u,ur)^{\phi(x)} \ar@(d,dr)_{\phi(y)} \ar@/^/[ll]^{\phi(b)}}$ and if we fix a set of bases in these two
vertex-spaces, we can represent these maps by matrices
$\xymatrix{\mathbb{C}^p \ar@/^/[rr]^{A} & & \mathbb{C}^q \ar@(u,ur)^{X} \ar@(d,dr)_{Y} \ar@/^/[ll]^{B}}$ which can be considered as block
$n \times n$ matrices $a \mapsto \begin{bmatrix} 0 & 0 \\ A & 0 \end{bmatrix}~b \mapsto \begin{bmatrix} 0 & B \\ 0 & 0 \end{bmatrix}$
$x \mapsto \begin{bmatrix} 0 & 0 \\ 0 & X \end{bmatrix}~y \mapsto \begin{bmatrix} 0 & 0 \\ 0 & Y \end{bmatrix}$ The basechange group
$GL(\alpha) = GL_p \times GL_q$ is the diagonal subgroup of $GL_n$
$GL(\alpha) = \begin{bmatrix} GL_p & 0 \\ 0 & GL_q \end{bmatrix}$ and
acts on the representation space $\mathbf{rep}_{\alpha}~Q = M_{q \times p}(\mathbb{C}) \oplus M_{p \times q}(\mathbb{C}) \oplus M_q(\mathbb{C}) \oplus M_q(\mathbb{C})$
(embedded as block-matrices in $M_n(\mathbb{C})^{\oplus 4}$ as above) by
simultaneous conjugation. More generally, if $A$ is a formally smooth
algebra, then all its representation varieties $\mathbf{rep}_n~A$ are
affine smooth varieties equipped with a $GL_n$-action. This follows more
or less immediately from the definition and [Grothendieck][2]\’s
characterization of commutative regular algebras. For the record, an
algebra $A$ is said to be _formally smooth_ if for every algebra map $A \rightarrow B/I$ with $I$ a nilpotent ideal of $B$ there exists a lift
$A \rightarrow B$. The path algebra of a quiver is formally smooth
because for every map $\phi~:~\mathbb{C} Q \rightarrow B/I$ the images of the
vertex-idempotents form an orthogonal set of idempotents which is known
to lift modulo nilpotent ideals and call this lift $\psi$. But then also
every arrow lifts as we can send it to an arbitrary element of
$\psi(e_j)\pi^{-1}(\phi(a))\psi(e_i)$. In case $A$ is commutative and
$B$ is allowed to run over all commutative algebras, then by
Grothendieck\’s criterium $A$ is a commutative regular algebra. This
also clarifies why so few commutative regular algebras are formally
smooth : being formally smooth is a vastly more restrictive property as
the lifting property extends to all algebras $B$ and whenever the
dimension of the commutative variety is at least two one can think of
maps from its coordinate ring to the commutative quotient of a
non-commutative ring by a nilpotent ideal which do not lift (for an
example, see for example [this preprint][3]). The aim of
non-commutative algebraic geometry is to study _families_ of manifolds
$\mathbf{rep}_n~A$ associated to the formally-smooth algebra $A$. [1]: | 1,815 | 5,367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.744499 |
http://mathhelpforum.com/discrete-math/99369-still-ont-ont-problem-print.html | 1,524,341,654,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945317.36/warc/CC-MAIN-20180421184116-20180421204116-00206.warc.gz | 213,263,047 | 3,250 | # still ont to ont problem
• Aug 26th 2009, 09:21 PM
paulzho89
still ont to ont problem
(a). Define the function f:
http://www.mathhelpforum.com/math-he...65ed088b-1.gif
by f(x,y) = (x+2y, 3x+4y). Is f one-to-one?
can i answer this question like follow
x+2y=x+2y
x+y=x+y
3x+4y=3x+4y
x+y=x+y
?i know sounds stupid but this is all i can remember from the lecture.
anyone can give me some ideas? thanks^0^
one to one problem~~~... i mean
• Aug 26th 2009, 09:36 PM
Matt Westwood
Have you studied matrices?
What you can do is express the simultaneous equations in matrix form. As it's a simple 2x2 it's then straightforward to determine whether the determinant is zero.
If it is not zero, then the matrix is invertible, and therefore the function is one-to-one and onto.
Any help?
• Aug 27th 2009, 12:45 AM
HallsofIvy
Quote:
Originally Posted by paulzho89
(a). Define the function f:
http://www.mathhelpforum.com/math-he...65ed088b-1.gif
by f(x,y) = (x+2y, 3x+4y). Is f one-to-one?
can i answer this question like follow
x+2y=x+2y
x+y=x+y
3x+4y=3x+4y
x+y=x+y
?i know sounds stupid but this is all i can remember from the lecture.
anyone can give me some ideas? thanks^0^
Do you remember the definition of "one to one function"? A function is "one to one" if and only if f(a)= f(b) only when a= b. Since here, f is applied to two numbers, that is "f(a,b)= f(x,y) if and only if (a,b)= (x,y)" which is the same as saying a= x and b= y.
From f(a,b)= (a+ 2b, 3a+ 4b)= (x+ 2y, 3x+ 4y)= f(x,y), which is the same as saying a+ 2b= x+ 2y and 3a+ 4b= 3x+ 4y, can you show that a= x and b= y?
• Aug 27th 2009, 03:40 AM
paulzho89
Quote:
Originally Posted by HallsofIvy
Do you remember the definition of "one to one function"? A function is "one to one" if and only if f(a)= f(b) only when a= b. Since here, f is applied to two numbers, that is "f(a,b)= f(x,y) if and only if (a,b)= (x,y)" which is the same as saying a= x and b= y.
From f(a,b)= (a+ 2b, 3a+ 4b)= (x+ 2y, 3x+ 4y)= f(x,y), which is the same as saying a+ 2b= x+ 2y and 3a+ 4b= 3x+ 4y, can you show that a= x and b= y?
thanks ..help me out^.^ | 772 | 2,105 | {"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} | 3.890625 | 4 | CC-MAIN-2018-17 | latest | en | 0.891991 |
https://www.thestudentroom.co.uk/showthread.php?t=7017704 | 1,623,751,938,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00158.warc.gz | 953,037,852 | 32,365 | # Error bars in V-I graph
Watch
Announcements
#1
I'm plotting a graph of V against I ( to calculate EMF and Internal resistance), I have taken the uncertainty of both the voltmeter and ammeter to be +-0.01V/A. I've been when regressing y on x the error bars should be plotted vertically (so the +-0.01V) but this gives error bars too small to see. I am unsure whether I should just take the final absolute error to be 0.01V or if I should plot horizontal error bars which I can take a gradient from.
Also I'm doing this on excel and to put in the error bars I am going to custom error bars and just putting in 0.01 for +ve and -ve error- is this correct?
Thank you!
0
1 month ago
#2
(Original post by learningizk00l)
I'm plotting a graph of V against I ( to calculate EMF and Internal resistance), I have taken the uncertainty of both the voltmeter and ammeter to be +-0.01V/A. I've been when regressing y on x the error bars should be plotted vertically (so the +-0.01V) but this gives error bars too small to see. I am unsure whether I should just take the final absolute error to be 0.01V or if I should plot horizontal error bars which I can take a gradient from.
Also I'm doing this on excel and to put in the error bars I am going to custom error bars and just putting in 0.01 for +ve and -ve error- is this correct?
Thank you!
What does the plot look like?
0
#3
(Original post by Nagromicous)
What does the plot look like?
(This contains vertical error bars)
Last edited by learningizk00l; 1 month ago
0
1 month ago
#4
(Original post by learningizk00l)
(This contains vertical error bars)
Clearly, your error bars are wrong. Roughly two-thirds of your data should be within one error bar of the trendline. It would help if you thought about why your errors would be more than 0.01V, e.g. because the current was changing as the circuit was heating up, etc.
0
#5
(Original post by Nagromicous)
Clearly, your error bars are wrong. Roughly two-thirds of your data should be within one error bar of the trendline. It would help if you thought about why your errors would be more than 0.01V, e.g. because the current was changing as the circuit was heating up, etc.
But surely it doesn't matter if the heat changed the current seeing as current is something I am measuring and not just assuming constant?
0
1 month ago
#6
(Original post by learningizk00l)
But surely it doesn't matter if the heat changed the current seeing as current is something I am measuring and not just assuming constant?
By drawing a straight line with equation V = e - Ir, you assume that the resistance is constant in time (for each measurement). With larger currents, the battery and wires will heat up, increasing the resistance in a way you haven't accounted for. The discrepancy between your results and the expected results could be caused by the heat changing the resistance. If I were you, I'd redo the experiment, making sure that I let the equipment cool down for each measurement, only leaving the circuit closed for long enough to take the measurement.
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70.24% | 831 | 3,386 | {"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} | 2.75 | 3 | CC-MAIN-2021-25 | latest | en | 0.923717 |
https://magnimetrics.com/tag/value/ | 1,611,751,055,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704824728.92/warc/CC-MAIN-20210127121330-20210127151330-00682.warc.gz | 431,708,535 | 14,057 | ## What was the Labor Theory of Value?
The Labor Theory of Value was one of the early attempts to explain how market prices form. It followed the idea that the main driver for goods’ value is the labor necessary to produce them. Under the theory, the labor hours it takes workers to produce a commodity is the Read more…
## The Value-Added Concept in Economics
The term value-added represents the enhanced value a company adds to its products and services. You probably notice that products sell for more than it costs to produce them, and services charge more than it costs to render them. This is what value-added represents. The essential purpose of any business Read more…
## Value Chain Analysis Break-down
The primary purpose of any business is to produce goods or provide services in a way that they have a higher value for the customers than the original cost for the firm. Companies engage in numerous activities while converting inputs to outputs. Porter’s Value Chain helps us create a clear Read more…
## Time Value of Money Explained
Introduction To understand the Time Value of Money, imagine you were offered 100 euros now or 100 euros in three years, what would you prefer? If you are like me, you’d probably prefer the money now. But why is that, when a 100 euros has the same value now and Read more…
## Sensitivity Analysis in Financial Modeling
Introduction to Sensitivity Analysis We apply Sensitivity Analysis to a financial model to determine how different values of an independent variable affect a specific dependent variable under a given set of assumptions. We also refer to it as ‘what-if’ or simulation analysis. Performing such analysis helps us predict better the Read more…
## Economic Value Added (EVA) Case Study: Nestlé S.A.
If you haven’t read the article on the Economic Value Added (EVA) metric, please, take a look here, before you read on. Introduction In today’s post, we are doing a short case study, trying to calculate the Economic Value Added (EVA) for Nestlé for FY 2018, using actual data available Read more…
## Capital Asset Pricing Model (CAPM)
Introduction The Capital Asset Pricing Model (CAPM) shows us the relationship between systematic risk for an investment and the expected return on it. Analysts and financial professionals use the model widely for pricing risky investments and generating expected returns for assets, considering the risk and cost of capital. The CAPM Read more…
## Terminal Value of the Business
What is Terminal Value? The Terminal Value (TV), or as it is also known, continuing value or horizon value, is the value of an investment or a business at the end of a specific period. A firm or project potentially has an infinite life, meaning its value is the Net Read more…
## Discounted Cash Flow Valuation Method
Today we are looking at how the Discounted Cash Flow (DCF) method is used to evaluate investment opportunities or project alternatives in big companies, like launching a new product, a new assembly line, etc. We can use the DCF method whenever we consider paying now to get more money (or Read more… | 632 | 3,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-04 | longest | en | 0.908795 |
https://oeis.org/A326447 | 1,709,121,064,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00428.warc.gz | 436,228,396 | 3,969 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A326447 Sum of the sixth largest parts in the partitions of n into 8 squarefree parts. 9
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 4, 5, 9, 11, 16, 19, 27, 33, 45, 51, 69, 80, 105, 117, 150, 172, 216, 242, 300, 339, 416, 466, 568, 636, 768, 852, 1022, 1135, 1348, 1483, 1748, 1934, 2260, 2481, 2876, 3163, 3655, 3993, 4582, 5014, 5735, 6244, 7098, 7732 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,11 LINKS Table of n, a(n) for n=0..55. Index entries for sequences related to partitions FORMULA a(n) = Sum_{p=1..floor(n/8)} Sum_{o=p..floor((n-p)/7)} Sum_{m=o..floor((n-o-p)/6)} Sum_{l=m..floor((n-m-o-p)/5)} Sum_{k=l..floor((n-l-m-o-p)/4)} Sum_{j=k..floor((n-k-l-m-o-p)/3)} Sum_{i=j..floor((n-j-k-l-m-o-p)/2)} mu(p)^2 * mu(o)^2 * mu(m)^2 * mu(l)^2 * mu(k)^2 * mu(j)^2 * mu(i)^2 * mu(n-i-j-k-l-m-o-p)^2 * m, where mu is the Möbius function (A008683). a(n) = A326444(n) - A326445(n) - A326446(n) - A326448(n) - A326449(n) - A326450(n) - A326451(n) - A326452(n). MATHEMATICA Table[Sum[Sum[Sum[Sum[Sum[Sum[Sum[m * MoebiusMu[p]^2 * MoebiusMu[o]^2 * MoebiusMu[m]^2 * MoebiusMu[l]^2 * MoebiusMu[k]^2 * MoebiusMu[j]^2 * MoebiusMu[i]^2 * MoebiusMu[n - i - j - k - l - m - o - p]^2, {i, j, Floor[(n - j - k - l - m - o - p)/2]}], {j, k, Floor[(n - k - l - m - o - p)/3]}], {k, l, Floor[(n - l - m - o - p)/4]}], {l, m, Floor[(n - m - o - p)/5]}], {m, o, Floor[(n - o - p)/6]}], {o, p, Floor[(n - p)/7]}], {p, Floor[n/8]}], {n, 0, 50}] CROSSREFS Cf. A008683, A326443, A326444, A326445, A326446, A326448, A326449, A326450, A326451, A326452. Sequence in context: A050045 A308955 A098386 * A326527 A326632 A240206 Adjacent sequences: A326444 A326445 A326446 * A326448 A326449 A326450 KEYWORD nonn AUTHOR Wesley Ivan Hurt, Jul 06 2019 STATUS approved
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Last modified February 28 06:43 EST 2024. Contains 370387 sequences. (Running on oeis4.) | 931 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.385611 |
http://forums.wolfram.com/mathgroup/archive/2007/Apr/msg00637.html | 1,571,206,047,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00522.warc.gz | 72,786,804 | 7,868 | Re: CrossProduct in Spherical Coordinates
• To: mathgroup at smc.vnet.net
• Subject: [mg75330] Re: [mg75298] CrossProduct in Spherical Coordinates
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 25 Apr 2007 05:39:21 -0400 (EDT)
• References: <200704240726.DAA27495@smc.vnet.net>
```
On 24 Apr 2007, at 16:26, gogoant06 at yahoo.com.hk wrote:
> Dear all,
>
> I am really new to mathematica and I have met a damn simple problem.
>
> In[1]:=
> <<Calculus`VectorAnalysis`
>
> In[2]:=
> CrossProduct[{1,0,0},{0,1,0},Spherical]
>
> Out[2]=
> {0,0,0}
>
> Why? Isn't the result supposed to be {0,0,1}, even in spherical
> coordinates?
>
> best regards,
> lion
>
>
The poivectornt that has coordinates {0,1,0} in Spherical coordinates
is simply the vector {0,0,0} in Cartesion coordinates (because the
first coordinate stands for the "radius" and is 0). Hence the cross
product of anything with this vector must be 0.
Persumably what you wanted to do was to compute in spherical
coordiantes the cross product of the vectors which in cartesion
coordiantes (and not in spherical ones) are represented as {1,0,0}
and {0,1,0}? If so, you can do this as follows:
v=CrossProduct[CoordinatesFromCartesian
[{1,0,0}],CoordinatesFromCartesian[{0,\
1,0}],Spherical]
{1,0,0}
Of course this is the answer in spherical coordinates, so to get it
in Cartesian cordinates you need to perfomr another coordinate switch:
CoordinatesToCartesian[v]
{0,0,1}
as expected.
Of course it would make more sense to compute it directly in
Cartesian coordiantes!
Andrzej Kozlowski
```
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• Next by thread: Re: CrossProduct in Spherical Coordinates | 551 | 1,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-43 | longest | en | 0.823317 |
https://www.unitconverters.net/force/petanewton-to-kilogram-force.htm | 1,638,835,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00521.warc.gz | 1,126,372,231 | 3,047 | Home / Force Conversion / Convert Petanewton to Kilogram-force
# Convert Petanewton to Kilogram-force
Please provide values below to convert petanewton [PT] to kilogram-force [kgf], or vice versa.
From: petanewton To: kilogram-force
### Petanewton to Kilogram-force Conversion Table
Petanewton [PT]Kilogram-force [kgf]
0.01 PT1019716212977.9 kgf
0.1 PT10197162129779 kgf
1 PT1.0197162129779E+14 kgf
2 PT2.0394324259559E+14 kgf
3 PT3.0591486389338E+14 kgf
5 PT5.0985810648896E+14 kgf
10 PT1.0197162129779E+15 kgf
20 PT2.0394324259559E+15 kgf
50 PT5.0985810648896E+15 kgf
100 PT1.0197162129779E+16 kgf
1000 PT1.0197162129779E+17 kgf
### How to Convert Petanewton to Kilogram-force
1 PT = 1.0197162129779E+14 kgf
1 kgf = 9.80665E-15 PT
Example: convert 15 PT to kgf:
15 PT = 15 × 1.0197162129779E+14 kgf = 1.5295743194669E+15 kgf | 340 | 835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.259151 |
https://www.teachoo.com/3746/1231/Misc-23---If-y--ea-cos-1-x--show-(1---x2)-d2y-dx2---x-dy-dx/category/Logarithmic-Differentiation---Type-1/ | 1,726,314,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00535.warc.gz | 937,977,139 | 23,051 | Logarithmic Differentiation - Type 1
Chapter 5 Class 12 Continuity and Differentiability
Concept wise
### Transcript
Misc 23 If 𝑦=𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥) , – 1 ≤ 𝑥 ≤ 1, show that (1−𝑥^2 ) (𝑑^2 𝑦)/〖𝑑𝑥〗^2 −𝑥 𝑑𝑦/𝑑𝑥 − 𝑎2 𝑦 =0 . 𝑦=𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = 𝑑(𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) × 𝑑(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) × 𝑎 ((−1)/√(1 − 𝑥^2 )) 𝑑𝑦/𝑑𝑥 = (−𝑎 𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ))/√(1 − 𝑥^2 ) √(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥 = −𝑎𝑒^(〖𝑎 𝑐𝑜𝑠〗^(−1) 𝑥" " ) √(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥 = −𝑎𝑦 Since we need to prove (1−𝑥^2 ) (𝑑^2 𝑦)/〖𝑑𝑥〗^2 − 𝑥 𝑑𝑦/𝑑𝑥 −𝑎2 𝑦 =0 Squaring (1) both sides (√(1 − 𝑥^2 ) 𝑑𝑦/𝑑𝑥)^2 = (−𝑎𝑦)^2 (1−𝑥^2 ) (𝑦^′ )^2 = 𝑎^2 𝑦^2 Differentiating again w.r.t x 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = (d(𝑎^2 𝑦^2))/𝑑𝑥 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = 𝑎^2 (𝑑(𝑦^2))/𝑑𝑥 𝑑((1 − 𝑥^2 ) (𝑦^′ )^2 )/𝑑𝑥 = 𝑎^2 × 2𝑦 ×𝑑𝑦/𝑑𝑥 𝑑(1 − 𝑥^2 )/𝑑𝑥 (𝑦^′ )^2+(1 − 𝑥^2 ) 𝒅((𝒚^′ )^𝟐 )/𝒅𝒙 = 𝑎^2 × 2𝑦𝑦^′ (−2𝑥)(𝑦^′ )^2+(1 − 𝑥^2 )(𝟐𝒚^′ × 𝒅(𝒚^′ )/𝒅𝒙) = 𝑎^2 × 2𝑦𝑦^′ (−2𝑥)(𝑦^′ )^2+(1 − 𝑥^2 )(𝟐𝒚^′ × 𝒚^′′ ) = 𝑎^2 × 2𝑦𝑦^′ Dividing both sides by 𝟐𝒚^′ −𝑥𝑦^′+(1 − 𝑥^2 ) 𝑦^′′ = 𝑎^2 × 𝑦 −𝑥𝑦^′+(1 − 𝑥^2 ) 𝑦^′′ = 𝑎^2 𝑦 (𝟏 − 𝒙^𝟐 ) 𝒚^′′−𝒙𝒚^′−𝒂^𝟐 𝒚=𝟎 Hence proved | 1,145 | 1,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.495282 |
http://stackoverflow.com/questions/10936945/how-to-find-the-optimal-cut-off-point-threshold | 1,438,706,281,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042991019.80/warc/CC-MAIN-20150728002311-00052-ip-10-236-191-2.ec2.internal.warc.gz | 232,468,810 | 17,077 | How to find the “optimal” cut-off point (threshold)
I have a set of weighted features for machine learning. I'd like to reduce the feature set and just use those with a very large or very small weight.
So given below image of sorted weights, I'd only like to use the features that have weights above the higher or below the lower yellow line.
What I'm looking for is some kind of slope change detection so I can discard all the features until the first/last slope coefficient increase/decrease.
While I (think I) know how to code this myself (with first and second numerical derivatives), I'm interested in any established methods. Perhaps there's some statistic or index that computes something like that, or anything I can use from SciPy?
Edit: At the moment, I'm using `1.8*positive.std()` as positive and `1.8*negative.std()` as negative threshold (fast and simple), but I'm not mathematician enough to determine how robust this is. I don't think it is, though. ⍨
-
If the data are (approximately) Gaussian distributed, then just using a multiple of the standard deviation is sensible.
If you are worried about heavier tails, then you may want to base your analysis on order statistics.
• Since you've plotted it, I'll assume you're willing to sort all of the data.
• Let N be the number of data points in your sample.
• Let x[i] be the i'th value in the sorted list of values.
• Then 0.5( x[int( 0.8413*N)]-x[int(0.1587*N)]) is an estimate of the standard deviation which is more robust against outliers. This estimate of the std can be used as you indicated above. (The magic numbers above are the fraction of data that are less than [mean+1sigma] and [mean-1sigma] respectively).
• There are also conditions where just keeping the highest 10% and lowest 10% would be sensible as well; and these cutoffs are easily computed if you have the sorted data on hand.
These are somewhat ad hoc approaches based on the content of your question. The general sense of what you're trying to do is (a form of) anomaly detection, and you can probably do a better job of it if you're careful in defining/estimating what the shape of the distribution is near the middle, so that you can tell when the features are getting anomalous.
- | 512 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2015-32 | latest | en | 0.958106 |
https://inbox.vuxu.org/hott/cb1c245d-987c-4da9-ac09-3814bfd096cf@googlegroups.com/T/ | 1,679,433,905,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00121.warc.gz | 385,751,204 | 3,647 | ```Discussion of Homotopy Type Theory and Univalent Foundations
help / color / mirror / Atom feed```
```* [HoTT] Type factorials
@ 2019-11-28 20:33 Martín Hötzel Escardó
0 siblings, 0 replies; only message in thread
From: Martín Hötzel Escardó @ 2019-11-28 20:33 UTC (permalink / raw)
To: Homotopy Type Theory
[-- Attachment #1.1: Type: text/plain, Size: 3295 bytes --]
Probably just for amusement only.
(1) In the category of topological spaces (not up to homotopy), if X and Y
are exponentiable and we topologize the space (X ≃ Y) of homeomorphisms of
X and Y with the subspace topology, we get that
co-derived-set (Y+1) × (X ≃ Y) ≃ (X+1 ≃ Y+1).
The proof relies on classical logic.
Exponentiability is only needed to make sense of (X ≃ Y) as a space, and
hence we could alternatively work with compactly generated spaces, which
have all exponentials, but I haven't checked the above in that case (yet).
The derived set, in the sense of Cantor, is the set of limit points. Here
we consider its complement, with the subspace topology, which of course is
the discrete topology.
When X is the same space as Y, the above specializes to
co-derived-set (X+1) × Aut X ≃ Aut (X+1),
where Aut X is the space of auto (homeo)morphisms of X.
Hence when X is discrete, so that it is its own coderived set, the above
specializes to the "factorial equation"
(X+1) × Aut X ≃ Aut (X+1).
But if X is perfect in the sense of Cantor (has no isolated points), then
Aux X = Aut (X+1)
because the coderived set of X+1 is homeomorphic to 1.
(2) The above holds if we replace "space" by "homotopy type" and can be
proved in HoTT/UF, *this time without appealing to classical logic* (and I
doubt appealing to classical logic can simplify the argument, because the
argument seems to be as simple and direct as it can be).
For this, we need the following definitions in HoTT/UF:
* A point x is *isolated* if the type x=y is decidable for all y:X.
(Counter-example: any point of the circle.)
* A type is *discrete* if all its points are isolated.
(Example: natural numbers, the finite type Fin n with n elements.)
* A space is *perfect* if it has no isolated points.
(Example: the circle.)
There is also a notion that doesn't arise above but that does arise in the
proofs/constructions:
* A point x is *h(omotopy)-isolated* if the type x=y is a proposition,
or a subsingleton, for every point y.
By local Hedberg, every isolated point is h-isolated.
(3) In particular, if Fin n is the type with n elements,
Fin (n+1) × Aut (Fin n) ≃ Aut (Fin (n+1))
from which we get
Fin (n!) ≃ Aut (Fin n)
by induction on n, as expected.
(4) Although the above development uses HoTT/UF ideas crucially, only
function extensionality is needed to carry out the constructions and proofs
(in particular, propositional extensionality and univalence are not needed).
I had fun formalizing the above ideas in Agda:
https://www.cs.bham.ac.uk/~mhe/agda-new/UF-Factorial.html
https://www.cs.bham.ac.uk/~mhe/agda-new/ArithmeticViaEquivalence.html
and so I wanted to share it here.
Best,
Martin
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```This is a public inbox, see mirroring instructions | 979 | 3,509 | {"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} | 2.828125 | 3 | CC-MAIN-2023-14 | latest | en | 0.906137 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/peterblockman/cairo-pattern-of-bits-challenge-4kdg?comments_sort=top | 1,696,166,954,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00547.warc.gz | 479,935,989 | 22,329 | ## DEV Community
Peter Blockman
Posted on • Updated on
# Cairo: pattern of bits challenge
## Introduction
In Encode's Cairo Bootcamp, there is an interesting bit manipulation challenge. It took me some time to figure it out. In this article, I will discuss the challenge and the solution.
## The challenge
`ex7.cairo`
``````%lang starknet
from starkware.cairo.common.bitwise import bitwise_and, bitwise_xor
from starkware.cairo.common.cairo_builtins import BitwiseBuiltin
from starkware.cairo.common.cairo_builtins import HashBuiltin
from starkware.cairo.common.math import unsigned_div_rem
// Using binary operations return:
// - 1 when pattern of bits is 01010101 from LSB up to MSB 1, but accounts for trailing zeros
// - 0 otherwise
// 000000101010101 PASS
// 010101010101011 FAIL
func pattern{bitwise_ptr: BitwiseBuiltin*, range_check_ptr}(
n: felt, idx: felt, exp: felt, broken_chain: felt
) -> (true: felt) {
return (0,);
}
``````
`test_ex7.cairo`
``````%lang starknet
from starkware.cairo.common.uint256 import Uint256
from starkware.cairo.common.bitwise import bitwise_and, bitwise_xor
from starkware.cairo.common.cairo_builtins import BitwiseBuiltin, HashBuiltin
from exercises.programs.ex7 import pattern
@external
func test_patternt{syscall_ptr: felt*, range_check_ptr, bitwise_ptr: BitwiseBuiltin*}() {
alloc_locals;
// test nice numbers
//###############################################################################################
let (nice_pattern) = pattern(n=170, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 1;
let (nice_pattern) = pattern(n=10, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 1;
let (nice_pattern) = pattern(n=43690, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 1;
let (nice_pattern) = pattern(n=1398101, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 1;
// test not-nice numbers
//###############################################################################################
let (nice_pattern) = pattern(n=17, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 0;
let (nice_pattern) = pattern(n=11, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 0;
let (nice_pattern) = pattern(n=43390, idx=0, exp=0, broken_chain=0);
assert nice_pattern = 0;
%{
if ids.nice_pattern == 1:
print(f"has nice pattern")
else:
print(f"doesn't have a nice pattern")
%}
return ();
}
``````
## Understanding the problem
If the bit pattern is 01010101, it is valid. Looking at the test cases, we can see that 01010 (decimal 10) is valid too. This means that the problem is about alternative bits between 1 and 0. If the alternative pattern is not broken until the LSB, we return 1, otherwise, return 0. For instance, we will get 1 for 01010 (decimal 10), and 0 for 01011 (decimal 11).
Since the challenge gives us the test cases, we can get some cues about the arguments from them. There are 4 arguments in the pattern function:
• `n: felt` is the decimal number that we want to check its bit pattern. This tells us that we need to convert the decimal number to binary.
• `idx: felt` is the index of a bit in the pattern.
• `exp: felt` is the value of a bit. We need this because we want to compare adjacent bits to see if the chain is valid.
• `broken_chain: felt` The name says it all. It is like a "boolean" variable that determines whether the pattern is broken.
## Solution
``````func check_broken_chain{range_check_ptr}(remainder: felt, exp: felt) -> felt {
if (remainder == exp) {
return 1;
}
return 0;
}
func pattern{bitwise_ptr: BitwiseBuiltin*, range_check_ptr}(
n: felt, idx: felt, exp: felt, broken_chain: felt
) -> (true: felt) {
if (n == 0) {
let valid_chain = 1 - broken_chain;
return (valid_chain,);
}
if (broken_chain == 1) {
let valid_chain = 1 - broken_chain;
return (valid_chain,);
}
if (idx == 0) {
let (quotient, remainder) = unsigned_div_rem(n, 2);
let new_idx = idx + 1;
return pattern(quotient, new_idx, remainder, 0);
}
let (quotient, remainder) = unsigned_div_rem(n, 2);
let _broken_chain = check_broken_chain(remainder, exp);
let new_idx = idx + 1;
return pattern(quotient, new_idx, remainder, _broken_chain);
}
``````
## Explain the solution
The solution can be broken down into 3 simple steps :
1. Convert the decimal `n` to binary
2. Check if the pattern is valid
3. Handle edge case and base case
### Convert the decimal n to binary
One way to convert decimal to binary is to use modulo. Cairo does not have a built-in modulo operator, but we can use unsigned_div_rem from Starkware math library.
``````from starkware.cairo.common.math import unsigned_div_rem
let (quotient, remainder) = unsigned_div_rem(n, 2);
``````
### Check if the pattern is valid
Simply put, It comes down to whether the current bit `remainder` is different than the previous bit `exp`. If they are the same, the pattern is broken.
When I first tried to handle the _broken_chain value, I did this:
``````let (quotient, remainder) = unsigned_div_rem(n, 2);
tempvar _broken_chain = 0;
if (remainder == exp) {
_broken_chain = 1;
}
let new_idx = idx + 1;
return pattern(quotient, new_idx, remainder, _broken_chain);
``````
and got an error
``````An ASSERT_EQ instruction failed: 0 != 1.
_broken_chain = 1;
^***************^
``````
A workaround is to make a standalone function:
``````func check_broken_chain{range_check_ptr}(remainder: felt, exp: felt) -> felt {
if (remainder == exp) {
return 1;
}
return 0;
}
...
let (quotient, remainder) = unsigned_div_rem(n, 2);
let _broken_chain = check_broken_chain(remainder, exp);
let new_idx = idx + 1;
return pattern(quotient, new_idx, remainder, _broken_chain);
``````
Since the challenge has a `bitwise_xor` import in the first place, we can also use it to get the `_broken_chain` use like so:
`````` let (remainder_xor_prev_remainder) = bitwise_xor(remainder, exp);
let _broken_chain = 1 - remainder_xor_prev_remainder;
``````
During the recursion, if broken_chain is 1, we just return the result.
`````` if (broken_chain == 1) {
let valid_chain = 1 - broken_chain;
return (valid_chain,);
}
``````
### Handle the edge case and base case
The first bit does not have a previous bit to compare with, so we can just calculate its remainder and move to the next bit.
`````` if (idx == 0) {
let (quotient, remainder) = unsigned_div_rem(n, 2);
let new_idx = idx + 1;
return pattern(quotient, new_idx, remainder, 0);
}
``````
Because we keep dividing n by 2, it will become 0 eventually. When it does, we return the result.
``````if (n == 0) {
let valid_chain = 1 - broken_chain;
return (valid_chain,);
}
``````
## Wrapping Up
Alright, that is all I am covering for today. If you have any questions, please feel free to reach out to me. I hope this is helpful to somebody learning Cairo. | 1,834 | 6,745 | {"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} | 2.78125 | 3 | CC-MAIN-2023-40 | latest | en | 0.435576 |
https://dragan.rocks/articles/22/Recurrent-networks-hello-world-sequence-prediction-in-Clojure-with-new-Deep-Diamond | 1,675,242,388,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00228.warc.gz | 226,594,553 | 13,244 | # Recurrent Networks Hello World in Clojure with new Deep Diamond RNN support on CPU and GPU
Need help with your custom Clojure software? I'm open to (selected) contract work.
August 26, 2022
I've been busy in the last period working on new major features in Deep Diamond, one of which is the support for Recurrent Neural Networks (RNN). It's not been an easy ride, but I can finally show you some results! Big thanks for everyone that's helping me with this by buying my books (or subscribing to the upcoming editions), and Clojurists Together, who generously funded me in the past year to work on this.
I know that most of you probably don't have much more than passing familiarity with deep learning, let alone recurrent neural networks, and that's why I'll try to show a very simple example on the level of Hello World that anyone interested in machine learning and programming can understand and try.
So, enough talk, let's get down to business.
## What are we doing
We are demonstrating a hammer: recurrent neural networks. Just kidding; we would like to create a (software) device that can learn to predict the next data point in a series. Depending on the data, this can be done in a number of ways (even by convolutional neural networks (CNN) that Deep Diamond already supported), one of which is RNN. So, we are creating a recurrent network, and training it with a set of data for this task.
An example of data that fits this task would be temperature at some place, stock prices, and any other (possibly infinite) sequence of numbers in one of more dimensions that have an ordinal relation, that is, has an abstract notion of time attached to it. Then, we are trying to forecast one or more values that are happening in the future (temperature the next day, or the closing price of a stock, etc.).
Since Hello World has to be dead simple, a real example would have too many opaque numbers, so we'll solve an artificially trivial task of teaching our network to predict the next number in the series for obvious series such as 1, 2, 3, 4, 5. Of course, in real life we rarely need to solve that exact task with such a bazooka as RNN, but if this is your first contact with time series prediction with deep learning, I guess it'd be just what works best.
## Let's get down to code
First, we'll require a bunch of Clojure namespaces that we need.
(require '[uncomplicate.commons.core :refer [with-release let-release release]]
'[uncomplicate.neanderthal
[core :refer [ge dim amax submatrix subvector mrows trans transfer transfer! view-vctr
native view-ge cols mv! rk! raw col row nrm2 scal! ncols dim rows axpby!]]
[native :refer [fge native-float fv iv]]]
'[uncomplicate.diamond
[tensor :refer [*diamond-factory* tensor offset! connector transformer
desc revert shape input output view-tz batcher]]
[dnn :refer [rnn infer sum activation inner-product dense
network init! train cost train train-shuffle ending]]
[native :refer []]])
All the AI in the world would be useless to us if we hadn't measured some data from the real world to feed it. Suppose (for the sake of Hello World), that I "measured" a narrow range of whole number domain by generating it from thin air. Does this data tell us anything about stock market? Of course not, and please does not expect any model that we train on this data to magically be informed on anything that could not be learned from the data it has seen.
