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Author: trintith [ Tue Feb 28, 2012 4:20 pm ] Post subject: Ccc 2012 How did everyone do? I made a really, really stupid mistake on S4, which means that I won't get any points for n=6,7 (so I doubt I made stage 2). The other questions seemed pretty straightforward. I don't think we're allowed to talk about specific solutions yet.
Author: Snario [ Tue Feb 28, 2012 5:02 pm ] Post subject: RE:Ccc 2012 What did you do for S4?
Author: trintith [ Tue Feb 28, 2012 5:04 pm ] Post subject: RE:Ccc 2012 We're not allowed to give specific solutions, if it's like last year.
Author: crossley7 [ Tue Feb 28, 2012 5:10 pm ] Post subject: RE:Ccc 2012 Yeah, probably safest to wait a day or 2 to post or discuss any actual solutions. 1,2,3,5 were all really easy and probably took me a little over an hour combined to do them all, but I didn't know how to do 4. I already know my solution times out so as I ran out of time I put a cap on it so that it might have a slightly better chance to get some cases. Hoping to get lucky enough to make round 2 with a score around 64-65.. Excited to see the proper solution for 4 (I already know mine is a bunch of crap. Couldn't get my head around an efficient solution and used about the worst possible algorithm) I was actually really disappointed with the difficulty of the contest and thought 5 should have been maybe #3 and then ditch number 3 altogether for a challenging final problem. There will be too many scores in mid 60's or higher.
Author: ultimatebuster [ Tue Feb 28, 2012 5:29 pm ] Post subject: RE:Ccc 2012 Thats what we said last year and stage 2 ends being around 62... remind you people in ib or ap can't go due to exams... like myself Found 1235 to be simple, though my solution for 3 may take too long for 2 million lines... python is pretty slow .. also I madethe mistake of looping through the entire set up one more time than I had to... it took about half a second with 2 million inputs 4 is interesting ... my solution is not efficient enough and I didn't have enough time to intergrate a star as I misread 4 and wasted about an half hour. Anyway it was fun... probably won't get much over 65, unfortunately
Author: ihsh [ Tue Feb 28, 2012 5:46 pm ] Post subject: RE:Ccc 2012 Yeah 1, 2, 3 and 5 were so easy that I finished them within 1-1.5 hours. But I just couldn't even think of an algorithm for problem 4, for I had trouble finding a good representation of each state.
Author: ProgrammingFun [ Tue Feb 28, 2012 5:49 pm ] Post subject: Re: RE:Ccc 2012 ultimatebuster @ Tue Feb 28, 2012 5:29 pm wrote:Thats what we said last year and stage 2 ends being around 62... remind you people in ib or ap can't go due to exams... like myselfWait, what?
Author: ultimatebuster [ Tue Feb 28, 2012 5:52 pm ] Post subject: Re: RE:Ccc 2012 ihsh @ Tue Feb 28, 2012 5:46 pm wrote:Yeah 1, 2, 3 and 5 were so easy that I finished them within 1-1.5 hours. But I just couldn't even think of an algorithm for problem 4, for I had trouble finding a good representation of each state. I didn't have time but i think A star search would have done it.
Author: mirhagk [ Tue Feb 28, 2012 6:41 pm ] Post subject: RE:Ccc 2012 A* is only as good as the heuristic you choose, which is hard in this scenario. PS: I did pretty well this year (got 1,2,3,5 and some cases for 4), but it sounds like that might not be good enough, which would suck. Last year was 45 for the certificate cut-off for senior)
Author: 102897108 [ Tue Feb 28, 2012 7:00 pm ] Post subject: RE:Ccc 2012 This year's difficulty is so low that I spent 1.5 hrs checking errors. My eyes were almost bleeding at the end. Q3 and Q5 are overly simple. I guess the cutoff is around the same as last year
Author: DanielKang [ Tue Feb 28, 2012 7:11 pm ] Post subject: Re: Ccc 2012 Hello, I did the junior one, and the same problem appeared as J5. Thankfully, after rocking my head trying to figure out how to do it, I was able to successfully analyze each case then use println. Notice that for S4, n<8, while in J5, n<5, which made manual work possible (33 cases). Can I get full marks with this approach? Where I just do the calculations by hand and just print the results out for each case? If I do, then I have high chance of getting 75/75 which would be awesome considering this is my very first compsci competition.
Author: crossley7 [ Tue Feb 28, 2012 7:11 pm ] Post subject: RE:Ccc 2012 Seems like most people got 4 and part solution. Hopefully that is just because the top people are all posting on here I get my results tomorrow during class when my teacher goes over them (He normally has a student help him with scoring programs) Q3 and 5 were far too easy for their question number as was already stated. Hoping the cut off is around where it was last year so I can hopefully make it
Author: RandomLetters [ Tue Feb 28, 2012 7:20 pm ] Post subject: RE:Ccc 2012 How did you manage to play 33 towers of hanoi games in under 3 hours?
Author: ihsh [ Tue Feb 28, 2012 7:22 pm ] Post subject: Re: Ccc 2012 DanielKang @ Tue Feb 28, 2012 7:11 pm wrote:Hello, I did the junior one, and the same problem appeared as J5. Thankfully, after rocking my head trying to figure out how to do it, I was able to successfully analyze each case then use println. Notice that for S4, n<8, while in J5, n<5, which made manual work possible (33 cases). Can I get full marks with this approach? Where I just do the calculations by hand and just print the results out for each case? If I do, then I have high chance of getting 75/75 which would be awesome considering this is my very first compsci competition. Wait, I thought for n<=5, there would be 5!=120 cases?
Author: DanielKang [ Tue Feb 28, 2012 7:23 pm ] Post subject: Re: Ccc 2012 n is less than 5, does not include n = 5
Author: crossley7 [ Tue Feb 28, 2012 7:29 pm ] Post subject: RE:Ccc 2012 I think that there are n! possible permutations so if n<5 then it would be 1!+2!+3!+4! = 33 possible starting positions with 3 of them being easy to solve. I won't go into more detail there because I think we are still supposed to wait on posting any solutions for longer (some people will have to write tomorrow or something) try not talking about problem specifics right now until we know it is ok to discuss
Author: DanielKang [ Tue Feb 28, 2012 7:31 pm ] Post subject: Re: RE:Ccc 2012 RandomLetters @ Tue Feb 28, 2012 7:20 pm wrote:How did you manage to play 33 towers of hanoi games in under 3 hours? There is some speculations, but i cannot say because we are not yet to discuss.
Author: Mamick [ Tue Feb 28, 2012 7:42 pm ] Post subject: Re: Ccc 2012 My teacher marked everyone's an I got perfect, though it hasn't been confirmed by Troy yet. I do agree with everyone else though that S4 was the hardest. My program would have failed for n=8 and above, and it took about a second for the n=7. Every other question was pretty trivial in my opinion. I really hope the cutoff is below 55 so other people from my school can make it. Wasn't it about there a few years ago?
Author: chengbin [ Tue Feb 28, 2012 7:49 pm ] Post subject: Re: Ccc 2012 Are they trying to make the cutoff a perfect score this year??? 1, 2, 3, and 5 were trivial. I was fairly close to solving 4. There has to be at least 20 people in Canada that can get S4 perfectly. It is not terribly hard, just tedious to code.
Author: crossley7 [ Tue Feb 28, 2012 7:50 pm ] Post subject: RE:Ccc 2012 It will be over 60 because of the difficulty. The last few years the problems have been easier (this year + last year) allowing much higher marks and therefore a higher cutoff. Also, I thought time limit was something like a minute (no idea why) which would allow for alightly weaker solutions. As it is, that still cutoff the most naive solutions for #4. Last year 54 was the lowest score for honourable mention and this year the problems were arguably easier than last year meaning that sub 55 score definitely won't make it on. Once we can post solutions, I would love to see how you solved #4 Mamick.
Author: chengbin [ Tue Feb 28, 2012 7:55 pm ] Post subject: RE:Ccc 2012 This year is really unfair for the good programmers. If S4 is not your type of question, which is a really bad exercise in recursion, then you're not making stage 2. There should be at least 2 really hard questions so the cutoff a near perfect score.
Author: Mamick [ Tue Feb 28, 2012 7:57 pm ] Post subject: Re: Ccc 2012 I actually wasted about half an hour because I thought it was n<=8, but then I reread the question. I feel that this year's was harder than last year's. In my opinion S1 was harder, S2 was about the same, S3 was about the same (though less tedious), S4 is about as hard as S5 from last year, and S5 requires more knowledge than S4 last year (though less logic is needed).
Author: DanielKang [ Tue Feb 28, 2012 7:57 pm ] Post subject: Re: Ccc 2012 chengbin wrote: This year is really unfair for the good programmers. If S4 is not your type of question, which is a really bad exercise in recursion, then you're not making stage 2. There should be at least 2 really hard questions so the cutoff a near perfect score. ? That didn't make sense.
Author: crossley7 [ Tue Feb 28, 2012 8:04 pm ] Post subject: RE:Ccc 2012 I felt that s1 and s2 were trivial both years so just discounted them, s3 was a bit easier than last year, s4 this year and s5 last year are comparable and s5 a bit easier than s4 was last year. Again that is my opinion on it, but that is also probably partly because I have an extra year of coding and so I pick up the problems a little bit quicker. And yeah, it kind of sucks that if you struggle with using recursion at all (like I do) then you are in for a challenge to make it to round 2 this year.
Author: chengbin [ Tue Feb 28, 2012 8:06 pm ] Post subject: Re: Ccc 2012 DanielKang @ Tue Feb 28, 2012 7:57 pm wrote:chengbin wrote: This year is really unfair for the good programmers. If S4 is not your type of question, which is a really bad exercise in recursion, then you're not making stage 2. There should be at least 2 really hard questions so the cutoff a near perfect score. ? That didn't make sense. You have to get perfect to make stage 2 since 4 of the problems are trivial. What if it is an insane ad hoc problem (which is the case this year) that you just can't get? As a senior contest, where are the true computer science problems involving hard dynamic programming, recursion, graph theory or computational geometry that I like (actually I don't like graph theory)?
Author: Mamick [ Tue Feb 28, 2012 8:10 pm ] Post subject: Re: Ccc 2012 chengbin @ Tue Feb 28, 2012 8:06 pm wrote:DanielKang @ Tue Feb 28, 2012 7:57 pm wrote:chengbin wrote: This year is really unfair for the good programmers. If S4 is not your type of question, which is a really bad exercise in recursion, then you're not making stage 2. There should be at least 2 really hard questions so the cutoff a near perfect score. ? That didn't make sense. You have to get perfect to make stage 2 since 4 of the problems are trivial. What if it is an insane ad hoc problem (which is the case this year) that you just can't get? As a senior contest, where are the true computer science problems involving hard dynamic programming, recursion, graph theory or computational geometry that I like (actually I don't like graph theory)? I would just like to say that I've never seen a problem anything like S4 before, yet I still managed to get it. I do agree that the other problems should be harder, or there should be more (difficult) questions.
Author: RandomLetters [ Tue Feb 28, 2012 8:11 pm ] Post subject: RE:Ccc 2012 I thought S4 last year was not that hard (although incredibly tedious). I didn't get S4 and S5 at all this year though. It would have been nice to see a more uniform progression in difficulty though, rather than 3 questions which everyone gets, and 2 questions which most people simply frustrate over
Author: 102897108 [ Tue Feb 28, 2012 8:14 pm ] Post subject: Re: Ccc 2012 chengbin @ Tue Feb 28, 2012 5:06 pm wrote:DanielKang @ Tue Feb 28, 2012 7:57 pm wrote:chengbin wrote: This year is really unfair for the good programmers. If S4 is not your type of question, which is a really bad exercise in recursion, then you're not making stage 2. There should be at least 2 really hard questions so the cutoff a near perfect score. ? That didn't make sense. You have to get perfect to make stage 2 since 4 of the problems are trivial. What if it is an insane ad hoc problem (which is the case this year) that you just can't get? As a senior contest, where are the true computer science problems involving hard dynamic programming, recursion, graph theory or computational geometry that I like (actually I don't like graph theory)? S4 is a "true computer science problem" and anyone that qualifies for stage 2 should be able to solve that without too much struggle
Author: ihsh [ Tue Feb 28, 2012 8:15 pm ] Post subject: RE:Ccc 2012 I think 1 and 2 are perfectly fine as they are (trivial). S3 is easy. It really doesn't require too much thought. S4 is difficult compared to previous years; I had solved every S4 except the one from 2010 and this one. S5 is definitely too easy; it's more like a S3 problem. There isn't really even a need to be efficient because of the small bounds.
Author: bbi5291 [ Tue Feb 28, 2012 8:21 pm ] Post subject: Re: Ccc 2012 chengbin, based on experience, I highly doubt that 20 contestants will get full marks on S4. The cutoff is not going to be a perfect score. Bright contestants usually overestimate the cutoff, because they expect everyone else to find the contest as easy as they do. S4 is not "a really bad exercise in recursion"; it has a straightforward solution. But even if it were, it should be pointed out that the early IOIs were full of such kinds of problems.
Author: Advecticity [ Tue Feb 28, 2012 8:31 pm ] Post subject: RE:Ccc 2012 What are your opinions on S4 as a contest question in general? A problem that involves a straightforward solution that works for n = 7 but that would likely fail for n = 9 or something seems to just invite dirty solutions. Unless the solutions that the people that solved S4 came up with are different then what I have in mind.
Author: Yves [ Tue Feb 28, 2012 8:54 pm ] Post subject: RE:Ccc 2012 The last case for #4 was surprisingly easy; my extremely slow solution just passed with 49 seconds on my computer, so I tentatively have a perfect. (:
Author: ihsh [ Tue Feb 28, 2012 8:56 pm ] Post subject: Re: RE:Ccc 2012 Yves @ Tue Feb 28, 2012 8:54 pm wrote:The last case for #4 was surprisingly easy; my extremely slow solution just passed with 49 seconds on my computer, so I tentatively have a perfect. (: The test-cases are out to the public already? If so, could you tell me where I could find them? Thanks.
Author: Drenn [ Tue Feb 28, 2012 8:57 pm ] Post subject: Re: Ccc 2012 I did 1-3 relatively quickly. My solution to #3 started working before I expected it to... which is really weird... I hope it works in all test cases. I spent the rest of my time on #4 expecting #5 to be way too hard for me. After giving it a closer look I realized it was far easier than I thought! Oh well. I couldn't get 4 and didn't have the time to start 5.
Author: Snario [ Tue Feb 28, 2012 9:49 pm ] Post subject: Re: Ccc 2012 DanielKang @ Tue Feb 28, 2012 7:11 pm wrote:Hello, I did the junior one, and the same problem appeared as J5. Thankfully, after rocking my head trying to figure out how to do it, I was able to successfully analyze each case then use println. Notice that for S4, n<8, while in J5, n<5, which made manual work possible (33 cases). Can I get full marks with this approach? Where I just do the calculations by hand and just print the results out for each case? If I do, then I have high chance of getting 75/75 which would be awesome considering this is my very first compsci competition. Same here. Was it just factorial business? I didn't see a pattern
Author: crossley7 [ Tue Feb 28, 2012 10:35 pm ] Post subject: RE:Ccc 2012 it wan't just factorials. n! is the number of possible inputs for input of size n but other than that I don't think the result had anything to do with factorials. No simple math that you could do to solve.
Author: d310 [ Tue Feb 28, 2012 10:35 pm ] Post subject: Re: Ccc 2012 I also got perfect! I do think that problem 5 should have been much harder.
Author: mirhagk [ Tue Feb 28, 2012 11:12 pm ] Post subject: RE:Ccc 2012 So I see 3 perfects here.... and a couple more that probably did better than me.... Dang looks like me 4.2/5 problems just ain't going to cut it.