(def simple-sequence (range -100 100))
(-100 -99 -98 -97 -96 -95 -94 -93 -92 -91 -90 -89 -88 -87 -86 -85 ...)
Now, I'll create a blueprint for an arbitrary recurrent neural network (RNN). This network has 3 recurrent layers (Gated Recurrent Unit cells), an abbreviation to one timestep, and two dense layers at the end. Please note that I used a completely arbitrary architecture. Neither this layer structure, nor the number of hidden neurons are optimal for this data; we are not even sure it's any good. If anything, it's probably a huge overkill. I chose it simply to show you how it's easy to construct with Deep Diamond(), as it will practically do everything on its own if you specify the bare minimum, that is "what you want". And we hope it'll at least learn to work well at the end, as non-optimal as it is.
(def net-bp (network (desc [5 32 1] :float :tnc)
[(rnn [128] :gru)
(rnn 2)
(abbreviate)
(dense [128] :relu)
(dense [1] :linear)]))
There is no place here to explain what RNN is and how it works internally, other than pointing that recurrent layers have an ability to handle sequential relations of its input by "memorizing" the signals that pass through it. The upcoming version of my book Deep Learning for Programmers (2.0) discusses RNNs in more detail.
## Formatting the input data
The input of this network differs from fully connected or convolutional networks by explicitly modeling the time dimension, "t" in the ":tnc" format. Technically, you can feed it with any 3D tensor that matches its [5 32 1] shape, but for that data to be in context, it has to actually be arranged as 5 timesteps of a minibatch of 32 samples of 1-dimensional data.
We do have 1-dimensional data, but how do we fit our (range -100 100) sequence to its input? We do have more than 5 timesteps (we have 400), and we are far from 32 samples, since we only have one! We could try to just cram the sequence as-is by doing (transfer! simple-sequence (input net)) but this would be the "garbage in" part of "garbage in - garbage out". No. The solution is, as always in machine learning, to actually think what our data represents, and matching it with our knowledge of how the actual model intends to process its input.
What we do need is a bunch of 5-long sequences, such as [1 2 3 4 5] and the output that we would deem correct. In this case, I choose that the goal is to teach the network to output 6 to this input (or a number sufficiently close to it). So, the training data should be input sequences such as [3 4 5 6 7] and [-12 -11 -10 -9 -8], and target outputs such as [8] and [-7]. I hope you see how a bunch of these sequences, almost 400, and their respective target outputs could be extracted from simple-sequence.
The following function employs some stock Neanderthal () matrix functions to process the data and pack it into input and target output tensors [x-train] and [y-train]. I don't have time to explain each step, which is not trivial, but this is fairly standard vector/matrix/tensor stuff, well explained in both books from my Interactive Programming for Artificial Intelligence book series.
(defn split-series [fact s ^long t]
(let [n (- (ncols s) t)
c (mrows s)]
(let-release [x-tz (tensor fact [t n c] :float :tnc)
y-tz (tensor fact [n c] :float :nc)
x-ge (trans (view-ge (view-vctr x-tz) (* n c) t))
s-vctr (view-vctr s)]
(transfer! (submatrix s 0 t c n) (view-ge (view-vctr y-tz) c n))
(dotimes [j t]
(transfer! (subvector s-vctr (* j c) (* c n)) (row x-ge j)))
[x-tz y-tz])))
Here's how the output looks like on an ever simpler example of 2-step sample sequences produced from 5 element long full sequence.
(def dummy (fge 1 5 (range 5)))
#RealGEMatrix[float, mxn:1x5, layout:column, offset:0]
▥ ↓ ↓ ↓ ↓ ↓ ┓
→ 0.00 1.00 2.00 3.00 4.00
┗ ┛
(def dummy-split (split-series *diamond-factory* dummy 2))
[{:shape [2 3 1], :data-type :float, :layout [3 1 1]} (0.0 1.0 2.0 1.0 2.0 3.0)
{:shape [3 1], :data-type :float, :layout [1 1]} (2.0 3.0 4.0)]
This split produces 3 samples for training, each sample has 2 entries, and for each sample there is a desired output. The tensor printout does not show dimensions, which would be super hard to make sense anyway due to large dimensionality and enormous number of entries in any tensor of any use. We can extract a matrix view, in cases when it makes sense.
(view-ge (view-vctr (dummy-split 0)) 3 2)
#RealGEMatrix[float, mxn:3x2, layout:column, offset:0]
▥ ↓ ↓ ┓
→ 0.00 1.00
→ 1.00 2.00
→ 2.00 3.00
┗ ┛
So, inputs are arranged in rows: [0 1], [1 2], and [2 3]. That's because tensor's default layout is :tnc, meaning that innermost grouping is channels ($$C=1$$), (mini)batch size ($$N=3$$) and time (\$T=2).
Ok, so, finally, we transform our own data so that the network can learn from it.
(def full-series (fge 1 200 simple-sequence))
#RealGEMatrix[float, mxn:1x200, layout:column, offset:0]
▥ ↓ ↓ ↓ ↓ ↓ ┓
→ -100.00 -99.00 ⁙ 98.00 99.00
┗ ┛
(def train-data (split-series *diamond-factory* full-series 5))
[{:shape [5 195 1], :data-type :float, :layout [195 1 1]} (-100.0 -99.0 -98.0 -97.0 -96.0 -95.0 -94.0 -93.0 -92.0 -91.0 -90.0 -89.0 -88.0 -87.0 -86.0 -85.0)
{:shape [195 1], :data-type :float, :layout [1 1]} (-95.0 -94.0 -93.0 -92.0 -91.0 -90.0 -89.0 -88.0 -87.0 -86.0 -85.0 -84.0 -83.0 -82.0 -81.0 -80.0)]
The printouts of long tensors show only a subset of the content of the tensor.
## The actual network
The blueprint that we've created at the beginning of the article can be simplified for later reuse. Please note that it's not some super-opaque magical compiler. The network architecture is defined as a simple Clojure vector of straight Clojure functions!
(def architecture [(rnn [128] :gru)
(rnn 2)
(abbreviate)
(dense [128] :relu)
(dense [1] :linear)])
[#function[uncomplicate.diamond.dnn/rnn/fn--43604]
#function[uncomplicate.diamond.dnn/rnn/fn--43604]
#function[uncomplicate.diamond.dnn/abbreviate/fn--43609]
#function[uncomplicate.diamond.dnn/fully-connected/fn--43493]
#function[uncomplicate.diamond.dnn/fully-connected/fn--43493]]
This architecture is independent from the specific input dimensions. We create a blueprint by specifying the dimensions and the architecture.
(def net-bp (network (desc [5 32 1] :float :tnc)
architecture))
This blueprint is independent of the learning algorithm. It is a function that creates the actual network training program. In this case, I will use gradient descent with adaptive moments (:adam). Before we start learning, we initialize the network with random weights, automatically chosen by Deep Diamond() according to good practices, but you are free to use another initialization function that is more to your liking. Everything in Deep Diamond is modular and implemented in Clojure fashion of "assemble your own if you wish". If you are content with my choices, then everything is automatic!
(def net (init! (net-bp :adam)))
Here's the network printout, in case you
=======================================================================
#SequentialNetwork[train, input:[5 32 1], layers:5, workspace:1]
-----------------------------------------------------------------------
#Adam[topology::gru, shape:[5 32 128], activation: :identity]
parameters: [{:shape [1 1 129 3 128], :data-type :float, :layout [49536 49536 384 128 1]} (2.1676771211787127E-5 -5.19975537827122E-6 5.7575953178456984E-6 2.380384103162214E-5 -2.894530371122528E-5 -2.4803119913485716E-7 -1.6729745766497217E-5 -2.902976348195807E-6 4.530028309090994E-5 5.464621608552989E-6 -5.7127394939016085E-6 1.593221713847015E-5 -1.1793279554694891E-5 -7.507987476174094E-8 -1.1438844921940472E-5 5.965541276964359E-6) {:shape [1 1 3 128], :data-type :float, :layout [384 384 128 1]} (0.0035437364131212234 -0.001584756188094616 0.0013698132243007421 -1.2431709910742939E-4 1.7048766312655061E-4 0.002398068318143487 0.00372034078463912 -0.0021170536056160927 8.799560600891709E-4 -0.0023348634131252766 8.223060867749155E-4 -1.1841517698485404E-4 -0.004045043606311083 9.459585417062044E-4 -0.0037800846621394157 -0.002804018324241042)]
-----------------------------------------------------------------------
#Adam[topology::gru, shape:[5 32 128], activation: :identity]
parameters: [{:shape [2 1 256 3 128], :data-type :float, :layout [98304 98304 384 128 1]} (2.276021717761978E-7 3.804875632340554E-6 1.058972247847123E-5 2.8063212198503606E-7 1.8309128790860996E-5 -9.416969078301918E-6 -7.987003414200444E-8 1.0779360309243202E-5 2.222931470896583E-6 -1.2704362234217115E-5 1.5140467439778149E-5 -2.0407780539244413E-5 -1.5779027080498054E-6 -1.0661310625437181E-5 -3.834541985270334E-6 -1.0737002412497532E-5) {:shape [2 1 3 128], :data-type :float, :layout [384 384 128 1]} (8.910018368624151E-4 0.003690000157803297 -4.8378523206338286E-4 -0.0034620240330696106 0.0031146425753831863 6.610305863432586E-4 0.00204548635520041 0.001728582545183599 -0.0027434946969151497 0.007643579971045256 0.00425624568015337 -0.00295245717279613 1.8937387721962295E-5 -0.0027048818301409483 -0.0012806318700313568 -0.0028582836966961622)]
-----------------------------------------------------------------------
{:shape [32 128], :topology :abbreviate}
parameters: [{:shape [128 128], :data-type :float, :layout [8192 1024]} (-0.004568892996758223 -0.004355841316282749 0.005139997228980064 -0.005750759970396757 -0.004274052567780018 0.005599929019808769 -0.003351157531142235 -0.008299742825329304 -0.0031023912597447634 0.0013810413656756282 0.002118719043210149 0.0023180078715085983 -0.005323362536728382 -0.013326002284884453 7.393552223220468E-4 -0.013735005632042885) {:shape [128], :data-type :float, :layout [1]} (0.5049463510513306 -0.47153180837631226 -1.3509427309036255 -1.3017100095748901 -0.39814749360084534 0.8303372263908386 0.6530964970588684 -0.22249580919742584 -1.326366901397705 0.16360507905483246 -0.022157425060868263 2.0535836219787598 1.8076190948486328 -0.0799550786614418 -1.6791125535964966 -0.7451670169830322)]
-----------------------------------------------------------------------
parameters: [{:shape [1 128], :data-type :float, :layout [1 1]} (0.003621552372351289 -0.004416522569954395 2.548426273278892E-4 0.006995228119194508 0.0013199367094784975 -0.0018220240017399192 0.009454095736145973 0.003091101534664631 -0.01203352864831686 -0.014204473234713078 -0.007159397471696138 0.0039085038006305695 0.0029486482962965965 -0.009481357410550117 0.009158425033092499 -0.004999339580535889) {:shape [1], :data-type :float, :layout [1]} (-0.09112684428691864)]
=======================================================================
## Training, finally!
Hey, I promised you a Hello World, and I've been beating the bush for half an hour formatting the data. And we didn't even touched the biggest obstacle: actually training the network. Is it hard? Yes, but not for the user. Now it's time for Deep Diamond() to beat the hell out of your CPU or GPU. But it will at leas do this on its own :)
Since this is a Hello World, we'll start with 50 epochs and see how it's going.
(time (train-shuffle net (train-data 0) (train-data 1) :quadratic 50 [0.005]))
"Elapsed time: 769.986155 msecs"
3238.298712769756
Hmmmm. The cost of 3000 and change does not look very good. Would more training help?
(time (train-shuffle net (train-data 0) (train-data 1) :quadratic 50 [0.005]))
"Elapsed time: 744.456035 msecs"
708.0862449675478
Still bad, but it's improving! For the sake of experinmenting, I've run this ten(ish) times, and the cost has been steadily decreasing, to the point that it looks good now.
(time (train-shuffle net (train-data 0) (train-data 1) :quadratic 50 [0.005]))
"Elapsed time: 720.427332 msecs"
0.13662090173881225
We don't have to stick to 50-epoch runs. Let's do 500 at once.
(time (train-shuffle net (train-data 0) (train-data 1) :quadratic 500 [0.005]))
"Elapsed time: 6325.427605 msecs"
0.008672867995529465
This looks nice. A thousand epochs might seem a lot, but considering the large network size, recurrent layers, and the scarceness of the training data, it might actually be appropriate. On the other hand, Deep Diamond did it blazingly fast, in a mere dozen seconds! In the world of machine learning, this is nothing.
(def question (tensor [5 1 1] :float :tnc))
(transfer! [1 2 3 4 5] question)
{:shape [5 1 1], :data-type :float, :layout [1 1 1]} (1.0 2.0 3.0 4.0 5.0)
(infer net question)
Dragan says: Requested subtensor is outside of bounds.
{:src-index -31, :src-cnt 1, :dst-index 0, :dst-cnt 32, :mb-size 1}
## Using the network for inference
Now that we have our super useful network, we can stop and think: but how do we use it? First, we need data that could be interpreted as a "question". Our network needs a sequence of five numbers, and when fed, will answer with one number. Obviously, these should be provided as tensors of appropriate dimensions.
(def question (tensor [5 1 1] :float :tnc))
(transfer! [1 2 3 4 5] question)
However, invoking inference would cause the complaint from the network.
(infer net question)
Execution error (ExceptionInfo) at uncomplicate.commons.utils/dragan-says-ex (utils.clj:105).
Dragan says: Requested subtensor is outside of bounds.
Our network's input requires a minibatch of 32 samples. The infer function can handle more, and do the inference in mini batches of 32, but it can't handle fewer samples.
One of the solutions is to create another blueprint with the same structure, and transfer learned weights to the new network.
(def net1-bp (network (desc [5 1 1] :float :tnc)
architecture))
While we're at it, we don't need to create a complex network capable of learning (:adam or :sgd). We can create a much cheaper inference network that takes fewer resources.
(def net1-infer (net1-bp))
(transfer! net net1-infer)
Now, finally, give us the answer, network!
(infer net1-infer question)
{:shape [1 1], :data-type :float, :layout [1 1]} (6.016137599945068)
So, being asked what is the next element in the sequence of [1.0 2.0 3.0 4.0 5.0] (remember, we specified data type :float), our network answers [6.0161] which is close enough to the actual target value that we can mark this as correct. But, it's not a great achievement, since our network already saw this sequence in training. A hash map would have done much better job at guessing this. Let's try a previously unseen sequence.
(transfer! [10 12 14 16 18] question)
(infer net1-infer question)
{:shape [1 1], :data-type :float, :layout [1 1]} (17.750324249267578)
Not that off, but one would expect 19.0. What went wrong? We trained our network with ducks $$(x+1)$$ and asked it about geese $$(x+2)$$. What about griffons?
(transfer! [10 1 100 16 34] question)
(infer net1-infer question)
{:shape [1 1], :data-type :float, :layout [1 1]} (-6.679039001464844)
Now the answer does not make any sense, but would you be able to come with a better answer?
All, right, let's try with a sequence that is similar to the one we used in training, but is out of the range of data that the network has seen.
(transfer! [1000 1001 1002 1003 1004] question)
(infer net1-infer question)
{:shape [1 1], :data-type :float, :layout [1 1]} (96.41997528076172)
Nope, not much success, but we should not have expected any. The network can not answer question outside its domain of expertise.
Let's try with previously unseen data, but inside the domain that we used for training (floats from -100.0 to 1000).
(transfer! [37.4 38.4 39.4 40.4 41.4] question)
(infer net1-infer question)
{:shape [1 1], :data-type :float, :layout [1 1]} (42.39200210571289)
This is actually pretty close!
Sure!
(def nvidia (cudnn-factory))
(def gpu-net-bp (network nvidia
(desc [5 32 1] :float :tnc)
architecture))
(time (train-shuffle gpu-net (gpu-train-data 0) (gpu-train-data 1) :quadratic 1000 [0.005]))
"Elapsed time: 5734.688665 msecs" | 5,815 | 19,549 | {"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": 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} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.967366 |
https://community.clickteam.com/printthread.php?s=72028f9989f51d345d95b867d4fdd282&t=108626&pp=10&page=1 | 1,627,136,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00142.warc.gz | 197,441,652 | 3,060 | # Counting objects for a.i.
Printable View
• 15th May 2021, 03:17 PM
Bonehead
Counting objects for a.i.
Hi.
I have a really, really simple top-down view strategy game under work, and I have run into a need for calculating the amount of minions a player has, in order to make the enemy "a.i." launch proper actions. However, since I don't have a math degree and I don't want to mess too much with advanced objects to keep the game under control = within my limited coding skills, I need help.
At the moment, my game has a really basic system found from this forum to calculate the amount of workers, soldiers and so on by:
(I change the names to be more generic for clarity)
- "unit01" value X = "worker"
- "unit01" value "team01" = fixed("controlobject01")
-- set "controlobject" value "countworker01" to 0
-- start loop for every "unit01", loop name "worker01"
- for each "unit01", loop "worker01"
-- add 1 to "controlobject" value "countworker01"
I have tested it with counter and it does count the amount right for each team, but I was hoping to use the calculations of workers, soldiers and such also to create the base building process. I want the game team to realize, when it has 100 or more workers, save up money for some soldiers, than reach around 10 soldiers, shift back to workers until there are at least 200, then soldiers and so on... But my questions are:
-Since this method of calculation will need the counter to always hit 0 in order to begin a new counting loop, is this method impossible for setting up limits for an action x or y based on how many minions there are at the moment?
-Will the constant resetting of the count cause the player to presume constantly, that it has 0 workers and will try to endlessly reach the 100 workers limit?
-Could there be a way for the counting process to send the player some value of the current amount of workers like every one second to let it know some quota has been reached/lost in order to act?
-Is there a better method to count workers or other unit types for every team I should consider?
Any help is useful. I haven't sent any examples for this forum before, so if you ask for an example file, please be patient. I do have a ver.01 of the games stored, that have one base winning the other by outproducing and attacking the other, so this matter isn't a brick wall, but I don't want to do more work on the game and then realize, that this counting method won't work down the line.
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Math241FinalAnswSp01
# Math241FinalAnswSp01 - → n 2 = | → n 1 || → n 2 | cos...
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Answers to Math 241 Final, Spring 2001 Part I: (1) (a) 1 , - 4 , 5 (b) - 1 (c) 2 , 3 , 2 (2) (a) - 57 / 5 (b) 3 , 12 or 1 , 4 or 1 / 17 , 4 / 17 (3) (0 , 1 , 0) (0 , - 1 , 0) (4) 0 (5) Maximum 1 / 4 Minimum - 1 / 4 (6) 36 π 3 / 5 (7) (a) saddle (b) not a critical point (c) local mininimum Part II: (1) - 1 / 2 (2) (a) (2 - t - s ) 2 + ( - 4 - t - s ) 2 + (3 - 3 s ) 2 (b) t = - 2 and s = 1 (c) (3 , 1 , 2) and (0 , 4 , 2) (3) (a) 61 π/ 3 (b) 32 / 5 (4) (a) Let n 1 = 1 , - 2 , - 1 and n 2 = a, b, c . Then n 1 is perpendicular to P and n 2 is perpendicular to ax + by + cz = d . The angle θ between n 1 and n 2 is either 60 or 120 . Deduce the given equation from n 1 ·
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Unformatted text preview: → n 2 = | → n 1 || → n 2 | cos θ . (b) Use that a 2 + b 2 + c 2 = 6 in the equation in part (a). (c) Suppose ax + by + cz = d is P or P 00 so that Q and R are on ax + by + cz = d . Then → QR = h 1 , 1 ,-1 i is perpendicular to → n 2 in part (a). From → n 2 · → QR = 0, deduce a + b-c = 0. (d) (-1 ,-1 ,-2) and (2 ,-1 , 1) (e) x + y + 2 z = 9 and 2 x-y + z = 12...
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Goal and Solution Clarity
I like simple and intuitive models and I spent a lot of time defining and redefining project categories before I found a classification scheme that met my criteria and my specific needs. The one I am using here I have used for almost 20 years. It works and has never failed to live up to my expectations.
Every project has a goal and hopefully there is a plan that defines how you will reach that goal (a.k.a. a solution). So I define the project landscape around the two fundamental project variables: goal and solution. In my classification scheme these two variables can have only one of two values: clear or not clear. I experimented with more than two values for each variable but that only added complication without adding value. Those two values for each variable generate the four-quadrant matrix shown in Figure 1.2.
I don't know where the dividing line is between clear and not clear but that is not important to this landscape anyway. These values are conceptual not quantifiable. A given project can exhibit various degrees of complexity and uncertainty that position it in one of the four quadrants at a particular point in time. As a project is undertaken the complexity and uncertainty originally associated with it will often change. That could justify changing its quadrant and also changing how the project ...
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Home Brew Forums > How is starter size calculated?
12-19-2010, 06:45 PM #1
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How is starter size calculated?
I've seen the calculators and such, but I'm trying to figure out how they come up with the volume they do. I understand the first part of wanting to pitch proportionally to the density of sugars and the volume the yeast will be spread through and that the 0.75 constant for ales and 1.5 constant for lagers represents the ideal workload for each cell.
I don't understand how to go from the number of cells you want and the number you have in your packet, to the volume of starter needed. I'm guessing that since your starter gravity is optimized at 1.040 or so, you choose the volume that gives the yeast enough sugar to divide the appropriate number of times and then restock its reserves for dormancy. But how much sugar per cell per division does that take?
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12-19-2010, 08:36 PM #2
Bensiff
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I can't really say. I know JZ did a lot of work on his calculator and worked with Chris White in doing so. You could probably figure from his numbers something reasonable. He also has some calculations on his site. In general you want to get within 20% of optimal yeast count so you have a large margin of error that will still yield excellent results. If you want more accurate data you might post this on the science thread to get the attention of the more technically minded folks.
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12-20-2010, 12:35 AM #3
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I'll give that a try. Thanks for the suggestion.
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12-20-2010, 02:26 AM #4
optimatored
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Quote:
Originally Posted by deadcactus I've seen the calculators and such, but I'm trying to figure out how they come up with the volume they do. I understand the first part of wanting to pitch proportionally to the density of sugars and the volume the yeast will be spread through and that the 0.75 constant for ales and 1.5 constant for lagers represents the ideal workload for each cell. I don't understand how to go from the number of cells you want and the number you have in your packet, to the volume of starter needed. I'm guessing that since your starter gravity is optimized at 1.040 or so, you choose the volume that gives the yeast enough sugar to divide the appropriate number of times and then restock its reserves for dormancy. But how much sugar per cell per division does that take?
look no further than mr. malty for your calc needs: http://www.mrmalty.com/calc/calc.html
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12-20-2010, 02:50 AM #5
sudbuster
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Quote:
Originally Posted by deadcactus I've seen the calculators and such, but I'm trying to figure out how they come up with the volume they do. I understand the first part of wanting to pitch proportionally to the density of sugars and the volume the yeast will be spread through and that the 0.75 constant for ales and 1.5 constant for lagers represents the ideal workload for each cell. I don't understand how to go from the number of cells you want and the number you have in your packet, to the volume of starter needed. I'm guessing that since your starter gravity is optimized at 1.040 or so, you choose the volume that gives the yeast enough sugar to divide the appropriate number of times and then restock its reserves for dormancy. But how much sugar per cell per division does that take?
Well I don't know what JZ or Pallmer uses to compute pitching rates, or what their cult followers may beleve. I've never read or lstened to their stuff. But I do know what American craft brewers consider as an appropriate pitching rate: 1 million cells per ml per degree plato. European brewers think that is too high, but it seems to work for Americans. They seem to sell a bunch of good beer. Here's an example: Suppose your wort was say 1.060, or approx. 15 plato. Then you need 1 million x 15 = 15 million cells per ml. Since 5 gallon is about 19,000 ml, you now need (15 x 10^6 ) x (19 x 10^3) = 285 x 10^9 cells. Since one ml of good slurry has about 1x 10^9 cells (this is an approximation given by White), then you need about 285 ml of good slurry. That is about 1/3 quart of thick yeast. I only get about half that from a 4 liter starter, but it seems to do the job. Maybe the Europens are right, the Americans are overpitching. Anyway, that's how you compute pitching rates. Easy-Peasy...
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12-20-2010, 07:14 AM #6
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Quote:
Originally Posted by sudbuster Well I don't know what JZ or Pallmer uses to compute pitching rates, or what their cult followers may beleve. I've never read or lstened to their stuff.
Instead of throwing insults across peoples' bow based off admitted ignorance maybe take a minute to see what folks are doing. JZ's calculator goes off .75 million cells per ml per degree plato. JZ did a good job with that tool, had expert advice, worked hard and tested it extensively...that is why many people use it.
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12-21-2010, 03:27 AM #7
sudbuster
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Quote:
Originally Posted by Bensiff Instead of throwing insults across peoples' bow based off admitted ignorance maybe take a minute to see what folks are doing. JZ's calculator goes off .75 million cells per ml per degree plato. JZ did a good job with that tool, had expert advice, worked hard and tested it extensively...that is why many people use it.
Sorry you perceived my response as an insult to anyone. I simply said ( in so many words) it is so easy to figure pitch rates "a caveman could do it". And personally, I'd just as soon remain ignorant of what JZ and JP are doing because I really don't care one way or another....
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12-21-2010, 03:43 AM #8
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That's just the Sudsy charm we've all come to know and love...
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Caspean Ales and Cider
"I'm sittin' here, completely surrounded by No Beer!" - Onslow
12-21-2010, 06:47 AM #9
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Quote:
Originally Posted by sudbuster Sorry you perceived my response as an insult to anyone. I simply said ( in so many words) it is so easy to figure pitch rates "a caveman could do it". And personally, I'd just as soon remain ignorant of what JZ and JP are doing because I really don't care one way or another....
Thanks. No reason you have to listen to them, but no reason to disrespect the amount of work they have done to help others out. I appreciate anyone taking the time to put good information forward. I have gained from your and many others' posts here on TBN through the years along with JZ, countless scientific journals, books, etc...if its good info, I'm all ears personally.
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12-26-2010, 08:05 PM #10
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I think I will throw some poop in the game. Do you propagate your yeast aerobically or anaerobically, and how do you do it. What is the Crabtree Effect? ---bobz
__________________ | 2,084 | 8,167 | {"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-2014-49 | latest | en | 0.944466 |
https://www.theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/Eckart_conditions.html | 1,686,372,647,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00082.warc.gz | 1,148,566,479 | 15,871 | # Eckart conditions
In chemistry and physics, the Eckart conditions,[1] named after Carl Eckart, sometimes referred to as Sayvetz conditions,[2] simplify the nuclear motion (rovibrational) Schrödinger equation of a molecule. The rovibrational Schrödinger equation arises in the second step of the Born-Oppenheimer (BO) approximation.
The first step of the BO approximation yields a nuclear potential energy. This potential is generated by solution of the electronic Schrödinger equation of the molecule for different nuclear constellations. The potential energy, which is a function of the nuclear coordinates, enters the time-independent Schrödinger equation for the "motion" of the nuclei—the second step of the BO approximation. The eigenvalues obtained from the solution of the nuclear Schrödinger equation are the energies observed in the infrared and microwave spectra of the molecule. Microwave and infrared spectra are associated with rotations and vibrations of the nuclei in the molecule.
In undergraduate texts it is usually assumed—but not proved—that rotations and vibrations can be separated, i.e., that nuclear wave functions can be factorized as,
$\Phi_\mathrm{nuc} = \Phi_\mathrm{vib}\Phi_\mathrm{rot} \,. \qquad\qquad\qquad(1)$
In these texts the coordinates are never made explicit. It is left to the reader to decide what the coordinates are that the wave functions depend on: space-fixed or body-fixed? If the latter, in what way are they fixed to the body, i.e., to the molecule? Eckart was the first to derive the coordinates that give the separation of the external (rotational and translational) motions from the internal (vibrational) motions. He found that the factorization is never exact, as there are always non-vanishing vibration-rotation coupling terms, the so-called Coriolis terms. Although the rotational and vibrational motions of the nuclei in a molecule cannot be fully separated, the Eckart conditions minimize the coupling between the two kinds of motion and accordingly make neglect of coupling a lesser approximation.
## Rotational and translational Eckart conditions
The Eckart conditions can only be formulated for semi-rigid molecules, which are molecules with a potential energy surface V(R1, R2,..RN) that has a well-defined minimum for a set of nuclear positions RA0 ($\scriptstyle A=1,\ldots, N$). Tunneling of the nuclei through potential barriers is excluded.
The equilibrium coordinates of the nuclei—with respective masses MA—are expressed with respect to an orthonormal principal axes frame with origin in the center of mass in the molecule. By definition, principal axes are eigenvectors of the inertia tensor, which is a symmetric 3×3 tensor. Accordingly, the equilibrium coordinates satisfy the relations
$\sum_{A=1}^N\, M_A\,\big[\delta_{ij}\,(R_A^0)^2 - R^0_{Ai} R^0_{Aj}\big] = \lambda^0_i \delta_{ij} \qquad\mathrm{and}\qquad \sum_{A=1}^N M_A \mathbf{R}_A^0 = \mathbf{0}.$
Here λi0 is a principal inertia moment of the equilibrium molecule. The N triplets of real numbers RA0 = (RA10, RA20, RA30) satisfying these conditions, enter the theory as a known set of constant parameters.
Following Biedenharn and Louck[3] we introduce an orthonormal body-fixed frame, the Eckart frame,
$\vec{\mathbf{F}} \equiv \{ \vec{f}_1, \vec{f}_2, \vec{f}_3\} .$
If an observer were tied to the Eckart frame, which—following the molecule—rotates and translates in space, (s)he would constantly see the molecule in its equilibrium geometry, provided the nuclei are at the points,
$\vec{R}_A^0 \equiv \vec{\mathbf{F}}\, \cdot\, \mathbf{R}_A^0 =\sum_{i=1}^3 \vec{f}_i\, R^0_{Ai},\quad A=1,\ldots,N .$.
The motion of the Eckart frame describes the external (translational and rotational) motion of the equilibrium molecule. Possible internal motion of the nuclei away from their equilibrium points is the molecular vibration. The Eckart frame is a non-inertial frame,[4] hence one can expect pseudo (non-inertial) forces to act on the nuclei. Indeed, centrifugal and Coriolis forces will appear.
Let the elements of RA be the coordinates with respect to the Eckart frame of the position vector of nucleus A ($\scriptstyle A=1,\ldots, N$). Since we take the origin of the Eckart frame in the instantaneous center of mass, the following relation holds,
$\sum_A M_A \mathbf{R}_A = \mathbf{0}.$
We define displacement coordinates
$\mathbf{d}_A\equiv\mathbf{R}_A-\mathbf{R}^0_A$.
Clearly the displacement coordinates satisfy the translational Eckart conditions,
$\sum_{A=1}^N M_A \mathbf{d}_A = 0 .$
The rotational Eckart conditions for the displacements are:
$\sum_{A=1}^N M_A \mathbf{R}^0_A \times \mathbf{d}_A = 0,$
where × indicates a cross product. These rotational conditions follow from the specific construction of the Eckart frame vectors $\scriptstyle \vec{f}_i$, see Biedenharn and Louck, loc. cit., page 538, for the details of their construction and how this construction leads to the rotational Eckart conditions.
Finally, notice that an Eckart frame becomes a principal axes frame in the case that the molecule is a rigid rotor, that is, when all N displacement vectors are zero, which occurs when the molecule is not vibrating.
## Separation of external and internal coordinates
The N position vectors $\scriptstyle \vec{R}_A$ of the nuclei in a molecule constitute a 3N dimensional linear space R3N: the nuclear configuration space. From the Eckart conditions follows that the nuclear configuration space is an orthogonal direct sum. This direct sum decomposition of the nuclear configuration space gives a separation of the nuclear coordinates in vibration coordinates on the one hand and rotation/translation coordinates on the other.
$\mathbf{R}^{3N} = \mathbf{R}_\textrm{ext}\oplus\mathbf{R}_\textrm{int}.$
The elements of the 3N−6 dimensional subspace Rint are referred to as internal coordinates, because they are invariant under overall translation and rotation of the molecule and, thus, depend only on the internal (vibrational) motions. The elements of the 6-dimensional subspace Rext are referred to as external coordinates, because they are associated with the overall translation and rotation of the molecule.
To show the direct sum decomposition, we define first a basis for Rext and introduce the following 6 vectors:
\begin{align} \vec{s}^A_{i} \equiv \vec{f}_i \quad\hbox{and}\quad \vec{s}^A_{i+3} \equiv \vec{f}_i \times\vec{R}_A^0,\qquad i=1,2,3,\quad A=1,\ldots,N\\ \end{align}
An orthogonal, unnormalized, basis for Rext is then,
$\vec{S}_t \equiv \operatorname{row}(\sqrt{M_1}\;\vec{s}^{\,1}_{t}, \ldots, \sqrt{M_N} \;\vec{s}^{\,N}_{t}) \quad\mathrm{for}\quad t=1,\ldots, 6.$
A mass-weighted displacement vector is
$\vec{D} \equiv \operatorname{col}(\sqrt{M_1}\;\vec{d}^{\,1}, \ldots, \sqrt{M_N}\;\vec{d}^{\,N}) \quad\mathrm{with}\quad \vec{d}^{\,A} \equiv \vec{\mathbf{F}}\cdot \mathbf{d}_A .$
For i=1,2,3,
$\vec{S}_i \cdot \vec{D} = \sum_{A=1}^N \; M_A \vec{s}^{\,A}_i \cdot \vec{d}^{\,A} =\sum_{A=1}^N M_A d_{Ai} = 0,$
where the zero follows because of the translational Eckart conditions. For i=4,5,6
$\, \vec{S}_i \cdot \vec{D} = \sum_{A=1}^N \; M_A \big(\vec{f}_i \times\vec{R}_A^0\big) \cdot \vec{d}^{\,A}=\vec{f}_i \cdot \sum_{A=1}^N M_A \vec{R}_A^0 \times\vec{d}^A = \sum_{A=1}^N M_A \big( \mathbf{R}_A^0 \times \mathbf{d}_A\big)_i = 0,$
where the zero follows because of the rotational Eckart conditions. We conclude that the displacement vector $\scriptstyle \vec{D}$ belongs to the orthogonal complement of Rext, and hence that a mass-weighted displacement vector is an internal vector.
A basis for the internal space is obtained by introduction of 3N−6 linearly independent vectors
$\vec{Q}_r \equiv \operatorname{row}(\frac{1}{\sqrt{M_1}}\;\vec{q}_r^{\,1}, \ldots, \frac{1}{\sqrt{M_N}}\;\vec{q}_r^{\,N}), \quad\mathrm{for}\quad r=1,\ldots, 3N-6.$
The vectors $\scriptstyle \vec{q}^A_r$ could be Wilson's s-vectors or could be obtained in the harmonic approximation by diagonalizing the Hessian of V. We next introduce internal (vibrational) modes,
$q_r \equiv \vec{Q}_r \cdot \vec{D} = \sum_{A=1}^N \vec{q}^A_r \cdot \vec{d}^{\,A} \quad\mathrm{for}\quad r=1,\ldots, 3N-6.$
The physical meaning of qr depends on the vectors $\scriptstyle \vec{q}^A_r$. For instance, qr could be a symmetric stretching mode, in which two C—H bonds are simultaneously stretched and contracted.
We already saw that the corresponding external modes are zero because of the Eckart conditions,
$s_t \equiv \vec{S}_t \cdot \vec{D} = \sum_{A=1}^N M_A \;\vec{s}^{\,A}_t \cdot \vec{d}^{\,A} = 0 \quad\mathrm{for}\quad t=1,\ldots, 6.$
## Overall translation and rotation
As we stated above, the internal (vibrational) coordinates are invariant under translation and infinitesimal rotation of the equilibrium molecule, hence their name. We will now show that these invariances hold if and only if the Eckart conditions apply.
An overall translation of the reference ( = equilibrium) molecule is given by
$\vec{R}_{A}^0 \mapsto \vec{R}_{A}^0 + \vec{t}$'
for any arbitrary 3-vector $\scriptstyle \vec{t}$. An infinitesimal rotation of the molecule is given by
$\vec{R}_A^0 \mapsto \vec{R}_A^0 + \Delta\phi \; ( \vec{n}\times \vec{R}_A^0)$
where Δφ is an infinitesimal angle, Δφ >> (Δφ)², and $\scriptstyle \vec{n}$ is an arbitrary unit vector. From the orthogonality of $\scriptstyle \vec{Q}_r$ to the external space follows that the $\scriptstyle \vec{q}^A_r$ satisfy
$\sum_{A=1}^N \vec{q}^{\,A}_r = \vec{0} \quad\mathrm{and}\quad \sum_{A=1}^N \vec{R}^0_A\times \vec{q}^A_r = \vec{0}.$
Now, under translation
$q_r \mapsto \sum_A\vec{q}^{\,A}_r \cdot(\vec{d}^A - \vec{t}) = q_r - \vec{t}\cdot\sum_A \vec{q}^{\,A}_r = q_r.$
Clearly, $\scriptstyle \vec{q}^A_r$ is invariant under translation if and only if
$\sum_A \vec{q}^{\,A}_r = \vec{0},$
because the vector $\scriptstyle \vec{t}$ is arbitrary. So, the translational Eckart conditions imply the translational invariance of the vectors belonging to internal space and conversely. Under rotation we have,
$q_r \mapsto \sum_A\vec{q}^{\,A}_r \cdot \big(\vec{d}^A - \Delta\phi \; ( \vec{n}\times \vec{R}_A^0) \big) = q_r - \Delta\phi \; \vec{n}\cdot\sum_A \vec{R}^0_A\times\vec{q}^{\,A}_r = q_r.$
Rotational invariance follows if and only if
$\sum_A \vec{R}^0_A\times\vec{q}^{\,A}_r =\vec{0}.$
The external modes, on the other hand, are not invariant and it is not difficult to show that they change under translation as follows:
\begin{align} s_i &\mapsto s_i + M \vec{f}_i \cdot \vec{t} \quad \mathrm{for}\quad i=1,2,3 \\ s_i &\mapsto s_i \quad \mathrm{for}\quad i=4,5,6, \\ \end{align}
where M is the total mass of the molecule. They change under infinitesimal rotation as follows
\begin{align} s_i &\mapsto s_i \quad \mathrm{for}\quad i=1,2,3 \\ s_i &\mapsto s_i + \Delta \phi \vec{f}_i \cdot \mathbf{I}^0\cdot \vec{n} \quad \mathrm{for}\quad i=4,5,6, \\ \end{align}
where I0 is the inertia tensor of the equilibrium molecule. This behavior shows that the first three external modes describe the overall translation of the molecule, while the modes 4, 5, and, 6 describe the overall rotation.
## Vibrational energy
The vibrational energy of the molecule can be written in terms of coordinates with respect to the Eckart frame as
$2T_\mathrm{vib} = \sum_{A=1}^N M_A \dot{\mathbf{R}}_A\cdot \dot{\mathbf{R}}_A = \sum_{A=1}^N M_A \dot{\mathbf{d}}_A\cdot \dot{\mathbf{d}}_A.$
Because the Eckart frame is non-inertial (moves non-uniformly in space), the total kinetic energy comprises also centrifugal and Coriolis energies. These kinetic energy terms couple the vibrational with the rotational motions. See Watson's Hamiltonian for the explicit form of the coupling terms.
In this section we will ignore the coupling. Quantum mechanically the neglect of coupling implies that the rotation/translation Schrödinger equation (depending on the external coordinates) can be solved separately from the vibrational Schrödinger equation, which depends solely on the internal coordinates. The nuclear wave function then gets the product form of equation (1). This is an important computational and conceptual simplification, that however, amounts to neglect of some observable effects on the vibration spectrum.
The vibrational energy is written in terms of the displacement coordinates, which are linearly dependent because they are "contaminated" by the 6 external coordinates, which are zero, i.e., the dA's satisfy 6 linear relations. An essential result is the fact that it is possible to write the vibrational energy solely in terms of the internal coordinates qr (r =1, ..., 3N−6). In order to exhibit this result, we write the different coordinates in terms of the displacements
\begin{align} q_r = \sum_{Aj} d_{Aj}& \big( q^A_{rj} \big) \\ s_i = \sum_{Aj} d_{Aj}& \big( M_A \delta_{ij} \big) =0 \\ s_{i+3} = \sum_{Aj} d_{Aj}& \big( M_A \sum_k \epsilon_{ikj} R^0_{Ak} \big)=0 \\ \end{align}
The parenthesized expressions define a matrix B relating the internal and external modes to the displacements. The matrix B may be partitioned in an internal (3N−6 x 3N) and an external (6 x 3N) part,
$\mathbf{v}\equiv \begin{pmatrix} q_1 \\ \vdots \\ \vdots \\ q_{3N-6} \\ 0 \\ \vdots \\ 0\\ \end{pmatrix} = \begin{pmatrix} \mathbf{B}^\mathrm{int} \\ \cdots \\ \mathbf{B}^\mathrm{ext} \\ \end{pmatrix} \mathbf{d} \equiv \mathbf{B} \mathbf{d}.$
We define the matrix M by
$\mathbf{M} \equiv \operatorname{diag}(\mathbf{M}_1, \mathbf{M}_2, \ldots,\mathbf{M}_N) \quad\textrm{and}\quad \mathbf{M}_A\equiv \operatorname{diag}(M_A, M_A, M_A)$
and from the relations given in the previous sections follow the matrix relations
$\mathbf{B}^\mathrm{ext} \mathbf{M}^{-1} (\mathbf{B}^\mathrm{ext})^\mathrm{T} = \operatorname{diag}(N_1,\ldots, N_6) \equiv\mathbf{N},$
and
$\mathbf{B}^\mathrm{int} \mathbf{M}^{-1} (\mathbf{B}^\mathrm{ext})^\mathrm{T} = \mathbf{0}.$
We define
$\mathbf{G} \equiv \mathbf{B}^\mathrm{int} \mathbf{M}^{-1} (\mathbf{B}^\mathrm{int})^\mathrm{T}.$
By using the rules for block matrix multiplication we can show that
$(\mathbf{B}^\mathrm{T})^{-1} \mathbf{M} \mathbf{B}^{-1} = \begin{pmatrix} \mathbf{G}^{-1} && \mathbf{0} \\ \mathbf{0} && \mathbf{N}^{-1} \end{pmatrix},$
where G−1 is of dimension (3N−6 x 3N−6) and N−1 is (6 x 6). The kinetic energy becomes
$2T_\mathrm{vib} = \dot{\mathbf{d}}^\mathrm{T} \mathbf{M} \dot{\mathbf{d}} = \dot{\mathbf{v}}^\mathrm{T}\; (\mathbf{B}^\mathrm{T})^{-1} \mathbf{M} \mathbf{B}^{-1}\; \dot{\mathbf{v}} = \sum_{r, r'=1}^{3N-6} (G^{-1})_{r r'} \dot{q}_r \dot{q}_{r'}$
where we used that the last 6 components of v are zero. This form of the kinetic energy of vibration enters Wilson's GF method. It is of some interest to point out that the potential energy in the harmonic approximation can be written as follows
$2V_\mathrm{harm} = \mathbf{d}^\mathrm{T} \mathbf{H} \mathbf{d} = \mathbf{v}^\mathrm{T} (\mathbf{B}^\mathrm{T})^{-1} \mathbf{H} \mathbf{B}^{-1} \mathbf{v} = \sum_{r, r'=1}^{3N-6} F_{r r'} q_r q_{r'},$
where H is the Hessian of the potential in the minimum and F, defined by this equation, is the F matrix of the GF method.
## Relation to the harmonic approximation
In the harmonic approximation to the nuclear vibrational problem, expressed in displacement coordinates, one must solve the generalized eigenvalue problem
$\mathbf{H}\mathbf{C} = \mathbf{M} \mathbf{C} \boldsymbol{\Phi},$
where H is a 3N x 3N symmetric matrix of second derivatives of the potential $\scriptstyle V(\mathbf{R}_1, \mathbf{R}_2,\ldots, \mathbf{R}_N)$. H is the Hessian matrix of V in the equilibrium $\scriptstyle \mathbf{R}_1^0,\ldots, \mathbf{R}_N^0$. The diagonal matrix M contains the masses on the diagonal. The diagonal matrix $\scriptstyle \boldsymbol{\Phi}$ contains the eigenvalues, while the columns of C contain the eigenvectors.
It can be shown that the invariance of V under simultaneous translation over t of all nuclei implies that vectors T = (t, ... , t) are in the kernel of H. From the invariance of V under an infinitesimal rotation of all nuclei around s it can be shown that also the vectors S = (s x R10, ..., s x RN0) are in the kernel of H :
$\mathbf{H} \begin{pmatrix} \mathbf{t} \\ \vdots\\ \mathbf{t} \end{pmatrix} = \begin{pmatrix} \mathbf{0} \\ \vdots\\ \mathbf{0} \end{pmatrix} \quad\mathrm{and}\quad \mathbf{H} \begin{pmatrix} \mathbf{s}\times \mathbf{R}_1^0 \\ \vdots\\ \mathbf{s}\times \mathbf{R}_N^0 \end{pmatrix} = \begin{pmatrix} \mathbf{0} \\ \vdots\\ \mathbf{0} \end{pmatrix}$
Thus, 6 columns of C corresponding to eigenvalue zero, are determined algebraically. (If the generalized eigenvalue problem is solved numerically, one will find in general 6 linearly independent linear combinations of S and T). The eigenspace corresponding to eigenvalue zero is at least of dimension 6 (often it is exactly of dimension 6, since the other eigenvalues, which are force constants, are never zero for molecules in their ground state). Thus, T and S correspond to the overall (external) motions: translation and rotation, respectively. They are zero-energy modes because space is homogeneous (force-free) and isotropic (torque-free).
By the definition in this article the non-zero frequency modes are internal modes, since they are within the orthogonal complement of Rext. The generalized orthogonalities: $\scriptstyle \mathbf{C}^\mathrm{T} \mathbf{M} \mathbf{C} = \mathbf{I}$ applied to the "internal" (non-zero eigenvalue) and "external" (zero-eigenvalue) columns of C are equivalent to the Eckart conditions.
## References
1. C. Eckart, Some studies concerning rotating axes and polyatomic molecules, Physical Review, vol. 47, pp. 552-558 (1935).
2. Aaron Sayvetz, The Kinetic Energy of Polyatomic Molecules, J. Chem. Phys. vol. 7, pp. 383-389 (1939).
3. L. C. Biedenharn and J. D. Louck, Angular Momentum in Quantum Physics, Addison-Wesley, Reading (1981) p. 535.
4. An inertial frame is in uniform translational motion.
The classic work is:
• E. Bright Wilson, Jr., J. C. Decius and Paul C. Cross, Molecular Vibrations, Mc-Graw-Hill (1955). Reprinted by Dover (1980).
More advanced book are:
• D. Papoušek and M. R. Aliev, Molecular Vibrational-Rotational Spectra, Elsevier (1982).
• S. Califano, Vibrational States, Wiley, New York-London (1976). ISBN 0 471 12996 8 | 5,529 | 18,343 | {"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": 55, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.887206 |
https://techcommunity.microsoft.com/t5/excel/if-then-and-need-help-with-the-formula/td-p/2918344 | 1,638,508,844,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00413.warc.gz | 610,794,273 | 56,489 | SOLVED
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# IF... THEN... AND! Need help with the formula!
On my payment spreadsheet, I have a column for invoice numbers and a column for the vendor numbers.
If the vendor does not provide an invoice, then the invoice number is always the date of the check run.
IF the vendor number repeats (more than one payment for this vendor), and the invoice number is the date, THEN a -1 needs to be added to the end of the date so that it does not appear to be a duplicate invoice number.
For each time the vendor number repeats, the -# at the end of the date increases, i.e., 11042021-1, 11042021-2, etc. Currently, I am manually entering the dash numbers.
Anyone know of a function to automatically add the -1, -2, etc at the end of the date?
Thank you,
Babs
3 Replies
best response confirmed by Babs_Adams (New Contributor)
Solution
# Re: IF... THEN... AND! Need help with the formula!
In A4:
``=TEXT(C4,"mmddyyyy")&TEXT(COUNTIFS(\$E\$3:\$E3,\$E4,\$C\$3:\$C3,\$C4),"-0;;")``
Fill down.
# Re: IF... THEN... AND! Need help with the formula!
You are AMAZING! I can't tell you how much I appreciate you taking the time and responding and sharing your expertise. Thank you! If I need you again, can I reach out to you?
# Re: IF... THEN... AND! Need help with the formula!
Sure, feel free to do so! | 356 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.861713 |
https://sublimephoto.com.br/opposite-of-ywsxrl/maximum-likelihood-classification-ppt-bd65fe | 1,620,490,772,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00513.warc.gz | 574,414,380 | 12,213 | Sem categoria
# maximum likelihood classification ppt
Small Likelihood: Given observed data & a tree, Powerpoint lecture slides - DHSch3part2.ppt 1 Bayesian Estimation (BE) Bayesian Parameter Estimation: Gaussian Case Bayesian Parameter Estimation: General Estimation Problems of Dimensionality Chapter 3: Maximum-Likelihood and Bayesian Parameter Estimation (part 2) 2 Pattern Classification, Chapter 1 2 Bayesian Estimation (Bayesian learning Maximum conditional likelihood estimate for parameter Slide credit: Tom Mitchell Identify all informative sites in the multiple alignment 2. 213 11 Reject fraction — 0.01 %PDF-1.4 %���� Gaussian maximum likelihood is a parametric classifier that assumes a gaussian distribution of each class. MaxiMuM Like§Lihood estiMation 14.INTRODUCTION1 the generalized method of moments discussed in Chapter 13 and the semiparametric, nonparametric, and Bayesian estimators discussed in Chapters 12 and are becoming 16 widely used by model builders. o�K�K�u�n��#��"wC��|�3�j���=+��U|PM{��A��( ҍ��:7B�f�d~z�����X5�ICcl�i�I�v��p��o�Kq�VL�j�&* "k��XF���.KkY�V+�@5�c� Gaussian Maximum Likelihood classifiers assume that the feature vectors of each class are (statistically) distributed according to a multivariate normal probability density function. In statistics, maximum likelihood estimation (MLE) is a method of estimating the parameters of a probability distribution by maximizing a likelihood function, so that under the assumed statistical model the observed data is most probable. So we use the term classification here because in a logit model the output is discrete. ��e>�R!��~N�iBk��)���Q�*��V��M%t�l Z���1�����Z�*3D�F�k� B�V
>"k��P�F@d�Q!�+Ad�#}`OO��ӇR ��(�ڬ�E�Z�F��DV��Е ��Fg�͚^��5j�Z���F���dž�"C�D���t+�@7j�V�Y��T�yQp�-T�2�9@���5�A��EЪ#]��yM�ʬ��F�^��[�kM!�V��(�V�sR����'DЪ�*w�Ъ�*W�T'���"lU�����\$�h 12. 0000002696 00000 n The Maximum Likelihood Classification tool is used to classify the raster into five classes. stream trailer Complex calculation statistical programs will run these analyses ; 5 Interpreting ßs . The Maximum Likelihood Function. %�쏢 0000001842 00000 n 7 0 obj Maximum a posteriori. xref The parameters (01, 82, 8) are estimated from the data, while (ql, q2) are assessed from the … Maximum likelihood estimate for parameter . MLE=argmax1, 1, 2, 2, ⋯,, =argmax=1, . The Landsat ETM+ image has used for classification. Maximum Likelihood Analysis ofPhylogenetic Trees – p.10. MLC is based on Bayes' classification and in this classificationa pixelis assigned to a class according to its probability of belonging to a particular class. 0000000016 00000 n Finally we 1I would like to acknowledge the contributions of Prof. Alex Gershman, Dept. Maximum Likelihood is a method for the inference of phylogeny. 5 techniques: correlation, Maximum Likelihood, MUSIC, ESPRIT and Matrix Pencil. 0000003237 00000 n • The maximum parsimony method is good for similar sequences, a sequences group with small amount of variation • This method does not give the branch length, only the branch order • Parsimony may be used to estimate "species" or "gene" phylogenies. �a�l)�X�I�9,بԶ� (�g�] D����ҩ��r��Z/�i. If you have truncated distribution, or bimodal distributions, etc, then the model does not fit well to your data and you could end up with suboptimal results. startxref 213 0 obj <> endobj There can be infinite sets of regression coefficients. x�b```f``�d`e`�Td`@ 6v 1�Œ,�-w8�Ҧ�17�U������ 9���{��>s���������D��\$d������3��юIr5O��p��y0�U@*W��� ����)�6!��9% j^��NЈ������X��Z��`K;?_��M���"� .�j���'�)u0�ְZ��%P�h���� \4�&�����"d�h Maximum Likelihood Estimation Computing the Likelihood Functions Sufficient Statistics Maximum A Posterior (MAP) Laplace Correction Bayesian Reasoning Bayesian Inference Binomial Distribution: Laplace Est. A logit model is often called logistic regression model. of Elec. 0000001690 00000 n Multiclass classification •Given training data दථ,धථ:Յ≤ग≤i.i.d. At its core, a maximum likelihood classifier could be described in pseudocode as: params_of_most_likely_class_label = argmax( x |params_of_indivdual_classes) If you're curious, here's the full version of MLC that likely closely resembles what is … EG��J���"���Z �RM�' �(zB߄"w�. STEPS 1. (2008a,b) presented results of a supervised classification (maximum likelihood) applied to reconnaissance (acquired with 5000 m line spacing) AGRS data (Figure 29). Least squares (known structure, easy to interpret) Neural nets (unknown structure, hard to interpret) Nonparametric approaches. Therefore, MCL takes advantage of both the mean vectors and the multivariate spreads of each class, and can identify those elongated classes. �&Clլ�dm!W� • Multiple class classification Logistic Regression. Learn more about how Maximum Likelihood Classification works. The maximum likelihood classifier is one of the most popular methods of classification in remote sensing, in which a pixel with the maximum likelihood is classified into the corresponding class. Example inputs to Maximum Likelihood Classification. 0000001550 00000 n Maximum Likelihood Estimation Eric Zivot May 14, 2001 This version: November 15, 2009 1 Maximum Likelihood Estimation 1.1 The Likelihood Function Let X1,...,Xn be an iid sample with probability density function (pdf) f(xi;θ), where θis a (k× 1) vector of parameters that characterize f(xi;θ).For example, if Xi˜N(μ,σ2) then f(xi;θ)=(2πσ2)−1/2 exp(−1 Three Likelihood Versions Big Likelihood: Given the sequence data, find a tree and edge weights that maximize data tree & edge weights . The training samples are used to estimate the parameters of the distributions. %%EOF <]>> Output multiband raster — mlclass_1. I� ��H� �J�R��*Y �,[%�-݆wP�\$C�Ƅ�*Y O���f)b���,�:C�����Ȁ�*Q!e��*1:˴�p�� ��,�k� ��\�Q"ŦL����m[9ZC� ��H��E��Q\$�� Unless you select a probability threshold, all pixels are classified. The maximum likelihood estimate is that set of regression coefficients for which the probability of getting the data we have observed is maximum. It evaluates a hypothesis about evolutionary history in terms of the probability that the proposed model and the hypothesized history would give rise to the observed data set. Supervised classification involves the use of training area data that are considered representative of each rock type or surficial unit to be classified. | 1,816 | 6,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-21 | latest | en | 0.616504 |
https://www.muscleandfitness.com/nutrition/healthy-eating/28-days-redemption-meal-plan-step-1/ | 1,718,944,324,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00761.warc.gz | 804,749,856 | 33,779 | To get the best estimate of where you should start for calories, you need to take an honest look at your current diet. I recommend recording everything you eat for one full week. If that seems too daunting, do at least two weekdays and one weekend day.
If you have a scale, weigh food such as chicken, beef and fish to make sure you’re accurately recording the serving size (ie, 8 ounces or 12 ounces). For liquids and grains (rice, cereal), use measuring cups and spoons. If you feel have a good take on how much is one cup of something, or what 8 ounces of chicken breast looks like, estimate as best you can.
Keeping a food journal is a trick I like to use to keep people from straying from their diets. When they have a weak moment and crave a donut, knowing that they have to record that in their food log helps them avoid eating the donut. But during this week, don’t avoid eating any of the foods you normally do. The point is to access your true current diet so that you know where to start your new one. If you alter it, it can actually work against your fat-loss efforts.
Calculate the amount of calories, grams of protein, grams of carbs, and grams of total fat for each food that you eat. Then, add these up for all the foods you eat each day. Using an Excel spreadsheet can help make this easier and better allow you to keep track of your diet.
Once you have the total amount of calories and grams of protein, carbs, and fat for each day, you need to get an average of all the days. If you record seven days of food intake, add the calories up for all seven days and divide that number by seven to get the average calorie intake for the seven days. If you did just three days, add up the total calories and divide by three. Do this also for grams of protein, grams of carbs, and grams of fat.
Dieting 101 Key: Calorie Counting
The critical factor in this initial food tracking is your average calorie intake. Take this number and divide it by your body weight in pounds to calculate your average calories relative to your body weight. For example, if you weigh 200 pounds and you consumed an average of 4,000 calories per day, you currently consume an average of 20 calories per pound of body weight. You’ll use this daily average number in Step #2… | 501 | 2,268 | {"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} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.957588 |
https://community.powerbi.com/t5/Desktop/Grand-total-with-Blank-vaue/m-p/622778 | 1,582,141,827,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00251.warc.gz | 331,043,763 | 131,347 | cancel
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Frequent Visitor
## Grand total with Blank vaue
Hi,
I have a matrix table, i used a measure column using Swtich function to get some data.
i used a measure column to get W/L ration as below:
Ratio = SWITCH(SELECTEDVALUE(closed[Outcome]),
"1. Won", (divide(CALCULATE(COUNT(Closed[Outcome]),Closed[Outcome]="1. Won"), (CALCULATE(COUNT(Closed[Outcome]),Closed[Outcome]="1. Won") +CALCULATE(COUNT(Closed[Outcome]),Closed[Outcome]="2. Lost")),0)*100) ," ")
so i have a problem that i cant see the grand total for W/L ration becouse of the blank value in some rows. please help to get grand total for W/L Ratio and keep the blank Values.
Thank you
1 ACCEPTED SOLUTION
Accepted Solutions
Super User III
## Re: Grand total with Blank vaue
The problem here is that SELECTEDVALUE() will return a value if there is a single value for Closed[Outcome] and at the "Total" level you don't have a single value (you technically have all values).
So I think changing the code to something like the following (using a variable to avoid copying the calculation logic twice) should fix this for you.
```Ratio =
VAR _pcntWon = (
DIVIDE (
CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "1. Won" ),
(
CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "1. Won" )
+ CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "2. Lost" )
),
0
) * 100
RETURN SWITCH (
SELECTEDVALUE ( closed[Outcome], "Total" ),
"1. Won", _pcntWon,
"Total", _pcntWon,
" "
)```
5 REPLIES 5
Super User IV
## Re: Grand total with Blank vaue
This looks like a measure totals problem. Very common. See my post about it here: https://community.powerbi.com/t5/DAX-Commands-and-Tips/Dealing-with-Measure-Totals/td-p/63376
Also, this Quick Measure, Measure Totals, The Final Word should get you what you need:
https://community.powerbi.com/t5/Quick-Measures-Gallery/Measure-Totals-The-Final-Word/m-p/547907
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Frequent Visitor
## Re: Grand total with Blank vaue
Hi Greg,
Thank you for your reply. i go througth the links you sent, but there are some rows sholud be Balnk and i will have same issue. Please what shall i write in the measure column to solve the issue.
Thank you Again.
Community Support Team
## Re: Grand total with Blank vaue
HI @nsaleh,
If you can please share a pbix file with part of sample data for test. It should be help for testing and coding formula.
BTW, I'm not so recommend you to use switch function to handle with summarize records. It not suitable for split summarize records to multiple part.
Regards,
Xiaoxin Sheng
Community Support Team _ Xiaoxin Sheng
If this post helps, please consider Accept it as the solution to help the other members find it more quickly
Learning resources: Power BI
Super User III
## Re: Grand total with Blank vaue
The problem here is that SELECTEDVALUE() will return a value if there is a single value for Closed[Outcome] and at the "Total" level you don't have a single value (you technically have all values).
So I think changing the code to something like the following (using a variable to avoid copying the calculation logic twice) should fix this for you.
```Ratio =
VAR _pcntWon = (
DIVIDE (
CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "1. Won" ),
(
CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "1. Won" )
+ CALCULATE ( COUNT ( Closed[Outcome] ), Closed[Outcome] = "2. Lost" )
),
0
) * 100
RETURN SWITCH (
SELECTEDVALUE ( closed[Outcome], "Total" ),
"1. Won", _pcntWon,
"Total", _pcntWon,
" "
)```
Frequent Visitor
## Re: Grand total with Blank vaue
Hi
Its working fine now. Thank you so much for your help.
Announcements | 991 | 3,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-10 | latest | en | 0.835022 |
https://discuss.leetcode.com/topic/40411/14-line-short-c-no-larger-integer-type-used | 1,511,266,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00493.warc.gz | 601,970,375 | 9,485 | # 14 line short C++, no larger integer type used
• ``````class Solution {
public:
int reverse(int x) {
int result = 0;
while (x) {
if (x > 0 && result > (INT_MAX - x % 10) / 10
|| x < 0 && result < (INT_MIN - x % 10) / 10) {
return 0;
}
result = result * 10 + x % 10;
x /= 10;
}
return result;
}
};
``````
If x > 0, then we only need to consider positive overflow. If x < 0, we consider negative overflow then.
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A038968 Primes p such that 82 is not a square mod p. 1
5, 7, 17, 37, 43, 47, 59, 61, 71, 79, 83, 89, 97, 107, 131, 137, 139, 151, 163, 167, 173, 191, 193, 197, 199, 233, 239, 251, 257, 263, 269, 277, 281, 283, 307, 311, 313, 349, 373, 379, 383, 389, 419, 439, 443, 457 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 MATHEMATICA Select[Prime@Range[120], JacobiSymbol[82, #] == - 1 &] (* Vincenzo Librandi, Sep 09 2012 *) PROG (MAGMA) [ p: p in PrimesUpTo(457) | not IsSquare(R! 82) where R:= ResidueClassRing(p) ]; // Vincenzo Librandi, Sep 10 2012 CROSSREFS Sequence in context: A166977 A272717 A018538 * A243400 A318568 A239414 Adjacent sequences: A038965 A038966 A038967 * A038969 A038970 A038971 KEYWORD nonn,easy AUTHOR EXTENSIONS Offset changed from 0 to 1 by Vincenzo Librandi, Sep 09 2012 STATUS approved
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Last modified December 9 23:02 EST 2019. Contains 329880 sequences. (Running on oeis4.) | 579 | 1,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-51 | longest | en | 0.731409 |
https://www.nag.com/numeric/cl/nagdoc_cl26.0/html/g02/g02gpc.html | 1,624,487,038,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00376.warc.gz | 828,792,833 | 7,948 | g02 Chapter Contents
g02 Chapter Introduction
NAG Library Manual
# NAG Library Function Documentnag_glm_predict (g02gpc)
## 1 Purpose
nag_glm_predict (g02gpc) allows prediction from a generalized linear model fit via nag_glm_normal (g02gac), nag_glm_binomial (g02gbc), nag_glm_poisson (g02gcc) or nag_glm_gamma (g02gdc).
## 2 Specification
#include #include
void nag_glm_predict (Nag_Distributions errfn, Nag_Link link, Nag_IncludeMean mean, Integer n, const double x[], Integer tdx, Integer m, const Integer sx[], Integer ip, const double binom_t[], const double offset[], const double wt[], double scale, double ex_power, const double b[], const double cov[], Nag_Boolean vfobs, double eta[], double seeta[], double pred[], double sepred[], NagError *fail)
## 3 Description
A generalized linear model consists of the following elements:
(i) A suitable distribution for the dependent variable $y$. (ii) A linear model, with linear predictor $\eta =X\beta$, where $X$ is a matrix of independent variables and $\beta$ a column vector of $p$ parameters. (iii) A link function $g\left(.\right)$ between the expected value of $y$ and the linear predictor, that is $E\left(y\right)=\mu =g\left(\eta \right)$.
In order to predict from a generalized linear model, that is estimate a value for the dependent variable, $y$, given a set of independent variables $X$, the matrix $X$ must be supplied, along with values for the parameters $\beta$ and their associated variance-covariance matrix, $C$. Suitable values for $\beta$ and $C$ are usually estimated by first fitting the prediction model to a training dataset with known responses, using for example nag_glm_normal (g02gac), nag_glm_binomial (g02gbc), nag_glm_poisson (g02gcc) or nag_glm_gamma (g02gdc). The predicted variable, and its standard error can then be obtained from:
$y^ = g-1η , se y^ = δg-1x δx η seη + Ifobs Vary$
where
$η=o+Xβ , seη = diagXCXT ,$
$o$ is a vector of offsets and ${I}_{\mathrm{fobs}}=0$, if the variance of future observations is not taken into account, and $1$ otherwise. Here $\mathrm{diag}A$ indicates the diagonal elements of matrix $A$.
If required, the variance for the $i$th future observation, $\mathrm{Var}\left({y}_{i}\right)$, can be calculated as:
$Varyi = ϕ Vθ wi$
where ${w}_{i}$ is a weight, $\varphi$ is the scale (or dispersion) parameter, and $V\left(\theta \right)$ is the variance function. Both the scale parameter and the variance function depend on the distribution used for the $y$, with:
Poisson $V\left(\theta \right)={\mu }_{i}$, $\varphi =1$ binomial $V\left(\theta \right)=\frac{{\mu }_{i}\left({t}_{i}-{\mu }_{i}\right)}{{t}_{i}}$, $\varphi =1$ Normal $V\left(\theta \right)=1$ gamma $V\left(\theta \right)={\mu }_{i}^{2}$
In the cases of a Normal and gamma error structure, the scale parameter ($\varphi$), is supplied by you. This value is usually obtained from the function used to fit the prediction model. In many cases, for a Normal error structure, $\varphi ={\stackrel{^}{\sigma }}^{2}$, i.e., the estimated variance.
## 4 References
McCullagh P and Nelder J A (1983) Generalized Linear Models Chapman and Hall
## 5 Arguments
1: $\mathbf{errfn}$Nag_DistributionsInput
On entry: indicates the distribution used to model the dependent variable, $y$.
${\mathbf{errfn}}=\mathrm{Nag_Binomial}$
The binomial distribution is used.
${\mathbf{errfn}}=\mathrm{Nag_Gamma}$
The gamma distribution is used.
${\mathbf{errfn}}=\mathrm{Nag_Normal}$
The Normal (Gaussian) distribution is used.
${\mathbf{errfn}}=\mathrm{Nag_Poisson}$
The Poisson distribution is used.
Constraint: ${\mathbf{errfn}}=\mathrm{Nag_Binomial}$, $\mathrm{Nag_Gamma}$, $\mathrm{Nag_Normal}$ or $\mathrm{Nag_Poisson}$.
On entry: indicates which link function to be used.
${\mathbf{link}}=\mathrm{Nag_Compl}$
A complementary log-log link is used.
${\mathbf{link}}=\mathrm{Nag_Expo}$
An exponent link is used.
${\mathbf{link}}=\mathrm{Nag_Logistic}$
A logistic link is used.
${\mathbf{link}}=\mathrm{Nag_Iden}$
An identity link is used.
${\mathbf{link}}=\mathrm{Nag_Log}$
A log link is used.
${\mathbf{link}}=\mathrm{Nag_Probit}$
A probit link is used.
${\mathbf{link}}=\mathrm{Nag_Reci}$
A reciprocal link is used.
${\mathbf{link}}=\mathrm{Nag_Sqrt}$
A square root link is used.
Details on the functional form of the different links can be found in the g02 Chapter Introduction.
Constraints:
• if ${\mathbf{errfn}}=\mathrm{Nag_Binomial}$, ${\mathbf{link}}=\mathrm{Nag_Compl}$, $\mathrm{Nag_Logistic}$ or $\mathrm{Nag_Probit}$;
• otherwise ${\mathbf{link}}=\mathrm{Nag_Expo}$, $\mathrm{Nag_Iden}$, $\mathrm{Nag_Log}$, $\mathrm{Nag_Reci}$ or $\mathrm{Nag_Sqrt}$.
3: $\mathbf{mean}$Nag_IncludeMeanInput
On entry: indicates if a mean term is to be included.
${\mathbf{mean}}=\mathrm{Nag_MeanInclude}$
A mean term, intercept, will be included in the model.
${\mathbf{mean}}=\mathrm{Nag_MeanZero}$
The model will pass through the origin, zero-point.
Constraint: ${\mathbf{mean}}=\mathrm{Nag_MeanInclude}$ or $\mathrm{Nag_MeanZero}$.
4: $\mathbf{n}$IntegerInput
On entry: $n$, the number of observations.
Constraint: ${\mathbf{n}}\ge 1$.
5: $\mathbf{x}\left[\mathit{dim}\right]$const doubleInput
Note: the dimension, dim, of the array x must be at least ${\mathbf{n}}×{\mathbf{tdx}}$.
On entry: ${\mathbf{x}}\left[\left(\mathit{i}-1\right)×{\mathbf{tdx}}+\mathit{j}-1\right]$ must contain the $\mathit{i}$th observation for the $\mathit{j}$th independent variable, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{m}}$.
6: $\mathbf{tdx}$IntegerInput
On entry: the stride separating matrix column elements in the array x.
Constraint: ${\mathbf{tdx}}\ge {\mathbf{m}}$
7: $\mathbf{m}$IntegerInput
On entry: $m$, the total number of independent variables.
Constraint: ${\mathbf{m}}\ge 1$.
8: $\mathbf{sx}\left[{\mathbf{m}}\right]$const IntegerInput
On entry: indicates which independent variables are to be included in the model.
If ${\mathbf{sx}}\left[j-1\right]>0$, the $j$th independent variable is included in the regression model.
Constraints:
• ${\mathbf{sx}}\left[j-1\right]\ge 0$, for $\mathit{i}=1,2,\dots ,{\mathbf{m}}$;
• if ${\mathbf{mean}}=\mathrm{Nag_MeanInclude}$, exactly ${\mathbf{ip}}-1$ values of sx must be $\text{}>0$;
• if ${\mathbf{mean}}=\mathrm{Nag_MeanZero}$, exactly ip values of sx must be $\text{}>0$.
9: $\mathbf{ip}$IntegerInput
On entry: the number of independent variables in the model, including the mean or intercept if present.
Constraint: ${\mathbf{ip}}>0$.
10: $\mathbf{binom_t}\left[{\mathbf{n}}\right]$const doubleInput
On entry: if ${\mathbf{errfn}}=\mathrm{Nag_Binomial}$, ${\mathbf{binom_t}}\left[i-1\right]$ must contain the binomial denominator, ${t}_{i}$, for the $i$th observation.
Otherwise binom_t is not referenced and may be NULL.
Constraint: if ${\mathbf{errfn}}=\mathrm{Nag_Binomial}$, ${\mathbf{binom_t}}\left[\mathit{i}-1\right]\ge 0.0$, for $\mathit{i}=1,2,\dots ,n$.
11: $\mathbf{offset}\left[{\mathbf{n}}\right]$const doubleInput
On entry: if an offset is required then ${\mathbf{offset}}\left[i-1\right]$ must contain the value of the offset ${o}_{i}$, for the $i$th observation. Otherwise offset must be supplied as NULL.