Author: crossley7 [ Tue Feb 28, 2012 11:18 pm ] Post subject: RE:Ccc 2012 I'm hoping that 4.5/5 will work (although that depends based on test cases for 4. May be higher/lower if they like 6 and 7 that have solutions) I just output impossible for 6 and 7 because it took my program too long to compute. It will be frustrating for me not knowing my score until last period at school tomorrow and seeing all these high scores raising the bar for round 2.
Author: Velocity [ Wed Feb 29, 2012 12:14 am ] Post subject: RE:Ccc 2012 4.9/5 ... Just when i thought it was too easy. Its never perfect!!!!!!!! How do i know if i qualify for round 2
Author: 102897108 [ Wed Feb 29, 2012 12:17 am ] Post subject: Re: RE:Ccc 2012 Velocity @ Tue Feb 28, 2012 9:14 pm wrote:4.9/5 ... Just when i thought it was too easy. Its never perfect!!!!!!!! How do i know if i qualify for round 2 Haha I'm really curious where did you lose that 0.1 mark?
Author: DanielKang [ Wed Feb 29, 2012 12:38 am ] Post subject: Re: Ccc 2012 Snario @ Tue Feb 28, 2012 9:49 pm wrote:DanielKang @ Tue Feb 28, 2012 7:11 pm wrote:Hello, I did the junior one, and the same problem appeared as J5. Thankfully, after rocking my head trying to figure out how to do it, I was able to successfully analyze each case then use println. Notice that for S4, n<8, while in J5, n<5, which made manual work possible (33 cases). Can I get full marks with this approach? Where I just do the calculations by hand and just print the results out for each case? If I do, then I have high chance of getting 75/75 which would be awesome considering this is my very first compsci competition. Same here. Was it just factorial business? I didn't see a pattern A pattern was obvious, i will say it later
Author: ultimatebuster [ Wed Feb 29, 2012 8:30 am ] Post subject: RE:Ccc 2012 Just confirmed that my solution for S4 takes about 1 min to run for n=8 n=7 around 2 seconds.. Damn python. lol
Author: saltpro15 [ Wed Feb 29, 2012 10:20 am ] Post subject: RE:Ccc 2012 When will the senior problems be released? I'm undergrad this year but I'd still like to take a shot at some of them. I'm assuming from the posts here that the difficulty was lowered a bit from previous years?
Author: danc [ Wed Feb 29, 2012 2:01 pm ] Post subject: RE:Ccc 2012 I got 60... so maybe not good enough to advance. I just didn't get an at all working solution for 4... though I know how to do it algorithmically. It's just difficult to represent. I was a little sad that recursion would run in time for S5. And with the difficulty in general.
Author: crossley7 [ Wed Feb 29, 2012 5:06 pm ] Post subject: RE:Ccc 2012 saltpro, hit me up on fb and I will send you the problems. I have my copy and perfect solutions for 4 of them.
Author: A.J [ Wed Feb 29, 2012 5:46 pm ] Post subject: RE:Ccc 2012 I'm glad most of you found the contest enjoyable and/or easy. I must agree that the contest was easier than most years, however I don't think that the cutoff will be close to perfect (maybe mid to high 60's). Please refrain from discussing the solutions for another day or so. I can post a copy of the contest after a couple of days (if people want this).
Author: crossley7 [ Wed Feb 29, 2012 6:38 pm ] Post subject: RE:Ccc 2012 Crossing my fingers that it is low-mid 60's so I can make it on. Either way, looking forward to seeing other solutions for the problems (specifically 3,4,5)
Author: Velocity [ Wed Feb 29, 2012 6:50 pm ] Post subject: RE:Ccc 2012 S3, i could not find the correct base value for 500, some of the values worked but 500 didnt.
Author: NeilV [ Wed Feb 29, 2012 8:56 pm ] Post subject: RE:Ccc 2012 Yeah I finished S1,S2,S3, and S5 within an hour. Then I spend a solid 2 hours on S4 and still didn't figure it out. But very shortly afterwards I came up with a solution that would have gotten at least 12/15, possibly 15. The frustrating part is that when I implemented this new solution, 90% of my code was copied and pasted from my solution during the contest. Anyway, final score is 63.
Author: ultimatebuster [ Thu Mar 01, 2012 8:02 am ] Post subject: RE:Ccc 2012 Any idea what's the Max execution time allowed? Also can we talk about solutions now?
Author: danc [ Thu Mar 01, 2012 11:41 am ] Post subject: RE:Ccc 2012 max execution time is 60 seconds...
Author: ultimatebuster [ Thu Mar 01, 2012 3:06 pm ] Post subject: RE:Ccc 2012 My script for 4 gets it after 48 sec for n=8 on cpython on my computer ... 16 sec on pypy. Given that my marker 's laptop is probably slower... maybe u should convince her to get pypy?
Author: crossley7 [ Thu Mar 01, 2012 3:56 pm ] Post subject: RE:Ccc 2012 I ended up with 66 and probably still wait to comment on solutions. I think AJ will have an idea of when we can start discussing them and will let us know when that is the case
Author: Yves [ Thu Mar 01, 2012 4:36 pm ] Post subject: Re: RE:Ccc 2012 ultimatebuster @ Thu Mar 01, 2012 3:06 pm wrote:My script for 4 gets it after 48 sec for n=8 on cpython on my computer ... 16 sec on pypy. Given that my marker 's laptop is probably slower... maybe u should convince her to get pypy? It doesn't really matter, as every program with an unofficial score of at least 40 gets remarked on the same computer for the official score.
Author: 102897108 [ Thu Mar 01, 2012 7:53 pm ] Post subject: RE:Ccc 2012 Oh yeah!!!! I got perfect 75~ Isn't 1 min of execution time too long?? The contests I have participated before mostly had a time limit of 1 sec or 2 sec. It will take a lot of time to mark if everyone's program runs 1 min for each test case= = At least my solution for Q4 gives answer well within 1s for n<8
Author: Annology [ Thu Mar 01, 2012 8:09 pm ] Post subject: Re: Ccc 2012 102897108 wrote: Oh yeah!!!! I got perfect 75~ Woah, congrats This is my first computing contest....I got 60. Would that be enough to make it to honor roll or group 3?
Author: danc [ Thu Mar 01, 2012 8:26 pm ] Post subject: Re: Ccc 2012 Annology @ Thu Mar 01, 2012 8:09 pm wrote:102897108 wrote: Oh yeah!!!! I got perfect 75~ Woah, congrats This is my first computing contest....I got 60. Would that be enough to make it to honor roll or group 3? I guess we won't know 'til the results come out, eh?
Author: crossley7 [ Thu Mar 01, 2012 8:27 pm ] Post subject: RE:Ccc 2012 It gives 1 min since some programs take very long to execute and it has some of the lowest restrictions of any contest I know of. Also, since it's your first contest, did you do Junior or Senior? If you did senior it would be some level of the honour role at the very least.
Author: ihsh [ Thu Mar 01, 2012 8:36 pm ] Post subject: Re: Ccc 2012 The unofficial solutions site has posted the test cases and problems, so I think we can freely discuss the problems now. Anyway, from the test cases on the site, I have an unofficial score of 63. I hope this would somehow give me a rank of 20th... (though I highly doubt it)... And yeah, I am very curious about how you guys solved #4.
Author: danc [ Thu Mar 01, 2012 9:12 pm ] Post subject: RE:Ccc 2012 Well, me being a noob, I didn't know that in C++ you can store vectors inside pairs... so I didn't get it until after I found that out (after the contest D: ) but (if we're allowed to post solutions?) it's BFS. You can represent the game as a graph, with edge weights of 1 and edges between any given state and all the possible moves from there. (If you don't think I should have posted that, yell at me and I'll take it down)
Author: crossley7 [ Thu Mar 01, 2012 9:50 pm ] Post subject: RE:Ccc 2012 bfs is 1 way to solve it although the naive implementation is expensive for memory (though how expensive I don't know). There is also the recursive solution (basically a dfs i guess) and most any search algorthm would work if you knew the correct way to implement it. I used the naive recursion with an inefficient memoization that times out for 3 of the 5 test cases but does at least get the correct answer eventually
Author: danc [ Thu Mar 01, 2012 10:17 pm ] Post subject: RE:Ccc 2012 Hmmm, my solution uses 54 Mb of memory and runs in 10s on the largest case... but that's because I use and compare dynamic arrays representing the states... Of course, it can also be done by representing the state in a string, but that's more difficult but that makes it much faster and better for memory.
Author: 102897108 [ Thu Mar 01, 2012 10:34 pm ] Post subject: RE:Ccc 2012 I used a number(indicating the No. of a state) to represent a state and some minor tricks to avoid unnecessary calculations. Memory used is 14mb
Author: crossley7 [ Thu Mar 01, 2012 11:01 pm ] Post subject: RE:Ccc 2012 also curious how people went about number 3 with the giant input. I had a frequency array of 1 - 1000 that would slowly increment values as it reads in the 2 million inputs then naively finds how many max values, second max values there are and then checked the 3 cases for the output (multiple max, 1 max, multiple second max, 1 max, 1 second max). My solution runs in about half a second in the worst case in c++. What were other ways of going about that on (ideally more efficient)
Author: trintith [ Thu Mar 01, 2012 11:41 pm ] Post subject: crossley7 @ Thu Mar 01, 2012 11:01 pm wrote:also curious how people went about number 3 with the giant input. I had a frequency array of 1 - 1000 that would slowly increment values as it reads in the 2 million inputs then naively finds how many max values, second max values there are and then checked the 3 cases for the output (multiple max, 1 max, multiple second max, 1 max, 1 second max). My solution runs in about half a second in the worst case in c++. What were other ways of going about that on (ideally more efficient) My C++ solution runs in 70 ms on a solid state, or 200 ms on a normal hard drive on the worst case. I used sort(...) from on a vector filled with elements of a class which stores value (pH) and count (frequency). You need to define a member function in this class (lets call the class 'C'): 'bool operator<(const C& other) const' based on frequency. Once it's sorted by frequency, I did the trickery you described with the max frequency and second-to-max frequency. This is dominated by the hard drive, but is there a better way? Also, when grading solutions, are all test cases for a question worth the same? Are all questions worth the same. My wonderful teacher promised to finish marking by the end of next week and I'm impatient.
Author: danc [ Thu Mar 01, 2012 11:44 pm ] Post subject: RE:Ccc 2012 Test cases are all worth the same (3), except for number 5 I think, which has 7 test cases with between 1-3 points.
Author: crossley7 [ Thu Mar 01, 2012 11:55 pm ] Post subject: RE:Ccc 2012 Nice, I don't like using classes for contest problems (I generally feel they are overkill), but it would have cut down the time for the second aspect of the program (after assigning input). Nice solution there. I haven't checked my exact runtime, but I do know that considering it does over 4 million commends (2 million read, 2 million increments) + finds the max/min in under half a second I feel good about that. Now for my teacher to send my code to Waterloo for official marking and find out when round 2 qualifiers get posted
Author: trintith [ Fri Mar 02, 2012 12:23 am ] Post subject: RE:Ccc 2012 I got 63. I did a DFS instead of a BFS for 4. I even knew I needed to do a BFS. A two-line change (which I thought I made), and I would have gotten 72. Je suis idiot.
Author: VK_eipi [ Fri Mar 02, 2012 4:02 am ] Post subject: Re: Ccc 2012 For S3, after building the frequency table, you can iterate through the possible readings 1 to 1000 just once, keeping track of 6 variables: frequency, lower bound, and upper bound of the max-frequency and second-max-frequency readings. Then at the end, you check which case it is and work it out. E.g. If second max frequency is different from max (only one reading with max, so max_low=max_high), then take the bigger of (max_low - max2_low) and (max2_high - max_low), if one is negative the other's absolute value is bigger anyways. Because you iterate in order, simplifications can be made such as only setting lower bound when a higher frequency is found and setting upper bound every single time the frequency matches. (You can even share one variable for the two upper bounds, since they are never simultaneously relevant, but it makes things confusing). For S4, my friend came up with the idea to represent the state as an n-length array of n-bit binary numbers (integers), which might be the same as what user 102897108 means by "a number." Essentially, the coins are considered to be worth powers of two and each pile is represented by its total value, with moving a coin just equal to adding and subtracting the relevant entries. The smallest coin in a pile is equal to the smallest significant 1-bit, which can be calculated by the trick x&(-x).
Author: 102897108 [ Fri Mar 02, 2012 4:11 am ] Post subject: Re: Ccc 2012 VK_eipi @ Fri Mar 02, 2012 1:02 am wrote:For S3, after building the frequency table, you can iterate through the possible readings 1 to 1000 just once, keeping track of 6 variables: frequency, lower bound, and upper bound of the max-frequency and second-max-frequency readings. Then at the end, you check which case it is and work it out. E.g. If second max frequency is different from max (only one reading with max, so max_low=max_high), then take the bigger of (max_low - max2_low) and (max2_high - max_low), if one is negative the other's absolute value is bigger anyways. Because you iterate in order, simplifications can be made such as only setting lower bound when a higher frequency is found and setting upper bound every single time the frequency matches. (You can even share one variable for the two upper bounds, since they are never simultaneously relevant, but it makes things confusing). For S4, my friend came up with the idea to represent the state as an n-length array of n-bit binary numbers (integers), which might be the same as what user 102897108 means by "a number." Essentially, the coins are considered to be worth powers of two and each pile is represented by its total value, with moving a coin just equal to adding and subtracting the relevant entries. The smallest coin in a pile is equal to the smallest significant 1-bit, which can be calculated by the trick x&(-x). Haha that's what i meant. But i guess what you wanted to say was that the coins are worth powers of N instead of powers of 2 )
Author: VK_eipi [ Fri Mar 02, 2012 5:02 am ] Post subject: Re: Ccc 2012 102897108 @ Fri Mar 02, 2012 1:11 am wrote:VK_eipi @ Fri Mar 02, 2012 1:02 am wrote:For S3, after building the frequency table, you can iterate through the possible readings 1 to 1000 just once, keeping track of 6 variables: frequency, lower bound, and upper bound of the max-frequency and second-max-frequency readings. Then at the end, you check which case it is and work it out. E.g. If second max frequency is different from max (only one reading with max, so max_low=max_high), then take the bigger of (max_low - max2_low) and (max2_high - max_low), if one is negative the other's absolute value is bigger anyways. Because you iterate in order, simplifications can be made such as only setting lower bound when a higher frequency is found and setting upper bound every single time the frequency matches. (You can even share one variable for the two upper bounds, since they are never simultaneously relevant, but it makes things confusing). For S4, my friend came up with the idea to represent the state as an n-length array of n-bit binary numbers (integers), which might be the same as what user 102897108 means by "a number." Essentially, the coins are considered to be worth powers of two and each pile is represented by its total value, with moving a coin just equal to adding and subtracting the relevant entries. The smallest coin in a pile is equal to the smallest significant 1-bit, which can be calculated by the trick x&(-x). Haha that's what i meant. But i guess what you wanted to say was that the coins are worth powers of N instead of powers of 2 ) I actually do mean powers of 2, because I have a separate number for each position. The coins can only be "there"/"not there". I considered using powers of N (base-N number with digits equal to pile #) since it is the most compact way to represent the state, but I didn't know how to determine the smallest coin of each pile efficiently, since the piles are all jumbled together. What is your method for doing this?
Author: trishume [ Fri Mar 02, 2012 8:09 am ]
Post subject: Re: Ccc 2012
I was using ruby to write the contest so I thought that it would be too slow for S3 but it wasn't.
The first thing I did is write a script to generate a huge test case so I could constantly check speed.
It turns out that it was fast enough and that ruby's massive method chains helped me write a really short solution
For S4 I only had an hour left and I am a really slow programmer, so I had to make the decision between taking the whole hour to write an A* solution, or to look for a simpler mathematical solution.
Unfortunately I never found a mathematical solution. So my final submission was
code: puts "IMPOSSIBLE" # for me to find a mathematical solution...