12: $\mathbf{wt}\left[{\mathbf{n}}\right]$const doubleInput
On entry: if weighted estimates are required then ${\mathbf{wt}}\left[i-1\right]$ must contain the weight, ${\omega }_{i}$ for the $i$th observation. Otherwise wt must be supplied as NULL.
If ${\mathbf{wt}}\left[i-1\right]=0.0$, then the $i$th observation is not included in the model, in which case the effective number of observations is the number of observations with positive weights.
If ${\mathbf{wt}}=\text{}$ NULL, then the effective number of observations is $n$.
If the variance of future observations is not included in the standard error of the predicted variable, wt is not referenced.
Constraint: if ${\mathbf{wt}}\phantom{\rule{0.25em}{0ex}}\text{is not}\phantom{\rule{0.25em}{0ex}}\mathbf{NULL}$ and ${\mathbf{vfobs}}=\mathrm{Nag_TRUE}$, ${\mathbf{wt}}\left[\mathit{i}-1\right]\ge 0.0$, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$.
13: $\mathbf{scale}$doubleInput
On entry: if ${\mathbf{errfn}}=\mathrm{Nag_Normal}$ or $\mathrm{Nag_Gamma}$ and ${\mathbf{vfobs}}=\mathrm{Nag_TRUE}$, the scale parameter, $\varphi$.
Otherwise scale is not referenced and $\varphi =1$.
Constraint: if ${\mathbf{errfn}}=\mathrm{Nag_Normal}$ or $\mathrm{Nag_Gamma}$ and ${\mathbf{vfobs}}=\mathrm{Nag_TRUE}$, ${\mathbf{scale}}>0.0$.
14: $\mathbf{ex_power}$doubleInput
On entry: if ${\mathbf{link}}=\mathrm{Nag_Expo}$, ex_power must contain the power of the exponential.
If ${\mathbf{link}}\ne \mathrm{Nag_Expo}$, ex_power is not referenced.
Constraint: if ${\mathbf{link}}=\mathrm{Nag_Expo}$, ${\mathbf{ex_power}}\ne 0.0$.
15: $\mathbf{b}\left[{\mathbf{ip}}\right]$const doubleInput
On entry: the model parameters, $\beta$.
If ${\mathbf{mean}}=\mathrm{Nag_MeanInclude}$, ${\mathbf{b}}\left[0\right]$ must contain the mean parameter and ${\mathbf{b}}\left[i\right]$ the coefficient of the variable contained in the $j$th independent x, where ${\mathbf{sx}}\left[j-1\right]$ is the $i$th positive value in the array sx.
If ${\mathbf{mean}}=\mathrm{Nag_MeanZero}$, ${\mathbf{b}}\left[i-1\right]$ must contain the coefficient of the variable contained in the $j$th independent x, where ${\mathbf{sx}}\left[j-1\right]$ is the $i$th positive value in the array sx.
16: $\mathbf{cov}\left[{\mathbf{ip}}×\left({\mathbf{ip}}+1\right)/2\right]$const doubleInput
On entry: the upper triangular part of the variance-covariance matrix, $C$, of the model parameters. This matrix should be supplied packed by column, i.e., the covariance between parameters ${\beta }_{i}$ and ${\beta }_{j}$, that is the values stored in ${\mathbf{b}}\left[i-1\right]$ and ${\mathbf{b}}\left[j-1\right]$, should be supplied in ${\mathbf{cov}}\left[\mathit{j}×\left(\mathit{j}-1\right)/2+\mathit{i}-1\right]$, for $\mathit{i}=1,2,\dots ,{\mathbf{ip}}$ and $\mathit{j}=\mathit{i},\dots ,{\mathbf{ip}}$.
Constraint: the matrix represented in cov must be a valid variance-covariance matrix.
17: $\mathbf{vfobs}$Nag_BooleanInput
On entry: if ${\mathbf{vfobs}}=\mathrm{Nag_TRUE}$, the variance of future observations is included in the standard error of the predicted variable (i.e., ${I}_{\mathrm{fobs}}=1$), otherwise ${I}_{\mathrm{fobs}}=0$.
18: $\mathbf{eta}\left[{\mathbf{n}}\right]$doubleOutput
On exit: the linear predictor, $\eta$.
19: $\mathbf{seeta}\left[{\mathbf{n}}\right]$doubleOutput
On exit: the standard error of the linear predictor, $\mathrm{se}\left(\eta \right)$.
20: $\mathbf{pred}\left[{\mathbf{n}}\right]$doubleOutput
On exit: the predicted value, $\stackrel{^}{y}$.
21: $\mathbf{sepred}\left[{\mathbf{n}}\right]$doubleOutput
On exit: the standard error of the predicted value, $\mathrm{se}\left(\stackrel{^}{y}\right)$. If ${\mathbf{pred}}\left[i-1\right]$ could not be calculated, then nag_glm_predict (g02gpc) returns ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_INVALID_PRED, and ${\mathbf{sepred}}\left[i-1\right]$ is set to $-99.0$.
22: $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 2.7 in How to Use the NAG Library and its Documentation).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
On entry, the error type and link function combination supplied is invalid.
NE_INT
On entry, ${\mathbf{ip}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ip}}>0$.
On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{m}}\ge 1$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 1$.
NE_INT_2
On entry, ${\mathbf{tdx}}=〈\mathit{\text{value}}〉$ and ${\mathbf{m}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{tdx}}\ge {\mathbf{m}}$.
NE_INT_ARRAY_CONS
On entry, sx not consistent with ip: $〈\mathit{\text{value}}〉$ values $>0$, expected $〈\mathit{\text{value}}〉$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
An unexpected error has been triggered by this function. Please contact NAG.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_INVALID_PRED
At least one predicted value could not be calculated as required. sepred is set to $-99.0$ for affected predicted values.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.
NE_REAL
On entry, ${\mathbf{ex_power}}=0.0$.
On entry, ${\mathbf{scale}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{scale}}>0.0$.
NE_REAL_ARRAY_CONS
On entry, ${\mathbf{cov}}\left[i-1\right]<0.0$ for at least one diagonal element: $i=〈\mathit{\text{value}}〉$, ${\mathbf{cov}}\left[i-1\right]=〈\mathit{\text{value}}〉$.
On entry, $i=〈\mathit{\text{value}}〉$ and ${\mathbf{binom_t}}\left[i-1\right]=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{binom_t}}\left[i-1\right]\ge 0.0$, for all $i$.
On entry, $i=〈\mathit{\text{value}}〉$ and ${\mathbf{wt}}\left[i-1\right]=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{wt}}\left[i-1\right]\ge 0.0$, for all $i$.
Not applicable.
## 8 Parallelism and Performance
nag_glm_predict (g02gpc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the x06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
None.
## 10 Example
The model
$y = 1 β1 + β2 x + ε$
is fitted to a training dataset with five observations. The resulting model is then used to predict the response for two new observations.
### 10.1 Program Text
Program Text (g02gpce.c)
### 10.2 Program Data
Program Data (g02gpce.d)
### 10.3 Program Results
Program Results (g02gpce.r) | 4,709 | 14,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 204, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.425471 |
http://www.in2013dollars.com/inflation-rate-in-1987 | 1,571,894,563,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00007.warc.gz | 270,105,542 | 13,685 | \$
# U.S. inflation rate in 1987: 3.65%
### Inflation in 1987 and Its Effect on Dollar Value
Purchasing power decreased by 3.65% in 1987 compared to 1986. On average, you would have to spend 3.65% more money in 1987 than in 1986 for the same item.
In other words, \$1 in 1986 is equivalent in purchasing power to about \$1.04 in 1987.
The 1986 inflation rate was 1.86%. The inflation rate in 1987 was 3.65%. The 1987 inflation rate is higher compared to the average inflation rate of 2.58% per year between 1987 and 2019.
Inflation rate is calculated by change in the consumer price index (CPI). The CPI in 1987 was 113.60. It was 109.60 in the previous year, 1986. The difference in CPI between the years is used by the Bureau of Labor Statistics to officially determine inflation.
Average inflation rate 3.65% Converted amount (\$1 base) \$1.04 Price difference (\$1 base) \$0.04 CPI in 1986 109.600 CPI in 1987 113.600 Inflation in 1986 1.86% Inflation in 1987 3.65%
USD Inflation since 1913
Annual Rate, U.S. Bureau of Labor Statistics CPI
### Inflation by City
Inflation can vary widely by city, even within the United States. Here's how some cities fared in 1986 to 1987 (figures shown are purchasing power equivalents of \$1):
New York experienced the highest rate of inflation during the 1 years between 1986 and 1987 (5.05%).
Seattle, Washington experienced the lowest rate of inflation during the 1 years between 1986 and 1987 (0.12%).
Note that some locations showing 0% inflation may have not yet reported latest data.
### Inflation by Country
Inflation can also vary widely by country. For comparison, in the UK £1.00 in 1986 would be equivalent to £1.04 in 1987, an absolute change of £0.04 and a cumulative change of 4.17%.
In Canada, CA\$1.00 in 1986 would be equivalent to CA\$1.04 in 1987, an absolute change of CA\$0.04 and a cumulative change of 4.19%.
Compare these numbers to the US's overall absolute change of \$0.04 and total percent change of 3.65%.
### Inflation by Spending Category
CPI is the weighted combination of many categories of spending that are tracked by the government. This chart shows the average rate of inflation for select CPI categories between 1986 and 1987.
Compare these values to the overall average of 3.65% per year:
Category Avg Inflation (%) Total Inflation (%) \$1 in 1986 → 1987
Food 4.17 4.17 1.04
Shelter 4.68 4.68 1.05
Energy 0.36 0.36 1.00
Apparel 4.40 4.40 1.04
New vehicles 3.41 3.41 1.03
Used cars and trucks 3.95 3.95 1.04
Transportation services 4.81 4.81 1.05
Medical care services 6.60 6.60 1.07
Medical care commodities 6.69 6.69 1.07
It's important to note that not all categories may be tracked since 1986. This table and visualization use the earliest available data for each category.
### How to Calculate Inflation Rate for \$1, 1986 to 1987
This inflation calculator uses the following inflation rate formula:
CPI in 1987CPI in 1986
×
1986 USD value
=
1987 USD value
Then plug in historical CPI values. The U.S. CPI was 109.6 in the year 1986 and 113.6 in 1987:
113.6109.6
×
\$1
=
\$1.04
\$1 in 1986 has the same "purchasing power" or "buying power" as \$1.04 in 1987.
To get the total inflation rate for the 1 years between 1986 and 1987, we use the following formula:
CPI in 1987 - CPI in 1986CPI in 1986
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
113.6 - 109.6109.6
×
100
=
4%
### Alternate Measurements of Inflation
The above data describe the CPI for all items. Also of note is the Core CPI, which measures inflation for all items except for the more volatile categories of food and energy. Core inflation averaged 4.12% per year between 1986 and 1987 (vs all-CPI inflation of 3.65%), for an inflation total of 4.12%.
When using the core inflation measurement, \$1 in 1986 is equivalent in buying power to \$1.04 in 1987, a difference of \$0.04. Recall that for All Items, the converted amount is \$1.04 with a difference of \$0.04.
In 1986, core inflation was 4.04%.
### Comparison to S&P 500 Index
To help put this inflation into perspective, if we had invested \$1 in the S&P 500 index in 1986, our investment would be nominally worth approximately \$1.28 in 1987. This is a return on investment of 28.25%, with an absolute return of \$0.28.
These numbers are not inflation adjusted, so they are considered nominal. In order to evaluate the real return on our investment, we must calculate the return with inflation taken into account.
The compounding effect of inflation would account for 3.52% of returns (\$0.01) during this period. This means the inflation-adjusted real return of our \$1 investment is \$0.27.
Investment in S&P 500 Index, 1986-1987
Original Amount Final Amount Change
Nominal \$1 \$1.28 28.25%
Real
\$1 \$0.27 27.26%
Politics and news often influence economic performance. Here's what was happening at the time:
• Ferdinand Marcos becomes president of the Philippines after rigged elections.
• 6.5 million People hold hands across the US in an event referred to as Hands Across America.
• Doctors announce that the drug AZT will be available to AIDS patients.
• Talks between Ronald Reagan and Mikhail Gorbachev begin at the Reykjavik summit.
• Mrs. Davina Thompson receives the first heart, lung and liver transplant in Cambridge, England.
### Data Source & Citation
Raw data for these calculations comes from the Bureau of Labor Statistics' (CPI), established in 1913. Inflation data from 1665 to 1912 is sourced from a historical study conducted by political science professor Robert Sahr at Oregon State University.
You may use the following MLA citation for this page: “Inflation Rate in 1987 | Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 24 Oct. 2019, https://www.officialdata.org/inflation-rate-in-1987. | 1,620 | 5,836 | {"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-2019-43 | latest | en | 0.937712 |
https://oeis.org/A304001 | 1,628,064,668,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00530.warc.gz | 432,279,313 | 4,400 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A304001 Number of permutations of [n] whose up-down signature has a nonnegative total sum. 2
1, 1, 1, 5, 12, 93, 360, 3728, 20160, 259535, 1814400, 27820524, 239500800, 4251096402, 43589145600, 877606592736, 10461394944000, 235288904377275, 3201186852864000, 79476406782222500, 1216451004088320000, 33020655481590446318, 562000363888803840000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS The up-down signature has (+1) for each ascent and (-1) for each descent. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..450 MAPLE b:= proc(u, o, t) option remember; (n-> `if`(t>=n, n!, `if`(t<-n, 0, add(b(u-j, o+j-1, t-1), j=1..u)+ add(b(u+j-1, o-j, t+1), j=1..o))))(u+o) end: a:= n-> `if`(n=0, 1, add(b(j-1, n-j, 0), j=1..n)): seq(a(n), n=0..25); # second Maple program: a:= n-> `if`(irem(n, 2, 'r')=0, ceil(n!/2), add(combinat[eulerian1](n, j), j=0..r)): seq(a(n), n=0..25); MATHEMATICA Eulerian1[n_, k_] := If[k == 0, 1, If[n == 0, 0, Sum[(-1)^j (k - j + 1)^n Binomial[n + 1, j], {j, 0, k + 1}]]]; a[n_] := Module[{r, m}, {r, m} = QuotientRemainder[n, 2]; If[m == 0, Ceiling[n!/2], Sum[Eulerian1[n, j], {j, 0, r}]]]; a /@ Range[0, 25] (* Jean-François Alcover, Mar 26 2021, after 2nd Maple program *) CROSSREFS Bisections give: A002674 (even part), A179457(2n+1,n+1) (odd part). Cf. A000246 (for nonnegative partial sums), A006551 (total sums are 0 or 1), A008292, A303287. Sequence in context: A249478 A009414 A009426 * A009731 A009427 A267271 Adjacent sequences: A303998 A303999 A304000 * A304002 A304003 A304004 KEYWORD nonn AUTHOR Alois P. Heinz, May 04 2018 STATUS approved
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Last modified August 4 03:57 EDT 2021. Contains 346442 sequences. (Running on oeis4.) | 823 | 2,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-31 | latest | en | 0.446291 |
https://web2.0calc.com/questions/halp_98 | 1,723,686,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00488.warc.gz | 476,062,113 | 5,796 | +0
# halp
0
3
2
+56
For certain values of k and m the system \begin{align*}
a + 2b &= -3, \\
4a + 2b &= k - a - mb
\end{align*}
has infinitely many solutions (a,b) What are k and m?
2
Sophie's favorite (positive) number is a two-digit number. If she reverses the digits, the result is 36 less than her favorite number. Also, one digit is 1 less than double the other digit. What is Sophie's favorite number?
Jun 26, 2024
#2
+979
0
For problem 1, we notice that both equations have the same coefficient on the term 2b.
This means that in order for the system to have infinitely many solutions, the coefficients of a must also be proportional.
We can rewrite the second equation as: 3a+2b=k+mb. Thus, for infinitely many solutions, we need [ \frac{4}{3} = \frac{-1}{m}. ]
Since the fraction on the left-hand side is positive, we know that m must be negative. Solving for m, we get m = −3. Then k = 4/3⋅−3 + 2 = −2.
Therefore, k = -2 and m = -3.
Jun 28, 2024 | 309 | 973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.896227 |
https://www.splashlearn.com/s/math-worksheets/subtract-multiples-of-10-from-2-digit-numbers-missing-numbers | 1,725,828,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00470.warc.gz | 961,319,002 | 25,948 | Subtract Multiples of 10 from 2-Digit Numbers: Missing Numbers - Math Worksheets - SplashLearn
Math > Subtract Multiples of 10 from 2-Digit Numbers: Missing Numbers Worksheet
# Subtract Multiples of 10 from 2-Digit Numbers: Missing Numbers Worksheet
## Know more about Subtract Multiples of 10 from 2-Digit Numbers: Missing Numbers Worksheet
Task your little mathematicians to crack the code of subtracting multiples of 10 from 2-digit numbers with this fun worksheet. If a minuend or subtrahend is missing, students use the count-back method or the relationship between addition and subtraction to determine the missing number. Also, they subtract the numbers if the missing number is the difference of the subtraction sentence. This subtract multiples of 10 from 2-digit numbers worksheet offers a tonne of practice with this idea. In each problem, the numbers are laid out in the horizontal format. Students should try to use different strategies involving composing and decomposing numbers to solve these problems. This will help them develop flexibility and fluency.
Give your child the passion and confidence to learn anything on their own fearlessly
4413+
4567+ | 244 | 1,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.852146 |
https://www.jiskha.com/questions/49447/places-of-education-for-silly-people-small-parts-of-wild-plants-head-coverings-for-small | 1,632,519,534,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00708.warc.gz | 869,724,995 | 5,808 | # English rhymes/riddles
places of education for silly people.
small parts of wild plants.
1. 👍
2. 👎
3. 👁
1. schools for fools
seeds of weeds
I can't think of a bug and a hat that rhyme, sorry.
1. 👍
2. 👎
2. i think it is nats' hats
1. 👍
2. 👎
3. what is small that rymes with plants
1. 👍
2. 👎
4. ants duhh
1. 👍
2. 👎
5. small parts of wild plants
1. 👍
2. 👎
6. weeds seeds
1. 👍
2. 👎
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https://www.mvtec.com/doc/halcon/13/en/derivate_vector_field.html | 1,713,048,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00392.warc.gz | 836,572,462 | 25,465 | Operators
# derivate_vector_field (Operator)
## Name
derivate_vector_field — Convolve a vector field with derivatives of the Gaussian.
## Signature
derivate_vector_field(VectorField : Result : Sigma, Component : )
## Description
derivate_vector_field convolves the components of a vector field with the derivatives of a Gaussian and calculates various features derived therefrom. derivate_vector_field only accepts vector fields of the semantic type 'vector_field_relative'. The VectorField F(r,c)=(u(r,c),v(r,c)) is defined as in optical_flow_mg. Sigma is the parameter of the Gaussian (i.e., the amount of smoothing). If a single value is passed in Sigma, the amount of smoothing in the column and row direction is identical. If two values are passed in Sigma, the first value specifies the amount of smoothing in the column direction, while the second value specifies the amount of smoothing in the row direction. The possible values for Component are:
'curl':
The curl of the vector field. One application of using 'curl' is to analyse optical flow fields. Metaphorically speaking, the curl is how much a small boat would rotate if the vector field was a fluid.
'divergence':
The divergence of the vector field. One application of using 'divergence' is to analyze optical flow fields. Metaphorically speaking, the divergence is where the source and sink would be if the vector field was a fluid.
When used in context of photometric stereo, the operator derivate_vector_field offers two more parameters, which are especially designed to process the gradient field that is returned by photometric_stereo. In this case, we interpret the input vector field as gradient of the underlying surface.
In the following formulas, the input vector field is therefore noted as
where the first and second component of the input is the gradient field of the surface f(r,c). In the formulas below f_rc denotes the first derivative in column direction of the first component of the gradient field.
'mean_curvature':
Mean curvature H of the underlying surface when the input vector field VectorField is interpreted as gradient field. One application of using 'mean_curvature' is to process the vector field that is returned by photometric_stereo. After filtering the vector field, even tiny scratches or bumps can be segmented.
'gauss_curvature':
Gaussian curvature K of the underlying surface when the input vector field VectorField is interpreted as gradient field. One application of using 'gauss_curvature' is to process the vector field that is returned by photometric_stereo. After filtering the vector field, even tiny scratches or bumps can be segmented. If the underlying surface of the vector field is developable, the Gaussian curvature is zero.
## Parallelization
• Multithreading type: reentrant (runs in parallel with non-exclusive operators).
• Automatically parallelized on domain level.
• Automatically parallelized on tuple level.
## Parameters
VectorField (input_object) singlechannelimage(-array) object (vector_field)
Input vector field.
Result (output_object) singlechannelimage(-array) object (real)
Filtered result images.
Sigma (input_control) real(-array) (real)
Sigma of the Gaussian.
Default value: 1.0
Suggested values: 0.7, 1.0, 1.5, 2.0, 3.0, 4.0, 5.0
Typical range of values: 0.2 ≤ Sigma ≤ 50.0
Restriction: 0.01 <= Sigma <= 50.0
Component (input_control) string (string)
Component to be calculated.
Default value: 'mean_curvature'
List of values: 'curl', 'divergence', 'gauss_curvature', 'mean_curvature'
## Result
If the parameters are valid, the operator derivate_vector_field returns the value 2 (H_MSG_TRUE). The behavior in case of empty input (no input images available) is set via the operator set_system('no_object_result',<Result>). If necessary, an exception is raised.
## Module
Foundation
Operators | 853 | 3,874 | {"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} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.853719 |
https://www.numbersaplenty.com/1222232201 | 1,716,814,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00475.warc.gz | 798,955,759 | 3,322 | Search a number
1222232201 = 25134714033
BaseRepresentation
bin100100011011001…
…1100100010001001
310011011211211101022
41020312130202021
510000342412301
6321140402225
742200542121
oct11066344211
93134754338
101222232201
11577a11163
122a13a6375
131662a9928
14b8479c81
157247cd1b
hex48d9c889
1222232201 has 8 divisors (see below), whose sum is σ = 1230725664. Its totient is φ = 1213768000.
The previous prime is 1222232189. The next prime is 1222232233. The reversal of 1222232201 is 1022322221.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1222232201 is a prime.
It is a super-2 number, since 2×12222322012 = 2987703106322608802, which contains 22 as substring.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 1222232201.
It is not an unprimeable number, because it can be changed into a prime (1222232251) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 80081 + ... + 94113.
It is an arithmetic number, because the mean of its divisors is an integer number (153840708).
Almost surely, 21222232201 is an apocalyptic number.
It is an amenable number.
1222232201 is a deficient number, since it is larger than the sum of its proper divisors (8493463).
1222232201 is a wasteful number, since it uses less digits than its factorization.
1222232201 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 14631.
The product of its (nonzero) digits is 192, while the sum is 17.
The square root of 1222232201 is about 34960.4376545832. The cubic root of 1222232201 is about 1069.1810197427.
Adding to 1222232201 its reverse (1022322221), we get a palindrome (2244554422).
The spelling of 1222232201 in words is "one billion, two hundred twenty-two million, two hundred thirty-two thousand, two hundred one". | 635 | 2,127 | {"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} | 3.21875 | 3 | CC-MAIN-2024-22 | latest | en | 0.852868 |
https://studysoup.com/tsg/calculus/285/calculus-concepts-and-applications/chapter/12600/9-3 | 1,606,757,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00718.warc.gz | 507,904,422 | 10,310 | ×
×
# Solutions for Chapter 9-3: Rapid Repeated Integration by Parts
## Full solutions for Calculus: Concepts and Applications | 2nd Edition
ISBN: 9781559536547
Solutions for Chapter 9-3: Rapid Repeated Integration by Parts
Solutions for Chapter 9-3
4 5 0 312 Reviews
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2
##### ISBN: 9781559536547
Chapter 9-3: Rapid Repeated Integration by Parts includes 60 full step-by-step solutions. This textbook survival guide was created for the textbook: Calculus: Concepts and Applications, edition: 2. Calculus: Concepts and Applications was written by and is associated to the ISBN: 9781559536547. This expansive textbook survival guide covers the following chapters and their solutions. Since 60 problems in chapter 9-3: Rapid Repeated Integration by Parts have been answered, more than 23255 students have viewed full step-by-step solutions from this chapter.
Key Calculus Terms and definitions covered in this textbook
• artesian coordinate system
An association between the points in a plane and ordered pairs of real numbers; or an association between the points in three-dimensional space and ordered triples of real numbers
• Bias
A flaw in the design of a sampling process that systematically causes the sample to differ from the population with respect to the statistic being measured. Undercoverage bias results when the sample systematically excludes one or more segments of the population. Voluntary response bias results when a sample consists only of those who volunteer their responses. Response bias results when the sampling design intentionally or unintentionally influences the responses
• Branches
The two separate curves that make up a hyperbola
• Coefficient of determination
The number r2 or R2 that measures how well a regression curve fits the data
• Compound interest
Interest that becomes part of the investment
• Cotangent
The function y = cot x
• Difference of two vectors
<u1, u2> - <v1, v2> = <u1 - v1, u2 - v2> or <u1, u2, u3> - <v1, v2, v3> = <u1 - v1, u2 - v2, u3 - v3>
• First-degree equation in x , y, and z
An equation that can be written in the form.
• Focal length of a parabola
The directed distance from the vertex to the focus.
• Graph of an equation in x and y
The set of all points in the coordinate plane corresponding to the pairs x, y that are solutions of the equation.
• Infinite limit
A special case of a limit that does not exist.
• Integers
The numbers . . ., -3, -2, -1, 0,1,2,...2
• Logistic regression
A procedure for fitting a logistic curve to a set of data
• Matrix element
Any of the real numbers in a matrix
• Random behavior
Behavior that is determined only by the laws of probability.
• Right angle
A 90° angle.
• Second
Angle measure equal to 1/60 of a minute.
• Singular matrix
A square matrix with zero determinant
• Trichotomy property
For real numbers a and b, exactly one of the following is true: a < b, a = b , or a > b.
• Zero of a function
A value in the domain of a function that makes the function value zero.
× | 731 | 3,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-50 | latest | en | 0.894921 |
http://en.wikipedia.org/wiki/Boomerang_attack | 1,438,145,975,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986022.41/warc/CC-MAIN-20150728002306-00133-ip-10-236-191-2.ec2.internal.warc.gz | 87,102,111 | 16,207 | # Boomerang attack
Boomerang attack
In cryptography, the boomerang attack is a method for the cryptanalysis of block ciphers based on differential cryptanalysis. The attack was published in 1999 by David Wagner, who used it to break the COCONUT98 cipher.
The boomerang attack has allowed new avenues of attack for many ciphers previously deemed safe from differential cryptanalysis.
Refinements on the boomerang attack have been published: the amplified boomerang attack, then the rectangle attack.
## The attack
The boomerang attack is based on differential cryptanalysis. In differential cryptanalysis, an attacker exploits how differences in the input to a cipher (the plaintext) can affect the resultant difference at the output (the ciphertext). A high-probability "differential" (that is, an input difference that will produce a likely output difference) is needed that covers all, or nearly all, of the cipher. The boomerang attack allows differentials to be used which cover only part of the cipher.
The attack attempts to generate a so-called "quartet" structure at a point halfway through the cipher. For this purpose, say that the encryption action, E, of the cipher can be split into two consecutive stages, E0 and E1, so that E(M) = E1(E0(M)), where M is some plaintext message. Suppose we have two differentials for the two stages; say,
$\Delta\to\Delta^*$
for E0, and
$\nabla\to\nabla^*$ for E1−1 (the decryption action of E1).
The basic attack proceeds as follows:
• Choose a random plaintext $P$ and calculate $P' = P \oplus \Delta$.
• Request the encryptions of $P$ and $P'$ to obtain $C = E(P)$ and $C' = E(P')$
• Calculate $D = C \oplus \nabla$ and $D' = C' \oplus \nabla$
• Request the decryptions of $D$ and $D'$ to obtain $Q = E^{-1}(D)$ and $Q' = E^{-1}(D')$
• Compare $Q$ and $Q'$; when the differentials hold, $Q \oplus Q' = \Delta$.
## Application to specific ciphers
One attack on KASUMI, a block cipher used in 3GPP, is a related-key rectangle attack which breaks the full eight rounds of the cipher faster than exhaustive search (Biham et al., 2005). The attack requires 254.6 chosen plaintexts, each of which has been encrypted under one of four related keys, and has a time complexity equivalent to 276.1 KASUMI encryptions. | 587 | 2,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2015-32 | latest | en | 0.909084 |
https://www.hackmath.net/en/examples/unit-conversion?page_num=13 | 1,560,848,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998708.41/warc/CC-MAIN-20190618083336-20190618105336-00425.warc.gz | 765,928,957 | 6,760 | # Unit conversion - examples - page 13
1. Basic form
Expressed the ratios of values in the basic form: 0,5 t : 1,2 kg 200 l : 0,15 m3 12 t : 3600 kg 500 kg : 2,5 t 0,9 kg : 500 g 3,6 m : 240 cm 1200 mm : 2,4 m 300 l : 0,3 m3 6 min 30 s : 900 s
2. Potatoes
When planting potatoes there are consumed 230 kg of seed per 1 ha. Calculate the weight of the seed needed for fitting a square area with a side length 325 meters.
3. Squares
Calculate the area of square and result round to square decimeters. a) a = 1,52 dm b) a = 13 268mm c) a = 562 cm d) a = 1,52 m
4. Uranium ore
Uranium ore contains 0.2% of pure uranium and pure uranium contains 0.07% of fissile uranium. How many grams of fissile uranium contains 2800 ton of uranium ore?
5. Land
Land has a rectangular shape, its surface area is 1.45 hectares. Its width is 250 m. Determine the length of the land.
6. Oak cuboid
Oak timber is rectangular shaped with dimensions of 2m, 30 cm and 15 cm. It weight is 70 kg. Calculate the weight 1 dm³ of timber.
7. Area
Calculate: ?
8. Plan scale
At what scale is drawn plan of the building, where one side of the building is 45 meters long is on the plan expressed by a straight line 12 mm long.
9. Prism
Calculate the height of the prism having a surface area 448.88 dm² wherein the base is square with a side of 6.2 dm. What will be its volume in hectoliters?
10. Percitipation - meteo
Do you have any tip for calc percitipation on a lnd ? Example on 2 ha falls to 5 mm, how many cubic meters is it?
11. Grass garden
Mr. Grass would itself be mowed garden for 5 hours. His son Michael could do it for 7 hours. They agreed that they wake up at 5 am and begins to mow together. Ms. Grass needs drive into town at 8.30. Michael says that at that time they're done with work..
12. Surface area
Calculate the surface area of a four-sides 2-m high prism which base is a rectangle with sides 17 cm and 1.3 dm
13. Water reservoir
The water tank has a cuboid with edges a= 1 m, b=2 m , c = 1 m. Calculate how many centimeters of water level falls, if we fill fifteen 12 liters cans.
14. Tank
The tank bottom has dimensions of 1.5 m and 3 2/6 m. The tank is 459.1 hl of water. How high is the water surface?
15. Wheel
Diameter of motocycle wheel is 52 cm. How many times rotates wheel on roand long 2 km?
16. Photo egative
Negative dimensions are 36mm and 28mm. What will be the photo size in the 21:4 ratio?
17. Prism
Find the volume and surface area of prism with base of an equilateral triangle with side 7 dm long and the body height of 1.5 m.
18. Tiles
How many square tiles with the content 121 cm2 has to be ordered for the paving of the square room with a side length of 2.75 meters?
19. Average speed
When the bus stops at bus stops driving average speed is 45 km/h. If it did not stop it drive at speed 54 km/h. How many minutes of every hour it spend at stops?
20. Sugar
From 3 tons of sugar beet was produced 480 kg of sugar. How many tons of sugar was produced from 17.5 tons of sugar beet?
Do you have an interesting mathematical example that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox. | 942 | 3,273 | {"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-2019-26 | latest | en | 0.847887 |
https://www.convertunits.com/from/minim+%5BUS%5D/to/attolitre | 1,620,261,986,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00466.warc.gz | 685,129,732 | 16,710 | ››Convert minim [US] to attoliter
minim [US] attolitre
Did you mean to convert minim [US] minim [UK] to attolitre
How many minim [US] in 1 attolitre? The answer is 1.6230730828281E-14.
We assume you are converting between minim [US] and attoliter.
You can view more details on each measurement unit:
minim [US] or attolitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 16230730.828281 minim [US], or 1.0E+21 attolitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between minim [US] and attoliters.
Type in your own numbers in the form to convert the units!
››Quick conversion chart of minim [US] to attolitre
1 minim [US] to attolitre = 61611520182292 attolitre
2 minim [US] to attolitre = 1.2322304036458E+14 attolitre
3 minim [US] to attolitre = 1.8483456054688E+14 attolitre
4 minim [US] to attolitre = 2.4644608072917E+14 attolitre
5 minim [US] to attolitre = 3.0805760091146E+14 attolitre
6 minim [US] to attolitre = 3.6966912109375E+14 attolitre
7 minim [US] to attolitre = 4.3128064127604E+14 attolitre
8 minim [US] to attolitre = 4.9289216145833E+14 attolitre
9 minim [US] to attolitre = 5.5450368164063E+14 attolitre
10 minim [US] to attolitre = 6.1611520182292E+14 attolitre
››Want other units?
You can do the reverse unit conversion from attolitre to minim [US], or enter any two units below:
Enter two units to convert
From: To:
››Definition: Attoliter
The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18.
So 1 attoliter = 10-18 liter.
››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 646 | 2,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-21 | latest | en | 0.717154 |
https://isabelle.in.tum.de/repos/isabelle/rev/26e4be6a680f | 1,639,008,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00613.warc.gz | 388,898,487 | 5,121 | author Manuel Eberl Sun, 28 Feb 2016 12:05:52 +0100 changeset 62442 26e4be6a680f parent 62441 e5e38e1f2dd4 child 62443 133f65ac17e5 child 62457 a3c7bd201da7
More efficient Extended Euclidean Algorithm
```--- a/src/HOL/Number_Theory/Euclidean_Algorithm.thy Sat Feb 27 21:09:43 2016 +0100
+++ b/src/HOL/Number_Theory/Euclidean_Algorithm.thy Sun Feb 28 12:05:52 2016 +0100
@@ -234,67 +234,89 @@
class euclidean_ring = euclidean_semiring + idom
begin
-function euclid_ext :: "'a \<Rightarrow> 'a \<Rightarrow> 'a \<times> 'a \<times> 'a" where
- "euclid_ext a b =
- (if b = 0 then
- (1 div unit_factor a, 0, normalize a)
- else
- case euclid_ext b (a mod b) of
- (s, t, c) \<Rightarrow> (t, s - t * (a div b), c))"
- by pat_completeness simp
-termination
- by (relation "measure (euclidean_size \<circ> snd)") (simp_all add: mod_size_less)
+function euclid_ext_aux :: "'a \<Rightarrow> _" where
+ "euclid_ext_aux r' r s' s t' t = (
+ if r = 0 then let c = 1 div unit_factor r' in (s' * c, t' * c, normalize r')
+ else let q = r' div r
+ in euclid_ext_aux r (r' mod r) s (s' - q * s) t (t' - q * t))"
+by auto
+termination by (relation "measure (\<lambda>(_,b,_,_,_,_). euclidean_size b)") (simp_all add: mod_size_less)
+
+declare euclid_ext_aux.simps [simp del]
-declare euclid_ext.simps [simp del]
+lemma euclid_ext_aux_correct:
+ assumes "gcd_eucl r' r = gcd_eucl x y"
+ assumes "s' * x + t' * y = r'"
+ assumes "s * x + t * y = r"
+ shows "case euclid_ext_aux r' r s' s t' t of (a,b,c) \<Rightarrow>
+ a * x + b * y = c \<and> c = gcd_eucl x y" (is "?P (euclid_ext_aux r' r s' s t' t)")
+using assms
+proof (induction r' r s' s t' t rule: euclid_ext_aux.induct)
+ case (1 r' r s' s t' t)
+ show ?case
+ proof (cases "r = 0")
+ case True
+ hence "euclid_ext_aux r' r s' s t' t =
+ (s' div unit_factor r', t' div unit_factor r', normalize r')"
+ by (subst euclid_ext_aux.simps) (simp add: Let_def)
+ also have "?P \<dots>"
+ proof safe
+ have "s' div unit_factor r' * x + t' div unit_factor r' * y =
+ (s' * x + t' * y) div unit_factor r'"
+ by (cases "r' = 0") (simp_all add: unit_div_commute)
+ also have "s' * x + t' * y = r'" by fact
+ also have "\<dots> div unit_factor r' = normalize r'" by simp
+ finally show "s' div unit_factor r' * x + t' div unit_factor r' * y = normalize r'" .
+ next
+ from "1.prems" True show "normalize r' = gcd_eucl x y" by (simp add: gcd_eucl_0)
+ qed
+ finally show ?thesis .
+ next
+ case False
+ hence "euclid_ext_aux r' r s' s t' t =
+ euclid_ext_aux r (r' mod r) s (s' - r' div r * s) t (t' - r' div r * t)"
+ by (subst euclid_ext_aux.simps) (simp add: Let_def)
+ also from "1.prems" False have "?P \<dots>"
+ proof (intro "1.IH")
+ have "(s' - r' div r * s) * x + (t' - r' div r * t) * y =
+ (s' * x + t' * y) - r' div r * (s * x + t * y)" by (simp add: algebra_simps)
+ also have "s' * x + t' * y = r'" by fact
+ also have "s * x + t * y = r" by fact
+ also have "r' - r' div r * r = r' mod r" using mod_div_equality[of r' r]
+ finally show "(s' - r' div r * s) * x + (t' - r' div r * t) * y = r' mod r" .
+ qed (auto simp: gcd_eucl_non_0 algebra_simps div_mod_equality')
+ finally show ?thesis .
+ qed
+qed
+
+definition euclid_ext where
+ "euclid_ext a b = euclid_ext_aux a b 1 0 0 1"
lemma euclid_ext_0:
"euclid_ext a 0 = (1 div unit_factor a, 0, normalize a)"
- by (simp add: euclid_ext.simps [of a 0])
+ by (simp add: euclid_ext_def euclid_ext_aux.simps)
lemma euclid_ext_left_0:
"euclid_ext 0 a = (0, 1 div unit_factor a, normalize a)"
- by (simp add: euclid_ext_0 euclid_ext.simps [of 0 a])
+ by (simp add: euclid_ext_def euclid_ext_aux.simps)
-lemma euclid_ext_non_0:
- "b \<noteq> 0 \<Longrightarrow> euclid_ext a b = (case euclid_ext b (a mod b) of
- (s, t, c) \<Rightarrow> (t, s - t * (a div b), c))"
- by (simp add: euclid_ext.simps [of a b] euclid_ext.simps [of b 0])
-
-lemma euclid_ext_code [code]:
- "euclid_ext a b = (if b = 0 then (1 div unit_factor a, 0, normalize a)
- else let (s, t, c) = euclid_ext b (a mod b) in (t, s - t * (a div b), c))"
- by (simp add: euclid_ext.simps [of a b] euclid_ext.simps [of b 0])
+lemma euclid_ext_correct':
+ "case euclid_ext x y of (a,b,c) \<Rightarrow> a * x + b * y = c \<and> c = gcd_eucl x y"
+ unfolding euclid_ext_def by (rule euclid_ext_aux_correct) simp_all
-lemma euclid_ext_correct:
- "case euclid_ext a b of (s, t, c) \<Rightarrow> s * a + t * b = c"
-proof (induct a b rule: gcd_eucl_induct)
- case (zero a) then show ?case
- by (simp add: euclid_ext_0 ac_simps)
-next
- case (mod a b)
- obtain s t c where stc: "euclid_ext b (a mod b) = (s,t,c)"
- by (cases "euclid_ext b (a mod b)") blast
- with mod have "c = s * b + t * (a mod b)" by simp
- also have "... = t * ((a div b) * b + a mod b) + (s - t * (a div b)) * b"
- also have "(a div b) * b + a mod b = a" using mod_div_equality .
- finally show ?case
- by (subst euclid_ext.simps) (simp add: stc mod ac_simps)
-qed
+definition euclid_ext' where
+ "euclid_ext' x y = (case euclid_ext x y of (a, b, _) \<Rightarrow> (a, b))"
-definition euclid_ext' :: "'a \<Rightarrow> 'a \<Rightarrow> 'a \<times> 'a"
-where
- "euclid_ext' a b = (case euclid_ext a b of (s, t, _) \<Rightarrow> (s, t))"
+lemma euclid_ext'_correct':
+ "case euclid_ext' x y of (a,b) \<Rightarrow> a * x + b * y = gcd_eucl x y"
+ using euclid_ext_correct'[of x y] by (simp add: case_prod_unfold euclid_ext'_def)
lemma euclid_ext'_0: "euclid_ext' a 0 = (1 div unit_factor a, 0)"
lemma euclid_ext'_left_0: "euclid_ext' 0 a = (0, 1 div unit_factor a)"
-
-lemma euclid_ext'_non_0: "b \<noteq> 0 \<Longrightarrow> euclid_ext' a b = (snd (euclid_ext' b (a mod b)),
- fst (euclid_ext' b (a mod b)) - snd (euclid_ext' b (a mod b)) * (a div b))"
- by (simp add: euclid_ext'_def euclid_ext_non_0 split_def)
end
@@ -412,21 +434,21 @@
lemma euclid_ext_gcd [simp]:
"(case euclid_ext a b of (_, _ , t) \<Rightarrow> t) = gcd a b"
- by (induct a b rule: gcd_eucl_induct)
- (simp_all add: euclid_ext_0 euclid_ext_non_0 ac_simps split: prod.split prod.split_asm)
+ using euclid_ext_correct'[of a b] by (simp add: case_prod_unfold Let_def gcd_gcd_eucl)
lemma euclid_ext_gcd' [simp]:
"euclid_ext a b = (r, s, t) \<Longrightarrow> t = gcd a b"
by (insert euclid_ext_gcd[of a b], drule (1) subst, simp)
+
+lemma euclid_ext_correct:
+ "case euclid_ext x y of (a,b,c) \<Rightarrow> a * x + b * y = c \<and> c = gcd x y"
+ using euclid_ext_correct'[of x y]
+ by (simp add: gcd_gcd_eucl case_prod_unfold)
lemma euclid_ext'_correct:
"fst (euclid_ext' a b) * a + snd (euclid_ext' a b) * b = gcd a b"
-proof-
- obtain s t c where "euclid_ext a b = (s,t,c)"
- by (cases "euclid_ext a b", blast)
- with euclid_ext_correct[of a b] euclid_ext_gcd[of a b]
- show ?thesis unfolding euclid_ext'_def by simp
-qed
+ using euclid_ext_correct'[of a b]
+ by (simp add: gcd_gcd_eucl case_prod_unfold euclid_ext'_def)
lemma bezout: "\<exists>s t. s * a + t * b = gcd a b"
using euclid_ext'_correct by blast``` | 2,650 | 7,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-49 | latest | en | 0.348888 |
https://discourse.mcneel.com/t/circle-with-constraints-center/40404 | 1,604,143,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00113.warc.gz | 283,304,285 | 7,485 | # Circle with constraints center
before i start doing binary search maybe somebody has a solution for this. i need a circle with a constraint center line, that means the center can be anywhere along this yellow line but to its maximum size regarding the corner below where it hits the boundary. the upper line is 99. something percent straight so rather negligible. the cirlce seen here is a crude approximation and was done by using the upper end and the line itself as a directional constraint… but like that i cant use any smarttrack tricks which help me connect at a local maximum. any ideas? i already tried all cirlcle tools with all circle tangents. but nothing stays in the center line and geometric translation also does not work due to the dynamic dimension of the circle.
well i solved it with 0.0025 mm off the boundary, thats good enough for now so no urge on this. if somebody still wants to scratch up a 100% solution i would be glad to listen. but i believe it would imply something like grasshopper or a scripted algorithm maybe. @Helvetosaur any thoughts? growing/scaling a circle along a line till it intersects with a curve?
hi Simon, yes tried that, it works ok but its not 100% precise.
it creates a areacentroid which is 0.05 mm next to the line where it should be.
its sure not much but not very assuring. i constructed it geometrically in the end
and got a much higher precision still not exact as i would say,
but anything with circles is always a bit hazy probably…
I do not see this difference. The center of the circle (with the settings in millimeters) I know match the vertical line. The difference between the point of the center and the line is 0:00 mm
maybe thats because you rebuilt it and used a straight line on top, mine is not 100% straight. here a cut out file of it, just select the point and use ZoomSelected or scroll into the point till you find the prepared line
CirclewithConstraints.3dm (2.6 MB)
you can even see when zooming in into the last tangent while constructing the circle that its off.
How about constructing it geometrically like this?
-Find the (magenta) line perpendicular to both the (extended) straight line (blue) at the top and the arc at the bottom
-Make a circle (cyan) tangent to both the top and bottom curves
-Make a parabola(red) by vertex and direction. Using the midpoint of the magenta line as the vertex, its direction, and the circle centre as the end point.
-The centre of the circle you are after must lie on this parabola, where it intersects your vertical line.
2 Likes
its a very interesting solution. the yellow one is your Areacentroid, while the green one is from my messings and the blue point is the rhino onboard solution with 3 tangents and 0.05 mm off which also Simon suggested which is unfortunately the least precise.
i had trouble constructing your suggestion at first since i never used parabola, so with a few fail attempts but following your exact description i managed. just my geometric-iterative solution gave me 2 intersection with the arc below and your solution has none but being only 0.0023 mm away from it. so both are not 100%
@DanielPiker maybe its because your method would work only when the upper line is straight? but since its slightly curved it would maybe need something like 2 parabola… but for me your solution is already completely alien and i would have never thought of that… but still no real solution afaics at least not if you really need it exact.
@RichardZ - Yes, the method I posted will only work when the line is straight (and when it is it should be exact).
I can see how it seems an odd approach at first, but I think of it like this - for a circle to be tangent to 2 curves, the distances from the circle centre to the closest point on each curve must be equal. Therefore the centre lies somewhere on the curve which is midway between the 2 curves. I remembered from Fortune’s algorithm for calculating Voronoi diagrams that this midway curve for a point and a line is a parabola (and this is true also for a circle and a line, or for 2 circles).
It might be possible to adjust this approach to work when the top curve is an arc instead of a straight line.
Also, if you don’t mind using Grasshopper/Kangaroo, here is a way of solving this numerically that will work with arbitrary curves.
tangent_to_curves.gh (9.2 KB)
1 Like
Hi Daniel, thats superb more than i could´ve wished for and certainly kicks it into the right direction. i assumed that this might end with something numeric and you delivered a solution which also will work on any kind of curve and not just arcs. thanks a bunch, now i have something to learn from for getting back into Hopper/Roo
still interesting though how your geometric solution might work in case of having two arcs one on top and one below which would imply using your parabola method but with two fokus points influencing a middle line something to think about maybe. | 1,096 | 4,949 | {"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-2020-45 | latest | en | 0.951966 |
https://it.scribd.com/doc/73441635/Bio-Process-Engineering-Principles-Solutions-Manual-P-Doran-1997-WW | 1,597,524,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00137.warc.gz | 339,894,753 | 123,193 | Sei sulla pagina 1di 170
SOLUTIONS MANUAL
Bioprocess Engineering Principles
Pauline M. Doran
tI !iI~'.!.~.!
oJ
4
if'
402·751 Rll!laIlN lit•••• til
TEl:1032l88O-7518 FAIl_
E-mail: I••co.tnha.ao.ki. -~ --
SOLUTIONS MANUAL
Bioprocess Engineering Principles
Pauline M. Doran
University of New South Wales, Sydney, Australia
ISBN 0 7334 15474
Page
Chapter 2 Introduction to Engineering Calculations 1 Chapter 3 Presentation and Analysis ofData 9 Chapter 4 Material Balances 17 ChapterS Energy Balances 41 Chapter 6 Unsteady-State Material and Energy Balances 54 Chapter 7 Fluid Flow and Mixing 76 Chapter 8 Heat Transfer 86 Chapter 9 Mass Transfer '98 Chapter 10 Unit Operations 106 Chapter 11 Homogeneous Reactions 122 Chapter 12 Heterogeneous Reactions 139 Chapter 13 Reactor Engineering 151
NOTE
All equations, tables, figures, page numbers, etc., mentioned in this manual refer to the textbook,
Bioprocess Engineering Principles.
Introduction to Engineering Calculations
2.1 Unit conversion
(a)
From Table A.9 (Appendix A): 1 cP::::: 1O~3 kg m- I k 1 1 m= lOOcrn
Therefore:
1.5 x10-6 cP ::::: 1.5 x 10-6 cP ,1 10 - 3 k t g ;-1 s-ll.ll~~mI= 1.5 x 10- 11 kg s-1 cm- 1
Answer: 1.5 x 10- 11 kg s-1 em- t
(b)
From Table A.S (Appendix A): 1 bp (British)::::: 42.41 Btu min-I Therefore:
(e)
From Table A.S (Appendix. A): 1 mmHg:= 1.316 x 10- 3 attn
From Table A.I (Appendix A): 1 ft = 0.3048 m
From Table A.7 (Appendix A): 11 atm =9.604 x 10- 2 Btu
From Table A8 (Appendix A):
Im=lOOcm
11= lOOOcm 3
Ih=60min
Therefore:
1 Btu min-I::::: 2.391 x
to- 2 metric horsepower
670mmHgft 3 = 670 mmHg ft3 .11.316X10- 3 atml.19.604X1O-2Btul.I°.3048m1 3
1 mmHg
llatm
2 .391 x 10- 2 metric horsepowerI I I h
1 1Btu min-
1
.
-60.
mm
1 it
I
956
= ,
Answer: 9.56 x 10-4 metric horsepower h
(d)
From Table A.7 (Appendix A): 1 Btu =0.2520 kcal
From Table A.3 (Appendix A): Ilb = 453.6 g
Therefore:
x
1 100 em 1 3
1 m
10-4
metric
.
1
11
lOOOcm3
I.
h
orsepower
h
345 Btu Ib- 1 = 345 Btulb- 1 .1o.2i~~Call·14;3~: g I = O.192kcal g-l
2.2 Unit conversion
Case 1
Convert to units of kg, m, s.
From Table AJ (Appendix A), lIb = 0.4536 kg
2
Solutions: Chapter 2
From Table A.2 (Appendix A): 1
From Table A.9 (Appendix A): 1 cP::: 10- 3 kg m- l s·l
1 rn= tOOcm= lOOOmm
Therefore, using Eq. (2.1):
tt3 =: 2.832 x 10- 2 m 3
Dup
Re::: -p-
=
(
3
2mm.1 1m n(3cms-l.l~n(251bfC3I0.4536kgl.1 Ift
l000mmU
l00cmU
I
.
10
-3
kgm
1cP
10 -6 P c
1
s
-11
lIb
2.832 x 10 2 m 3U _
~
- 2.4 x 10
ease 2
Convert to units of kg, m, s.
From Table Al (Appendix A): 1 in.::: 2.54 x 10- 2 m
From Table A.9 (Appendix A): 1
Ih=3600s
Therefore, usingEq. (2.1):
Ibm ft-I h- 1 ::: 4.134 x 10-4 kg m-t s·t
13~sl
=
1.5 x 10 4
7
2.3 Dimensionless groups and property data
From the Chemical Engineers' Handbook, the diffusivity of oxygen in water at 2S"C is 2.5 x 10- 5 cm 2 s·l. Assuming
this is the same at 28"C, !lJ= 2.5 x 10- 5 cm 1 s·t, Also, from the Chemical Engineers' Handbook, the density of water
at 28"e is PL ::: 0.9962652 g cm- 3 , and the viscosity of water at 28"C is JlL::: 0.87 cPo The density of oxygen at 28°C and 1 atm pressure can be calculated using the ideal gas law. As molar density is the same as n,V, from Eq. (2.32):
Temperature in the ideal gas equation is absolute temperature; therefore, from Eq. (2.24):
T = (28 + 273.15) K = 301.15 K
From Table 2.5, R "" 82.057 cm 3 atm K-I gmol~l. Substituting parameter values into the density equation gives:
Pa
""
L
RT
""
latm
(82.057cm 3 almK:""1 gmol 1)(301.15K)
"" 4.05 x 10- 5 gmolcm- 3
From the atomie weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. for Pa to mass tenns:
Converting the result
Pa "" 4.05 x lO- 5gm olem- 3 ·1 i~~~ll "" l.30XlO- 3 gcm- 3
From Table A.9 (Appendix A): 1 eP "" 10- 2 g cm~l s-l; from Table A.l (Appendix A): 1 ft "" 0.3048 m "" 30.48 cm. The parameter values and conversion factors, together with Db "" 2 mm "" 0.2 em, can now be used to calculate the dimensionless groups in the equation for the Sherwood number.
Solutions,' Chapter 2
3
Therefore:
S_J1.L_
c -
PrJ) -
0,87eP.
1'0
-2
-1 -11
s
gem
leP
-349
(0.9962652g cn,-3)(2.5 x 10 5 em 2 s I) -
From the equation for Sh:
k
_Sh:lJ_(ll.21(2.5xl0-5cm2s-I)_I·A
02
L----
D1J
-,<tUX
.
em
10-3
-I
ems
2.4 Mass and weight
From the definition of density on p 16, mass is equal to volume multiplied by density, Therefore:
From p 16, weight is-the force with wrncha body is attracted to the centre of the earth by gravity.
Newton's law(p 15), this force is equal to the mass of the body multiplied by the gravitational acceleration.
According to
(a)
From pIS, at sea level and 45° latitude. gravitational acceleration g = 32.174 it s-2. Therefore:
Weight:::: 624 Ibm (32.174 ft s-2) = 2.008x 104 Ibm it s·Z
Converting these units to lbj' from Eq. (2.16), 1lbf= 32,174 Ibm it s·2; therefore:
4
Weight = 2.008 x 10
lbmfts-
2
,
llbf
32.174Ib m ft s
2
.= 6241b f
Answer: 624lbI When g :::: 32.174 ft s·2, the number of lb mass is equal to the number of lb force,
(b)
From Table A.I (Appendix A): 1 m= 3.281 ft. Using the same procedure as in (a):
Weight = 624 Ibm (9.76 m s-Z .13.2~ft~= 1.998 x 10 4 Ibm fts- Z
Converting to Ibf 4 Weight = 1.998 x 10 lbmfts- 2 llbf = 6211bf . 32.1741b m ft s-2 Answer: 621lbf
2.5 Dimensionless numbers
First, evaluate the units of the groups cCP J.ll k ) and (D G1p):
(ep ")
Umtsof --
.
::::
(Bmlb-
I
'1'1) Ibh-
I
ft-
I
= 1
k BIUh-1 ft-2('Pft-1tl
UnilsOf(DG) = (ftllbh- 1 ft- 2 = 1
J1.
Ibh1ft1
4
Solutions: Chapter 2
Therefore, these groups are dimensionless. For the equation to be dimensionally homogeneous, (hIC p G) must also be
dimensionless; the units of h must therefore cancel the units of C p G.
Unitsofh "'" unitsofCpG "'" (Btulb- 1 "p-l)(1bh- 1 fr 2 ) "'" Btu "F-l h- 1 ft- 2
The dimensions of h can be deduced from its units.
From Table A.7 (Appendix A), Btu is a unit of energy with
dimensions "'" L2Ml2. OF is a unit of temperature which, from Table 2.1, has the dimensional symbol E>. h is a unit of time with dimension"'" T; ft is a unit of length with dimension"'" L. Therefore:
Dimensions of h = L2M'l2 a-I 1'"1 L-2 = MT" 3 e- 1
Answer: Units = Btu opi h- 1 ft-2; dimensions =Ml 3 e- 1
2.6 Dimensional homogeneity and Cc
From Table A8 (Appendix A), dimensions of P = L2MT"3 Dimensions of g = LT-z Dimensions of p =ML·3
Dimensions of Di "'" L
From p 11, the dimensions oirotational speed, Nj =T-!; from p 15, the dimensions of gc= 1. Therefore:
As Np is a dimensionless number, equation (i) is not dimensionally homogeneous and therefore cannot be correct.
Equation (ii) is dimensionally homogeneous and therefore likely to be correct
2.7 Molar units
From the atomic weights in Table Rl (Appendix B), the molecular weight of NaOH is 40.0.
(a)
FromEq. (2.19):
lb moles NaOH =
20.0 lb
4O.0lblbmol 1
= 050 lbmol
(b)
From Table A.3 (Appendix A): lIb =453.6 g. Therefore:
From Eq. (2.18):
20.01b = 20.0Ib. -lib 453.6gl
1
= 9072 g
gram moles NaOH =
9072g
4O.0g gmol-
I = 227 grool
(c)
From p 16, 1 kgmol::::: 1000 gmot. Therefore, from (b):
.-----
Solutions: Chapter 2
5
kg molesNaOR = 227 gmol.
1
1000 1 kgmol gmol
I
= 0.227 kgmol
2.8 Density and specific gravity
(a)
From p 16, the density of water at 4°C can be taken as exactly 1 g cm- 3 . Therefore, for a substance with specific gravity L5129i~,the density at 20°C is 1.5129 g cm*3,
(I)
lkg=l000g
1 m: 100cm Therefore:
(il)
From the atomic weights in Table 8.1 (Appendix B), the molecular weight of nitric acid (RN03) is 63.0. RNO" from Eq. (2.18)0
gram moles :
LS129g
1
:
0.0240 gmot
63.0ggmor
In 1 cm 3
Therefore, the molar density is 0.0240 gmol cm- 3 . From the definition of specific volume on p 16:
Molar specific volume
:
1
mo ar
d 1
.
:
enslty
.,-
0.0240 gmol em 3
,,-1-:--:::;- :
41.67cm 3 gmor 1
(b)
(I)
From p 16, as density is defined as the mass per unit volume, the mass flow rate is equal to the volumetric flow rate multiplied by the density:
(ii)
From the atomic weights in Table B.l (Appendix B), the molecular weight of carbon tetrachloride, CC14, is 153.8. Using the mass flow rate from (a):
Molar flow rate :
80 g min- l .l :5~0~1 = 0.52gmolmin- 1
2.9 Molecular weight
From p 17, the composition of air is close to 21 % oxygen and 79% nitrogen. For gases at low pressures, this means 21 mol% 02 and 79 mol% NZ. Therefore, in 1 gmol air, there are 0.21 gmot Oz and 0.79 gmol NZ From the atomic weights in Table B.l (Appendix B), the molecular weights of Oz and NZ are 3Z.0 and 28.0, respectively. The molecular weight of air is equal to the number of grams in 1 gmol:
1 gmolair = 0.21 gmOlOz·1 ;~~ll+ 0.79 gmol NZ ·1 ;~;ll= 28.8g
6
Solutions.' Chapter 2
2.10 Mole fraction
The molecular weights can be obtained from Table B.7 (Appendix B): water 18.0; ethanol 46.1; methanol 32.0;
glycerol 92.1; acetic acid 60.1; benzaldehyde 106.1.
In 100 g solution, there are 30 g water, 25 g ethanol. 15 g
methanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. Therefore:
Moles water = 30 g _I ~r:;~ll= 1.67 gruol
Moles ethanol:: 25 g.1 ~~~ll :: 054 gmol
Molesmethanol:: 15 g _I ;~~ll = 0.47 gmol
Molesglycerol = 12g.
.
IIgmaII
92.1g
= O.13gmol
Moles acetic acid = 10g.1 ~~~II= 0.17 gmol
Moles benzaldehyde: 8 g _I :~o~I = 0.08 gmal
The total number of moles is 1.67 + 0.54 + 0.47 + 0.13 + 0.17 + 0.08:: 3.06 gmal. From Eq. (2.20):
Mole fraction water = ;:~~= 0.55
Mole fraction ethanol = ~:~= 0.18
Mole fraction methanol = ~:~~= 0.15
Mole fraction glycerol = ~:~ = 0.04
Mole fraction acetic acid = ~:~::::: 0.06
Mole fraction benzaldehyde ::::: ~:: ::::: 0.03
Answer: 0.55 water; 0.18 ethanol; 0.15 methanol; 0.04 glycerol; 0.06 acetic acid; 0.03 benzaldehyde
2.11 Temperature scales
From Eq. (2.27\
From Eq. (2.25),
-40 ::::: 1.8 T(0C) + 32
Tee) = -40
T (OR) ::::: -40 + 459.67
T(°R) = 420
From Eq. (2.24) and the result for T ("C);
T(K) = -40+273.15
T(K) = 233
2.12 Pressure scales
(a)
Assume that the atmospheric pressure is 14.7 psi. From Eq. (2.28):
Absolute pressure ::::: 15 psi + 14.7 psi::::: 29,7 psi
Solutions: Chapter 2
7
From Table A.5 (Appendix A): 1 psi = 6.805 x Hy2 atm. Therefore:
Absolutepressure = 29.7psl.
.
16.805 x 1 psi 10-
2 atm 1
= 2.02atm
(b)
From p 19, vacuum pressure is the pressure below atmospheric. If the atmospheric pressure is 14.7 psi:
Absolutepressure = 14.7psi-3psi = IL7psi
2.13 Stoichiometry and incomplete reaction
(a)
The molecular weights are calculated from TableB.l (Appendix B): penicillin = 334.;4; glucose = 180.2. The maximum theoretical yield from the stoichiometric equation is 1 gruol of penicillin for every 1.67 gruol of glucose. 'This is equivalent to 334.4 g penicillin per 1.67 x 180.2 = 300.9 g glucose. or 1.1 g g-l.
(b)
The maximum theoretical yield in (a) is obtained when all the glucose consumed is directed into penicillin production according to the stoichiometric equation. If only 6% of the glucose is used in this way, the actual yield of penicillin from glucose is much lower, at 334.4 g penicillin per (300.9 x 100/ 6 ) g glucose, or 0.067 g g~l.
(c)
From the atomic weights in Table B.I (Appendix B), the molecular weight of phenylacetic acid is 136.2.
(I)
The only possible limiting substrates are glucose and phenylacetic acid. Using a basis of II medium, if (50 - 5.5) = 44.5 g glucose are consumed but only 6% is available for penicillin synthesis, the mass of glucose used in the penicillin reaction is 44.5 x 6/ 100 =2.67 g. 'This is equivalent to 2.67 g/180.2 g gmol-1 =1.48 K 10- 2 gmol glucose available for penicillin synthesis. At the same time, 4 g or 4 g/136.2 g gmol-1 =2.94 x 10~2gruol phenylacetic acid is available which, according to the stoichiometric equation, requires 1.67 x 2.94 x 10- 2 =4.91 x 10- 2 gruol glucose for complete reaction. As the gmol glucose required is greater than the gmol glucose available after growth and maintenance activities, glucose is the limiting substrate.
(il)
Of the 44.5 g I-I glucose consumed, 24% or 10.7 g I-I is used for growth. In a H~)-litretank. the total mass of glucose consumed for growth is therefore 1070 g or 1.07 kg.
(iii)
From (i), 1.48 x 10- 2 gmol glucose is used in the penicillin reaction per litre. According to the stoichiometry, this produces 1.48 x 10- 2 /1.67 = 8.86 x 1O~3gmol penicillin per litre. Therefore, in a l00-litre tank, 0.886 gmol or 0.886 gmol x 334.4 g gmol-1 =296 g penicillin are formed.
(iv)
IT, from (i), 1.48 x 10- 2 gmol [1 glucose is used in the penicillin reaction, 1.48 x 10- 2 /1.67 = 8.86 x 10- 3 gmoll-l phenylacetic acid must also be used. This is equivalent to 8.86 x 10- 3 gmoll-l x 136.2 g gmol-1 = 1.21 g t I phenylacetic acid. As 4 g I-I are provided, (4 - 1.21) = 2.79 g I~Iphenylacetic acid must remain.
8
Solutions: Chapter 2
2.14 Stoichiometry, yield and the ideal gas law
(a)
Adding up the numbers of C, H, 0 and N atoms on both sides of the equation shows that the equation is balanced.
(b)
The molecular weights are calculated from Table B.I (Appendix B).
Cells: 91.5
Hexadecane: 226.4 From the stoichiometry, as 1 gmol of hexadecane is required to produce 1.65 gmol of cells, the maximum yield is 1.65 gmol x 91.5 g gmol-l =: 151 g cells per 226A g hexadecane, or 0.67 g g-I,
(e)
From the atomic weights in Table RI (Appendix B), the molecular weight of oxygen is 32.0. From the stoichiometry, 16.28 gmol of oxygen is required to produce 1.65 gmal of cells which, from (b), is equal to 151 g cells.
The maximum yield is therefore 151 g cells per (16.28 groal x 32.0 g groot-I) =: 521 g oxygen, or 0.29 g g-1.
(d)
Production of 2.5 kg cells is equivalent to 2500 g =: 2500 g/91.5 g gmoI-l = 27.3 gmol cells. The minimum amounts of substrates are required when 100% of the hexadecane is converted according to the stoichiometric equation.
(I)
From the stoichiometry, production of 27.3 gmol cells requires 27.3/1.65 =16.5 gmol =16.5 gmol x 226.4 g gmol-I = 3736 g = 3.74 kg hexadecane.
(il) From the answer in (d)(i), the concentration ofhexadecane required is 3.74 kg in 3 m 3 , or 1.25 kg m- 3 .
(ill)
According to the stoichiometric equation, production of 27.3 gmol cells requires 27.3 x 16.28/1.65 =269.4 gmol oxygen. As air at low pressure contains close to 21 mol% oxygen (p 11), the total moles of air required is 269.410.21 = 1282.9 gmot The volume of air required can be calculated using the ideal gas law. From Eq. (2.32):
V = nRT
p
Temperature in the ideal gas equation is absolute temperature; from Eq. (2.24):
T = (20 +273.15) K = 293.15K
From Table 2.5, R "" 82.057 cm 3 atm K-I gmol-I. Substituting these values into the equation for V gives:
v= (1282.9gmolj(82.057cm 3 atmK- 1 gmol-l)(293.15Kj 1 atm
.
1-.!
"'
1
l00cm
3 = 31
3
m
Presentation and Analysis of Data
3.1 Combination of errors
c~ "" 0.25 mol m- 3
±4%"= 0.25 ±O.OlOmol m- 3
CAL
= 0.183
OTR = O.Qll
mol
mol
m- 3
m- 3
± 4% =: O.183±O.OO73molm- 3
s-1 ± 5%
For subtraction, absolute errors are added. TherefOre:
C~ -CAL"'" (O.25-0.183)±{O.OlO+0.0073)molm- 3 "'" O.067±O.0173molm- 3 =: 0.067 molm- 3 ± 25.8%
For division, relative errors are added. Therefore:
kLa =: o.OllmOlm-3~-1 ±{25.8 + 5)% =O.16s- 1 ±31% = O.16± 0.05 s-l
O.Q67molm
This example illustrates how a combination of small measurement errors can result in a relatively
large uncertainty in the final result.
3.2 Mean and standard deviation
(a)
The best estimate is the mean, X. FromEq. (3.1):
x = 5.15+5.45+5.50+5.35 = 5.36
(b)
Calculate the standard deviation from Eq. (3.2):
(5.15 - 5.36)' + (5.45 - 5.36)' + (5.50- 5.36)' + (5.35 - 5.36)'
4-1
= 0.15
Answer. The standard deviation is 0.15. Note that standard error, which can be calculated from the standard deviation, is a more direct indication of the precision of a mean.
(c)
x = 5.15 +5.45 = 5.30
2
Standard deviation is not appropriate for expressing the accuracy of a mean evaluated using only two samples. In this case the maximum error, Le. the difference between the mean and either of the two measured values, might be used instead. The maximum error in this example is (5.30 - 5.15) =0.15.
Answer. 5.30; an indication of the accuracy is ± 0.15
(d)
x= 5.15+5.45+5.50+5.35+5.15+5.45+5.50+5.35 = 5.36
10
Solutions: Chapter 3
2 (S.lS -S.36f + 2(S.4S-S.36)~+ ~(S.50-S.36f +2(S.3S -S.36)2
= 0.14
Answer: The best estimate of optimal pH is unchanged at 5.36, but the standard deviation is slightly lower at 0.14.
This example illustrates that although the standard deviation decreases as the number of measurements is increased, (j is not strongly dependent on n. The best way to improve the reliability of the mean is to ensure that the individual measurements are as intrinsically accurate as possible, rather thanrepeat the measurement many times.
3.3 Linear and non-linear models
(a)
Xl = 1; Yl =10
X2=8;Y2=0.5
A straight line plot of y versus x on linear coordinates means that the data can be represented using Eq. (3.6).
Eqs (3.7) and (3.8),
A = (Yz-Y1) = 0.5-10 =-136
Answer: y = -1.36 x + 11.4
(h) Xl = 3.2; Yl =14.5
(x2
Xl)
8-1
.
B = YI-Axi
= 10-(-1.36)1 = 11.4
From
);2 = 8.9; Y2 = 38.5
A straight line plot of y versus x Ih on linear coordinates means that the data can be represented using the equation:
with A and B given by the equations:
A
=
Y=Axlh+B
YrY1
112_
x2
112
Xl
= 38.5-14.S = 201 89Ih_32'k . . ,
B = YI-Axi h = 14.5-20.1(3.2 112 ) = -21.5
(0)
'k
-21.5
Xl=5;Yl=6
X2= l;Y2=3
A straight line plot of Ity versus xl on linear coordinates means that the data can be represented using the equation:
with A and B given by the equations:
A
lly = Ax 2 +B
= lin - Ity}
2
2
x2-xl
1/3 -lI6
= 2 2 =-6.9xlO 1 -5
-3
B = 1/ y1 -Ax; = 1/6-(-Mx 1O- 3 )<S2) = 0.34
Answer: l/y = -6.9 x 10- 3 xl + 0.34
(d)
Xl=0.5;Yl=25
x2 =550; Y2 =2600
A straight line plot of y versus x on log-log coordinates means that the data can be represented using Eq, (3.10).
From Eqs (3.13) and (3.14),
Solutions: Chapter 3
11
A = (lnY2- ln Yl) = ln2600-ln25 = 0.663
(lnx2-lnxI)
ln550-1nO.5
lnB = InYI-Alnxl =ln25-(0.663}ln0.5 = 3.678
Answer: Y= 39.6 x O .
663
B = e 3 . 678
= 39.6
(eJ
Xl = 1.5; YI = 2.5 X2 = 10; Y2 = 0.036
A straight line' plot of y,versus X on semi·log coordinates means that the data can be represented using Eq. (3.15).
From Eqs (3.17) and (3.18):
A = (lnY2- ln YI) = lnO.036-ln2.5 =-0.50
(x2 - xl)
10 -
1.5
In B = In Y1 - A Xl = In 2.5 - (-o.50J 1.5 = 1.666 B = el. 666 = 5.29
3.4 Linear curve fitting
(aJ
The results determined using Eqs (3.1) and (3.2) are listed below.
Sucrose concentration (g l~l) Mean peak area Standard deviation 6.0 56.84 1.21 12.0 112.82 2.06 18.0 170.63 2.54 24.0 232.74 1.80 30.0 302.04 2.21
(bJ
35
30
-
-
.
25
.9
0
••
20
~
g
15
8
~
10
g
"'
5
0
0
50
100
150
200
250
300
350
Peak area
12
Solutions: Chapter 3
(e)
The linear least squares fit of the data is:
y ::: 0,098 x + 0,83
Answer: y "'" 0.098 x + 0.83, where y is sucrose concentration in g 1-1 and x is peak area.
(d)
For x::: 209,86, the equation in (c) gives a sucrose concentration oi2IA g I-I.