Author: NeilV [ Fri Mar 02, 2012 5:54 pm ] Post subject: RE:Ccc 2012 see this is why I love python. For S4 I just kept the state as a list and stored it in a dictionary, which has constant lookup time. In other words, I let python hash it for me
Author: bl0ckeduser [ Fri Mar 02, 2012 6:42 pm ] Post subject: Re: Ccc 2012 Can anyone explain how the grade is calculated ? Someone seems to have mentionned "full 75", but I only see 27 test files on the Robart website. BTW my unoptimal S1, S2, S5 C solutions are here: http://sites.google.com/site/bl0ckeduserssoftware/ccc-2012-partial-sols.zip (I messed up S3, didn't know what to do for S4)
Author: Yves [ Fri Mar 02, 2012 6:52 pm ] Post subject: Re: Ccc 2012 bl0ckeduser @ Fri Mar 02, 2012 6:42 pm wrote:Can anyone explain how the grade is calculated ? Someone seems to have mentionned "full 75", but I only see 27 test files on the Robart website. BTW my unoptimal S1, S2, S5 C solutions are here: http://sites.google.com/site/bl0ckeduserssoftware/ccc-2012-partial-sols.zip (I messed up S3, didn't know what to do for S4) For S1 to S4, there are 5 cases each with 3 points awarded per case. Partial marks are not awarded for partially correct cases. For S5, there are 7 cases. The first four are worth 2 each, the second-to-last worth 1, and the remaining two 3 each.
Author: ihsh [ Fri Mar 02, 2012 10:27 pm ]
Post subject: Re: Ccc 2012
I finally coded a working solution for S4, which involves a DFS, and states represented like this:
code: 1344 //coin '1' is in location 1, coin '2' is in location 3, coin '3' is in location 4, and coin '4' is in location 4
And since it's a brute-force DFS, I had to limit the the number of steps to 50 for it to run in time...
Author: danc [ Fri Mar 02, 2012 10:34 pm ] Post subject: RE:Ccc 2012 But why not just use BFS with the same state? BFS, I believe, would run much faster and be not much more difficult to code? But I probably don't know best.
Author: ihsh [ Fri Mar 02, 2012 10:51 pm ] Post subject: Re: RE:Ccc 2012 danc @ Fri Mar 02, 2012 10:34 pm wrote:But why not just use BFS with the same state? BFS, I believe, would run much faster and be not much more difficult to code? But I probably don't know best. Because I don't know how to implement a BFS. Oh, and I just noticed that my java IDE doesn't support the Queue class. Hmm, it seems like I should really switch to another IDE; no Scanner, no ArrayList, no HashMap, and now no Queue. Also, please allow me to ask a very ignorant question: without queues, is it possible to (easily) code a BFS?
Author: crossley7 [ Fri Mar 02, 2012 11:19 pm ] Post subject: RE:Ccc 2012 create a manual queue with an array and index of where you started. Ideally a flexible array/vector would be best to use for this purpose or at least something that you can quickly remove elements/indexes from. In C++ I normally use vectors for my queue in a bfs, but you can use makeshift ways of creating a queue.
Author: VK_eipi [ Sat Mar 03, 2012 1:26 am ] Post subject: Re: RE:Ccc 2012 NeilV @ Fri Mar 02, 2012 2:54 pm wrote:see this is why I love python. For S4 I just kept the state as a list and stored it in a dictionary, which has constant lookup time. In other words, I let python hash it for me To clarify, were you referring to a tuple instead of an actual Python list? Lists in Python are not hashable and cannot be used as dictionary keys.
Author: danc [ Sat Mar 03, 2012 7:52 am ] Post subject: Re: RE:Ccc 2012 crossley7 @ Fri Mar 02, 2012 11:19 pm wrote:create a manual queue with an array and index of where you started. Ideally a flexible array/vector would be best to use for this purpose or at least something that you can quickly remove elements/indexes from. In C++ I normally use vectors for my queue in a bfs, but you can use makeshift ways of creating a queue. So do you usually just leave all the elements in the vector and simply shift your index of access variable? Why not just use a queue? And BFS is not too hard. Sounds like maybe you do know how to implement it if you're asking about a queue. You push your starting state onto the queue, and loop until some condition is true (in this case, whether you win, or queue is empty). Inside the loop, you access the next state, and then find all the possible moves from there and push them onto the back of the queue. It is also necessary to keep track of which states you have already examined, in order to prevent pushing and repushing useless states back onto the queue. Maybe you didn't need that explanation, but I hope that helps.
Author: crossley7 [ Sat Mar 03, 2012 8:47 am ] Post subject: RE:Ccc 2012 I normally index the value at the end of a depth then remove the first element as I check it. Once it reaches that value that I know is the last one of a depth, I increase my depth counter and reset the index to the end of the vector.
Author: danc [ Sat Mar 03, 2012 9:19 am ] Post subject: RE:Ccc 2012 How much faster is that than actually using a queue?
Author: crossley7 [ Sat Mar 03, 2012 11:17 am ] Post subject: RE:Ccc 2012 Not sure since I never have used Java and I'm not sure of exact efficiency. It is more that I know how a vector works and haven't used a queue type if that even exists in C++. My knowledge of the language is relatively primitive and so I make do with what I do know
Author: danc [ Sat Mar 03, 2012 11:23 am ] Post subject: RE:Ccc 2012 Oh. I was referring to the queue in C++. Makes life pretty easy. You can store pairs in it etc.
Author: NeilV [ Sat Mar 03, 2012 11:33 am ] Post subject: Re: RE:Ccc 2012 VK_eipi @ Sat Mar 03, 2012 1:26 am wrote:NeilV @ Fri Mar 02, 2012 2:54 pm wrote:see this is why I love python. For S4 I just kept the state as a list and stored it in a dictionary, which has constant lookup time. In other words, I let python hash it for me To clarify, were you referring to a tuple instead of an actual Python list? Lists in Python are not hashable and cannot be used as dictionary keys. Yeah actually I did convert it to a tuple for the dictionary. But then I worked with it as a list since tuples are immutable.
Author: crossley7 [ Sat Mar 03, 2012 12:49 pm ] Post subject: RE:Ccc 2012 ok, danc. Well, now after the last dwite and this, I have learned 2 new types in c++ Didn't know about pairs or queue. Both of those would make some problems much easier
Author: ultimatebuster [ Sat Mar 03, 2012 1:06 pm ]
Post subject: RE:Ccc 2012
Ah this is embarassing.. I have a 54 according to my teachers. I failed at question 3 and 4, where question 3 it's a matter of just being stupid: all of my own test cases are really the same case.. and here's the incorrect line:
Where as the correct is
... Can't believe I made a stupid error especially since i instantly saw my problem when i ran through the first test case.. damnit.
Something went wrong in my BFS algorithm or my hashing algorithm for S4. It looks like all of the test cases gave incorrect answers for my solution in S4, even though all my own tests passed.
It's probably going to turn out to be something really really dumb as well.. so I don't even want to spend hours debugging my BFS algorithm..
Author: danc [ Sat Mar 03, 2012 1:12 pm ] Post subject: Re: RE:Ccc 2012 crossley7 @ Sat Mar 03, 2012 12:49 pm wrote:ok, danc. Well, now after the last dwite and this, I have learned 2 new types in c++ Didn't know about pairs or queue. Both of those would make some problems much easier haha, and i guess while you're at it, i'll mention priority queues too! useful for dijkstra's, but top element is always the largest value, so you have to negate them. both are in . useful stuff, yo!
Author: Advecticity [ Sat Mar 03, 2012 8:37 pm ] Post subject: RE:Ccc 2012 What's the O() running time for the solution of S4?
Author: smaxd [ Sun Mar 04, 2012 12:36 pm ] Post subject: Re: Ccc 2012 I did the Junior competition during the CCC. Got all but J5 perfect. Yesterday did a mock run through of the senior competition. Got all but S4 perfect. fml
Author: bbi5291 [ Sun Mar 04, 2012 1:31 pm ] Post subject: Re: Ccc 2012 I've put the CCC 2012 problem descriptions and test data up on the PEG Judge, so you can try solving problems you didn't get during the contest, or whatever. You're limited to 60 seconds of execution time and 1 GiB of memory.
Author: Snario [ Sun Mar 04, 2012 2:01 pm ] Post subject: RE:Ccc 2012 Did anyone write it in Turing? lol
Author: D. McSmurfin [ Mon Mar 05, 2012 9:16 pm ] Post subject: RE:Ccc 2012 I hear CCC Stage 2 is going to be easier than Stage 1 this year.
Author: ihsh [ Mon Mar 05, 2012 10:37 pm ] Post subject: Re: RE:Ccc 2012 D. McSmurfin @ Mon Mar 05, 2012 9:16 pm wrote:I hear CCC Stage 2 is going to be easier than Stage 1 this year. Where did you hear this from? If this is true, then everyone who advances to stage two will be able to solve (almost) all of the problems!
Author: VK_eipi [ Wed Mar 07, 2012 1:11 am ] Post subject: Re: RE:Ccc 2012 Advecticity @ Sat Mar 03, 2012 5:37 pm wrote:What's the O() running time for the solution of S4? For naive breadth-first search with a set storing visited states, checking expanded neighbours for membership in the visited set totals to O(E log V) and adding new vertices to the set is O(V log V) so running time is O((E+V) log V). From my own data for n=3 to 7, it looks like, in the worst cases, V approaches n^n and E approaches n^(n+1). So that is O((n^(n+1)+n^n) log(n^n)) = O(n^(n+1) n log n) = O(n^(n+2) log n) There is probably a much more efficient way of doing it, though. Does anyone have one to share?
Author: d310 [ Wed Mar 07, 2012 8:06 pm ] Post subject: Re: Ccc 2012 I would like to correct you. BFS has a runtime of O(V+E), you've mixed it up with Dijkstra's runtime of O((E+V) log V) (for a binary heap). I agree with the n^n vertices statement, but for the number of edges... not so much. Firstly n^(n+1) assumed that each vertex is connected to n other neighbours, but evidently that is sometimes impossible (e.g. all coins are in one stack). Additionally the graph is definitely not completely connected (as seen with the IMPOSSIBLE test cases), and you forgot to account for the same edge being used twice in your estimate. So an overestimate for the number of edges is n^(n+1)/2. I also disagree with the O(V log V) runtime for state checking. This can be reduced to O(Vn) (see ihsh's comment on 3110, meaning coin 0 is in stack 3, coin 1 is in stack 1, and so on), with the use of computer number systems. Thus: O(V+E+Vn) =O(n^n+n^(n+1)/2+n^(n+1)) Simplified to O(n^(n+1)) The runtime based on the unsimplified expression is about 9.5 million operations in the worst case of n=7. This translates to roughly 1 second on a modern computer, for each game.
Author: A.J [ Wed Mar 07, 2012 8:10 pm ] Post subject: Re: RE:Ccc 2012 D. McSmurfin @ Mon Mar 05, 2012 9:16 pm wrote:I hear CCC Stage 2 is going to be easier than Stage 1 this year. I can assure you that this won't be the case. Don't take Stage 2 too lightly. A good way to prepare for it is to take a look at some of the past Stage 2 problems, and maybe some other national olympiads.
Author: mirhagk [ Wed Mar 07, 2012 8:26 pm ] Post subject: RE:Ccc 2012 does anyone know what the cut-off marks will be around?
Author: VK_eipi [ Thu Mar 08, 2012 2:37 am ] Post subject: Re: Ccc 2012 d310 @ Wed Mar 07, 2012 5:06 pm wrote:I would like to correct you. BFS has a runtime of O(V+E), you've mixed it up with Dijkstra's runtime of O((E+V) log V) (for a binary heap). I agree with the n^n vertices statement, but for the number of edges... not so much. Firstly n^(n+1) assumed that each vertex is connected to n other neighbours, but evidently that is sometimes impossible (e.g. all coins are in one stack). Additionally the graph is definitely not completely connected (as seen with the IMPOSSIBLE test cases), and you forgot to account for the same edge being used twice in your estimate. So an overestimate for the number of edges is n^(n+1)/2. I also disagree with the O(V log V) runtime for state checking. This can be reduced to O(Vn) (see ihsh's comment on 3110, meaning coin 0 is in stack 3, coin 1 is in stack 1, and so on), with the use of computer number systems. Thus: O(V+E+Vn) =O(n^n+n^(n+1)/2+n^(n+1)) Simplified to O(n^(n+1)) The runtime based on the unsimplified expression is about 9.5 million operations in the worst case of n=7. This translates to roughly 1 second on a modern computer, for each game. My mistake. The operations I thought would use logarithmic time are constant time on average (with hash table or similar data structure), so like you said, log V does not play a part and it's just O(V+E)=O(n^(n+1)). Thank you for the corrections (I didn't even account for state checking against the solution). For the number of edges, I agree that n neighbours per vertex is an overestimate; in fact, the maximum for any vertex is n-1 neighbours (for each pair of adjacent piles). However, it turns out that this maximum occurs surprisingly often, such that the average neighbours per vertex is usually between n-1 and n-2, leading to the same order of magnitude as my original estimate for large n. The reason is that a number of neighbours below (n-1) occurs only when there are 2 or more adjacent empty locations, which is rarely required for larger n where the coins are generally spread out. I also doubt the significance of the IMPOSSIBLE cases, since I only found one ("2 1") but I'm open to counter-examples. As for dividing by two for each edge, I did not keep track of parent states in my algorithm so I had to think about both directions. Perhaps that was a mistake, but I didn't want to store to essentially double the size of my states.
Author: crossley7 [ Tue Mar 20, 2012 9:51 pm ] Post subject: RE:Ccc 2012 We are now 3 weeks removed from doing the contest. Anyone have an idea of when round 2 competitors come out? I don't remember from last year and it isn't up on the website yet. Still 2 months until round 2, but it would be nice to know if I qualified
Author: d310 [ Tue Mar 20, 2012 10:12 pm ] Post subject: Re: Ccc 2012 Well, a date listed on last year's results was March 29, 2011 (on Comments, page 2). So expect results to come out around the end of March.
Author: crossley7 [ Tue Mar 27, 2012 6:03 pm ] Post subject: RE:Ccc 2012 Quote:Result for the 2012 CCC will be posted by Friday morning. Results are finally almost here. Not sure why, but that is about the only thing that has been in my mind when I'm not completely zoned in on work this past week. I'll be glad to finally know how I did so I can stop worrying about it
Author: D. McSmurfin [ Thu Mar 29, 2012 9:22 pm ] Post subject: Re: RE:Ccc 2012 Quote:Result for the 2012 CCC will be posted by Friday morning. How do you know?
Author: ihsh [ Thu Mar 29, 2012 9:28 pm ] Post subject: Re: RE:Ccc 2012 D. McSmurfin @ Thu Mar 29, 2012 9:22 pm wrote:Quote:Result for the 2012 CCC will be posted by Friday morning. How do you know? It's at the bottom of the CCC contest page.
Author: ihsh [ Thu Mar 29, 2012 10:06 pm ] Post subject: RE:Ccc 2012 Oh, they changed the date: Quote:Results for the 2012 CCC will be posted on Monday, April 2, 2012.
Author: crossley7 [ Fri Mar 30, 2012 6:35 am ] Post subject: RE:Ccc 2012 crap, I was looking forward to hopefully seeing them before I went to school today Oh well, 3 more days isn't all that bad.
Author: Yves [ Mon Apr 02, 2012 8:58 am ] Post subject: RE:Ccc 2012 Scores are up! 19 people advanced this year; the cutoff was 64 for Stage 2.
Author: trintith [ Mon Apr 02, 2012 10:21 am ] Post subject: RE:Ccc 2012 Of course I get 63... Don't they usually take more than 20 in the case of a tie?