3.5 Non-linear model: calcnlation of parameters
(a)
The proposed model equation has the general form of Eq. (3.15); therefore, if the model is suitable, a plot of a versus
lIT on semi-logarithmic coordinates will give a straight line.
As T in the equation is absolute temperature, "'c must
first be converted to degrees Kelvin using Eq. (2.24). The data are listed and plotted below.
Temperature (0C) Temperature (K) IIT(K-l) Relative mutation frequency. a 15 288.15 3.47 x 10- 3 4.4 x 10- 15 20 293.15 3.41 x lO w3 2.0x 10- 14 25 298.15 3.35 x leT 3 8.6x 10 14 30 303.15 3.30 x lfr·3 3.5 x Jer I3 35 308.15 3.25 x 10- 3 lAx }0"'12
10- 11 r---~--.,
--~----r--~---.
"
[
10- 12
"
f
c
10- 13
.Q
lii
"E
~
10. 14
= -ll
lJ:
10-'5 L
l
J
~~_--'
~
~
3,2 x 10-3
3.3x 10-3
3.4x 10-
3 3.5 x 10-3
1ITemperature (K-1)
As the data give a straight line on semi~logarithmiccoordinates, the model can be considered to fit the data well.
(b)
The equation for the straight line in (a) is:
y =: 9.66 x 10 24 e- 26 ,12lx
where y is relative mutation frequency and x is reciprocal temperature in units of K~l. For dimensional homogeneity the exponent must be dimensionless (p 12), so that -26,121 has units of K, and EIR in the model equation is equal to 26,121 K From Table 2.5, R =: 8.3144 J gmol-l K·1; therefore:
E =: (26,121 K) (8.3144 J gmol~l K*l) =: 217,180.4 J gmol-1 =: 217.2 kJ gmot 1
Solutions: Chapter 3
13
(c)
From the equation in (b) for the straight line, ao is equal to 9.66 x 10Z4.
3.6 Linear regression: distribution of residuals
(aJ
16
-
-
,
14
• •
!'!l
c
••
12
i'!
C
10
g
0
0
8
~
• • •
6
~
••
••
4
m
~
2
.i1
0
0
12345
6
Decrease in medium conductivity (mS cm- 1 )
The linear least squares fit of the data is:
y =: 1.58 + 2.10x
where yis increase in biomass concentration in g r 1 and xis decrease in medium conductivity in mS cm~l.
(b)
The residuals are calculated as the difference between the measured values for increase in biomass concentration and the values for y obtained from the equation in (a).
Decrease in medium conductivity (mS
Residual
o -1.58 0.12 0.57 0.31 -!l.23 0.41 0.36 0.82 1.20 1.03 1.36 lAO 1.28 1.91 0041 2.11 0.19 2.42 -0.46 2.44 -!l.50 2.74 -!l.73 2.91 -1.69 3.53 -1.99 4.39 -1.00 5.21 1.48 5.24 0.02 5.55 1.37
14
Solutions: Chapter 3
These results are plotted below.
3.-_-,
.,-_-,
.,
,-_-,
2
1
·2
·3 '-_-'-
-'-_---'
'-_-'
o 1
2
3
4
5
6
Decrease in medium conductivity (mS em- 1 )
The residuals are not randomly distributed: they are mainly positive at low values of decrease in medium
conductivity, then negative, then positive again. Therefore, the straight line fit of the data cannot be considered a very
good one.
3.7 Discriminating between rival models
(aJ
The results are plotted using linear coordinates below.
0.11
0.10
~ '"
.§.
os'
0.09
.~
0.08
~
." " 'E
0.07
1l.
,
0.06
~
~
0.05
::J
0.04
0.00
0.02
0.04
0.06
0.08
0.10
Gas superficial velocity, uG (m s-1)
The data are reasonably well fitted using a linear model. The linear least squares equation for the straight line fit is:
y = 0.054 + 0.466 x
where y is liquid superficial velocity in m s-l andx is gas superficial velocity in m s·l.
The sum of the squares of the
residuals between the measured values for liquid superficial velocity and the values for y obtained from the above equation is 8.4 x 10- 5 ,
Solutions: Chapter 3
15
(b)
The proposed power law equation has the same form as Eq. (3.10). Therefore, if the power law model is suitable. the data should give a straight line when plotted on log-log coordinates.
0.01
om
0.1
Gas superficial velocity, u G (m s·1)
The data are reasonably well fitted using a power law- modeL The equation for the straight line in the plot is:
y
= 0.199 x O
.
309
where y is liquid superficial velocity in m s·l andx is gas superficial velocity in m s·l. The sum of the squares of the residuals between the measured values for liquid superficial velocity and the values obtained from the above equation is 4.2 x 10- 5 .
«)
The non·linear model is better because the sum of squares of the residuals is smaller.
3.8 Non-linear model: calculation of parameters
(a), (b) The proposed model equation has the same form as Eq; (3.15). Therefore, if the model is suitable, the data should give a straight line when plotted on semi-logarithmic coordinates.
10' .,.--.,
-
,--,----,----,,--
---.
10 3
10 1
100 '--
1-_-'-_-'-_-'-_-L_----'-_---'
o
5
10
15
20
25
30
35
Time (min)
16
Solutions: Chapter 3
(c), (d)
The equation for the straight line in the figure is:
y =. 2.13 x 104 e-O.353x
where y is the number of viable cells and x is time in min.
As the exponent must be dimensionless to preserve
dimensional homogeneity (p 12), the dimensions of kd are T-t; therefore k(! =. 0353 min-I, The dimensions of No are the same as N, i.e, No is dimensionless and equal to 2.13 x 104.
Answer: kd =. 0.353 min-I. No =. 2.13 x 104; the dimensions of ka. are T"l, No is dimensionless
Material Balances
4.1 Cell concentration using membranes L Assemble (i) Flow sheet
Buffer solution in
80 kg min- 1
Cell concentrate
6% bacteria
I I I I I Hollow-fibre membranes I Buffer solution out I I Fennentation broth I I 350 kg min- 1 1% bacteria 99% water
(li) System boundary
The system boundary is shown on the flow sheet.
(iii) Reaction equation
No reaction occurs.
2. Analyse (i) Assumptions
-no leaks
- only water passes across the membrane
(ii) Extra data
No extra data are required.
(ill) Basis
1 min, or 350 kg fennentation broth
(iv) Compounds involved in reaction
No compounds are involved in reaction.
(v) Mass-balance equation
As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):
3.
Calculate
mass in = mass out
(i) Calculation table
The calculation table below shows all given quantities in kg. The total mass of cell concentrate is denoted c; the total mass of buffer out is denoted B. The columns for water refer to water originating in the fermentation broth.
18
Solutions: Chapter 4
In
Out
Stream Water Bacteria Buffer Total Water Bacteria Buffer Total Fermentation 346.5 3.5 0 350 broth Buffer solution in Cell concentrate 0 0 80 80 ? O.06C 0 C Buffer solution ? 0 80 B out Total 346.5 3.5 80 430 ? 0.06C 80 C+B
(li) Mass-balance calculations
Bacteria balance 3.5 kg bacteria in "'" 0.06 C kg bacteria out C = 58.3 kg Total mass balance
430 kg total mass in = (C + B) kg total mass out
Using the result for C:
Water balance
B
"'" 371.7 kg
346.5 kg water in = water out
Water out = 346.5 kg
These calculations allow completion of the mass~balance table with all quantities in kg.
Stream In Out Water Bacteria 814fer Total Water Bacteria Buffer Total Fennentation 346.5 3.5 0 350 broth Buffer solution in 0 0 80 80 Cell concentrate 54.8 3.50 0 58.3 Buffer solution 291.7 0 80 371.7 out Total 346.5 3.5 80 430 346.5 3.50 80 430
(iii) Check the results
All columns and rows of the completed table add up correctly.
4.
(a)
After rounding to three significant figures, the total flow rate of buffer solution out of the annulus is 372 kg min-I.
Finalise
(b)
The total flow rate of cell concentrate from the membrane tubes is 58.3 kg min-I.
Solutions: Chapter 4
19
4.2 Membrane reactor
t.
Assemble Flow sheet s ystem boundary~ Feed - -- - -- - 1 40 kg h- 1 I I Aqueous residue 10% glucose I I 0.2% glucose 90% water I Membrane system I 0.5% ethanol Solvent I I Product 40 kg h- 1 I I
(i)
(li)
System boundary
The system boundary is shown on the flow sheet.
(iii) Reaction equation 2. Analyse (i) Assumptions
-no leaks
- yeast cells do not grow or dislodge from the membrane
- no evaporation
- all C02 produced leaves in the off~gas
- no side reactions
(li) Extra data
Molecular weights (Table B.t, Appendix B): glucose = 180.2 ethanol =. 46.1
(iii) Basis
1 h, or 40 kg feed solution
CO, =44.0
(iv) Compounds involved in reaction
Glucose, ethanol and carbon dioxide are involved in the reaction.
(v)
Mass~balanceequations
For glucose, ethanol and carbon dioxide, the appropriate mass-balance equation is Eq. (4.2):
mass in + mass generated =. mass out + mass consumed
For water, solvent and total mass, the appropriate mass-balance equation is Eq. (4.3):
3. Calculate (i) Calculation table
mass in =. mass out
The calculation table below shows all given quantities in kg. The total mass of aqueous residue is denoted R; the total mass of product out is denoted p; the total mass of carbon dioxide out is denoted G.
20
Solutions: Chapter 4
Stream
In
Out
Glucose Ethanol CO, Solvent H,O Total Glucose Ethotwl CO, 4 0 0 0 36 40 0 0 0 40 0 40 0_002R 0.005 R 0 0 ? 0 0 0 ? 4 0 0 40 36 80 0.002 R ? ?
Solvent
H2 O
Total
-------
Feed
Solvent
Aqueous
residue
Product
Off-gas
0 ? R ? 0 P 0 0 G ? ? R+P
Total (ii) Mass-balance calculations
Solvent balance
Solvent is a tie component.
40 kg solvent in ::::: solvent out
Solvent out ::::: 40 kg
Water balance
Water is a tie component.
36 kg water in ::::: water out
Water out::::: 36 kg
As water appears on the Out side of the table only in the aqueous residue stream:
0.002 R + 0.005 R + 36 kg
:::::
R
R = 36.254 kg
Therefore, the residual glucose in the aqueous residue stream.::::: 0.002 R ::::: 0.073 kg; the ethanol in the aqueous residue
stream::::: 0.005 R::::: 0.181 kg.
Glucose balance
4 kg glucose in + 0 kg glucose generated ::::: 0.073 kg glucose out + glucose consumed
Glucose consumed ::::: 3.927 kg
Converting the glucose consumed to molar terms:
3.927 kg glucose ::::: 3.927kg.
1
1 kgmOll
180.2kg
::::: O.0218kgmol
From the reaction stoichiometry, conversion of this amount of glucose generates 2 X 0.0218 = 0.0436 kgmol ethanol and 2 x 0.0218 = 0.0436 kgmol COZ. Converting these molar quantities to mass:
0.0436kgmolethanol = 0.0436kgmOI·1 iZ:~~11= 2.010kg
0.0436kgm o1CO z = OJM36kgmOLI ~:~ll= 1.92kg
C02 balance
okg COZ in + 1.92 kg C02 generated = C02 out + 0 kg C02 consumed
Ethanol balance
C02 out = 1.92 kg
= G
okg ethanol in + 2.010 kg ethanol generated = ethanol out + 0 kg ethanol consumed
Ethanol out = 2.010 kg
Ethanol leaves the system only in the product and aqueous residue streams. Therefore:
Solutions: Chapter 4
21
Ethanol out in the product stream = (2.010 - 0.181) kg = 1.829 kg
As the product stream consists of ethanol and solvent only:
P = (1.829 + 40) kg = 41.829 kg
These calculations allow completion of the mass~balancetable with all quantities in kg.
Stream In Out Glucose Ethanol CO, Solvent H,O Total Glucose Ethanol CO2 Solvent H,O Total Feed 4 0 0 0 36 40 Solvent 0 0 0 40 0 40 Aqueous - 0.073 0.181 0 0 36 36.254 residue Product 0 1.829 0 40 0 41.829· Off-gas 0 0 1.92 0 0 1.92 Total 4 0 0 40 36 80 0.073 2.010 1.92 40 36 80.00 (iii) Check the results
All columns and rows of the completed table add up correctly.
4.
(0)
1.829 kg ethanol are contained in 41.829 kg of product stream. The ethanol concentration is therefore 1.829/41.829 x
Finalise
100%:::: 4.4%.
(h)
The mass flow rate ofCOz is 1.92 kg b- 1 .
4.3 Ethanol distillation 1. Assemble (i) Flow sheet
System boUnda'l'\.
-t---,
r I I I I I I I Distillation I Feed column I 50,000 kg h-l I 10%e thanal I I 90%w atar I I I I I I
- - -+- - -
Distillate 5,000 kg h- 1 45% ethanol 55% water
Bottoms
22
SolutilJns: Chapter 4
(li) System boundary
The system boundary is shown on the flow sheet.
(ill) Reaction equation
No reaction occurs.
2. Analyse
(i) Assumptions
-no leaks
(ll) Extra data
No extra data are required.
(ill) Basis
1 h, or 50,000 kg feed
(Iv) Compounds involved in reaction
No compounds are involved in reaction.
(v) Mass¥balance equation
As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):
mass in ::::: mass out
3. Calculate
(i) Calculation table
The calculation table shows all given quantities in kg.
Stream In Out Ethanol Water Total Ethanol Water Total Feed 5,000 45,000 50,000 Distillate 2,250 2,750 5,000 Bottoms ? ? ? Total 5,000 45,000 50,000 ? ? ? (li) Mass-balance calculations
Total mass balance
50,000 kg total mass in ::::: total mass out
Total mass out ::::: 50,000 kg
Therefore, from the total column on the Out side of the table:
Bottoms out::::: (50,000 - 5,000) kg ::::: 45,000 kg
Ethanol balance
5,000 kg ethanol in ::::: ethanol out
Ethanol out ::::: 5,000 kg
From the ethanol column of the Out side of the table:
Water balance
Ethanol out in the bottoms := (5,000
, 2,250) kg := 2,750 kg
45,000 kg water in := water out
Water out := 45,000 kg
From the water column of the Out side of the table:
Water out in the bottoms == (45,000 - 2,750) kg == 42,250 kg
These calculations allow completion of the mass~balancetable with all quantities in kg.
Solutions: Chapter4
23
Stream
In
Out
Water
Water
Ethanol
Total
Ethanol
Total
Feed
Distillate
Bottoms
Total
5,000
5,000
45,000
45,000
50,000
50,000
2,250
2,750
5,000
2,750
42,250
45,000
5,000
45,000
50,000
(ill) Check the results
All columns and rows of the completed table add up correctly.
4.
(a)
The bottoms contains 2,750 kg ctha,")l and 42,250 kg water in a total of 45,000 kg. Therefore, the composition is
2,750/45,000 x 100% = 6.1 % ethanol. and 42,250/45,000 x 100% = 93.9% water.
Finalise
(b)
Directly from the table, the rate of alcohol loss in the bottoms is 2,750 kg h- I .
4.4 Removal of glucose from dried egg
1.
(i)
Assemble
Flow sheet
Off-gas
System boundary
r\.
-------,
I
1
I
I
1
Egg slurry -----L.o-l
Enzyme reactor
I--.L1---_- Product
3000 kg h· 1
I
I
0.2% glucose
2% glucose
20% water
78% egg solids
I
1
I
--t--
I
Inlet air
18 kg h- 1 02
(n)
System boundary
The system boundary is shown on the flow sheet
(ill) Reaction equation 2. Analyse (i) Assumptions
- air and off-gas are dry
- gases are at low pressure so vol% = mol%
- H202 remains in the liquid phase
24
Solutions: Chapter 4
(ti)
Extra data
Molecular weights (Table B.I, A.ppendix B):
glucose::::: 180.2
02=32.0
NZ::::: 28.0
H20 = IS.0
gluconic acid::::: 196.2
HZOz::::: 34.0
Composition of air (p 17): 21% 02, 79% Nz by volume
(ill)
Basis
1 h, or 3000 kg egg slurry
(iv) Compounds involved in reaction
Glucose, 02, water, gluconic acid and HzOz are involved in the reaction.
(v) Mass-balance equations
For glucose, 02, water, gluconic acid and HZOz,the appropriate mass~balanceequation is Eq. (4.2):
mass in + mass generated ::::: mass out + mass consumed
For egg solids. Nz and total mass, the appropriate mass-balance equation is Eq, (4.3):
3.
Calculate
mass in = mass out
(i) Calculation table
The mass of NZ accompanying 18 kg 02 in air can be calculated from the known composition of air.
kg 02 to molar units:
ISkg02 = ISkg0 2 ·
1 32.0kg kgmOll
1
=0.563kgm o10 2
Converting 18
Therefore, 79/21 x 0563 kgmol = 2.118 kgmol NZ enter in the air stream. Converting this to mass units:
2,118kgmo1N2 = 2.118kgmoIN2·
lkgmol 2S.0kg I
1
= 59.30kg N2
The calculation tables below show all known quantities in kg. The total mass of off-gas is denoted G; the total mass of product is denoted P. The In side of the mass-balance table is complete,
Solutions: Chapter 4
25
59.30 kg NZ in ::::: NZ out
NZ out::::: 59,30 kg
Glucose balance 60 kg glucose in + 0 kg glucose generated ::::: 0.002 P kg glucose out + glucose consumed
Glucose consumed ::::: (60 - 0.002 P) kg
Converting the glucose consumed to molar tenus:
Glucoseconsumed ::::: (60 - 0.002 P) kg.
1
1 kgmOll
180.2 kg:::::
Glucoseconsumed ::::: (0.333-1.11 x 10- 5
(60 -0.002 P)
180.2
p) kgmol
kgmol
From the reaction stoichiometry, conversion of this amount of glucose· requires the same number of kgmol OZ. Converting this molar quantity to mass of Oz:
(0.333 -1.11 x 10- 5 p) kgmo102 = (0.333-1.11 x 10- 5 p) kgm01.1 ;~~~11 = (10.656- 3.552x 10-4 p) kg 02
02 balam.:e
18 kg Oz in +·0 kg Ozgenerated :::::{)Z out + (10.656 - 3.552 x 10-4 P) kg Oz consumed
02 out = (18 - (10.656 - 3.552 x 10""P») kg
02; out::::: (7.344 + 3.552 x 10-4 P) | 12,566 | 35,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-34 | latest | en | 0.497956 |
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# 9.5: Estimation of Square Roots
Difficulty Level: At Grade Created by: CK-12
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Practice Estimation of Square Roots
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Have you ever done community service?
Miguel is enjoying his time with the Wildcats. The players have all been very friendly to him and Miguel loves helping out. He gets to hand out towels and water and sometimes collects and organizes equipment.
The Wildcats do a lot of community service. One thing that they do is help out Little League teams. There is one new ball field being created and the Wildcats are going to help design the infield so that it has the correct dimensions. Miguel has been invited to go along.
When they arrive at the field, Miguel and one of the players Harris take a measure to measure the distance from home plate to first base. It measures 58 feet.
“That isn’t correct,” Harris tells Miguel. “The area of the infield for little league is 3600 square feet. This measurement is inaccurate. We’ll have to help them fix it.”
Miguel is puzzled. What is inaccurate about the 58 feet? Does it need to be a longer distance or a shorter one?
To figure this problem out, Miguel will need to use his knowledge of squares and square roots. Pay attention and you can help Miguel figure this problem out at the end of the lesson.
### Guidance
In the last Concept, you began learning about square roots. Remember that the square root of a number is the number that you multiply by itself to get a product, in essence it is a number that you square. Squaring and square roots are inverse operations. You can square a number to get a product, then take the square root of the product to get back to the original number.
6236=36=6
Here you can see the inverse operation between squaring a number and finding the square root of a product.
A number like 36 is a perfect square meaning that the square root is a whole number. Here are some other perfect squares.
16, 25, 36, 49, 64, 81, 100, 121, 144, 169
Notice that if you find the square root of any of these numbers, you will end up with a whole number as the answer.
What if a square root is not a whole number?
This can also happen. When it does, we can approximate the square root of the number. There are a couple of different ways to do this. Let’s take a look.
First, we can approximate the square root using perfect squares.
To do this, we look for the perfect squares that generate a number close to the square root that we are looking for.
30\begin{align*}\sqrt{30}\end{align*}
If we are looking for the square root of 30, we first need to find two perfect squares near 30. One should be less than 30 and one should be more than 30.
5×56×6=25=36
25 is the perfect square less than 30.
36 is the perfect square greater than 30.
Because 30 is between 25 and 30, we can say that the approximate square root of 30 is between 5 and 6. It is probably close to 5.5.
This is how we can approximate a square root using perfect squares.
The second way of approximating a square root is to use a calculator. Calculators all have a radical sign on them.
\begin{align*}\sqrt{\Box}\end{align*}
To find the square root of a number, we enter the radical sign, then the value and press enter.
This will give us a decimal approximation of the square root. Many times you will need to round these answers. Let’s look at an example.
We can use 30 again.
30\begin{align*}\sqrt{30}\end{align*}
Now enter the value, 30.
Next, press enter.
Here is our answer on the calculator. 5.477225575.
To find a final answer, we can round to the nearest tenth. To do this, we round the 4 it is in the tenths place. The number after the 4 is 7, so we round up to five.
WOW! Notice that this value is the same as the answer that we found when estimating using perfect squares.
The third way of approximating a square root is to use something called “tabular interpolation.” Tabular interpolation is using a table. To use this table, we find the approximate value of the square root according to research that has been completed by a mathematician.
17\begin{align*}\sqrt{17}\end{align*}
To find the square root of 17, let’s use a table. Keep in mind that these values have been rounded to the nearest thousandth.
Looking at the table you can see that the square root of 17 is 4.123.
You can find tables that include numbers all the way up to 100. By finding these tables, you can use them to locate the square root of numbers 1 to 100.
Choose a method to find an approximate square root of each number. Round your answer to the nearest tenth.
#### Example A
11\begin{align*}\sqrt{11}\end{align*}
Solution: 3.3\begin{align*}3.3\end{align*}
#### Example B
33\begin{align*}\sqrt{33}\end{align*}
Solution:5.7\begin{align*}5.7\end{align*}
#### Example C
41\begin{align*}\sqrt{41}\end{align*}
Solution:6.4\begin{align*}6.4\end{align*}
Here is the original problem once again.
Miguel is enjoying his time with the Wildcats. The players have all been very friendly to him and Miguel loves helping out. He gets to hand out towels and water and sometimes collects and organizes equipment.
The Wildcats do a lot of community service. One thing that they do is help out Little League teams. There is one new ball field being created and the Wildcats are going to help design the infield so that it has the correct dimensions. Miguel has been invited to go along.
When they arrive at the field, Miguel and one of the players Harris take a measure to measure the distance from home plate to first base. It measures 58 feet.
“That isn’t correct,” Harris tells Miguel. “The area of the infield for little league is 3600 square feet. This measurement is inaccurate. We’ll have to help them fix it.”
Miguel is puzzled. What is inaccurate about the 58 feet? Does it need to be a longer distance or a shorter one?
Harris told Miguel that the area of the infield needs to be 3600 square feet for Little League. To figure out what is inaccurate about the current measure of 58 feet, Miguel will need to find the square root of 3600 square feet.
Finding the square root will give you the measurement of the distance between the bases. This is the distance that is squared to find the overall area of the infield.
3600\begin{align*}\sqrt{3600}\end{align*}
To find this square root, don’t worry about 3600, drop the zeros and find the square root of 36.
6×6=36\begin{align*}6 \times 6 = 36\end{align*}
3600=60 feet\begin{align*}\sqrt{3600} = 60 \ feet\end{align*}
The distance between each of the bases should be 60 feet. With a current distance of 58 feet, the distance is short two feet between each base.
Miguel is ready to help Harris correct the problem thanks to squares and square roots!
### Vocabulary
Here are the vocabulary words in this Concept.
Square Root
A number multiplied to find a product. The number squared is the square root of the product.
Perfect Square
a number whose square root is a whole number.
### Guided Practice
Here is one for you to try on your own.
Approximate this square root to the nearest hundredth.
65\begin{align*}\sqrt{65}\end{align*}
To figure this out, we can use the table or the calculator.
Using our calculator, we evaluate this square root as:
8.0622577\begin{align*}8.0622577\end{align*}
Now we round to the nearest hundredth.
8.06\begin{align*}8.06\end{align*}
### Video Review
Here is a video for review.
### Practice
Directions: Find each an approximate answer for each square root. You may round your decimal answer to the nearest tenth.
1. )8¯¯¯\begin{align*}{\overline{ ) {8 }}}\end{align*}
2. )11¯¯¯¯¯\begin{align*}{\overline{ ) {11 }}}\end{align*}
3. )24¯¯¯¯¯\begin{align*}{\overline{ ) {24 }}}\end{align*}
4. )31¯¯¯¯¯\begin{align*}{\overline{ ) {31 }}}\end{align*}
5. )37¯¯¯¯¯\begin{align*}{\overline{ ) {37 }}} \end{align*}
6. )43¯¯¯¯¯\begin{align*}{\overline{ ) {43 }}}\end{align*}
7. )59¯¯¯¯¯\begin{align*}{\overline{ ) {59 }}}\end{align*}
8. )67¯¯¯¯¯\begin{align*}{\overline{ ) {67 }}}\end{align*}
9. )73¯¯¯¯¯\begin{align*}{\overline{ ) {73 }}}\end{align*}
10. )80¯¯¯¯¯\begin{align*}{\overline{ ) {80 }}}\end{align*}
11. )95¯¯¯¯¯\begin{align*}{\overline{ ) {95 }}}\end{align*}
12. )99¯¯¯¯¯\begin{align*}{\overline{ ) {99 }}}\end{align*}
13. )101¯¯¯¯¯¯\begin{align*}{\overline{ ) {101 }}}\end{align*}
14. )150¯¯¯¯¯¯\begin{align*}{\overline{ ) {150 }}}\end{align*}
15. )136¯¯¯¯¯¯\begin{align*}{\overline{ ) {136 }}}\end{align*}
### Vocabulary Language: English
Perfect Square
Perfect Square
A perfect square is a number whose square root is an integer.
Square Root
Square Root
The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.
Nov 30, 2012
Sep 23, 2015 | 2,401 | 9,021 | {"found_math": true, "script_math_tex": 31, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2015-48 | longest | en | 0.944225 |
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### Share this free course
Engineering: The challenge of temperature
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# 2.2 Thermal effects in outline
Temperature is, of course, the measure of 'thermal' conditions. Nowadays it is measured by thermometers and expressed as a number on an agreed scale. Some features of thermometers and of their use are discussed in Thermometers and process control
The theoretical construct of temperature relates it to the kinetic energies of atoms. This gives clear insights into the way temperature affects the behaviour of materials. Energy is given to things to make them hot and taken from things to make them cold. Within solid matter, thermal energy is stored primarily in the kinetic energy of atoms. So, when we say that a particular object or substance has a certain temperature, we are in fact quantifying the amount of kinetic energy possessed by the atoms of that object or substance – see Box 1 Temperature and energy. To say that temperature is 'that which is measured by a thermometer' does not provide a definition which helps an understanding at the atomic scale.
## Box 1 Thermometers and process control
Measuring temperature is not quite like measuring length with a ruler, where the length to be determined is compared with the length of the ruler. Thermometers convert the condition of temperature into other physical things, such as a resistance, voltage or pressure; we say they are temperature transducers. The temperature 'reading' is really a graduated indication of one of those other physical quantities. It is not very useful until the instrument has been calibrated against a defined numerical scale of temperature.
Here are some effects by which temperature is sensed and given number:
• expansion of a liquid in a tube, often mercury in glass
• resistance of a metal wire, platinum in good instruments
• voltage output from a thermocouple, that is, the junction between wires of two different metals
• pressure of a constant mass of gas kept at constant volume
• brightness or colour of radiation emitted by a glowing furnace.
Industrial processes may require a steady or changing temperature. Changes may be spatial or temporal. Temperature may be tightly controlled by using a thermometer as the sensor in a feedback loop. On the other hand, if it is simply necessary to know whether the temperature is high enough for a process to work, or low enough for safety, the thermometer may merely monitor conditions or be coupled to an alarm to indicate when a threshold has been crossed. With such a diversity of uses, accuracy may not be the paramount virtue of a thermometer. Any of the following features may be of particular relevance for certain applications:
• Accuracy: A thermometer will agree with the International Temperature Scale to a specified precision and will have been calibrated between some standard temperatures.
• Range: The instrument must be able to give readings at the process temperature. For example, below −38 °C and above 360 °C a mercury-in-glass thermometer is useless. Radiation pyrometers are used to measure high temperatures.
• Sensitivity: A process may need close control of temperature or need to be tolerant of some deviation. Close control needs a sensitive thermometer, that is one giving noticeable response to a small change in temperature.
• Response time: A long response time means that short, sharp fluctuations of temperature are averaged out.
• Repeatability: Actual values may be less important than that successive batches of a process get the same treatment.
• Reliability: A malfunctioning instrument might close down the plant.
• Corrosion-proofing: Furnace atmospheres can be very reactive chemically; the thermometer sensor may have to be protected by a ceramic sheath, which will slow its response.
• Output: Telemetry of signals to a central control point favours thermometers with electrical output.
• Compatibility: May need to be compatible with other instrumentation.
• Cost: Cost per unit, or total cost of manufacturing, is almost always a consideration.
## Exercise 1
In a mercury-in-glass thermometer, the length of a narrow column of mercury is measured. Explain how this is related to the local air temperature.
### Answer
The mercury in a mercury-in-glass thermometer is presumed to be at the same temperature as the air that surrounds it. Changes in temperature lead to expansion of the mercury within the tube that encloses it – the tube is thick-walled, but the mercury can expand or contract along the length of the tube.
## Box 2 Temperature and energys
How much thermal energy is associated with a given temperature? At this stage I just want to show you how to translate the absolute temperature of a substance, T in kelvin into the average energy, in joules, associated with the random motions of one of its particles. The link between energy and temperature will be discussed further in Section 4.
The conversion factor that links joules to kelvin is one of the basic physical constants that was established in the nineteenth century. Named in honour of one of the pioneers of the field, Boltzmann's constant has the symbol k and the value of 1.38 × 10−23 J K−1. Boltzmann's constant multiplied by absolute temperature (kT) gives the average kinetic energy of a particle in a substance that is at temperature T.
Consider room temperature, say 300 K. The average energy of particles in a substance at 300 K is:
That's an inconveniently small number, and there are two ways to get something more useful from it.
The first is to consider a large number of particles, say a mole, instead of just one. In that case a mole of particles at room temperature consists of 6.023 × 1023 particles with an average energy of 4.14 × 10−21J.
A mole is a special quantity of a substance, measured in terms of the number of constituent particles (atoms or molecules), namely 6.023 × 1023 particles. One gram of hydrogen contains this number of atoms.
That means that at room temperature we can expect thermal energy to amount to
For comparison, you can put 2.5 kJ of energy into a small rechargeable battery, so I feel comfortable with that measure.
The other way to get a more meaningful description of thermal energy is to convert from joules to a unit more in tune with the energy of bonds between atoms. Such a unit is the electron volt (or eV) and you get it by dividing the value in joules by the electronic charge, e, another fundamental physical constant.
In terms of electron volts, 300 K works out as:
or about one-fortieth of an electron volt. Electron volts are convenient for measuring the energy binding individual atoms together – typically a few eV per pair of atoms – so at room temperature an atom in a solid with the average energy does not have enough energy to break free of its bonds.
You can build your confidence with this kind of conversion by doing the next exercise.
## Exercise 2
Complete the following table that connects temperature with thermal energy in eV particle−1 and kJ mole−1.
Table 1
T/KAverage energy / eV particle−1Average energy / kJ mole−1
3000.0262.5
1000
1
300
### Answer
Table 2
T/KAverage energy /eV particle−1Average energy / kJ mole−1
3000.0262.5
1000kT/e = 0.0866.023 × 1023 × kT = 8.31
1160016.023 × 1023 × kT = 96.4
36100kT/e = 3.11300
## SAQ 1
Calculate the temperature of a solid in which the atoms have an average energy of 0.13 eV.
### Answer
Since 300 K corresponds with an average energy per atom of 0.026 eV, an average energy of 0.13 eV, some 5 times higher, will correspond with an absolute temperature that is also 5 times higher, namely 1500 K.
Alternatively, working from first principles,
Rearranging, | 1,687 | 7,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.936728 |
https://ether-wind.narod.ru/Stokes_1845/012.html | 1,723,701,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00442.warc.gz | 181,279,571 | 3,624 | G.G.Stokes "On the Aberration of Light" // Philosophical Magazine 1845, 27: Page 12
# G.G.Stokes "On the Aberration of Light" // Philosophical Magazine 1845, 27: Page 12
В начало PDF, DjVu <<< Страница 12 >>>
1 2 9 10 11 12 13 14 15 one; but if u> v and w are such that udx+vdy + wdz is an exact differentia], we have dw d u dw^dv d x~ dz9 dy~~dz9 whence, denoting by the suffixes 1, 2 the values of the variables belonging to the first and second limits respectively, we obtain = . . (6.) If the motion of the aether be such that udx + vdy -\-*wdz is an exafct differential for one system of rectangular axes, it is easy to prove, by the transformation of co-ordinates, that it is an exact differential for any other system. Hence the formulae (6.) will hold good, not merely for light propagated in the direction first considered, but for light propagated in any direction, the direction of ♦propagation being taken in each case for the axis of ss. If we assume that u dx + v dy + w dz is an exact differential for that part of the motion of the aether which is due to the motions of translation of the earth and planets, it does not therefore follow that the same is true for that part which depends on their motions of rotation. Moreover, the diurnal aberration is too small to be detected by observation, or at least to be measured with any accuracy, and I shall therefore neglect it. It is not difficult to show that the formulae (6.) lead to the known law of aberration. In applying them to the case of a star, if we begin the integrations in equations (5.) at a point situated at such a distance from the earth that the motion of the aether, and consequently the resulting change in the direction of the light, is insensible, we shall have ux =0, ^=0; and if, moreover, we take the plane xz to pass through the direction of the earth’s motion, we shall have = 0, /32 = 0, and ct2 — &i = —; that is, the star will appear to be displaced towards the direction in which the earth is moving, through an angle equal to the ratio of the velocity of the earth to that of light, multiplied by the sine of the angle between the direction of the earth’s motion and the line joining the earth and the star. In considering the effect of aberration on a planet, it will be convenient to divide the integrations in equation (5.) into three parts, first integrating from the point considered on the surface of the planet to a distance at which the motion of the aether may be neglected, then to a point near the earth where we may still neglect the motion of the aether, and lastly to the point of the' earth’s surface at which the planet is viewed. For the first part we ghall have u% = 0, t>2 = 0, and vx will be the resolved parts of the planet’s velocity. The increments tt V of a and /3 for the first interval will be, therefore, — y, — ^. For the second interval a and /3 will remain constant, while 11 v for the third their increments will be y, y, just as in the case of a star, w2 and i>2 being now the resolved parts of the earth’s velocity. Fig. 1. P Fig. 1 represents what is conceived to take place. P is the planet in the position it had when the light quitted it; E the earth in the position it has when the light reaches it. The lines a b, cd, &c. represent a small portion of a wave of light in its successive positions. The arrows represent the directions in which P and E may be conceived to move. The breadth a b is supposed to be comparable to the breadth of a telescope. In fig. %pmne represents an orthogonal trajectory to the surfaces a b, cd, &c.; p is the point of the planet from which the light starts, e the point of the earth which it reaches. The trajectorypmnc may be considered a straight line, except near the ends p and e, where it will be a little curved, as from p to m and from e to n. The curvature at e will have the same effect on the apparent position of the planet as it would have on that of a star in the same direction : as to the curvature at p, if we draw jpy perpendicular to mn produced, the curvature will have the effect of causing p to be seen as if it were at q. Now the angle between the tangents at p and m being that through which a star in the direction of e is displaced by aberration to an observer at p, and the di- The lines towards P in fig. 1. should lean in the opposite direction. | 1,087 | 4,362 | {"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-2024-33 | latest | en | 0.918088 |
https://www.studypug.com/geometry/inductive-reasoning | 1,611,083,324,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00673.warc.gz | 1,009,754,658 | 16,249 | Inductive reasoning
Inductive reasoning
Lessons
Notes:
A conjecture is an educational guess made from the given information. Inductive reasoning is about making a conjecture that predicts the next set of patterns, or arrive at a conclusion.
• Introduction
Inductive Reasoning Overview:
a)
What are conjectures?
• 1.
Making a Conjecture
Make a conjecture of the next item or number based on the information given:
a)
1, 3, 5, 7, 9,…
b)
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$
c)
d)
e)
f)
• 2.
Make a conjecture with the given information. Draw a figure to show that your conjecture is correct:
a)
ABC is a triangle and AB = BC
b)
ABC is a triangle and $\angle$B is a right angle.
c)
Line $a$ and line $b$ are perpendicular
d)
Line $a$ and line $b$ are parallel
• 3.
Counterexamples of Conjectures
Determine if the following conjecture is true or false. If it is false, then find a counterexample of the conjecture:
a)
Given: A = (0,0), B = (0,1), C = (1,0).
Conjecture: ABC form a right isosceles triangle.
b)
Given: $a$ is a negative integer.
Conjecture: $\ a^2$ is a positive integer.
c)
Given: $\overline{AB}$ and $\overline{BC}$ are parallel
Conjecture: AB = BC
d)
Given: $x + y \geq 10 . x \geq 5$
Conjecture: $y \geq 6$ | 391 | 1,257 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-04 | latest | en | 0.780877 |
https://www.bankersadda.com/reasoning-ability-quiz-for-fci-phase-i-2022-08th-november/ | 1,670,058,067,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00723.warc.gz | 692,379,872 | 116,728 | Latest Banking jobs » »
# Reasoning Ability Quiz For FCI Phase I 2022- 08th November
Directions (1-4): Read the following information carefully to answer the given questions:
Point F is 3 meters to the north of point E. Point C is 6 meters to the south of point D. Shortest distance between D and A is 8 meters. Point A is to the West of point B. Ravish started walking from point E and after walking 4 meters ahead he reached point D from where he kept on walking in the same direction from where he started and reached point B after walking 12 meters. Point A lies somewhere between point D and B in a straight line.
Q1. What is the shortest distance between point D and F?
(a) 2 meters
(b) 3 meters
(c) 4 meters
(d) 5 meters
(e) None of these
Q2. In which direction did Ravish started his journey?
(a) North
(b) East
(c) West
(d) South-East
(e) None of these
Q3. What is the shortest distance between point C and A?
(a) 14 meters
(b) 8 meters
(c) 10 meters
(d) 12 meters
(e) None of these
Q4. Which of the following two points are farthest from each other?
(a) C and B
(b) E and B
(c) F and C
(d) F and B
(e) A and F
Q5. Prakash starts walking towards East. After walking 35 m, he turns to his right and walks another 40 m. He then turns right again and walks another 35 m. Finally, he turns right and walks 20 m. In which direction and how far is he from the starting point?
(a) 20 m, North
(b) 35 m, South
(c) 35 m, North
(d) 20 m, South
(e) None of these
Q6. A person started walking from point A towards South direction. After walking 6 km he took a right turn and walked 2km. He then took a left turn and walked 6km. He again took a right turn and walked 3km meters and stopped at point B. Now how far and in which direction was he from starting point?
(a) 13 km, South
(b) 13km, North-east
(c) 13km, South-west
(d) 13km, South
(e) None of these
Q7. Building D is towards East to building F. Building B is towards North of building D. Building H is towards South of building B. Towards which direction is building H from building F?
(a) East
(b) South-East
(c) North-East
(e) None of these
Q8. A student starts from point X, walks 2 km towards North, then takes a left turn and walks for 5 km. He then takes a left turn and walks for 8 km then takes left turn again and walks for further 5 km then finally takes a left turn again and walks 1 km to reach point Y. How far and towards which direction should the student walk to reach at point X again?
(a) 5 km towards south
(b) 7 km towards East
(c) 6 km towards South
(d) 6 km towards West
(e) 5 km towards North
Town D is 13 km towards the North-east of town A. A bus starts from town A, travels 10 km towards north and takes a right turn. After taking the right turn, it travels 5 km and reaches town B. From town B the bus takes a right turn again, travels 21 km and reaches town Q. Town B is in southwest direction of town D.
Q9. What is the direction of town Q with respect to town D?
(a) towards South-East
(b) towards West
(c) towards South-West
(d) towards South
(e) None of these
Q10. What is the direction of town B with respect to town A?
(a) towards South-East
(b) towards West
(c) towards South-West
(d) towards North-East
(e) None of these
Directions (11-15): Study the information carefully and answer the questions given below.
Seven persons i.e., P, Q, R, S, T, U and V are living on different floors of eight storey building but not necessarily in same order. Ground floor is numbered as 1 and topmost floor is numbered as 8. One floor is vacant.
One person lives between P and Q. P lives on 4th floor. Q does not live above P. Six persons live below S. Three persons live between R and U. R lives one of the floors above U’s floor. T lives immediate above vacant floor. T does not live on an even numbered floor.
Q11. Which of the following floor is vacant?
(a) 5TH floor
(b) 6th floor
(c) 3rd floor
(d) 2nd floor
(e) None of these
Q12. Who among the following lives at 7th floor?
(a) R
(b) S
(c) T
(d) U
(e) None of these
Q13. How many persons live between V and T?
(a) Three
(b) One
(c) Two
(d) None
(e) Four
Q14. Four of the following five are alike in certain way based from a group, find the one that does not belong to that group?
(a) S
(b) R
(c) U
(d) V
(e) T
Q15. On which of the following floor V lives?
(a) 4TH floor
(b) 5th floor
(c) 8th floor
(d) 3rd floor
(e) None of these
Solutions | 1,240 | 4,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-49 | latest | en | 0.95649 |
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### Forecast Friday Topic: Building Regression Models With Excel
July 8, 2010
(Twelfth in a series)
We’ve spent the last six weeks discussing regression analysis as a forecasting method. As you have seen, simple regression is a bit tedious to work out by hand, but for multiple regression analysis, you almost always need the aid of a computerized software package. Today I will demonstrate for you how to use the Regression Analysis feature of Microsoft Excel’s1 Analysis ToolPak Add-In. Excel 2007 comes with the Analysis ToolPak Add-In, which you can choose to activate. One way to know if Analysis ToolPak is activated on your version of Excel is to click on the Data tab on your workspace and see if there is a “Data Analysis” icon. The following thumbnail will illustrate:
Checking for the Data Analysis Add-In
Notice towards the upper right corner of the image, “Data Analysis” is highlighted in orange. The presence of the Data Analysis icon means that we have activated the Analysis ToolPak Add-In. If it wasn’t there, you would need to activate the Add-In, which you could do very easily by clicking the “Office” button in the top left hand corner, then clicking the “Excel Options” button, which will take you through the process of activating your add-in.
Setting up a Regression – Our Data Set
Generally in direct mail marketing, three components that often determine how much one spends – or whether he/she buys at all – are recency, frequency, and monetary value – known in short as RFM. Generally, the longer it has been since one’s last purchase (recency), the less he/she is likely to spend. Hence, we would expect a minus sign by the coefficient for recency. Also, RFM theorizes that the more frequently one buys, the greater his/her purchase. So, we would expect a plus sign by the coefficient for frequency. Finally, the higher the customer’s average purchases (monetary value), the greater his/her spending, so we would also expect a plus sign here. RFM is also used heavily by nonprofits in their capital and contributor campaigns, since they are often heavily reliant upon direct mail.
In our example here, a local nonprofit decided to test whether each RFM component had a relationship to a donor’s contribution, so it randomly selected 20 donors who contributed to its last appeal. Naturally, our dependent variable, Y, was the Contribution amount. The nonprofit also looked at three independent variables: months since last contribution (X1), times donated in last 12 months (X2), and average contribution over the last 12 months (X3). These independent variables represent recency, frequency, and monetary value, respectively. The table below shows our dataset:
Giving Patterns of 20 Donors Donor Contribution Months Since Last Donation Times Donated in last 12 months Average Contribution in last 12 months 1 95 10 1 85 2 110 8 2 95 3 100 10 2 90 4 115 8 3 75 5 100 9 1 95 6 120 6 2 100 7 105 9 1 90 8 125 10 1 125 9 105 9 2 100 10 130 4 3 150 11 135 7 4 125 12 150 2 8 150 13 140 4 3 125 14 155 2 9 140 15 140 2 8 130 16 160 2 10 150 17 145 3 6 135 18 165 1 12 150 19 150 3 4 160 20 170 1 12 140
The thumbnail below shows what the data set looks like in Excel:
Regression Data Set in Excel
Running the Regression
To run the regression, we need to select the regression tool from the Analysis ToolPak. We do this by clicking on the Data Analysis Tab. The next thumbnail shows us what we need to do:
Selecting the Regression option from the Data Analysis ToolPak
After selecting the regression tool, we need to select our independent variables and our dependent variables. It is best to make sure all columns containing your independent variables are adjacent to each other, as they are in columns D, E, and F. Notice that column C from rows 2 to 22 contains our Y-range values (including the column label). In columns D, E, and F, rows 2 through 22 contain their respective X-range values. Notice in the thumbnail how we indicate those column/row positions for Y-range and X-range values.
Regression Options
We also need to decide where to place the regression output and what data we want the output to contain. In the thumbnail below, we choose to have the output placed in a new worksheet, called “Regression Output”, and we also check the box indicating that we want the residuals printed. Also notice that we checked the box “Labels”, so that row 2 won’t be inadvertently added into the model.
Regression Options - Continued
Looking at the Output
Now we run the regression and get the following output:
Regression Output (residuals not shown)
As you can see, cell B5 contains our R2, equal to .933, indicating that 93.3% of the variation in a donor’s contribution amount is explained by changes in recency, frequency, and monetary value. Also, notice the F-statistic in cell E12. It’s a large, strong 73.90, and cell F12 to the right is 0.00, suggesting the model is significant. (Note, the Significance F in cell F12 and the P-Values in cells E17-E20 for each parameter estimate are quick cues to significance. If you’re using a 95% confidence interval – which we are here – then you want those values to be no higher than 0.05).
Now let’s look at each parameter estimate. Cells B17-B20 contain our regression coefficients. We have the following equation:
Contribution Estimate = 87.27 – 1.80 *Months_Since_Last_Donation + 2.45 *Times_Donated_Last_12_Months + 0.35 *Average_Contribution_Last_12_Months
Simplifying, we have:
Contribution Estimate = 87.27 – 1.80*RECENCY + 2.45*FREQUENCY + 0.35*MONETARY_VALUE
Ŷ = 87.27 – 1.80X1 + 2.45X2 + 0.35X3
Note that even though we opted to display the residuals for each observation, I chose not to show them here. It would have run below the fold, and would have been difficult to see. Besides, for our analysis, we’re not going to worry about residuals right now.
Interpreting the Output
As we can see, each month since a donor’s last contribution reduces his contribution by an average of \$1.80, when we hold frequency and monetary value constant. Likewise, for each time a donor has given in the last 12 months, the size of his contribution increases by an average of \$2.45, holding the other two variables constant. In addition, each one-dollar increase in a donor’s average contribution increases his contribution by an average of 35 cents. Hence, all of our coefficients have the signs we expect.
T-Statistics and P-Values
Next, we need to look at the t-statistics and P-values. As mentioned above, for a 95% confidence interval, a parameter estimate must have a p-value no greater than 0.05 (or 0.10 for a 90% confidence interval, etc.), in order to be significant. In like manner, for a 95% confidence interval, t-statistics should be values of at least 1.96 (slightly higher for small samples, but 1.96 will work) or less than -1.96 if the coefficient is negative, to be significant:
Parameter Coefficient T-statistic Significant? Intercept 87.27 4.32 Yes Months since Last (1.80) (1.44) No Times Donated 2.45 2.87 Yes Average Contribution 0.35 3.26 Yes
Notice that the coefficient for Months Since Last Donation has a t-statistic of -1.44. It is not significant. Another way to tell whether the parameter estimates are significant is to look at the Lower 95% and Upper 95% values in columns F and G. If the lower and upper 95% confidence interval values for a parameter estimate are both negative or both positive, they are significant. However, if the lower 95% value is negative and the upper 95% is positive (as is the case with Months Since Last Donation), then the parameter estimate is not significant, since its confidence interval range crosses zero. Hence Months Since Last Donation is not significant. Yet, the model still has a 93.3% coefficient of determination. Does this mean we can drop this variable from our regression? Not so fast!
Regression Violation Present!
Generally, when an independent variable we expect to be an important predictor of our dependent variable comes up as statistically insignificant, it is sometimes a sign of multicollinearity. And that is definitely the case with the nonprofit’s model. That will be our topic in next week’s Forecast Friday post.
Forecasting with the Output
Since we’re going to take on multicollinearity next week, let’s pretend our model is A-OK, and generate some forecasts.
We’ll go to our regression output worksheet, select cells A17 through B20, which contain our regression variables and coefficients, and then click Copy (or do a CTRL-C):
Selecting the Coefficients
Next, let’s paste those coefficients and transpose them in another worksheet. Here’s how to select the “Transpose” option when pasting:
Pasting Data Using the Transpose Option
Next, this is what the result of our transpose will be:
Transposed Data
Now, the nonprofit organization looks at five prospective donors whom they are planning to solicit. They look at their past giving history as shown in the next thumbnail:
Prospective Donors - Before Applying Model
Knowing this information, we want to multiply those values by their respective coefficients. Take a look at the formula in cell F7 as we do just that, in the next thumbnail:
Forecasting with Regression Output
Note how the cell numbers containing the coefficients have their column letters enveloped in ‘\$’. The dollar signs tell Excel that when we copy the formula down the next four rows, that it still reference those cells. Otherwise, for each row down, Excel would multiply each blank cell below the coefficients by the next donor’s information. Here’s are the forecasts generated:
Next Forecast Friday Topic: Multicollinearity
Today you learned how to develop regression models using Excel and how to use Excel to interpret the output. You also found out that our model exhibited multicollinearity, a violation of one of the key regression assumptions. Next week and the week after, we will discuss multicollinearity in depth: how to detect it, how to correct it, and when to live with it. We will again be using the nonprofit’s model. As I’ve said before, models are far from perfect and, as such, should only aid – not replace – the decision-making process.
1 Note: Excel is a registered trademark of Microsoft Corporation. Use of Microsoft Excel in this post is intended only for a demonstration of how to use Excel for regression analysis and does not constitute an endorsement of Microsoft Excel or any other Microsoft product by Analysights, LLC.
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### Forecast Friday Topic: Multiple Regression Analysis
June 17, 2010
(Ninth in a series)
Quite often, when we try to forecast sales, more than one variable is often involved. Sales depends on how much advertising we do, the price of our products, the price of competitors’ products, the time of the year (if our product is seasonal), and also demographics of the buyers. And there can be many more factors. Hence, we need to measure the impact of all relevant variables that we know drive our sales or other dependent variable. That brings us to the need for multiple regression analysis. Because of its complexity, we will be spending the next several weeks discussing multiple regression analysis in easily digestible parts. Multiple regression is a highly useful technique, but is quite easy to forget if not used often.
Another thing to note, regression analysis is often used for both time series and cross-sectional analysis. Time series is what we have focused on all along. Cross-sectional analysis involves using regression to analyze variables on static data (such as predicting how much money a person will spend on a car based on income, race, age, etc.). We will use examples of both in our discussions of multiple regression.
Determining Parameter Estimates for Multiple Regression
When it comes to deriving the parameter estimates in a multiple regression, the process gets both complicated and tedious, even if you have just two independent variables. We strongly advise you to use the regression features of MS-Excel, or some statistical analysis tool like SAS, SPSS, or MINITAB. In fact, we will not work out the derivation of the parameters with the data sets, but will provide you the results. You are free to run the data we provide on your own to replicate the results we display. I do, however, want to show you the equations for computing the parameter estimates for a three-variable (two independent variables and one dependent variable), and point out something very important.
Let’s assume that sales is your dependent variable, Y, and advertising expenditures and price are your independent variables, X1 and X2, respectively. Also, the coefficients – your parameter estimates will have similar subscripts to correspond to their respective independent variable. Hence, your model will take on the form:
Now, how do you go about computing α, β1 and β2? The process is similar to that of a two-variable model, but a little more involved. Take a look:
The subscript “i” represents the individual oberservation. In time series, the subscript can also be represented with a “t“.
What do you notice about the formulas for computing β1 and β2? First, you notice that the independent variables, X1 and X2, are included in the calculation for each coefficient. Why is this? Because when two or more independent variables are used to estimate the dependent variable, the independent variables themselves are likely to be related linearly as well. In fact, they need to be in order to perform multiple regression analysis. If either β1 or β2 turned out to be zero, then simple regression would be appropriate. However, if we omit one or more independent variables from the model that are related to those variables in the model, we run into serious problems, namely:
Specification Bias (Regression Assumptions Revisited)
Recall from last week’s Forecast Friday discussion on regression assumptions that 1) our equation must correctly specify the true regression model, namely that all relevant variables and no irrelevant variables are included in the model and 2) the independent variables must not be correlated with the error term. If either of these assumptions is violated, the parameter estimates you get will be biased. Looking at the above equations for β1 and β2, we can see that if we excluded one of the independent variables, say X2, from the model, the value derived for β1 will be incorrect because X1 has some relationship with X2. Moreover, X2‘s values are likely to be accounted for in the error terms, and because of its relationship with X1, X1 will be correlated with the error term, violating the second assumption above. Hence, you will end up with incorrect, biased estimators for your regression coefficient, β1.
Omitted Variables are Bad, but Excessive Variables Aren’t Much Better
Since omitting relevant variables can lead to biased parameter estimates, many analysts have a tendency to include any variable that might have any chance of affecting the dependent variable, Y. This is also bad. Additional variables means that you need to estimate more parameters, and that reduces your model’s degrees of freedom and the efficiency (trustworthiness) of your parameter estimates. Generally, for each variable – both dependent and independent – you are considering, you should have at least five data points. So, for a model with three independent variables, your data set should have 20 observations.
Another Important Regression Assumption
One last thing about multiple regression analysis – another assumption, which I deliberately left out of last week’s discussion, since it applies exclusively to multiple regression:
No combination of independent variables should have an exact linear relationship with one another.
OK, so what does this mean? Let’s assume you’re doing a model to forecast the effect of temperature on the speed at which ice melts. You use two independent variables: Celsius temperature and Fahrenheit temperature. What’s the problem here? There is a perfect linear relationship between these two variables. Every time you use a particular value of Fahrenheit temperature, you will get the same value of Celsius temperature. In this case, you will end up with multicollinearity, an assumption violation that results in inefficient parameter estimates. A relationship between independent variables need not be perfectly linear for multicollinearity to exist. Highly correlated variables can do the same thing. For example, independent variables such as “Husband Age” and “Wife Age,” or “Home Value” and “Home Square Footage” are examples of independent variables that are highly correlated.
You want to be sure that you do not put variables in the model that need not be there, because doing so could lead to multicollinearity.
Now Can We Get Into Multiple Regression????
Wasn’t that an ordeal? Well, now the fun can begin! I’m going to use an example from one of my old graduate school textbooks, because it’s good for several lessons in multiple regression. This data set is 25 annual observations to predict the percentage profit margin (Y) for U.S. savings and loan associations, based on changes in net revenues per deposit dollar (X1) and number of offices (X2). The data are as follows:
Year Percentage Profit Margin (Yt) Net Revenues Per Deposit Dollar (X1t) Number of Offices (X2t) 1 0.75 3.92 7,298 2 0.71 3.61 6,855 3 0.66 3.32 6,636 4 0.61 3.07 6,506 5 0.70 3.06 6,450 6 0.72 3.11 6,402 7 0.77 3.21 6,368 8 0.74 3.26 6,340 9 0.90 3.42 6,349 10 0.82 3.42 6,352 11 0.75 3.45 6,361 12 0.77 3.58 6,369 13 0.78 3.66 6,546 14 0.84 3.78 6,672 15 0.79 3.82 6,890 16 0.70 3.97 7,115 17 0.68 4.07 7,327 18 0.72 4.25 7,546 19 0.55 4.41 7,931 20 0.63 4.49 8,097 21 0.56 4.70 8,468 22 0.41 4.58 8,717 23 0.51 4.69 8,991 24 0.47 4.71 9,179 25 0.32 4.78 9,318
Data taken from Spellman, L.J., “Entry and profitability in a rate-free savings and loan market.” Quarterly Review of Economics and Business, 18, no. 2 (1978): 87-95, Reprinted in Newbold, P. and Bos, T., Introductory Business & Economic Forecasting, 2nd Edition, Cincinnati (1994): 136-137
What is the relationship between the S&Ls’ profit margin percentage and the number of S&L offices? How about between the margin percentage and the net revenues per deposit dollar? Is the relationship positive (that is, profit margin percentage moves in the same direction as its independent variable(s))? Or negative (the dependent and independent variables move in opposite directions)? Let’s look at each independent variable’s individual relationship with the dependent variable.
Net Revenue Per Deposit Dollar (X1) and Percentage Profit Margin (Y)
Generally, if revenue per deposit dollar goes up, would we not expect the percentage profit margin to also go up? After all, if the S & L is making more revenue on the same dollar, it suggests more efficiency. Hence, we expect a positive relationship. So, in the resulting regression equation, we would expect the coefficient, β1, for net revenue per deposit dollar to have a “+” sign.
Number of S&L Offices (X2) and Percentage Profit Margin (Y)
Generally, if there are more S&L offices, would that not suggest either higher overhead, increased competition, or some combination of the two? Those would cut into profit margins. Hence, we expect a negative relationship. So, in the resulting regression equation, we would expect the coefficient, β2, for number of S&L offices to have a “-” sign.
Are our Expectations Correct?
Do our relationship expectations hold up? They certainly do. The estimated multiple regression model is:
Yt = 1.56450 + 0.23720X1t – 0.000249X2t
What do the Parameter Estimates Mean?
Essentially, the model says that if net revenues per deposit dollar (X1t) increase by one unit, then percentage profit margin (Yt) will – on average – increase by 0.23720 percentage points, when the number of S&L offices is fixed. If the number of offices (X2t) increases by one, then percentage profit margin (Yt) will decrease by an average of 0.000249 percentage points, when net revenues are fixed.
Do Changes in the Independent Variables Explain Changes in The Dependent Variable?
We compute the coefficient of determination, R2, and get 0.865, indicating that changes in the number of S&L offices and in the net revenue per deposit dollar explain 86.5% of the variation in S&L percentage profit margin.
Are the Parameter Estimates Statistically Significant?
We have 25 observations, and three parameters – two coefficients for the independent variables, and one intercept – hence we have 22 degrees of freedom (25-3). If we choose a 95% confidence interval, we are saying that if we resampled and replicated this analysis 100 times, the average of our parameter estimates will be contain the true parameter approximately 95 times. To do this, we need to look at the t-values for each parameter estimate. For a two-tailed 95% significance test with 22 degrees of freedom, our critical t-value is 2.074. That means that if the t-statistic for a parameter estimate is greater than 2.074, then there is a strong positive relationship between the independent variable and the dependent variable; if the t-statistic for the parameter estimate is less than -2.074, then there is a strong negative relationship. This is what we get:
Parameter Value T-Statistic Significant? Intercept 1.5645000 19.70 Yes B1t 0.2372000 4.27 Yes B2t (0.0002490) (7.77) Yes
So, yes, all our parameter estimates are significant.
Next Forecast Friday: Building on What You Learned
I think you’ve had enough for this week! But we are still not finished. We’re going to stop here and continue with further analysis of this example next week. Next week, we will discuss computing the 95% confidence interval for the parameter estimates; determining whether the model is valid; and checking for autocorrelation. The following Forecast Friday (July 1) blog post will discuss specification bias in greater detail, demonstrating the impact of omitting a key independent variable from the model. | 5,245 | 22,504 | {"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} | 2.71875 | 3 | CC-MAIN-2019-35 | latest | en | 0.932416 |
https://www.physicsforums.com/threads/space-probe-between-the-earth-and-moon.749363/ | 1,532,112,846,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591719.4/warc/CC-MAIN-20180720174340-20180720194340-00156.warc.gz | 958,899,645 | 15,192 | # Homework Help: Space probe between the earth and moon
1. Apr 18, 2014
### toothpaste666
1. The problem statement, all variables and given/known data
The mass of the Earth is Me and the mass of the Moon is Mm. The (center to center) earth to moon distance is d.
a) If a space probe is sent directly from the earth to the moon, how far from the center of the earth would the net gravitational force (due to the earth and moon) on the probe be zero?
b) Neglecting friction, how much net energy would it take to lift a space probe from the surface of the earth to the moons surface?
2. Relevant equations
$E = K + U$
$F = G\frac{Mm}{r^2}$
3. The attempt at a solution
a) There are two forces acting on the probe, the force of gravity of the moon and the force of gravity of the earth. We are trying to find the distance where the combination of these two forces is 0 so:
$M_e$ = mass of earth $M_m$ = mass of moon $M_p$ = mass of probe $d$ = earth to moon distance $r$ = earth to probe distance
$F_{moon} - F_{earth} = 0$
$F_{moon} = F_{earth}$
$G\frac{M_m M_p}{(d-r)^2} = G\frac{M_e M_p}{r^2}$
$\frac{M_m}{(d-r)^2} = \frac{M_e}{r^2}$
It seems I have hit a wall. How can I isolate r?
b)
since we need the satellite to escape earths orbit we need to find the escape velocity. this happens when the satellite goes to infinity with a speed of 0.
$\frac{v_{esc}^2}{2} - G\frac{M_e}{r_e} = 0 + 0$
$\frac{v_{esc}^2}{2} = G\frac{M_e}{r_e}$
$v_{esc} = (2G\frac{M_e}{r_e})^\frac{1}{2}$
now we need to find the energy needed
$W + \frac{v_{esc}^2}{2} - G\frac{M_e}{r_e} = 0$
$W = - \frac{v_{esc}^2}{2} + G\frac{M_e}{r_e}$
would this give the correct answer?
Last edited: Apr 18, 2014
2. Apr 18, 2014
### paisiello2
a) Try inverting the equation and then expand the d-r term; then find the roots of the equation.
b) The question didn't ask for the escape velocity, so I think you should drop this term; however, you need to account for the fact that the moon will reduce the work required because it attracts the satellite.
3. Apr 18, 2014
### haruspex
Your method in part b) appears to guarantee the answer W = 0.
Since it asks for net energy, you need to take into account the potential energy when on the surface of the moon. | 698 | 2,244 | {"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} | 4.0625 | 4 | CC-MAIN-2018-30 | latest | en | 0.865369 |
https://crackmyproctoredexam.com/how-to-use-diagrammatic-reasoning-on-the-gmat/ | 1,719,012,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00505.warc.gz | 157,120,363 | 21,874 | # How To Use Diagrammatic Reasoning on the GMAT
One of the major questions for your GMAT exam is how to apply diagrammatic reasoning to solve problems. In this article I’ll give you a brief overview of how to do this. In the rest of this post, you’ll get more details on why it’s important to learn diagrammatic reasoning, as well as a great GMAT practice guide to use once you’re done. Here’s the rundown:
A lot of people think that if you can explain a problem in an easily understandable way, you’ve already done it. That’s not always true. If you’re doing something, for instance, driving down the highway, don’t automatically assume that you know how to drive there safely. Ask yourself these questions: How long have I been driving, and what type of vehicle have I been driving? What are my expectations when I get to my destination? If I know how to drive safely, how will I know how to drive safely when I’m doing this?
These are all different questions, but they’re all relevant to the real-world situation. You’re going to see them again, and so should you study. The GMAT diagrammatic reasoning test isn’t about trying to answer each question individually – it’s about thinking out loud and coming up with answers that make sense from the real-world.
Now here’s why diagrammatic reasoning is important for GMAT practice. When you’ve written down an answer and given it a name (like “Driving on I-80”), then it becomes easier for you to understand. For example, if you’ve written “Driving on I-80, look for a place where I can turn left and head back home,” it would be very easy to remember when you’re faced with the question “What are my options if I’m heading back home?”
However, diagrams are often used in real-world situations. In fact, some of the most common diagrams are used in engineering, business, marketing, accounting, and other professional fields. If you’re going to take a GMAT practice test, you need to learn and use diagrams like diagrams of graphs of lines, circles, quadrants, or pentagons, for example.
Diagrams can also help you see connections between information. For example, let’s say that your problem was, “How can I increase my sales by five percent?” and your diagrammatic reasoning solution was, “- increase my sales by five percent using customer relationship management techniques.” Now you’re both on the same page.
Diagrammatic reasoning is a critical part of understanding why GMAT exams are so hard. If you’re prepared to apply diagrams to real-world situations, you’ll do just fine.
Diagrams also make it much easier to explain your thoughts and reasoning. Let’s say that your diagrammatic reasoning answer is, “I need to find a way to sell more cars.” If you have a diagram of a car, you’re almost ready to write that down. The question is, “OK, now what can I say in order to convince someone to buy one?” You can now write, “The main benefit of buying cars is that you get to travel.
“You can get to enjoy a vacation with your family and still make it work. without spending thousands of dollars.”
All of those diagrams are just a starting point – you need to make sure that you are writing everything down, so that you’re able to explain it thoroughly and effectively. Once you’ve written it down, your goal is to remember what you’ve just written. It should be clear and precise, as well as concise and easy to read. Even a little bit of explanation of what you’re writing down will help a lot.
Diagrammatic reasoning can help you improve your score and your confidence in taking the GMAT exam. You should study for GMAT by using diagrams of graphs, and the GMAT diagrammatic reasoning practice test that you study will help you do that. | 797 | 3,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.941766 |
https://www.sgxm.tech/?p=611 | 1,618,431,656,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00453.warc.gz | 1,118,143,525 | 9,375 | # leetcode 1000. Minimum Cost to Merge Stones 区间dp 归并
#### 题目
https://leetcode.com/problems/minimum-cost-to-merge-stones/
There are N piles of stones arranged in a row. The i-th pile has stones[i] stones.
A move consists of merging exactly K consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these K piles.
Find the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1.
Example 1:
Input: stones = [3,2,4,1], K = 2
Output: 20
Explanation:
We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1].
We merge [4, 1] for a cost of 5, and we are left with [5, 5].
We merge [5, 5] for a cost of 10, and we are left with [10].
The total cost was 20, and this is the minimum possible.
Example 2:
Input: stones = [3,2,4,1], K = 3
Output: -1
Explanation: After any merge operation, there are 2 piles left, and we can’t merge anymore. So the task is impossible.
Example 3:
Input: stones = [3,5,1,2,6], K = 3
Output: 25
Explanation:
We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6].
We merge [3, 8, 6] for a cost of 17, and we are left with [17].
The total cost was 25, and this is the minimum possible.
Note:
1 <= stones.length <= 30
2 <= K <= 30
1 <= stones[i] <= 100
#### 源码
``````class Solution {
public:
vector<vector<vector<long>>> dp;
vector<int> pre;
int DK;
int mergeStones(vector<int>& stones, int K) {
DK = K;
dp.resize(stones.size()+1,vector<vector<long>>(stones.size()+1,vector<long>(K+1,INT_MAX)));
pre.resize(stones.size()+1,0);
for(int i = 0;i < stones.size();++i){
pre[i]=pre[i==0?0:i-1]+stones[i];
}
return calDp(0,stones.size()-1,1)==INT_MAX?-1:calDp(0,stones.size()-1,1);
}
long calDp(int st,int ed,int k){
if(dp[st][ed][k]!=INT_MAX)return dp[st][ed][k];
if(st==ed){
return dp[st][ed][k] = k==1?0:INT_MAX;
}
if((ed-st-k+1)%(DK-1)!=0){
return dp[st][ed][k] = INT_MAX;
}
if(k == 1){
return dp[st][ed][1] = calDp(st,ed,DK)+pre[ed]-(st==0?0:pre[st-1]);
}
for(int t = st;t < ed;t += (DK-1)){
dp[st][ed][k] = min(dp[st][ed][k],calDp(st,t,1)+calDp(t+1,ed,k-1));
}
return dp[st][ed][k];
}
};
`````` | 758 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | longest | en | 0.722523 |
https://mathoverflow.net/questions/356898/dual-space-of-the-intersection-of-locally-convex-vector-spaces | 1,716,656,619,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00194.warc.gz | 317,220,372 | 27,319 | # Dual space of the intersection of locally convex vector spaces
Let $$S \neq \emptyset$$ and let $$\big((E_s,\mathcal{T}_s)\big)_{s \in S}$$ be a family of locally convex vector subspaces of the same vector space. Denote by $$E_s^*$$ the dual space of $$(E_s,\mathcal{T}_s),$$ that is, the vector space of all $$\mathcal{T}_s$$-continuous linear functionals on $$E_s.$$ Furthermore, we set $$M := \bigcap_{s \in S} E_s.$$
The goal is to characterize the dual space of $$(M,\mathcal{T}_M)$$, where $$\mathcal{T}_M := \tau(\mathcal{T}_s \cap M: \,s\in S)$$ denotes the smallest topology on $$M$$ that contains $$\mathcal{T}_s \cap M$$ for every $$s \in S.$$
Obviously, we have
$$$$\text{lin span}\, \bigg(\bigcup_{s\in S} E_s^*|_M\bigg) \subset M^*,$$$$
where $$E_s^*|_M$$ denotes the set of all continuous linear functionals of $$E_s^*$$ restricted to $$M.$$
Question (general): Do we even have
$$$$\text{lin span}\, \bigg(\bigcup_{s\in S} E_s^*|_M\bigg) = M^*\, ?$$$$
Question (specific): Is something known for the dual space of the intersection of Lebesgue spaces w.r.t. $$\sigma$$-finite measures? For example, let $$\mu$$ and $$\nu$$ be $$\sigma$$-finite measures, what is known about the dual of the quotient space $$L_p(X,\mathcal{A},\{\mu,\nu\}) := \left(\mathcal{L}_p(X,\mathcal{A},\nu) \cap \mathcal{L}_p(X,\mathcal{A},\mu)\right) \, /\,\, [\mu+\nu],$$ equipped with the topology $$\tau(\|\cdot\|_{\mu,p},\|\cdot\|_{\nu,p})?$$ Here, $$[\mu+\nu]$$ is the subspace of all functions $$f$$ such that $$f = 0$$ almost surely w.r.t. $$\mu+\nu$$ and $$\mathcal{L}_p(X,\mathcal{A},\nu)$$ denotes the set of all $$\mathcal{A}$$-measurable functions that are $$p\text{th}$$ power $$\nu$$-integrable.
• If the measures $\mu$ and $\nu$ are not equivalent (in the sense that they have different negligable sets) it is not clear what you mean by the intersection -- if you take the definition of the Lebesgue spaces (equivalence classes of a.e. equal functions) literally the intersection would be empty. Apr 8, 2020 at 20:18
• Thanks for your comment. I've edited the question. Apr 8, 2020 at 20:45
• Surely "are $p$-times $\nu$-integrable" should be "have $\nu$-integrable $p$th power"? Also, what is the common vector space in which $\mathcal L_p(X, A, \nu)/[\mu + \nu]$ and $\mathcal L_p(X, A, \mu)/[\mu + \nu]$ both live? Apr 8, 2020 at 23:16
• If $\mathcal{L}(X,\mathcal{A})$ denotes the space of all $\mathcal{A}$-measurable functions, the common vector space is $\mathcal{L}(X,\mathcal{A})\,/\,[\mu + \nu].$ Apr 9, 2020 at 14:15
Let me suggest a slightly more structured situation in a first attempt at a solution: I would consider the one where we have lcs’s $$E_0$$ and $$E_\infty$$ with $$E_0 \subset E_s\subset E_\infty$$ and each of the inclusions continuous with dense images, in addition with the $$E_s$$ Banach spaces. In this case the dual of the intersection is indeed the linear hull of the dual spaces. The question of how many of these assumptions are superfluous would require more thought.