Author: crossley7 [ Mon Apr 02, 2012 11:06 am ] Post subject: RE:Ccc 2012 I think they only took the 19 since the tie at 63 is fairly large at 8 people. It would likely not be economical for them to take 27 people since they need to plan sessions and stuff for the week. It turns out my name isn't on the results anywhere (not even honourable mention) even with a 66 so my teacher is trying to find out why that is the case. Hoping to maybe get lucky and my files were skipped over somehow and they will add me to the list for round 2... possibly? EDIT: Just found my name under Group 1 for Junior competitors for some reason. At least I know they have my results but somehow scored it in the wrong section. *crosses fingers and hopes for the best*
Author: trintith [ Mon Apr 02, 2012 1:49 pm ] Post subject: RE:Ccc 2012 Hmm... I'm not in the result booklet at all. Funny...
Author: michaelp [ Mon Apr 02, 2012 8:33 pm ] Post subject: RE:Ccc 2012 I managed to get an honourable mention this year! Unfortunately my last name was spelt wrong, so I'll have get that fixed.
Author: crossley7 [ Mon Apr 02, 2012 9:00 pm ] Post subject: RE:Ccc 2012 Just get an e-mail from my teacher. I'm in!!! Couldn't be much more psyched right now. Our school got 4 people in the results booklet this year which I think is a school record (Beating the previous record of 1 that we set last year) between JCCC and CCC. Not bad for a small town school
Author: ihsh [ Mon Apr 02, 2012 10:20 pm ] Post subject: Re: RE:Ccc 2012 crossley7 @ Mon Apr 02, 2012 9:00 pm wrote:Just get an e-mail from my teacher. I'm in!!! Couldn't be much more psyched right now. Our school got 4 people in the results booklet this year which I think is a school record (Beating the previous record of 1 that we set last year) between JCCC and CCC. Not bad for a small town school Congratulations. So that means that there are exactly 20 people who advanced to stage 2 (and that a score of 63 is actually the 21st place)?
Author: crossley7 [ Mon Apr 02, 2012 10:52 pm ] Post subject: RE:Ccc 2012 I believe so. Provided that there aren't any other errors in the results.
Author: VK_eipi [ Tue Apr 03, 2012 1:33 am ] Post subject: Re: Ccc 2012 I got invited to Stage 2. Congratulations to all the other invitees, as well as other award and recognition recipients!
Author: calaveras [ Tue Apr 03, 2012 7:25 pm ] Post subject: Re: Ccc 2012 Hi, Anybody know how the 1 week schedule looks like for CCC stage 2 ? I mean which day (Monday? ... Friday ?) do the contest ? Which day do the practice session (programming and try to submit) ? What development software tools they provide (Linux/Windows)? Yours information was appreciated in advance.
Author: pierret [ Thu May 24, 2012 7:18 pm ] Post subject: RE:Ccc 2012 how was like CCC stage 2 this year?
Author: crossley7 [ Thu May 24, 2012 7:42 pm ] Post subject: RE:Ccc 2012 Overall the problems were lower difficulty than past years and 2 people got perfect scores overall (1 from China). It was loads of fun and overall everyone did well during the contest.
Author: three0s [ Thu Aug 30, 2012 2:59 pm ] Post subject: RE:Ccc 2012 After 6 month of learning.... I decided its time to try S4 again (I skipped the question), and yes I figured it out.
Author: crossley7 [ Thu Aug 30, 2012 3:48 pm ] Post subject: RE:Ccc 2012 That thing was a pain in the ass. I didn't ever run my program long enough to see if it would eventually get the correct answer and haven't even bothered to go back to it. I know it is a bfs now but I haven't tried to think about how to implement it.
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https://cs.stackexchange.com/questions/tagged/regular-languages?sort=newest&page=3 | 1,555,880,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532882.36/warc/CC-MAIN-20190421195929-20190421221929-00253.warc.gz | 368,880,019 | 32,525 | # Questions tagged [regular-languages]
Questions about properties of the class of regular languages and individual languages.
1,128 questions
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### Prove or disprove L is regular
There is question in one of my exercise but I couldn't prove or disprove anything about it. This is language $L$ which is introduced with grammar: $$S \to 0S1 | 1S0 | AA$$ $$A \to 0A | \lambda|A1$$ ...
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### Show that $R^{+} \equiv R \leftrightarrow L(RR) \subset L(R)$
Show that $R^{+} \equiv R \leftrightarrow L(RR) \subset L(R)$ sigma is any alphabet. R is a regular expression. How can L(RR) even be a subset or equal to L(R)?
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### What is the regular expression for the following language?
What is the regular expression for the following language? $$L = \{acbc: a,b,c \in \{0,1\}^+ \}$$ maybe we can say $$L = ((0 + 1)^+ 0 (0 + 1)^+ 0) + ((0 + 1)^+ 1 (0 + 1)^+ 1)$$ Is it true??
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### Formal algorithm to test whether two given regular expressions define equal/identical or unequal languages
I'm trying to create a formal algorithm in order to determine whether two given regular expressions $a$, $a'$ define identical/equal or unequal languages and if those languages are subsets of each ...
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### Is SAT known to be non-context-free or even non-regular?
We have seen various languages proven to be outside of REG and CFL by corresponding pumping lemmas. Has something similar been done for SAT?
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### Prove regular languages are closed under $f(n) = 2^n$ and $f(n) = n^2$
Suppose $R$ is a regular language, let $f(R) = \{ w |$ $\exists x \text{ such that } |x| = f(|w|) \land wx \in R\}$, prove that $f(R)$ is regular for $f(n) = 2^n$ and for $f(n) = n^2$. I've ...
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### The reverse DFA is not working as expected
Assume a regular language contains all the strings that are ended with "01". We can draw the following DFA for it: And I reversed the DFA according to this answer (designing a DFA and the reverse of ...
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### How to show that the language made up of strings with nlogn 0s is not regular with the pumping lemma?
How to show that the following language is not regular with the pumping lemma? $$L=\left\{0^{n\lceil\log_2 n\rceil} \,\middle|\, n\in \mathbb{N}-\{0\}\right\}.$$
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### Proving a language is non-regular using the Pumping Lemma for non-binary strings [duplicate]
I am unsure of how to prove this language is non-regular. I do not even know where to start to develop a string that would prove the language is non-regular by contradiction. Any help would be ...
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### Can this DFA be converted to a regular expression? [duplicate]
I want to make the regular expression of this language but I can't: I tried but the regular expression didn't match some strings that it should. Is it even possible?
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### Is DFA and Regular Expression equivalent?
The language of a DFA can be the empty set (by defining no final states), but can a Regular Expression do that? If Regular Expression cannot do that, does it mean that DFA and Regular Expression are ... | 824 | 3,100 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-18 | latest | en | 0.902221 |
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# pH Watch
1. of of HCl was mixed with of water. Determine the pH of the diluted acid.
the answer is 2.31 but how do i get to this?
Thanks
2. (Original post by master y)
of of HCl was mixed with of water. Determine the pH of the diluted acid.
the answer is 2.31 but how do i get to this?
Thanks
You can't -log(0.02) because the HCl concentration is more dilute when you add the water.
Because of this you need to do the following
1. c*v=n so 0.02*0.025= 0.0005 moles [this gives you the no. of moles in 25cm3]
2. n/v=c so (0.0005)/0.1= 0.005 [this gives you the 'new' conc. of HCl in 100cm3 ]
3. now simply -log(0.005)= 2.301
3. (Original post by freetown)
You can't -log(0.02) because the HCl concentration is more dilute when you add the water.
Because of this you need to do the following
1. c*v=n so 0.02*0.025= 0.0005 moles [this gives you the no. of moles in 25cm3]
2. n/v=c so (0.0005)/0.1= 0.005 [this gives you the 'new' conc. of HCl in 100cm3 ]
3. now simply -log(0.005)= 2.301
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# How Many Inches Are in 20.5 Cm?
If you are trying to figure out how many inches are in 20.5 cm, you’ve come to the right place. You’ll need to do some mental math to find out. First, you must remember that 1 inch is 2.54 centimeters. Using this formula, you can calculate how many inches are in 20.5 centimeters.
When converting between inches and centimeters, it is important to consider the length and width of your measurement. This can be done in several ways. One simple method is to look at a ruler. Alternatively, you can take the length of your hand and multiply it by 2.54. This will yield the number of inches.
Another simple way to convert between inches and centimeters is to use a conversion factor. One centimeter is equivalent to 0.3937 inches, so if you measure 20.5 centimeters, you’ll have to multiply it by 0.3937. This will give you an answer of 8.07 inches.
The conversion factor is usually more accurate, but you can use a conversion factor to get a good estimate if you don’t have a calculator handy. Another option is to use an online tool that will allow you to quickly and easily convert 20.5 cm to inches. This will ensure that you’re getting an accurate result every time. There are other ways to convert the same measurement, so take advantage of whatever works for you.
Alternatively, you can also use a conversion factor that lets you select whether to display 20.5 cm in feet or centimeters. You can also find the same height in feet or inches. This method is easy and convenient. Just make sure that you have a ruler or a tape measure handy so you can use a ruler.
Inches are one of the most popular units for measurement in the United States and the United Kingdom. Essentially, an inch is a tenth of a foot. This means that 20.5 centimeters is equal to 7.62 centimeters. In general, the tenth and thirteenth centimeters are equal.
To convert inches to centimeters, you need to divide the length in centimeters by 2.54. Then, you can round your result to a decimal or a whole number. Whenever you’re using a decimal, remember to round up the result to the nearest whole or sixteenth of an inch. | 500 | 2,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | latest | en | 0.920603 |
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# ansASSIGN1_2450MW - ANSWERS TO ASSIGNMENT 1 Section I...
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A NSWERS TO A SSIGNMENT 1 Section I: Multiple Choice Questions (Total Marks 50) 1. A 2. D 3. A 4. D 5. B 6. A 7. B 8. C 9. B 10. C 11. B 12. C 13. B 14. B Section II: Long Question (Total Marks 50) a. By definition, planned expenditure is: E = C + I + G + GX – IM Substitute in all the equations given in the question: E = C 0 + C y (1-t)Y + I 0 – I r r + G 0 + X F Y F + X ε ε 0 – X ε ε r r + X ε ε r r F – IM y Y Now separate each of the components above in those that directly depend on Y and those that do not: E = [ C 0 + I 0 – I r r + G 0 + X F Y F + X ε ε 0 – X ε ε r r + X ε ε r r F ] + [ C y (1-t) – IM y ]Y The component that does not depend on national income (or real GDP) is autonomous expenditure: A = C 0 + I 0 – I r r + G 0 + X F Y F + X ε ε 0 – X ε ε r r + X ε ε r r F And the coefficient on income is the marginal propensity to expend, i.e., it tells you how much planned expenditure will change if income rises by one dollar: MPE = C y (1-t) – IM y Then we can re-write the planned expenditure line as:
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E = A + MPE*Y b. The goods market is in equilibrium when planned expenditure (or aggregate demand) is equal to real GDP (or national income): Y = E Using this equilibrium condition and the behavioral relationship from part (a) you can solve for equilibrium real GDP: Y = A/(1-MPE) See Figure 1 for the income expenditure diagram. In Figure 1 I plot both the
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http://nrich.maths.org/public/leg.php?code=149&cl=2&cldcmpid=163 | 1,495,810,919,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608668.51/warc/CC-MAIN-20170526144316-20170526164316-00489.warc.gz | 330,566,592 | 9,985 | # Search by Topic
#### Resources tagged with Area similar to Wrapping Presents:
Filter by: Content type:
Stage:
Challenge level:
### There are 93 results
Broad Topics > Measures and Mensuration > Area
### Wrapping Presents
##### Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
### A Square in a Circle
##### Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Inside Seven Squares
##### Stage: 2 Challenge Level:
What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### How Random!
##### Stage: 2 Challenge Level:
Explore this interactivity and see if you can work out what it does. Could you use it to estimate the area of a shape?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
### Numerically Equal
##### Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Ribbon Squares
##### Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
### Through the Window
##### Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
### Tiles in the Garden
##### Stage: 2 Challenge Level:
How many tiles do we need to tile these patios?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Overlapping Squares
##### Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Uncanny Triangles
##### Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
### Rope Mat
##### Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
### Making Boxes
##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Geoboards
##### Stage: 2 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Appearing Square
##### Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
### The Big Cheese
##### Stage: 2 Challenge Level:
Investigate the area of 'slices' cut off this cube of cheese. What would happen if you had different-sized block of cheese to start with?
### Making Squares
##### Stage: 2
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Shaping It
##### Stage: 1 and 2 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### A Day with Grandpa
##### Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
### Triangle Island
##### Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
### Square Areas
##### Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . . | 2,172 | 9,545 | {"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} | 3.96875 | 4 | CC-MAIN-2017-22 | longest | en | 0.892699 |
https://questions.examside.com/past-years/gate/question/which-one-of-the-following-field-patterns-represents-a-tem-w-gate-ece-2014-set-2-marks-1-qebqsle3nrdmvvaq.htm | 1,716,726,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00676.warc.gz | 401,056,179 | 30,741 | 1
GATE ECE 2014 Set 2
+1
-0.3
Which one of the following field patterns represents a TEM wave traveling in the positive $$x$$ direction?
A
$$E = + 8\widehat y,\,\,H = - 4\widehat z$$
B
$$E = - 2\widehat y,\,\,H = - 3\widehat z$$
C
$$E = + 2\widehat y,\,\,H = + 2\widehat z$$
D
$$E = - 3\widehat y,\,\,H = + 4\widehat z$$
2
GATE ECE 2014 Set 1
+1
-0.3
A two - port network has scattering parameters given by $$[S]$$ $$= \left[ {\matrix{ {{s_{11}}} & {{s_{12}}} \cr {{s_{21}}} & {{s_{22}}} \cr } } \right].$$
If the port - 2 of the two - port is short circuited, the $${{s_{11}}}$$ parameter for the resultant one - port network is
A
$${{{s_{11}} - {s_{11}}{s_{22}} + {s_{12}}{s_{21}}} \over {1 + {s_{22}}}}$$
B
$${{{s_{11}} + {s_{11}}{s_{22}} - {s_{12}}{s_{21}}} \over {1 + {s_{22}}}}$$
C
$${{{s_{11}} + {s_{11}}{s_{22}} + {s_{12}}{s_{21}}} \over {1 - {s_{22}}}}$$
D
$${{{s_{11}} - {s_{11}}{s_{22}} + {s_{12}}{s_{21}}} \over {1 - {s_{22}}}}$$
3
GATE ECE 2012
+1
-0.3
The electric field of a uniform plane electromagnetic wave in free spce, along the positive x direction, is given by $$\vec E = 10\left( {{{\widehat a}_y} + j{{\widehat a}_z}} \right){e^{ - j25x}}.$$ The frequency and polarization of the wave respectively are
A
$$1.2 GHz$$ and left circular
B
$$4 Hz$$ and left circular
C
$$1.2 GHz$$ and right circular
D
$$4 Hz$$ and right circular
4
GATE ECE 2012
+1
-0.3
A plane wave propagating in air with $$\vec E = \left( {8{{\widehat a}_x} + 6{{\widehat a}_y} + 5{{\widehat a}_z}} \right){\mkern 1mu} {\mkern 1mu} {e^{j\left( {\omega t + 3x - 4y} \right)}}{\mkern 1mu} {\mkern 1mu} V/m$$ is incident on a perfectly conducting slab positioned at $$x \le 0$$. The $$\overrightarrow E$$ field of the reflected wave is
A
$$\left( { - 8{{\widehat a}_x} - 6{{\widehat a}_y} - 5{{\widehat a}_z}} \right){\mkern 1mu} {e^{j\left( {\omega t + 3x + 4y} \right)}}{\mkern 1mu} {\mkern 1mu} V/m$$
B
$$\left( { - 8{{\widehat a}_x} + 6{{\widehat a}_y} - 5{{\widehat a}_z}} \right){\mkern 1mu} {e^{j\left( {\omega t + 3x + 4y} \right)}}{\mkern 1mu} {\mkern 1mu} V/m$$
C
$$\left( { - 8{{\widehat a}_x} - 6{{\widehat a}_y} - 5{{\widehat a}_z}} \right){\mkern 1mu} {e^{j\left( {\omega t - 3x - 4y} \right)}}{\mkern 1mu} {\mkern 1mu} V/m$$
D
$$\left( { - 8{{\widehat a}_x} + 6{{\widehat a}_y} - 5{{\widehat a}_z}} \right){\mkern 1mu} {e^{j\left( {\omega t - 3x - 4y} \right)}}{\mkern 1mu} {\mkern 1mu} V/m$$
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Class 12 | 1,127 | 2,516 | {"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} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.628245 |
https://www.physicsforums.com/threads/fusion-reaction.431963/ | 1,544,972,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827769.75/warc/CC-MAIN-20181216143418-20181216165418-00095.warc.gz | 1,038,225,656 | 14,164 | # Homework Help: Fusion reaction
1. Sep 24, 2010
### ProPatto16
1. The problem statement, all variables and given/known data
A nuclear reaction being studied for its potential as a source of fusion power is:
21H + 31H -> 42He + 10n
1) calculate energy in MeV released when two atoms fuse according to the reaction above.
2) calculate the energy in joules released per mole of reactants. (ie. one mole of H-2 and one mole of H-3)
3. The attempt at a solution
1) the answer is 17.59 Mev
and i dunno how to go about 2)
i know combined they release that energy found in part 1)... but im not sure how to divide that energy up between them. is it done on a ratio of 2:3 by atomic mass? or is it done by mass in kg? i just dont know the connection there...
2. Sep 25, 2010
### rock.freak667
Well if you know the masses of 21H and 31H, since they are both currently 1 mole, you can just add the atomic masses, which would be in the units kg/mol.
E=mc2
Last edited: Sep 25, 2010
3. Sep 25, 2010
### ProPatto16
the mass of 21H is 2.014102 so E is just 2.014102c^2?
same for 31H.
4. Sep 25, 2010
### rock.freak667
It is 2.014102u where u = 1.66053886 × 10-27 kg.
5. Sep 25, 2010
### ProPatto16
yeah, i used that conversion, just didnt clarify.
E of 21H = 3.01*10^-10J
E of 31H = 4.507*10^-10J
6. Sep 25, 2010
### rock.freak667
Sorry, I thought your question was asking something else. Just convert the value you got to Joules, then you only have 2 moles of reactants so, that the energy per unit mole of reaction is just Ereactant/2.
7. Sep 25, 2010
### ProPatto16
ohh... so the atomic masses and whatnot of the elements isnt even relevant? the energy is just half anf half?
8. Sep 25, 2010
### rock.freak667
Well you are finding per mole of reactant and not something like per mole of H-2 which I think would work out to be different, but my reasoning of the problem could be wrong.
9. Sep 25, 2010
### ProPatto16
2) calculate the energy in joules released per mole of reactants. (ie. one mole of H-2 and one mole of H-3)
so its energy per H-3 and per H-2. | 650 | 2,083 | {"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} | 3.390625 | 3 | CC-MAIN-2018-51 | latest | en | 0.925968 |
https://www.doe.in.gov/earlylearning/framework/m12 | 1,600,494,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00096.warc.gz | 848,011,471 | 15,193 | # Mathematics Foundation: 1 - Numeracy
#### Topic: M1.2 - Demonstrate understanding of written numerals
Basic math and number concepts are the foundation for learning more advanced math skills.
Understanding written numerals leads to the ability to:
*Identify numerals as different from letters or other symbols
*Begin to recognize that numerals indicate quantity
*Begin to recognize different numerals indicate different quantities
*Match numerals with amounts
*Name and write numerals
Looking Ahead to Kindergarten Family Engagement Special Populations
In kindergarten, students will be able to find the number that is one more than or less than any whole number up to 20 (K.NS.3). Lastly, they will be able to write whole numbers form 0 to 20 and recognize number words from 0 to 10 (K.NS.2). Encourage families to:
*Provide their child with many opportunities for counting objects at home (e.g. bottle caps or buttons) then have their child match the objects to a written number.
*Provide opportunities for their child to explore writing numerals by tracing, painting, or creating numerals.
Educators can:
*Provide children with a variety of textured or tactile numbers to feel and use.
*Use children’s interests to discuss numbers (e.g. counting dinosaurs and making groups to have conversations about the amount.)
*For DLL, provide numbers in native language to support learning of the concept in both the native language and English.
Powerful Practices
Across all developmental stages, educators can:
*Provide access to a variety of types of writing materials throughout the environment.
*Point out numerals in the environment (e.g. when reading a book, going for a field trip or walk, or when children create something that looks like a numeral, etc).
*Have numeral books freely available for children.
*Provide opportunities where children can form numeral shapes out of pasta, rice, paint, other sensory materials, or “loose parts”.
*Incorporate different ways for children to practice writing numerals (e.g. writing numerals in shaving cream, sand, or other sensory materials).
*Model the practical use of written numerals (e.g. calendars, weather temperature, etc).
*Share with families the importance of effort (e.g attempting to write numerals and simply holding writing utensils).
Infant
Younger Toddler
Draw child’s attention to numbers naturally occurring in the environment
Offer play materials that provide exposure to written numerals (e.g. old cell phones, number stickers, keyboards, etc.) and discuss how letters and numbers have different meanings
While reading, point out pictures of numbers (1-5) and connect the numeral to the actual item (e.g. “There are two dogs. See the two. Let's count the dogs. One, two.”)
Older Toddler
Provide opportunities for children to participate in creating number signs and labels for the environment (e.g. labeling tables or chairs or indicating number of children who are present)
Play games where children identify a numeral and make or move the quantity (e.g. Chutes and Ladders)
Read books that incorporate numerals and encourage children to help name the numerals and count quantities
Younger Preschool
Offer a variety of materials and opportunities to practice writing numerals (e.g. white boards, easels and paper, etc.)
Provide opportunities where children can create number books for 1-3 (i.e. children freely illustrate their own number books)
Read books that incorporate numerals and encourage children to independently name the numerals and count quantities
Older Preschool
Offer a variety of materials and opportunities to practice writing numerals (e.g. white boards, easels and paper, etc.)
Provide opportunities where children can create number books for 1-10 (i.e. children freely illustrate their own number books)
Read books that incorporate numerals and encourage children to independently name the numerals and count quantities
Encourage children to identify what comes next in a counting series | 824 | 3,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-40 | latest | en | 0.888585 |
https://allder.co/review-1ap-calculus/ | 1,642,370,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00024.warc.gz | 177,304,396 | 5,386 | # Review 1ap Calculus
Play this game to review Calculus. Given piecewise function f(x) shown, over which interval is f continuous? Copyright © 2008 Laying the Foundation, Inc., Dallas, Texas. All rights reserved. These materials may be used for face-to-face teaching with students only.
### Unit 1 - Limits and Continuity
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A: If the graph of a function has a sharp point, the function is not differentiable at that point. This is because the slopes directly to the left and right of the point do not approach the same value.
An absolute value function has a sharp point at its vertex. The graph off(x)= x + 4 is a horizontal translation (to the left 4 units) of the standard absolute value function, y = x , which has vertex (0, 0). Thus, the vertex of fis (-4, 0), which means the function is not differentiable at x =-4. Choice (A) is correct.
D: The only discontinuities that are removable are holes and holes with a point above or below—this function has neither. The function has 2 jump discontinuities (gaps), at x=-2 and x= 0, but neither of these is removable.
D: Examine the values in the table: f(x) increases as x gets larger, which indicates that f′(x), the slope of the function, is positive. This means f′(x)> 0, so eliminate (A) and (B).
To choose between (C) and (D), take a closer look at the slopes. The slope between the first pair of points is 3, and the slope between the second pair of points is also 3, so f′(x) is constant. This means f′′(x), which is the derivative of f′(x), must be 0. Choice (D) is correct
C: To use a local linear approximation, you need to find the equation of the tangent line. You’ve been given all the information you need in the question stem; you just need to piece it all together. The point on the function is given by f(-1)= 5, which translates to the point (-1, 5). The slope of the tangent line at x=-1 is given by f'(-1)= 2, so the slope is 2. Now the point-slope form of the tangent line is:
y-5 = 2(x-(-1))
y = 5+2(x+1)
## Ap Calculus Unit 1 Review
Substituting x=-0.9 into the equation of the tangent line yields:
5+2(-0.9+1) = 5+2(0.1) = 5.2
## Ap Calculus Bc Semester 1 Review
That’s (C).
For more practice questions, check out our AP Calculus Prep Plus book. | 985 | 3,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.779089 |
https://www.hackerearth.com/ja/practice/math/combinatorics/basics-of-combinatorics/practice-problems/algorithm/permutaions-and-querys-52af7b15/ | 1,596,624,902,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00377.warc.gz | 690,617,437 | 27,443 | Permutations and queries
/
## Combinatorics, Combinatorics basics, Easy-Medium, Mathematics, dp
Problem
Editorial
Analytics
Define the value of a permutation of $n$ elements as the number of subintervals where the number of inversions do not exceed $K$. That is, $value(P)=\sum_{i=1}^n\sum_{j=i}^n[K\ge \sum_{a=i}^{j-1}\sum_{b=i+1}^j[P_a>P_b]]$.
For example, if $P=[2,3,1]$, and K=0, then $value(P)=4$ (the valid intervals are $[1,1],[2,2],[3,3],[1,2]$).
Now you are given $Q$ queries. Each query you will receive $n$ and $K$, and you need to calculate the sum of value of all permutations of $1\dots n$.
Input
First line contains an integer $Q$.
Each of the next $Q$ lines contain two integers, $n$ and $K$.
Output
$Q$ lines, $i_{th}$ line contains the answer of $i_{th}$ query, modulo $10^9+7$.
Constraints
$1\le Q\le 100000$
$1\le n\le 500$
$0\le K\le 250000$
SAMPLE INPUT
3
2 0
7 10
10 14
SAMPLE OUTPUT
5
137228
184025758
Explanation
In the first query, $n=2$ and $K=0$, so $value([1,2])=3$ and $value([2,1])=2$, thus the answer is 5.
Time Limit: 2.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
## This Problem was Asked in
Initializing Code Editor...
? | 434 | 1,206 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-34 | latest | en | 0.570357 |
https://www.zigya.com/engineering-entrance-exam/11-Conic+Section/MAENJE11216242/370 | 1,718,770,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00166.warc.gz | 978,043,036 | 16,218 | Let PQ be a diameter of the circle x2 + y2 = 9. If α and β are the lengths of the perpendiculars from P and Q on the straight line, x + y = 2 respectively, then the maximum value of αβ is ..... | Conic Section
## Previous Year Papers
Download Solved Question Papers Free for Offline Practice and view Solutions Online.
## Test Series
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## Test Yourself
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# Conic Section
#### Multiple Choice Questions
371.
The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x+ y2 – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :
• ( - 3, 6)
• ( - 3, 1)
372.
• 7x - 4y = 1
• 4x - 2y = 1
• 8x - 2y = 5
• 4x - 3y = 2
# 373.
Ans : 7
#### Multiple Choice Questions
374.
If the common tangent to the parabolas, y2 = 4x and x2 = 4y also touches the circle, x2 + y2 = c2, then c is equal to :
• $\frac{1}{\sqrt{2}}$
• $\frac{1}{2\sqrt{2}}$
• $\frac{1}{2}$
• $\frac{1}{4}$
375.
If the co–ordinates of two points A and B are respectively and P is any point on the conic, 9x2 + 16y2 = 144, then PA + PB is equal to :
• 8
• 16
• 9
• 6
376.
If the line y = mx + c is a common tangent to the hyperbola
• 4c2 = 369
• 5m = 4
• c2 = 369
• 8m + 5 = 0
377.
If the normal at an end of latus rectum of an ellipse passes through an extremity of the minor axis, the eccentricity e of the ellipse satisfies :
378.
The centre of the circle passing through the point (0, 1) and touching the parabola y = x2 at the point (2, 4) is: | 612 | 1,720 | {"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} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.653293 |
https://byjusexamprep.com/barc-2020-engineering-mechanics-and-dss-nuclear-quiz-2-i-eda59e40-56cd-11ea-8e89-78b2e95d4b62 | 1,716,256,066,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00067.warc.gz | 125,382,013 | 51,794 | Time Left - 12:00 mins
# BARC 2020: Engineering mechanics and DSS Nuclear Quiz 2
Attempt now to get your rank among 739 students!
Question 1
A beam ABC 9 m long and of uniform section is fixed at A and simply supported at B, 6 m from A. The beam carries a point load W at the mid point of AB and a point load at the free end C. What is the value of so that W is maximum corresponding to collapse state?
Question 2
A diamond section with unequal diagonals, the shorter being b and longer h, is used as a beam. If the shorter diagonal is placed parallel to the xx – axis, the shape factor of the section will be _______.
Question 3
A plate carrying tensile load is 150 mm wide and 10 mm thick. It is connected to a gusset plate using 20 mm diameter bolts as shown in figure. The block shear strength of the plate is _________ kN.
Question 4
A fillet welded joint of 8mm size is shown in the figure. The weld surfaces meet at standard angle. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa respectively. The welding is done in the workshop (ϒmw =1.25). Find the factored load that can be transmitted by the welded joint as IS 800:2007 is
Question 5
An I-section beam, made of structural steel has an overall depth of 30 cm. If the developed flange stresses at the top and bottom of the beam are 1200 kg/cm2 and 300 kg/cm2 respectively, then the depth of the neutral axis from the top of the beam would be
Question 6
A heavy prismatic bar AB is rotating at an angular speed of 6 rad/s . The mass centre C of bar, thus moves upward at 120m/s. At this instant find the velocity of the topmost point A of the bar.
Question 7
A thin string of negligible mass with one end fixed to the roof is wounded around a circular disc of radius 2m and mass 10 kg as shown in the figure. The disc rolls vertically down under the action of its own weight. Considering acceleration due to gravity=10m/s2, the tension in the string is
Question 8
A bullet of mass m1=20 gm fired horizontally with a velocity of v=200 m/s hits a wooden block of mass m2 =500 gm (take g=9.8m/s2) resting on a horizontal plane as shown in the figure and the bullet remains embedded in the block after the impact. If the coefficient between the surfaces in contact remains constant at 0.3, the distance the block will move before coming to rest is
Question 9
A block weighting 100 N is resting on surface. The coefficient of friction between the block and surface is 0.3. A horizontal force of 20 N is applied on the block. What is the magnitude of force of friction acting on the block?
Question 10
Which of the following statements are correct regarding the design of lacings?
1) The lacing members can be made of flats only.
2) Lacing members are mirror image of each other on the opposite sides of the columns.
3) The maximum slenderness ratio for a lacing member carrying tension is 180.
4) Lacing members do not transfer any lateral loads.
• 739 attempts
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Jun 2ESE & GATE CE
GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 bepstudentsupport@byjus.com | 793 | 3,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.898309 |
https://www.aqua-calc.com/convert/density/microgram-per-us-fluid-ounce | 1,722,802,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00522.warc.gz | 529,301,360 | 16,254 | # Select a table to convert micrograms per US fluid ounce
## micrograms/US fluid ounce [µg/fl.oz] of density conversion tables
Select units whose names contain any of the following checked keywords, where ∗ is a wildcard for partial matches:
What is a microgram per US fluid ounce?