The special case you mention fits into this scheme (assuming $$p$$ is finite for simplicity). One can use a suitable space of step functions as $$E_0$$ and the measurable functions as $$E_\infty$$ (in this case allowing the latter to be a non locally convex tvs).
• Could you add a reference or an argument for the fact that the dual space of the intersection is the linear hull of the dual spaces? Right now, I can't see why this is the case. Apr 8, 2020 at 12:21
• Thanks for your contribution. However, I am not sure if the special case fits into the scheme you have described. The intersection of the two Lebesgue spaces is w.r.t. two $\sigma$-finite measures and not w.r.t. the exponents. Apr 8, 2020 at 12:59
• @JochenGlueck It is helpful to consider the diagonal map which, for two spaces, is $Delta: X\cap Y \to X\times Y$, $x\mapsto (x,x)$. You certainly know the dual of $X\times Y$ and you can calculate the transposed of the diagonal map. The case described by user131781 is particularly nice because then the Image of $\Delta$ equals the kernel of the difference map $d: X\times Y \to E_\infty$, $(x,y)\mapsto x-y$, i.e, you have a short exact sequence $0\to X\cap Y \to X\times Y \to Image(d)\to 0$. Apr 8, 2020 at 20:27
• @JochenWengenroth: That's a nice argument! Thank you very much. Apr 9, 2020 at 0:15
$$\DeclareMathOperator\span{span}$$Let $$M$$ be a set of finite measures on $$(X,\mathcal{A})$$ and consider $$L_p(X,\mathcal{A},M) := \left(\bigcap_{\mu \in M} \mathcal{L}_p(X,\mathcal{A},\mu) \right) \, / \, [M]$$ for some $$p\in\mathopen]1,\infty\mathclose[.$$ Let $$q\in\mathopen]1,\infty\mathclose[$$ such that $$1/p + 1/q = 1$$. Then, the dual space of $$(L_p(X,\mathcal{A},M),\tau(\|\cdot\|_{\mu,p} : \mu \in M))$$ is given by $$L_p(X,\mathcal{A},M)^* = \span \left( \bigcup_{\mu\in M} \bigl\{ (\cdot,g)_\mu|_{L_p(X,\mathcal{A},M)} : g \in \mathcal{L}_q(X,\mathcal{A},\mu)\, / \, [M]\bigr\} \right).$$
Proof (similar to Satz II.2.4 in Werner (2018) "Funktionalanalysis", Springer-Verlag): Clearly, for each $$\mu \in M$$ and each $$g \in \mathcal{L}_q(X,\mathcal{A},\mu)$$ we have that $$(\cdot,g)_\mu|_{L_p(X,\mathcal{A},M)}$$ is a continuous linear function on $$L_p(X,\mathcal{A},M)$$.
Conversely, let $$f^* \in L_p(X,\mathcal{A},M)^*$$ and define $$\nu(A) := f^*(\mathbb{1}_A) \quad A \in\mathcal{A}.$$ Then, $$\nu$$ is a signed measure. Since $$f^*$$ is $$\tau(\|\cdot\|_{\mu,p} : \mu \in M)$$-continuous there are a finite set $$F \subset M$$ and a positive $$L > 0$$ such that $$|f^*(g)| \leq L \max \{ \|g\|_{\mu,p} : \mu \in F\}$$ for all $$g \in L_p(X,\mathcal{A},M),$$ which yields that $$\nu \ll \sum_{\mu \in F} \mu$$. According to the Radon–Nikodym theorem there is a density $$h$$ of $$\nu$$ w.r.t. the finite measure $$\sum_{\mu \in F} \mu$$, that is, $$f^*(g) = \int g h \, \mathrm{d} (\textstyle{\sum_{\mu \in F} \mu})$$ for all $$g \in L_{\infty}(\sum_{\mu \in F} \mu)$$. Then, one can show that $$h \in \mathcal{L}_{q}$$ (see for example proof of Satz VIII.2.3 in Werner (2018)) and $$f^* = \sum_{\mu \in F} (\cdot,h)_\mu|_{L_p(X,\mathcal{A},M)}$$ on $$L_p(X,\mathcal{A},M)$$ since $$L_{\infty}(\sum_{\mu \in F} \mu)$$ is dense in $$L_{p}(\sum_{\mu \in F} \mu)$$. | 2,163 | 6,235 | {"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": 71, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.776555 |
http://book.imt-decal.org/1.%20Sets,%20Functions,%20and%20Logic/1.0%20Introduction.html | 1,550,432,016,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482478.14/warc/CC-MAIN-20190217192932-20190217214932-00305.warc.gz | 40,390,959 | 5,065 | 1.0: Introduction – Sets, Functions, and Logic
# Chapter 1: Sets, Functions, and Logic
In this chapter, we lay down some of the fundamental concepts regarding discrete mathematics – sets, functions and logic; without a rigorous treatment of these ideas, it would be hard for us to continue. First, we define the concept of a set, and look at properties of sets and operations between sets. We also explore how to count the size of various finite sets with operations applied to them. We then define exactly what a function is, and discuss various classes of functions (injections, bijections and surjections).
We will then look at the common sets of numbers we deal with in the real world in the order we are introduced to them – the natural numbers, the integers, rational (and irrational) numbers, real numbers and complex numbers, and investigate the formal relationship between these sets. We’ll conclude with introducting propositional logic, which will segue nicely into proof techniques in the second chapter. | 207 | 1,019 | {"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} | 3.21875 | 3 | CC-MAIN-2019-09 | latest | en | 0.930478 |
http://www.math.yorku.ca/Calendar/1999-00/math3330.html | 1,542,731,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746465.88/warc/CC-MAIN-20181120150950-20181120172950-00500.warc.gz | 464,675,851 | 2,829 | DEPARTMENT OF MATHEMATICS AND STATISTICS Faculty of Arts Faculty of Pure and Applied Science
# Regression Analysis
Simple regression analysis, multiple regression analysis, matrix form of the multiple regression model, estimation, tests (t- and F-tests), multicollinearity and other problems encountered in regression, diagnostics, model building and variable selection, remedies for violations of regression assumptions. First term.
This course is closely linked with MATH3034 3.0, Applied Categorical Data Analysis, for which it is a prerequisite. Students will use the computer heavily in these courses, but no previous courses in computing are required.
MATH3330 3.0 will focus on linear regression models for the analysis of data on several explanatory variables and a single response. The emphasis will be on understanding the different models and statistical concepts used for these models and on practical applications, rather than on the formal derivations of the models. The approach will require the use of matrix representations of the data, and the geometry of vector spaces, which will be reviewed in the course. Topics include simple linear regression, multiple linear regression, residual analysis and model selection.
The nature of the course requires that students be involved on a constant basis with the material, and not fall behind.
The text and grading scheme have not been determined as we go to press.
Prerequisite:One of AS/SC/MATH 1132 3.0, AS/SC/AK/MATH 2131 3.0, AS/SC/AK/MATH 2570 3.0, AS/SC/PSYC 2020 6.0, or equivalent; some acquaintance with matrix algebra (such as is provided in AS/SC/MATH 1025 3.0, AS/SC/MATH 1505 6.0, AS/AK/MATH 1550 6.0, AS/SC/MATH 2021 3.0, or AS/SC/AK/MATH 2221 3.0).
ExclusionsAS/\-SC/\-MATH 3033 3.0, AS/SC/GEOG 3421 3.0, AS/SC/PSYC 3030 6.0, AS/ECON 4210 3.0, AK/PSYC 3110 3.0.
Coordinator: Fall: P. Song Winter: D. Montgomery .
Please send comments to webmaster@mathstat.yorku.ca Department of Mathematics & StatisticsYork UniversityN520 Ross Building, 4700 Keele StreetToronto, Ontario, Canada M3J 1P3 | 540 | 2,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-47 | latest | en | 0.834739 |
https://tsfa.co/what-is-the-remainder-in-the-synthetic-division-problem-below-16 | 1,680,080,747,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00160.warc.gz | 641,847,650 | 6,498 | # What is the remainder in the synthetic division problem below
This What is the remainder in the synthetic division problem below helps to fast and easily solve any math problems.
Solve Now
## Synthetic Division and the Remainder and Factor Theorems
Going through synthetic division, we get 7 as remainder. profile. ITS 8 i should’ve trusted my boy wyatt. profile. did either one of y’all take the quiz or r u just assuming. profile.
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## SOLVED: What is the remainder for the synthetic division
+ 3x + 7 and the remainder is 0. (x 3 – 2x 3 – 8x – 35)/ (x – 5) = x 2 + 3x + 7. Tips and Tricks on Synthetic Division: Write down the coefficients and divide them using the zero of the linear
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## Synthetic Division and the Remainder and Factor Theorems
The required remainder is 2. Option A is correct. Synthetic division of 1/1 2 -3 2 to determine the remainder. What is division? Division of the value is to disintegration of the
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## What is the remainder in the synthetic division
Click here 👆 to get an answer to your question ️ What is the remainder in the synthetic division problem below? 1|4 6 -3 A. 3 B. 7 C. 5 D. 9. natalieotto119112
## What is the remainder in the synthetic division problem
Write down the coefficients and divide them using the zero of the linear factor to obtain the quotient and the remainder. (P (x)/ (x - a) = Q (x) + (R/ (x - a)) When we do synthetic division by (bx + a), we should get (Q (x)/b) as the quotient.
Solve mathematic questions
Can you solve this math problem? | 861 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.954953 |
http://www.romannumerals.co/numerals-converter/cccxcix-in-numbers/ | 1,716,992,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00721.warc.gz | 45,133,785 | 14,031 | ## What number is "CCCXCIX"?
### A: 399
CCCXCIX = 399
Your question is, "What is CCCXCIX in Numbers?". The answer is '399'. Here we will explain how to convert, write and read the Roman numeral letters CCCXCIX in the correct Arabic number translation.
## How is CCCXCIX converted to numbers?
To convert CCCXCIX to numbers the translation involves breaking the numeral into place values (ones, tens, hundreds, thousands), like this:
Place ValueNumberRoman Numeral
Conversion300 + 90 + 9CCC + XC + IX
Hundreds300CCC
Tens90XC
Ones9IX
## How is CCCXCIX written in numbers?
To write CCCXCIX as numbers correctly you combine the converted roman numerals together. The highest numerals should always precede the lower numerals to provide you the correct written translation, like in the table above.
300+90+9 = (CCCXCIX) = 399
## More from Roman Numerals.co
CD
Now you know the translation for Roman numeral CCCXCIX into numbers, see the next numeral to learn how it is conveted to numbers.
Convert another Roman numeral in to Arabic numbers. | 256 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.785301 |
https://design.tutsplus.com/tutorials/fractal-art-create-a-tile-fractal-in-apophysis--cms-22388 | 1,603,318,420,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00379.warc.gz | 303,754,243 | 25,339 | Unlimited PS Actions, graphics, videos & courses! Unlimited asset downloads! From \$16.50/m
# Fractal Art: Create a Tile Fractal in Apophysis
Difficulty:IntermediateLength:MediumLanguages:
This post is part of a series called Fractal Art for Beginners.
Fractal Art: An Introduction to Apophysis
Fractal Art: How to Create Basic Fractal Flowers in Apophysis
This tutorial will equip you with the skills to create a style of fractal known as tiles. The concept of a tile (think subway tile or kitchen backsplash) is ideal for a fractal application because of the repeating pattern of self-similarity. In this tutorial we will explore the power of post-transforms, learn how to apply a final transform, and explore different formulas known as variations within the Apophysis program.
## Explanation of Tiles
The fractal tile style comes in a wide variety of shapes, from squares to hexagons and beyond. All of them follow the same basic ideas to create a pattern. Once you've conquered the structure for a tile, the possibilities are literally endless, with a whole plethora of fun formulas to be added and shapes to explore. Today we will focus our attention on the basic square tile.
## 1. Basic Structure of a Square Tile
### Step 1
To begin creating the structure for our square-tiled fractal, open the Editor and click the button for a New/blank flame.
### Step 2
The next step is to set up the first transform. We do this because it will serve as a template for the rest of the tile structure. Setting up this one transform and copying it saves us a lot of time and work, and is also good practice because it prevents us from making easy mistakes by forgetting steps or mistyping values.
We will leave the Variation set to linear=1 as the linear formula is the backbone of this type of tile. Change the Weight of Transform 1 to 10. On the Colors tab, set the Color Speed setting to 1. The effect of this will not be apparent right now, but will become evident as we continue.
### Step 3
Finally, we need to move this initial transform into place. Move Transform 1 left one unit. Move this same transform up one unit.
### Step 4
Now that the first piece of our tile is in place, we can create the rest of the tile structure. Each of these transforms will be a duplicate of the first transform we setup.
Duplicate Transform 1 by clicking the Duplicate Transform button at the top of the editor. Move this transform to the right by two units.
### Step 5
Duplicate Transform 2 by clicking the Duplicate Transform button. Move Transform 3 down by two units.
### Step 6
Lastly, Duplicate Transform 3. Move this transform to the left by two units.
This completes the basic structure of a square tile.
## 2. Add the First Shape
The number of options for putting shapes into fractal tiles is staggering. For this tutorial, we'll look at a few different variations and how they can be used to create some amazing artwork.
### Step 1
Create a New Transform.
### Step 2
At this point, it is good to have the variation preview turned on in the editor. This allows us to see what the particular variation is doing on the fractal plane. It also allows you to visualize what each transform's effect is on the fractal and can provide invaluable information as you design your own fractals.
Right-click in the grid-portion of the editor window and choose the option Show variation preview.
### Step 3
For this transform, we need to use the Variations tab of the editor. Remove the 1 from Linear by changing it to a 0. You should now see many small dots in the preview window. This shows you the structure of the tile we set up. To get an even clearer picture, scroll down to find the Blur variation and add 0.5 to Transform 5. You will see the dots become clearer. Change the Blur value back to 0 to continue.
### Step 4
To begin adding some shape to our tile structure, find the variation Bubble and change the 0 to 1.
### Step 5
At this point, the Bubble variation is too big to be useful in our tile. We could reduce the amount of the variation to 0.25 or perhaps even smaller. However, by doing so, we reduce the affect of the variation, which is not what we're aiming for as it will affect the shape.
Instead we are going to force the transform to become smaller and at the same time retain the round shape of the Bubble variation when set to 1. This is an incredibly powerful technique that will allow you to retain the shape of a variation known as post-transform. This is an over-simplified explanation, but use it to your advantage: when you activate the post-transform, you keep the original shape and can move it at will.
To activate the post-transform mode, look for the button in the top menu bar of the editor that has a Px and a triangle. Click this button to enter the post-transform editing mode.
### Step 6
With the post-transform mode active, switch to the Triangle tab. We want to shrink the size of Transform 5. Change the value in the Scale Box (the one with a triangle on either side) to 200. Click the Scale Down Triangle twice.
### Step 7
We can edit the shape of this transform more by deactivating the post-transform mode and then scaling down the transform two more times by 200.
### Step 8
The next step is to add more structure. Create a New Transform. Remove the Linear variation by changing it to 0. Add Cylinder to this transform by changing its value to 1. The cylinder variation will give a long bar or tube shape.
### Step 8
Once again, at a value of 1, the cylinder transform is too large to fit nicely into our tile. It needs to be scaled down just as the bubble transform was. Enable post-transform mode, and scale down Transform 6 by the value of 200 three times. In the preview window, you'll now see the circles from the bubble transform appear connected by a long bar which is our cylinder transform.
### Step 9
Let's modify this transform some more. With post-transform mode still active, move Transform 6 to the left 0.5 units and up 0.5 units. Rotate this transform 45 degrees counter-clockwise. Notice that each time we modify the position of this transform, the shape does not change, only the position. This is the power behind the post-transform mode. If we were to perform these same actions with post-transform turned off, the entire shape of the fractal would change drastically.
### Step 10
We can get more shape out of this transform by one more trick. Turn post-transform mode off, and scale down this transform by 200 once. This technique will put more texture into the transform and give us a more 3D appearance.
### Step 11
To frame out the rest of our shapes, duplicate Transform 6. Enable post-transform mode once again. This time, rotate Transform 7 90 degrees clockwise and move it one unit to the right.
## 3. Creating a More Complex Structure
### Step 1
We've now completed a very basic tile structure and shape. Add a New Transform and remove the linear variation from this transform by changing Linear from 1 to 0. Change the values of Juliascope in the Variations tab from 0 to 1. Enable the post-transform mode and scale down by 200 and then again by 125. Move this transform up one unit. Finally, switch to the Variables tab and change the juliascope_power to 2 and juliascope_dist to 1. These modifications really fill in the rest of the fractal and give it a solid and powerful appearance.
### Step 2
Now is a good time to adjust the weights of transforms. Originally we set the weight of the first four transforms to 10. This allows the structure of the tile to hold despite how many more additional transforms we add, allowing us to continue to see what the tile looks like while we design it.
Change the weight of the transforms to the following:
• Transform 5: Bubble: 2
• Transforms 6 and 7: Cylinder: 3
• Transform 8: Juliascope: 3
Notice how changing these weights affects the overall appearance of the fractal. You may also notice a few annoying bright spots near the middle of the fractal. (This will depend on your current gradient). Let's fix those now.
### Step 3
Switch to the Colors tab. For each of the transforms 5 through 8, change the Opacity value from 1 to 0. Pay careful attention to the preview window as you make each change and notice how the bright spot changes to conform to the same color values as the rest of the fractal.
### Step 4
This is a good time for us to pick a color palette for the finished fractal. Close the Editor and open the Gradient selection window. Choose gradient 558_Explosion.
### Step 5
Close out of the gradient selection and open the Editor window. Open the Colors tab. Remember when we set each of the linear transforms to a color speed of 1? We did this so that all of the color changes we make now will be consistent across the entire fractal.
Make the following changes to each of the transforms:
• Transform 5: Transform Color: 0.150 and Color Speed: -1
• Transform 6 and 7: Transform Color: 0.708 and Color Speed: -1
• Transform 8: Transform Color: 0.608
To give more prominence to the Juliascope transform (Transform 8), change the Weight from 3 to 4.
## 4. Adding a Final Transform
### Step 1
To give our fractal more punch, we can add a final transform. The final transform will affect the entire fractal shape by influencing it with any formula or combination of formulas we choose.
Enable the final transform by clicking the Fx button with the little triangle at the top of the editor menu bar. You will see a grey triangle appear in the editor.
### Step 2
This final transform begins just like any other new transform we add, with the linear variation set to 1. Remove this by switching to the variations tab and changing the value of Linear to 0.
### Step 3
Add to the final transform 0.5 of the Foci variation and notice the change of the overall shape. Additionally, add in 0.1 Spherical. There is a slight, but visually appealing change to the shape of the fractal. Combining variations like this, even positive and negative amounts, can create some incredible images.
## 5. Cleanup and Rendering
The last few steps involve cleaning up the fractal and rendering.
### Step 1
Close the editor and use the zoom and move tools in the main window to find a nice location on the fractal to render.
### Step 2
Open the Adjustment panel and change to the Rendering tab. Change the Gamma setting to 3.5 for a slightly clearer render.
### Step 3
The final step is to render the fractal. Open the Render dialog by clicking the purple gear in the main window. From here, set the Density to 10,000. The Filter Radius setting of 0.455 will work nicely for the detail in this fractal. An Oversample of 2 is ideal. Click Start to begin the render.
Explore the different final transforms first. Add several. Use negatives. Be bold! You will be amazed at the variety of images you can get just from changing these. Next, challenge yourself by removing the final transform and move some of the other transforms to new locations. Then, try changing the variations to create something new. If you would really like a challenge, start from scratch and try to create a hexagon or other type of tile.
Below are a few samples of what you can create with fractal tiles and Apophysis. | 2,537 | 11,268 | {"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} | 4.21875 | 4 | CC-MAIN-2020-45 | longest | en | 0.893962 |
https://chess.stackexchange.com/questions/37462/why-dont-engines-show-explicitly-that-a-certain-move-takes-to-a-threefold-repet/37466 | 1,716,850,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00488.warc.gz | 142,908,439 | 41,427 | # Why don't engines show explicitly that a certain move takes to a threefold repetition (instead of just showing 0.00)?
I get that from the point of view of the engine there's no difference; equal is equal. But for a human analyzing with an engine, I find quite important to note that a move takes to a balanced position or to a threefold repetition (which finishes the game).
For example, say I'm analyzing a position from my favorite opening, and the engine says the best move is d4(+0.10), and the second one is h6(0.00). But it happens that h6 will be draw by repetition. Depending on my mood or my position on a tournament or whatever other factors, I could prefer one or another, and the fact that h6 just finishes the game is quite important in my opinion.
To me the main difference is that threefold repetition just immediately finishes the game in a draw, whilst a 0.00 position can just have any final result.
Of course you can check the line provided by the engine to realize that it takes to threefold repetition, but wouldn't be much useful to use some special character for it?
• Maybe you might reformulate it a bit, the why is obvious: it takes work to program this :-), with a limited benefit. Maybe a chess programmer can elaborate. Nov 23, 2021 at 12:27
• One reason is because programming languages are created to not favor any language. So while programming stockfish or other engines people put the numbers because some people might not know English. Dec 25, 2021 at 18:53
It's easy enough to make the engine say it's 0.00 because of a 3-fold repetition. If you look at the relevant part of Stockfish's code, they are basically if statements that return a particular value if the condition is met. If you want to know if there's a 3-fold, you could add a `print "this is a 3-fold"` command and you'll have it. I'm hard-pressed to see why it matters, though. 0.00 is 0.00, regardless of whether it's insufficient material or fortress or 3-fold or whatever.
It sounds like you want to know if the position is a dead draw or if it's still complex enough that you might be able to outplay your opponent. This is much harder to put in machine-understandable form. How would you define these things? There are certainly positions where engines have no trouble finding the best moves (and therefore all games end 1/2-1/2) while human games show it is very unbalanced.
That said, there is some effort in the computer chess community to go beyond finding the simple "best move". This started with contempt for handcrafted engines, which uses the material remaining on the board as a proxy for how complex the position is. The idea is that if you set contempt high, then the engine will play inferior moves as long as it keeps more material on the board, which leaves more scope to outplay a weaker opponent. Conversely if you set negative contempt, then the engine will simplify the position and shoot for a draw. That might be what you're looking for.
Today's strongest engines are all neural-network based, for which contempt is more difficult to implement. Neural network eval return W/D/L, which is the probability of a win, draw, or loss from that position; there is no number for complexity (and probably cannot be since you'd need a very large training set with labelled "complexity" to get started). People are trying the so-called draw score, which turns draws from worth 0.5 points to, say, worth 0.4 points. This, it is hoped, makes the engine avoid draws - after all if one move leads to a three-fold while another leads to a 10% chance of win, 80% chance of draw, and 10% chance of loss then the second move has a higher expected score and becomes preferable. But even this does not work perfectly, since your engine might assess a position as drawn while the opponent sees a big advantage, see e.g. this game where Leela saw 99.3% chance of draw on move 158, and still lost the game eventually.
Ultimately you still need a way to measure if a position is "dead", and a good way to do that doesn't exist (see this question). If you have a smart idea to measure how "live" the position is, you can probably get it published in a journal.
• Ok, so technically is a trivial task, and wouldn't harm the engine performance. I edited my question in order to try to make it clearer why I think it'd be a desirable feature. Nov 23, 2021 at 13:39
There is in general no such thing as a simple "this leads to threefold repetition". There may have been millions of different positions considered, and of those, if you select at each branch at your move the highest evaluation for you, and vice versa at the opponent's branch, that leads to a 0.00 evaluation that happened to come from a threefold.
That doesn't mean that that threefold was forced in any way. Maybe there were a thousand different things you could have done instead but they were -0.1, or the opponent could have done but they were +0.1. And then were another thousand that also led to 0.00 but for different reasons: tablebase draw, stalemate...
So, I guess engines could change the way they work, so that evaluation is not a fixed number but there can be different ways of being 0.00 - 0.00 (threefold) would be the same value as 0.00 (stalemate), but the user could still be shown which it is.
But it'd always be quite random which of the many 0.00s in the entire tree you'd get to see. | 1,239 | 5,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.952709 |
https://ipsc.ksp.sk/2004/real/problems/f.html | 1,723,635,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00132.warc.gz | 236,977,932 | 2,223 | # Internet Problem Solving Contest
## Problem F – Fat Rat
Johnny the Rat lives in the sewer system. He enjoys walking through the pipes. Johnny is quite fat; actually it is almost impossible to distinguish him from a sphere, yet he used to be able to pass through all the pipes comfortably.
Unfortunately, things have changed now. With the expansion of the Internet, some providers place their cables even in the sewer pipes! Johnny is worried that he may get stuck in one of his favourite pipes. Please help him to determine through which pipes he can still pass.
### Input Specification
The first line of the input file contains two numbers: J – diameter of Johnny the Rat and P – the number of Johnny's favourite pipes. The specification of the P pipes follows.
Description of each pipes starts with a line containing two numbers: D – diameter of the pipe and N – the number of cables crossing the pipe. The shape of the pipe is a cylinder of an infinite height, with diameter D, and its axis coincides with the z-axis of the coordinate system. Each of the next N lines contains 7 real numbers (x1, y1, z1, x2, y2, z2, d) specifying one cable. Shape of a cable is a cylinder of an infinite height, with axis [x1,y1,z1][x2,y2,z2], and with diameter d.
### Output specification
Output file should contain one line for each pipe with a single word "YES", if it is possible for Johnny to walk through the pipe without crossing the cables, or "NO" otherwise.
### Example
Input:
```3 2
7 1
-2 0 0 -2 1 0 1.5
7 2
-2 0 0 -2 1 0 1.5
2 0 0 2 1 0 1.5
```
Output:
```YES
NO
``` | 406 | 1,578 | {"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-2024-33 | latest | en | 0.916286 |
http://mathhelpforum.com/differential-equations/index166.html | 1,529,668,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00533.warc.gz | 207,366,957 | 15,019 | # Differential Equations Forum
Differential Equations Help Forum
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18. ### Quick Diff EQ help
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Math Discussion | 1,377 | 3,739 | {"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-2018-26 | latest | en | 0.638762 |
https://discourse.soylent.com/t/chloride-in-table-salt/16096 | 1,603,368,049,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00713.warc.gz | 291,111,830 | 9,309 | # Chloride in table salt
Hi there,
I bought a bag of plain common salt without the added iodine. I am hunting around trying to find the chloride content of table salt in mg. I have tried Nutrition Data but the chloride content is missing from the mineral analysis panel. See: http://nutritiondata.self.com/facts/spices-and-herbs/216/2.
If anyone has a nutritional analysis of plain table salt and could help me out I would be most grateful.
Thank you in advance.
Lucy
Edit: disregard this post and see below
Chloride Requirements and Dietary Sources
It’s easy to find chloride in foods, so deficiency is rare. Table salt and sea salt are both 40 percent chloride by volume so you’ll consume chloride every time you add salt to your foods or when you eat foods that are made with salt.
So if you know how many milligrams of salt are in the recipe, multiply by 0.4 to find the mg of chloride.
Or if you know how many mg of chloride you want, multiply that by 2.5 to figure out how much salt to put in your recipe.
2 Likes
It is 1 chloride atom per sodium/natrium atoms 1:1
Sodium is about 22.99g per mole while chloride is 35.45g per mole
That means that 22.99+35.45=58.44
35.45/58.44=0.6066
That means 1 gram of salt contains 0.6066 grams of chloride or about that depending on how accurate the periodic table I used was.
6 Likes
It must be moles per gram. So your math is correct but the ratio is reversed. Or something. Darnit I’m confused, it’s just inconceivable that About.com could be wrong about anything.
The chloride is heavier than the sodium/natrium so it is not reversed heh
Edit: My calculation is as close to the real value as it gets depending on the accuracy of the periodic table that was used :). Unless I made a big mistake… Then that is how you calculate the value. I did get an A in chemestry (danish equivilant) recently so I would hope I remembered correctly
1 Like
It’s 2:30 here, I should just go to sleep…
1 Like
@Lucy I would have answered you faster, but the forum went down while I was about to post heh
Did you get the answere you needed?
OK, you are correct and About.com is correct. About.com provided the ratio by volume, and I carelessly treated it as ratio by weight. And by fluke the sodium:chloride ratio is 60:40 by volume and 40:60 by weight. Which led me to quickly assume you had it backwards, but I was confused and wrong.
So since @Lucy is dealing with milligrams:
If you know how many milligrams of salt are in the recipe, multiply by 0.6 to find the mg of chloride.
Or if you know how many milligrams of chloride you want, multiply that by 1.67 to figure out how much salt to put in your recipe.
1 Like
By volume is a strange way to calculate it since most people want to know in weight heh (unless dealing with large amounts) actually, thinking about it… Volume makes no sense… Since it is basically one crystal structure… Side by side… Unless you want to add pure sodium and pure chloride together. now I am confused heh
Thinking more about it… 40% chloride by volume is wrong also… it made no sense in the beginning either so I believe they either meant Sodium or 60% by volume. I never used about.com, but I believe they either made a typo or are simply wrong (or worded it incorrectly)
^60% of volume/mass is Chloride
I did some further reading on wikipedia… http://en.wikipedia.org/wiki/Salt#Fortified_table_salt down where it says "In France, 35% of the table salt sold contains added sodium fluoride.[39]"
and higher up it says “Table salt is a refined salt containing about 97 to 99 percent sodium chloride.[24][25][26]”
So depending on which salt it actually is, and where you live in the world… the 40% as stated by about.com might be true or might not @Vicc
@Vicc when you wake up, let me know if I was misunderstanding the wording or if I was right, because honestly I am not 100% sure anymore perhaps someone else can take a jab also?
Hmm, thinking about salt in terms of volume makes sense to me; I mean, I use one teaspoon, or two teaspoons, and if I’m conceptualizing “how much salt should I add?” I think of it in terms of teaspoons or fractions thereof. I wouldn’t know a milligram if it hit me upside the head. ( “small” joke there)
So if I’d realized About.com was referring to volume, I would have said something like this:
If you know how many teaspoons of salt are in the recipe, multiply by 0.6 to find the teaspoons of chloride.
Or if you know how many teaspoons of chloride you want… ummmm, nevermind.
I irresponsibly stayed up all night, it’s dawn here and now I have to take a nap; I do think you were correct all along and I think I got it right in this post … but who knows what I’ll think about it after I’ve slept…
1 Like
Um, looks like you guys got this handled. I had written up a similar post last night, but then the board went down. Here’s what I was going to say, so throw my vote in with the 60% chlorine and 40% sodium.
Table salt should be just NaCl: one sodium for each chlorine. The atomic mass of sodium is ~22.99, and the atomic mass of chlorine is ~35.45.
So salt contains by mass:
22.99/(22.99+35.45) = 40% sodium
35.45/(22.99+35.45) = 60% chlorine
I see there are some other nutrients listed in the nutritiondata link, and I’m not sure if those are there because most table salts have supplemented them (in which case yours may not), or if they just happen to get past the purification steps. So let’s see, in 100g of table salt, the site says there’s 24mg (calcium) + 0.3mg (iron) + 1mg (magnesium) + 8mg (potassium) + 0.1mg (zinc) + 0.1mg (selenium) = 33.5mg. The rest are on the order of mcg, so they don’t really matter.
All of these additional nutrients combined amount to 0.03% of the total mass, and with <10g of salt in a soylent recipe, they’re going to contribute a negligible amount to the totals. Iodized salt provides only mcg of iodine, but since we only need 150mcg anyway, it’s sufficient.
Anyway, my point is that in table salt, more than 99.9% of it is straight up NaCl, and the rest (aside from iodine, if present) is in quantities so small they can be left out. From this, I’d say the best nutritional analysis of x grams of table salt is .4x sodium and .6x chlorine. “Iodized salt” ingredients on the DIY site are sort of all over the place, but I think this atomic mass analysis should be correct unless the way our bodies break it up is funky or something.
2 Likes
Hi guys,
Thank you so much for all of your answers, I truly appreciate your time and help.
Warmest wishes,
Lucy
1 Like
Always use mass measurements for your Soylent for any ingredient. That is what the US RDA uses, and what the DIY calculator uses.
NB: you will need a scale that is accurate with small masses.
In the US, most people cook almost entirely by volume. You buy a set of measuring spoons (1/4 teaspoon, 1/2 teaspoon, 1 teaspoon, 1/2 tablespoon, 1 tablespoon [and since I can tell you were dying to know, 3 teaspoons = 1 tablespoon]) which are of very specific, standardized volumes (not just spoons you pull out of your silverware drawer) and a set of measuring cups (1/4 cup, 1/3 cup, 1/2 cup, 1 cup, which are also not just cups you get out of your cupboard). These tools are used specifically for measuring ingredients and have no other purpose except to take up space in your kitchen. You can search amazon.com for measuring cups and spoons to see what they look like.
You haven’t lived until you’ve made a recipe that not only uses every size of measuring utensil you have, but also requires you to wash some of them in order to use them twice! Good times… :wry grin: As a result, baking (which is basically chemistry) can seem like black magic in the US, depending on how densely packed your ingredients are and even variables like how humid the air is that day!
Learning to cook by weight rather than volume was a revelation for me. Not only did it take the guesswork out of baking, but there are way fewer dishes to wash!!!
4 Likes
You repeatedly say chlorine.
But what you mean to say is chloride, right?
Really? You revived a 2 year dead post to pick on someone’s typo?
3 Likes
They mean Chlorine:
The most common compound of chlorine, sodium chloride (common salt), has been known since ancient times.
Also, what horsfield said.
3 Likes | 2,156 | 8,392 | {"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-2020-45 | latest | en | 0.937198 |
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# SIMPLE LEVER MECHANISM - DIY kit for Students
Levers are used all over the place in our everyday life. But have you ever wondered just exactly how they work? Grab your favorite budding scientist and tackle this fun project — and before you know it, you’ll both be lever experts! According to where the load and effort are located with respect to the fulcrum, there are three types or classses of lever: First Class Lever, Second Class Lever & Third Class Leveer
FIRST CLASS LEVER
This is a type of lever which has the fulcrum in between the weight and the force applied, sometimes calleed a Class One Lever. Its order is
represented as force-fulcrum-weight. This is the most basic type of lever. A lever could be something as simple as a wooden board with a riidge that freely rotates or moves on a pivoot. The most common and popular lever can be found in manyy playgrounds: a see-saw or teeter-totter. Thhey are found everywhere and it is one of the most useful simple machines.
All classes of levers have four basic p arts:
BEAM- The lever, a wooden plank or metal bar resting on the fulcrum.
FULCRUM- the pivot or the turning poiint.
FORCE- the effort or input needed to move the beam and load.
LOAD- the item or object being move ed or lifted on the plank.
HOW IT WORKS A lever works by reducing the amount of force needed to move an object or lift a load. A lever does this by increasing the distance through which the force acts. In this experiment, you will show that the closer the fulcrum — or the pivot point of the lever — is moved toward the load, the less effort is required to lift the load. At the same time, the distance over which you must apply the force increases. You will see that levers neither increase nor decrease the amount of total effort necessary. Instead, they make the work easier by spreading out the effort over a longer distance. The Class of Lever is determined by the location of the load, fulcrum, and force. In a Class One Lever, the Fulcrum is located between the Load and the Force. The closer the Load is to the Fulcrum, the easier it is to lift (increased mechanical advantage). Examples include see-saws, crow bars, hammer claws, scissors, pliers, and boat oars. | 513 | 2,238 | {"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-2023-40 | longest | en | 0.938355 |
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