#### Foods, Nutrients and Calories
BONITO DEL NORTE IN ORGANIC EXTRA VIRGIN OLIVE OIL, UPC: 8411320282977 contain(s) 200 calories per 100 grams (≈3.53 ounces) [ price ]
107060 foods that contain Cholesterol. List of these foods starting with the highest contents of Cholesterol and the lowest contents of Cholesterol
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Cichlid, White Sand weighs 1 169.35 kg/m³ (73.00014 lb/ft³) with specific gravity of 1.16935 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Trisodium orthophosphate [Na3PO4 or Na3O4P] weighs 2 540 kg/m³ (158.56702 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-123, liquid (R123) with temperature in the range of -28.89°C (-20.002°F) to 93.34°C (200.012°F)
#### Weights and Measurements
A zeptonewton is a SI-multiple (see prefix zepto) of the force unit newton and equal to 0.000,000,000,000,000,000,001 newton
The electric potential φ(A) of a point A (in the electric field) specifies the work to be done by the force F = −Q × E in order to move the charge Q from a fixed reference point P to the point A.
g/dm³ to µg/fl.oz conversion table, g/dm³ to µg/fl.oz unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a fluid in a horizontal cylindrical tank | 516 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-33 | latest | en | 0.71828 |
http://speleotrove.com/decimal/decimalj/deccfiel.html | 1,539,819,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511314.51/warc/CC-MAIN-20181017220358-20181018001858-00417.warc.gz | 338,682,711 | 2,633 | ### Fields
These fields contain the context settings or define certain values they may take.
#### Public constants
DEFAULT
A MathContext object whose settings have their default values.
The default values of the settings are: "digits=9 form=SCIENTIFIC lostDigits=0 roundingMode=ROUND_HALF_UP"
ENGINEERING
A constant of type int that signifies that standard floating point notation (with engineering exponential format, where the power of ten is a multiple of 3, if needed) should be used for the result of a BigDecimal operation.
PLAIN
A constant of type int that signifies that plain (fixed point) notation, without any exponent, should be used for the result of a BigDecimal operation. A zero result in plain form may have a decimal part of one or more zeros.
SCIENTIFIC
A constant of type int that signifies that standard floating point notation (with scientific exponential format, where there is one digit before any decimal point, if needed) should be used for the result of a BigDecimal operation.
The remaining constants indicate rounding algorithms. Rounding is applied when a result needs more digits of precision than are available; in this case the digit to the left of the first discarded digit may be incremented or decremented, depending on the rounding algorithm selected.[1]
ROUND_CEILING
A constant of type int that signifies that if any of the discarded digits are non-zero then the result should be rounded towards the next more positive digit.
ROUND_DOWN
A constant of type int that signifies that all discarded digits are ignored (truncated). The result is neither incremented nor decremented.
ROUND_FLOOR
A constant of type int that signifies that if any of the discarded digits are non-zero then the result should be rounded towards the next more negative digit.
ROUND_HALF_DOWN
A constant of type int that signifies that if the discarded digits represent greater than half (0.5) the value of a one in the next position then the result should be rounded up (away from zero). Otherwise the discarded digits are ignored.
ROUND_HALF_EVEN
A constant of type int that signifies that if the discarded digits represent greater than half (0.5) the value of a one in the next position then the result should be rounded up (away from zero). If they represent less than half, then the result should be rounded down.
Otherwise (they represent exactly half) the result is rounded down if its rightmost digit is even, or rounded up if its rightmost digit is odd (to make an even digit).
ROUND_HALF_UP
A constant of type int that signifies that if the discarded digits represent greater than or equal to half (0.5) the value of a one in the next position then the result should be rounded up (away from zero). Otherwise the discarded digits are ignored.
ROUND_UNNECESSARY
A constant of type int that signifies that rounding (potential loss of information) is not permitted. If any of the discarded digits are non-zero then an ArithmeticException should be thrown.
ROUND_UP
A constant of type int that signifies that if any of the discarded digits are non-zero then the result will be rounded up (away from zero).
#### Shared fields
These fields are shared with the BigDecimal class (that is, only 'default access' from the same package is allowed) for efficient access; they are never changed directly from another class.
digits
A value of type int that describes the number of digits (precision) to be used for a BigDecimal operation. A value of 0 indicates that unlimited precision (as many digits as are required) fixed-scale arithmetic will be used.
The BigDecimal operator methods use this value to determine the precision of results. Note that leading zeros (in the integer part of a number) are never significant. digits will always be non-negative.
form
A value of type int that describes the format of results from a BigDecimal operation. The BigDecimal operator methods use this value to determine the formatting of results. form will be ENGINEERING, PLAIN, or SCIENTIFIC.
lostDigits
A value of type boolean that is true if checking for lost digits is enabled, or is false otherwise. The BigDecimal operator methods use this value to determine whether checking for lost digits should take place.
roundingMode
A value of type int that describes the rounding algorithm to be used for a BigDecimal operation. The BigDecimal operator methods use this value to determine the algorithm to be used when non-zero digits have to be discarded in order to reduce the precision of a result. roundingMode will have the value of one of the public constants whose name starts with ROUND_.
Footnotes:
[1] These constants have the same values as the constants of the same name in java.math.BigDecimal, which have been preserved to maintain compatibility with earlier versions of BigDecimal.
[previous | contents | next] | 968 | 4,864 | {"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-2018-43 | latest | en | 0.789445 |
http://www.mathworks.fr/fr/help/matlab/ref/unmkpp.html?s_tid=gn_loc_drop&nocookie=true | 1,405,054,717,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776425626.86/warc/CC-MAIN-20140707234025-00091-ip-10-180-212-248.ec2.internal.warc.gz | 366,527,722 | 7,555 | Accelerating the pace of engineering and science
• Trials
# unmkpp
Piecewise polynomial details
## Syntax
[breaks,coefs,l,k,d] = unmkpp(pp)
## Description
[breaks,coefs,l,k,d] = unmkpp(pp) extracts, from the piecewise polynomial pp, its breaks breaks, coefficients coefs, number of pieces l, order k, and dimension d of its target. Create pp using spline or the spline utility mkpp.
## Examples
This example creates a description of the quadratic polynomial
as a piecewise polynomial pp, then extracts the details of that description.
```pp = mkpp([-8 -4],[-1/4 1 0]);
[breaks,coefs,l,k,d] = unmkpp(pp)
breaks =
-8 -4
coefs =
-0.2500 1.0000 0
l =
1
k =
3
d =
1``` | 205 | 693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-23 | latest | en | 0.798533 |
https://curriculum.illustrativemathematics.org/HS/teachers/2/4/10/practice.html | 1,718,634,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00766.warc.gz | 174,377,103 | 26,449 | # Lesson 10
Solving Problems with Trigonometry
### Problem 1
Technology required. Find the area of the isosceles trapezoid.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
Technology required. The sun is 62 degrees above the horizon. A tree casts a shadow that is 12 feet long. How tall is the tree?
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
Technology required. A plane leaves the ground with an elevation angle of 6 degrees. The plane travels 10 miles horizontally.
1. How high is the plane at the time?
2. What is the distance of the plane’s path?
### Solution
For access, consult one of our IM Certified Partners.
### Problem 4
Technology required. Find the missing measurements.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 9.)
### Problem 5
Technology required. Ramps in a parking garage need to be both steep and safe. The maximum safe incline for a ramp is 8.5 degrees.
Is this a safe ramp? Explain or show your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 9.)
### Problem 6
Select all true equations.
A:
$$\cos(37)=\sin(53)$$
B:
$$\tan(37)=\tan(53)$$
C:
$$\sin(37)=\cos(53)$$
D:
$$\sin(37)=\sin(53)$$
E:
$$\cos(\theta)=\sin(90-\theta)$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 8.)
### Problem 7
Technology required. Clare is flying a kite. She gets tired, so she stakes the kite into the ground. The kite is on a string that is 30 ft long and makes a 27 degree angle with the ground. How high is the kite?
A:
30 ft
B:
13.6 ft
C:
26.7 ft
D:
15.3 ft
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 7.)
### Problem 8
What is the length of the diagonal?
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 2.) | 539 | 1,965 | {"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} | 4.3125 | 4 | CC-MAIN-2024-26 | latest | en | 0.786895 |
https://astronomy.stackexchange.com/questions/13742/why-when-we-look-through-a-telescope-in-space-do-the-billions-of-stars-not-bloc | 1,723,756,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00389.warc.gz | 83,205,835 | 37,953 | # Why when we look through a telescope in space, do the billions of stars not block our view from seeing further?
I know that we are able to "see back into time" by looking very far away through telescopes such as the Hubble telescope, but my question is, wouldn't you run into at least a couple of stars that were blocking your vision when you were trying to look even farther?
Imagine a big Boulder in front of another smaller Boulder and if we are looking from the side of the larger Boulder, we wouldn't be able to see the smaller one because the bigger one was in the way. Hopefully that simplifies my question. If it is as simple as that, wouldn't we miss stars and not be able to view them?
• This sounds half-way to Olbers's Paradox Commented Jun 27, 2017 at 17:40
OK, Imagine the stars at distance x block an area of the sky. At a distance of 2x there should be four times as many stars, but they would seem four times smaller in terms of area covered. Thus, the blocking grows linearly until a significant part of the sky is blocked. The stars within 20 light years (excluding the Sun), blocks approximately $4.3 \cdot 10^{-16}$ of the sky. Even if we scale that up to the radius of the universe, 45 billion light years, only about $2 \cdot 10^{-6}$ of the sky is covered.
• Great answer, but where do you have the $3\times10^{-19}$ from?
• @RobJeffries One can apparently not be careful enough with data. Manually checking the result of my table look-up, I found an error. The RMS stellar radius is 0.55 of the Sun and the representative distance 15 ly. The correct fraction is then $4.3 \cdot 10^{-16}$. Sorry about the previous error. Commented Feb 14, 2016 at 17:26 | 413 | 1,683 | {"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} | 3.25 | 3 | CC-MAIN-2024-33 | latest | en | 0.964655 |
http://tailieu.vn/tag/parallel-circuits.html | 1,521,282,058,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257644877.27/warc/CC-MAIN-20180317100705-20180317120705-00351.warc.gz | 300,419,362 | 18,116 | # Parallel circuits
Xem 1-20 trên 28 kết quả Parallel circuits
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• ### Lessons In Electric Circuits, Volume I DC
It was discovered centuries ago that certain types of materials would mysteriously attract one another after being rubbed together. For example: after rubbing a piece of silk against a piece of glass, the silk and glass would tend to stick together. Indeed, there was an attractive force that could be demonstrated even when the two materials were separated: Glass and silk aren't the only materials known to behave like this. Anyone who has ever brushed up against a latex balloon only to nd that it tries to stick to them has experienced this same phenomenon.
• ### CHAPTER 5 - SERIES AND PARALLEL CIRCUITS
Mạch bao gồm một pin và một trong kháng tải là rất đơn giản để phân tích, nhưng họ không thường được tìm thấy trong các ứng dụng thực tế. Thông thường, chúng ta thấy mạch nơi có hơn hai các thành phần được kết nối với nhau. Có hai cách cơ bản để kết nối nhiều hơn hai thành phần mạch: loạt song song.
• ### Ebook Introductory circuit analysis (10th edition): Part 2
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• ### Lecture Radio Communication Circuits: Chapter 1 & 2 - Đỗ Hồng Tuấn
Lecture "Radio Communication Circuits: Chapter 1 & 2" presents the following contents: Introduction to Communication Systems (Elements of Communication Systems, Radio Frequency Metrics, Parallel-Tuned Circuit,...), Radio Frequency (RF) Power Amplifiers (Class C Amplifier, Class D Amplifier). Invite you to consult.
• ### Lecture Electric circuit theory: Basic laws - Nguyễn Công Phương
Lecture Electric circuit theory - Basic laws presents the following content: Ohm’s law; nodes, branches, and loops; Kirchhoff’s laws; series subcircuits; parallel subcircuits.
• ### Lecture Electric circuit theory: Frequency response - Nguyễn Công Phương
The main contents of this chapter include all of the following: Transfer function, the decibel scale, bode plots, series resonance, parallel resonance, passive filters, active filters, scaling.
• ### Electronics and Circuit Analysis Using MATLAB P7
This chapter discusses the application of MATLAB for analysis of two-port networks. The describing equations for the various two-port network representations are given. The use of MATLAB for solving problems involving parallel, series and cascaded two-port networks is shown. Example problems involving both passive and active circuits will be solved using MATLAB. 7.1 TWO-PORT NETWORK REPRESENTATIONS A general two-port network is shown in Figure 7.1. I1 + V1 Linear two-port network I2 + V2 - Figure 7.1 General Two-Port Network I 1 and V1 are input current and voltage, respectively.
• ### CHAPTER 7 - SERIES-PARALLEL COMBINATION CIRCUITS
Mạch này không phải là hàng loạt đơn giản, cũng không có song song đơn giản. Thay vào đó, nó chứa các yếu tố của cả hai. các thoát khỏi hiện tại phía dưới của pin, chia tách đi du lịch thông qua R3 và R4, gia nhập lại, sau đó chia tách lên một lần nữa để đi du lịch thông qua R1 và R2, sau đó gia nhập lại một lần nữa để quay trở lại đầu của pin. có tồn tại nhiều hơn một con đường cho hiện tại để đi du lịch (series), nhưng vẫn có hơn hai bộ điện...
• ### Electrical Power - W. J. R. H. Pooler
Automatic Voltage Regulator adjusts the excitation so that at the system design power factor, the voltage is correct whatever the current. If however it adjusts the excitation to give the correct voltage at other power factors, then two machines will not run in parallel. One can supply a huge leading current and the other a huge lagging current. A “droop” is needed to give a lower voltage if the power factor lags by more than the system design. This is achieved by the compounding. Faulty Compounding causes unstable sharing of kVAr which can be quite violent....
• ### Electrical Engineering - Know It All
The system of units used in engineering and science is the Système International d’Unités (International system of units), usually abbreviated to SI units, and is based on the metric system. This was introduced in 1960 and is now adopted by the majority of countries as the offi cial system of measurement.
• ### Image processing
Modern digital technology has made it possible to manipulate multi-dimensional signals with systems that range from simple digital circuits to advanced parallel computers. The goal of this manipulation can be divided into three categories...
• ### Audio Electronics
The progress of 'audio' towards the still-distant goal of a perfect imitation of reality has been, and remains, inextricably bound up with the development of electronic components and circuit technology, and with the parallel progress in the various transducers and interface devices used to generate and reproduce electrical signals.
• ### Micro Electronic and Mechanical Systems
The miniaturization and performance improvement in semiconductor devices and integrated circuits (ICs) are expected to continue through leveraging of nanotechnologies and nanomaterials. This evolution should accelerate the System-on-a-Chip (SoC) trend, i.e., singlechip integration of multifunctional, mixed-signal electronic components, toward realizing embedded nanoelectronic systems.
• ### ZENER DIODES
VOLTAGE REGULATION A voltage regulator circuit automatically maintains the output voltage of a power supply constant, regardless of - a change in the load - a change in the source voltage ZENER DIODES The simplest of all voltage regulators is the zener diode voltage regulator. A zener diode is a special diode that p is optimized for operation in the g breakdown region. ZENER DIODE CIRCUIT The zener diode is typically connected reverse biased, in parallel with the load. Resistor Rs limits current to zener.
• ### VLSI DESIGN
This book provides some recent advances in design nanometer VLSI chips. The selected topics try to present some open problems and challenges with important topics ranging from design tools, new post-silicon devices, GPU-based parallel computing, emerging 3D integration, and antenna design.
• ### Power Electronic Handbook P5
DC-AC Conversion Basic DC-AC Converter Connections (Square-Wave Operation) • Control of the Output Voltage • Harmonics in the Output Voltage • Filtering of Output Voltage • Practical Realization of Basic Connections • Special Realizations (Application of Resonant Converter Techniques) István Nagy Budapest University of Technology and Economics Dariusz Czarkowski Polytechnic University, Brooklyn 5.3 Resonant Converters Survey of Second-Order Resonant Circuits • Load Resonant Converters • Resonant Switch Converters • Resonant DC-Link Converters with ZVS Michael E. | 1,721 | 8,182 | {"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} | 2.90625 | 3 | CC-MAIN-2018-13 | latest | en | 0.899949 |
https://metanumbers.com/68249 | 1,643,144,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00206.warc.gz | 447,808,690 | 7,353 | 68249 (number)
68,249 (sixty-eight thousand two hundred forty-nine) is an odd five-digits composite number following 68248 and preceding 68250. In scientific notation, it is written as 6.8249 × 104. The sum of its digits is 29. It has a total of 2 prime factors and 4 positive divisors. There are 67,620 positive integers (up to 68249) that are relatively prime to 68249.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 29
• Digital Root 2
Name
Short name 68 thousand 249 sixty-eight thousand two hundred forty-nine
Notation
Scientific notation 6.8249 × 104 68.249 × 103
Prime Factorization of 68249
Prime Factorization 139 × 491
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 68249 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 68,249 is 139 × 491. Since it has a total of 2 prime factors, 68,249 is a composite number.
Divisors of 68249
1, 139, 491, 68249
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 68880 Sum of all the positive divisors of n s(n) 631 Sum of the proper positive divisors of n A(n) 17220 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 261.245 Returns the nth root of the product of n divisors H(n) 3.96336 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 68,249 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 68,249) is 68,880, the average is 17,220.
Other Arithmetic Functions (n = 68249)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 67620 Total number of positive integers not greater than n that are coprime to n λ(n) 33810 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6786 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 67,620 positive integers (less than 68,249) that are coprime with 68,249. And there are approximately 6,786 prime numbers less than or equal to 68,249.
Divisibility of 68249
m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 6 1 2
68,249 is not divisible by any number less than or equal to 9.
Classification of 68249
• Arithmetic
• Semiprime
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
Other numbers
• LucasCarmichael
Base conversion (68249)
Base System Value
2 Binary 10000101010011001
3 Ternary 10110121202
4 Quaternary 100222121
5 Quinary 4140444
6 Senary 1243545
8 Octal 205231
10 Decimal 68249
12 Duodecimal 335b5
20 Vigesimal 8ac9
36 Base36 1gnt
Basic calculations (n = 68249)
Multiplication
n×y
n×2 136498 204747 272996 341245
Division
n÷y
n÷2 34124.5 22749.7 17062.2 13649.8
Exponentiation
ny
n2 4657926001 317898791642249 21696274630791852001 1480749047276913107216249
Nth Root
y√n
2√n 261.245 40.8663 16.1631 9.26444
68249 as geometric shapes
Circle
Diameter 136498 428821 1.46333e+10
Sphere
Volume 1.33161e+15 5.85332e+10 428821
Square
Length = n
Perimeter 272996 4.65793e+09 96518.7
Cube
Length = n
Surface area 2.79476e+10 3.17899e+14 118211
Equilateral Triangle
Length = n
Perimeter 204747 2.01694e+09 59105.4
Triangular Pyramid
Length = n
Surface area 8.06776e+09 3.74647e+13 55725.1
Cryptographic Hash Functions
md5 65d5109db25d217a9212bbdc04317918 d5f35a4db5187a782d1e2ed757dcaee820bc3620 6aaf8e468e200bd41dc703156ba0b0afc44482774b1078d50636d37c71070a01 b65d328a740baeb2aa5c2342ff10c6b1197bc1bfd1b234bcdcd7b6d1371bd813b4a92871f0ff02c0e18f920c6d18b1ed5878c3353ea3480a5af73a7d50a185fa 7a2085aafde3d4d3a716dc79c799c4d0f235220e | 1,450 | 4,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.808264 |
http://docplayer.net/10013191-Determining-optimum-insurance-product-portfolio-through-predictive-analytics-badm-final-project-report.html | 1,529,494,798,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00370.warc.gz | 94,490,242 | 28,924 | # Determining optimum insurance product portfolio through predictive analytics BADM Final Project Report
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1 2012 Determining optimum insurance product portfolio through predictive analytics BADM Final Project Report Dinesh Ganti( ), Gauri Singh( ), Ravi Shankar( ), Shouri Kamtala( ), Supreet Kaur( ) Vinayak Palankar( ) Section A Group 1 12/27/2012
2 Table of Contents Executive Summary... 2 Business Problem... 2 Data... 2 Data Analytics solution... 2 Recommendations... 2 Problem Description (Business and Analytics)... 3 Data Description... 3 Data Mining solution... 4 K-Nearest Neighbors... 4 Naïve Bayes... 5 Classification Tree... 5 Ensembles... 6 Conclusion and operational recommendations... 6 Appendix P a g e
3 EXECUTIVE SUMMARY Business Problem An entrepreneur has an idea for a new business venture, which, in a nutshell, is to offer insurance to customers on price drops of certain products. Registered customers get a certain multiple of the insurance fee or a certain percentage of the drop provided that the drop happens within a certain period after purchase. Insurance is offered on a new portfolio of products every day and, for the sake of simplicity, the period within which price needs to drop is taken as one day. The business problem is to understand from the data, if the business idea is theoretically and financially viable. Data The available data is a list of products sold across five e-commerce sites. The data gives us information about shipping, average ratings, brand, category, in stock etc. Some features of the data are (a) Categorical variables such as model name, brand etc. which cannot simply be converted as they would explode the number of categorical variables (b) columns like average rating and review count which had large number of missing values and were filled through imputation based on rest of the columns (c) textual data columns such as shipping period which were converted into numerical averages. Data such as this is available on the internet and can be crawled off various sites easily. Data Analytics solution To check if it is possible to construct a portfolio of products such that as high a percentage of the products selected in the portfolio are due for a price increase so that losses on price drops don t overwhelm the insurance fee gained where the price goes up or stays the same. The idea is to build various classification models that can predict if the price will go up or not and then construct a portfolio by selecting top x% of products that have the highest predicted probability that the price will go up next day among all the products considered. Observe that data mining only takes us towards making a decision and quite a bit of external information, analysis and assumptions are needed before one can come to a conclusion. To achieve the data mining goal, we would (a) Use available data of various products sold by the five online retailers as predictors to classify products as Price Up or not (b) Rank order the results in terms of probability of price increase (c) Select top x% to offer insurance on. Recommendations Our data analysis shows that if products are randomly selected to form a portfolio only 2.4% of products on an average would show a price increase. However, using our best model of Classification tree, this percentage can shoot up to as high as 100% based on what x is chosen. If the top x% is intelligently and variably chosen everyday and if the multiple on insurance fee paid to customers upon price drop is carefully chosen so that it doesn t overwhelm the insurance revenue, the business can be very much viable. The entrepreneur can always explore additional streams of revenue like advertisements on the site or building a user database. 2 P a g e
4 PROBLEM DESCRIPTION (BUSINESS AND ANALYTICS) A young and bright entrepreneur has a new business idea. Customers are usually left disgruntled if there is a price drop on products they have recently purchased. More recent the purchase, more disgruntled one can expect customers to be. The idea is to offer insurance to customers on price drops. So, in case, the price drops within a pre-specified time, the customer is paid a multiple of insurance fee collected or a certain percentage of the price drop. Customer forfeits the insurance fee if the price goes up or remains the same within the period. The scope is limited to electronic products purchased on e- commerce websites. We are playing the role of consultants to this entrepreneur. Every insurance company works on the principle of minority defaults. Similarly, in this case, insurance must be offered on a set of products such that the actual price drop compensations do not exceed the total insurance token amounts collected. The business problem is to understand if the model is viable and advise the entrepreneur accordingly. As consultants, we would also help the entrepreneur get an idea regarding the number of products he/she needs to pick up to form a portfolio on a day to day basis and also what sort of a multiple he can offer on the insurance fee originally paid in case of a price drop. The analytics goal is to mine the available data and see if one can come up with a classification model that can accurately predict price increases on products to such an extent that the product portfolio formed on the models predictions turns an overall profit for the entrepreneur. DATA DESCRIPTION Data has been collected from five e-commerce sites namely BuyThePrice, HomeShop18, Indiaplaza, Infibeam and Saholic. Information such as model, brand, color, price, shipping period, stock-out and whether or not the price has increased is available on various products offered by these sites. Data preparation for building the model included: Converting categorical data into dummies. E.g. freeshipping, sitename and InStock. For running the Naïve Bayes method, we would need them as categories but for K-Nearest Neighbours, we needed to create categorical dummies. Dummies were also created for Brand and Color variables - since there were many different brands and colors, they were grouped first based on the outcome variable (brands/colors that have a very close average in terms of the output variable were put together as one group) and then categorized through dummy variables. These variables ultimately don t feature in any of the models as they didn t seem too important from a 3 P a g e
5 price up or not perspective and they were making the models too complicated with little improvement in predictive accuracy. Predicting missing values of input variables using K-NN. E.g. avrating and reviewcount. These variables had many values missing. Some of the other predictors available such as brand, color, free shipping, web portal etc. were used to predict the average rating or the review count where this information was missing. Numerical transformation of textual data using averages. E.g. shippingperiod. All the text had to be filtered out and the mid-value within the range specified in the test was used here. There were a few missing values in this variable as well but they were filled in using the average value of the shipping period across all products. Derivation of values from existing variables. E.g. timelast from date Some columns such as name and group were ignored Please refer to the Exhibit 0 for a few data visualizations. DATA MINING SOLUTION K-Nearest Neighbors The original dataset given has 7,765 rows which were partitioned into training and validation datasets. We also had a separate hold-out set on which we tested the model once we built it. The best k that minimizes error in the validation set turned out to be K = 5. Please refer table below. Value of k % Error Training % Error Validation <--- Best k Please refer to Exhibit 1A for the classification matrices on the training and validation datasets. Please refer to Exhibit 1B for the misclassification matrices based on the naïve rule. The KNN model built was used to score the test data. The lift chart is the ideal metric to use since we are concerned with picking up the top x% of products. The lift chart obtained for KNN on validation and test data is in Exhibit 1C. The following table demonstrates how one must select x based on performance on test dataset. # Cumulative PriceUp using Cumulative PriceUp when sorted Products average using predicted values % of PriceUp cases % % % % % 4 P a g e
6 Once products are sorted in descending order of predicted probabilities from the model based on test data, the number of products must be chosen such that the percentage of products where the price has actually gone up is highest. For example, in the above case, the number of products that should be chosen to offer insurance on is 14 out of a possible 350 considered in the test dataset or top 4% of products. We recommend that all the price change data from the previous day be used to retrain and validate the models every day. The revised models should then be run on the next day s consideration set of products and a similar top x% based on predicted probabilities must be picked up to offer insurance on. The single most important decision criterion among the various models is the identification of a top x% of products that has the maximum number of possible price increases that happened the last day. Data of price increases must be crawled and the modeling exercise done every day to keep the models fresh. Once x is determined, the percentage of products that might show a price rise is known and the return on insurance fee paid in case of price drop can be determined. For example, in the above case, since ~70% of the products chosen are likely to show a price increase, the entrepreneur may, at worst, end up paying a multiple of insurance fee on only 30% of products chosen. We say at worst because some products among the 30% will not show a price change and no money is lost on them. The insurance multiple that can be safely offered then in 70/30 = 2.33 (assuming simply that an equal number of people purchased insurance on all the products in the portfolio). Please note that the insurance multiple must be changed every day. Naïve Bayes The lift charts based on running this method on the test data set can be seen in Exhibit 2A. The table below shows why this model was not chosen as best. The best possible percentage is 41.18% if top 17 of 350 or top 4.86% of the products are chosen (not comparable to 71.43% from KNN) # Products Cumulative PriceUp using average Cumulative PriceUp when sorted using predicted values % of PriceUp cases % % % % % Classification Tree The lift charts based on running this method on the test data set can be seen in Exhibit 3A. The table below shows why this model is the best. The best possible percentage is 100% if top 10 of 350 or top 2.86% of the products are chosen (compared to 71.43%-KNN and 41.18%-NB.) # Cumulative PriceUp using Cumulative PriceUp when sorted Products average using predicted values % of PriceUp cases % 5 P a g e
7 % % % % Ensembles The only motivation to try ensembles after achieving 100% would be to increase the number of products on which to offer insurance and thus widen the net. We looked at a weighted combination of the probabilities obtained from the 3 models described above but we were not able to increase the number of products to offer insurance on. Weights were decided as inverse ratio of the error percentages on the test dataset. We end up with the exact same table above even for ensembles. CONCLUSIONS AND OPERATIONAL RECOMMENDATIONS Challenges in data collection: The model predicts price up and price down based on product model, brand, shipping and other meta-information. Collecting such data would require a webcrawler engine which parses these web sites dynamically and captures all the required metatags. The crawling software has to be built or acquired. Data needs to be collected daily. Data-driven: It might be tempting as an insurer to deviate from the product portfolio suggested by the model and include products where there is a general expectation of price increase. While such manual massaging works in the short term, it is a non-scalable approach. Automatic, datadriven portfolios are the way to go and for this. Time and money must be invested on improving data sources and data collection. Legal bindings: One should ensure that the insurance licenses cover the offered products. Also, the web crawling approach could be sued by the e-commerce sites and this must be taken into account in the planning stages. Review: The models and the web-crawling software should be reviewed and revised daily. Deployment: At the end of a day, all instances of price changes/non-changes must be captured and the models retrained, validated and tested on this data. Once finalized, the models need to be run on the consideration set for the next day and the x% and multiple details need to be figured out based on the test data set performance of the models. It is quite possible that the same model doesn t come out on top every day. If insurance is being offered on price drops over a certain period rather than a day, the problem becomes a lot more complex and historical information will need to be tracked. 6 P a g e
8 APPENDIX Exhibit 0: Data Visualizations Effect of list price on the PriceUp variable Effect of average rating on the PriceUp variable Effect of TimeLast on the PriceUp variable Effect of TimeNextPrice on the PriceUp variable 7 P a g e
9 Exhibit 1A: Performance of KNN on the training and validation data Training Data scoring - Summary Report (for k=5) Cut off Prob.Val. for Success (Updatable) 0.5 Classification Confusion Matrix Predicted Class Actual Class Error Report Class # Cases # Errors % Error Overall Validation Data scoring - Summary Report (for k=5) Cut off Prob.Val. for Success (Updatable) 0.5 Classification Confusion Matrix Predicted Class Actual Class Error Report Class # Cases # Errors % Error Overall Exhibit 1B: Performance of Naïve rule on the training and validation datasets created for KNN Training Data scoring - Naïve Rule Classification Confusion Matrix Predicted Class Actual 1 0 Class Error Report Class # Cases # Errors % Error Overall Validation Data scoring - Naïve Rule Classification Confusion Matrix Predicted Class Actual 1 0 Class Error Report Class # Cases # Errors % Error Overall P a g e
10 Exhibit 1C: Lift charts for KNN on test data Lift Charts of KNN on validation data 9 P a g e
11 Exhibit 2A: Lift charts on the test dataset based on Naïve Bayes Lift Charts of Naïve Bayes on validation data 10 P a g e
12 Exhibit 3A: Lift charts on the test dataset based on Classification Tree Lift Charts of Classification Tree on Validation dataset 11 P a g e
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© 2018 InstantÄ—, UAB. Visos teisÄ—s saugomos. | 2,454 | 11,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.902412 |
https://oasisacademyfallon.us/blog/math-update-jan-14-17/ | 1,620,439,866,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00403.warc.gz | 451,822,274 | 11,548 | array(1) { [0]=> string(1) "1" }
# Math Update: Jan. 14-17
January 13, 2019 by Kim Sorensen & Rebecca Farley
4th Grade Math: The 4th graders worked so hard this week! We had planned to continue multiplicative comparisons this week, but everyone was so solid last week we went ahead and tested on Thursday and our class rocked the test! This week we are looking forward to working with lines and angles.
IXLs that work with this week’s lessons: W.4, W.5, Z.1, Z.2, Z.5
3rd Grade Math: This past week the 3rd graders have worked hard at exploring the properties of multiplication. They did well at practicing the distributive property, which is when you take an equation like 5 x 14 and make it more friendly by distributing the 14 into 10 + 4 and multiplying both of those numbers by 5, then adding the products together.
5 x 14 = 5 x 10 + 5 x 4
This week we will review the properties of multiplication, identity, zero, commutative, associative, and distributive and test on Wednesday.
• Ask them to define the identity, zero, commutative, associative properties of multiplication. There is a page on the test where they will write out the definitions in their own words and the more chance they have to explain it, the more successful they will be!
• Ask them how to solve, 3 x 14 or 9 x 7 using the distributive property.
IXLs that match this week’s instruction: N.5 | 359 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-21 | latest | en | 0.94807 |
https://www.mathvids.com/browse/college/calculus/derivatives-and-integrals/derivatives-rules-of-derivatives/1037-derivatives-part-1 | 1,701,533,056,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00470.warc.gz | 1,001,526,584 | 9,471 | # Derivatives Part 1
Sick of ads? Sign up for MathVids Premium
Taught by MrRutter
• Currently 3.0/5 Stars.
5648 views | 2 ratings
Part of video series
Meets NCTM Standards:
Lesson Summary:
This lesson on derivatives provides an introduction to the concept and definition of a derivative. The lesson begins with an explanation of the tangent line to a curve and how to approximate its slope using two points. As the points get closer and closer together, the approximation gets better and better, allowing for a more accurate representation of the tangent line. The lesson then moves on to using the idea of a limit to calculate the slope of the tangent line, ultimately leading to the definition of a derivative as the slope of the tangent line.
Lesson Description:
An introduction to the concept and definition of a derivative.
Questions answered by this video:
• What is a derivative?
• What is the formal definition of a derivative?
• How do you estimate the slope of the tangent line to a curve at a point?
• What does delta mean in Calculus?
• What is delta x?
• What does it mean to find a derivative?
• #### Staff Review
• Currently 4.0/5 Stars.
This lesson is a very basic and simple explanation of what a derivative is, explained using the slope of the tangent line to a curve at a point. | 290 | 1,304 | {"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} | 3.296875 | 3 | CC-MAIN-2023-50 | latest | en | 0.930273 |
https://mapleprimes.com/questions/213420-From-Python--Into-Maple | 1,511,560,991,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808972.93/warc/CC-MAIN-20171124214510-20171124234510-00253.warc.gz | 663,695,267 | 23,247 | Question:From Python into Maple
Question:From Python into Maple
Maple
how to translate python code which use scipy, numpy to maple code
```import numpy as np
from scipy.sparse.linalg import svds
from functools import partial
def emsvd(Y, k=None, tol=1E-3, maxiter=None):
"""
Approximate SVD on data with missing values via expectation-maximization
Inputs:
-----------
Y: (nobs, ndim) data matrix, missing values denoted by NaN/Inf
k: number of singular values/vectors to find (default: k=ndim)
tol: convergence tolerance on change in trace norm
maxiter: maximum number of EM steps to perform (default: no limit)
Returns:
-----------
Y_hat: (nobs, ndim) reconstructed data matrix
mu_hat: (ndim,) estimated column means for reconstructed data
U, s, Vt: singular values and vectors (see np.linalg.svd and
scipy.sparse.linalg.svds for details)
"""
if k is None:
svdmethod = partial(np.linalg.svd, full_matrices=False)
else:
svdmethod = partial(svds, k=k)
if maxiter is None:
maxiter = np.inf
# initialize the missing values to their respective column means
mu_hat = np.nanmean(Y, axis=0, keepdims=1)
valid = np.isfinite(Y)
Y_hat = np.where(valid, Y, mu_hat)
halt = False
ii = 1
v_prev = 0
while not halt:
# SVD on filled-in data
U, s, Vt = svdmethod(Y_hat - mu_hat)
# impute missing values
Y_hat[~valid] = (U.dot(np.diag(s)).dot(Vt) + mu_hat)[~valid]
# update bias parameter
mu_hat = Y_hat.mean(axis=0, keepdims=1)
# test convergence using relative change in trace norm
v = s.sum()
if ii >= maxiter or ((v - v_prev) / v_prev) < tol:
halt = True
ii += 1
v_prev = v
return Y_hat, mu_hat, U, s, Vt```
| 471 | 1,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-47 | latest | en | 0.521343 |
https://questions.examside.com/past-years/gate/question/the-particular-solution-for-the-differential-equation-d2y-ov-gate-me-1996-marks-1-jju23q4er3u0jfce.htm | 1,719,017,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00576.warc.gz | 411,979,559 | 29,944 | 1
GATE ME 1996
+1
-0.3
The particular solution for the differential equation $${{{d^2}y} \over {d{t^2}}} + 3{{dy} \over {dx}} + 2y = 5\cos x$$ is
A
$$0.5\,Cos\,x + 1.5\,Sin\,x$$
B
$$1.5\,Cos\,x + 0.5\,Sin\,x$$
C
$$1.5\,Sin\,x$$
D
$$0.5\,Cos\,x$$
2
GATE ME 1996
+1
-0.3
The one dimensional heat conduction partial difference equation $$\,\,{{\partial T} \over {\partial t}} = {{{\partial ^2}T} \over {\partial {x^2}}}\,\,$$ is
A
parabolic
B
hyperbolic
C
elliptic
D
mixed
3
GATE ME 1995
+1
-0.3
The solution to the differential equation $$\,{f^{11}}\left( x \right) + 4{f^1}\left( x \right) + 4f\left( x \right) = 0$$
A
$${f_1}\left( x \right) = {e^{ - 2x}}$$
B
$${f_1}\left( x \right) = {e^{2x}},\,\,{f_2}\left( x \right) = {e^{ - 2x}}$$
C
$${f_1}\left( x \right) = {e^{ - 2x}},\,\,{f_2}\left( x \right) = x{e^{ - 2x}}$$
D
$${f_1}\left( x \right) = {e^{ - 2x}},\,\,{f_2}\left( x \right) = {e^{ - x}}$$
4
GATE ME 1995
True or False
+1
-0
A differential equation of the form $${{dy} \over {dx}} = f\left( {x,y} \right)\,\,$$ is homogeneous if the function $$f(x,y)$$ depends only on the ratio $${y \over x}$$ (or) $${x \over y}$$
A
TRUE
B
FALSE
GATE ME Subjects
EXAM MAP
Medical
NEET | 568 | 1,180 | {"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} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.576192 |
https://www.tenforums.com/drivers-hardware/27662-how-determine-what-cpu-better.html | 1,539,917,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00179.warc.gz | 1,109,403,167 | 12,266 | ## How to determine what CPU is better
For anyone looking here, this is just a guide I thought up and I think this is pretty useful.
If you are deciding between 2 or more CPUs, use this equation:
A= Number of cores
B= Clock speed of each core
C= The cache of the CPU in MB, also make it as, for example, 6MB cache, that would be 0.6. | 91 | 335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-43 | latest | en | 0.95717 |
http://gmatclub.com/forum/application-review-59048.html?fl=similar | 1,484,990,144,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00552-ip-10-171-10-70.ec2.internal.warc.gz | 122,426,902 | 57,799 | Application review... : The B-School Application
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# Application review...
Author Message
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Current Student
Joined: 07 Aug 2007
Posts: 1062
Followers: 4
Kudos [?]: 31 [2] , given: 0
### Show Tags
25 Jan 2008, 15:50
2
KUDOS
Found this thread on the s2s forums:
http://s2s.wharton.upenn.edu/wh-wharton/messages/?msg=20838
Basically a R1 Wharton reapplicant who got dinged put up his application and wanted feedback on why he was dinged. I think the responses to this thread is very informative and would highly recommend everyone thinking about applying next year to take a look at this thread.
GMAT Club Legend
Affiliations: HHonors Diamond, BGS Honor Society
Joined: 05 Apr 2006
Posts: 5926
Schools: Chicago (Booth) - Class of 2009
GMAT 1: 730 Q45 V45
Followers: 314
Kudos [?]: 2039 [2] , given: 7
### Show Tags
26 Jan 2008, 05:13
2
KUDOS
1) Resume is very weak - it fails completely to describe what he did and what the results of those actions are.
2) His essay attempts to recast the core question - the question asks how do you expect Wharton to help you achieve these goals - not "Why Wharton". It's a small difference, but there is a difference. Regardless, his why Wharton is exactly the kind of mindless paragraph that says nothing of consequence. I mean heck, his first point is that he has "no doubt" Wharton is "credible".... Its terrible any way you dice it, but its especially terrible since it's "qualified".. It's like saying to a friend "I have no doubt that you are capable" vs "You are really good at this". I don't advocate pandering, but if you are going to do it, at least make it sound like you actually believe it. He actually qualifies a lot of his words... things like "I think my exposure"? You think? You don't know? I also hate the 4th grader approach of providing headlines. Use transitional sentences. In the grand scheme of things, it doesn't matter, but it just makes for choppy reading.
3) Most damaging of all, its completely unclear what he wants from an MBA. He fails to answer that. Raise your hand if you think he forgot to tell you his long term goal? I raised mine.... but actually, its there... it just took me a while to realize it. The problem is, I thought his short term goal was to join a utility company or the World Bank. As it turns out, he lists his short term goal as getting an MBA, and the utility company thing is a long term goal! I think its pretty clear that "getting an MBA" is a horribly weak short term goal. He didn't understand the question, and thus, he didn't really answer it. This alone probably did him in.
4) I have to point this out too... The World Bank is a pretty HUGE leap from a utility company. The two have almost nothing in common. Immediately, it seems clear he has no clue what his goals are.
5) His why now is basically "I want it now cause my career isn't going anywhere". Not especially convincing.
6) The essays lack passion or introspection -- as a result they are entirely forgettable.
7) Differences from last application... This should have been the wow moment - the differentiator - but there's nothing here. He got promoted. He ran a few meetings. He raised $7k (I'd rather know what he did specifically than how much he made). He "managed large teams"... etc There's just nothing here but a laundry list of things, completely unconnected and most of them say nothing of consequence. So he ran meetings... with what goal? What was achieved? Why does it matter? How is that important for graduate school? What did he learn? Etc. etc. etc. I could go on and on.... In short, If I had to pick the top 3: 1) He failed to answer what his goals are / why he needs an mba. 2) Completely and totally fails to explain why Wharton 3) Whether its through lack of introspection or just a mediocre experience thus far, nothing in here is at all memorable. How did he get an interview? Its possible last years essays fill in some of these gaps. Or maybe, they decided to interview him to better understand the answers to these questions. Director Joined: 25 Dec 2007 Posts: 524 Schools: Harvard '11 Followers: 9 Kudos [?]: 64 [0], given: 28 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 06:20 That was very helpful for me to read. He seems qualified, but I agree with Rhyme - his essays are too generic. Thanks for the link dosa! Current Student Joined: 07 Aug 2007 Posts: 1062 Followers: 4 Kudos [?]: 31 [1] , given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 06:38 1 This post received KUDOS Thanks rhyme- that was a really detailed evaluation from your side. Definitely helps to get some expert opinions from you msday-when you visit chicago and attend a gsb live event, they share an application similar to this and discuss what they look for as well. i thought this was a good example on what NOT to do in an app. Senior Manager Joined: 26 Sep 2007 Posts: 264 Followers: 1 Kudos [?]: 4 [0], given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 07:15 Nice dosa! Uno kudos for you! Current Student Joined: 01 Aug 2007 Posts: 157 Followers: 1 Kudos [?]: 13 [0], given: 1 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 07:52 Now, we all can say what was wrong with his application. But the starnge thing is, if those flaws in essays were so blatant and easy to identify ( assuming nobody of the commentors in that thread went over his app for more than 20-30 minutes), then why Wharton decided to interview him ( after their stringent process of reading and re-reading and stacking of app files)? Did they not read his essays? And according to Wharton website, they care about the highest GMAT, then why in the feedback they mentioned the first GMAT score? This point is having an implication beyond that guy's application. This pokes me to think , how much should we trust the schools' to follow their own policies? Director Joined: 25 Dec 2007 Posts: 524 Schools: Harvard '11 Followers: 9 Kudos [?]: 64 [0], given: 28 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 08:35 It's entirely possible that he had very good recommendations...I think I value recommendations very highly in the process, as it is the only thing that is entirely out of your control. If you can get 2-3 other people to say very nice things about you (in the professional sense), then I think that'd be very important. filmcity wrote: Now, we all can say what was wrong with his application. But the starnge thing is, if those flaws in essays were so blatant and easy to identify ( assuming nobody of the commentors in that thread went over his app for more than 20-30 minutes), then why Wharton decided to interview him ( after their stringent process of reading and re-reading and stacking of app files)? Did they not read his essays? And according to Wharton website, they care about the highest GMAT, then why in the feedback they mentioned the first GMAT score? This point is having an implication beyond that guy's application. This pokes me to think , how much should we trust the schools' to follow their own policies? VP Joined: 10 Jun 2007 Posts: 1459 Followers: 7 Kudos [?]: 255 [0], given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 09:05 Great post. My redundant comments... The guy certainly has great GMAT score, great GPA from reputable engineering school, and nice salary. His work can potentially make significant impact in our deteriorating environment. In addition, he has shown the admission committee that he truly want to attend the school by being a re-applicant and desire to improve weaknesses by obtaining feedback. These maybe the reasons he was interviewed. On the other hand, it is unfortunate that his goal/passion/potentials doesn't come across in the essays. If I am the CEO of a company, I would not hire him simply from his writing. To me, his essays proof that 88%tile in verbal and 5.5 AWA in GMAT do not equate good writing. A lot of claims he made did not sound convincing, no specific details, no passion, confused, not logical, which ultimately led the readers to believe that he did not give enough thoughts on his essays. So much potentials lost Intern Joined: 26 Jan 2008 Posts: 9 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 10:15 In general, I would NEVER try to explain a low GMAT score in the optional essay or anywhere else (whether 1st or 2nd try). It just is what it is. Trying to explain it makes you look weak in my opinion. _________________ http://mbavolunteers.weebly.com - Application feedback in exchange for a donation to a non-profit Manager Joined: 23 Aug 2007 Posts: 136 Followers: 1 Kudos [?]: 8 [0], given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 10:17 I find it hard to criticise since I am not a good writer either, but I can provide my comments on the facts. The reasons mentioned to select CIVIL engineering are not convincing to me and the statement where he mentions "The idea of drinking water straight out of a tap without boiling it was alien to me until I traveled to Hungary as a 12 year old." would raise question mark if even one person at adcomm knew about India because most of the middle and lower class drinks water right out the tap without boiling it. "However, at EWBLA I felt that most members were treating chapter activities either as opportunities for having social gatherings or as a fashion or status symbol." Part judgmental, may be honest but schools will not like this if next thing you say is, "To avoid further conflict within the team I gradually reduced my inputs and started looking for other avenues to volunteer." If there was so much concern about victims as to make judgment that was made in the first statement, one must never give up the efforts for few people not collaborating. SVP Joined: 05 Aug 2007 Posts: 1502 Schools: NYU Stern '11 Followers: 15 Kudos [?]: 211 [0], given: 22 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 10:41 Useful comments but hindsight vision is always 20/20. I'm sure a lot of us will make the same mistakes when we actually write our essays on our own under the pressure of deadlines. As his interview call might indicate, a 'weak' essay isn't necessarily a dealbreaker. Maybe he just wasn't able to make the best use of his face to face time with the adcom during the interview. Current Student Joined: 22 Apr 2007 Posts: 1097 Followers: 5 Kudos [?]: 26 [0], given: 0 Re: Application review... [#permalink] ### Show Tags 26 Jan 2008, 10:49 Rhyme has done a good job evaluating the application. I agree with most of the posters that the biggest problem seems to be his somewhat unclear goals. I have had the same problem in my goals essays .. I haven't been able to make them very clear. hbs.aspirant wrote: "However, at EWBLA I felt that most members were treating chapter activities either as opportunities for having social gatherings or as a fashion or status symbol." Part judgmental, may be honest but schools will not like this if next thing you say is, "To avoid further conflict within the team I gradually reduced my inputs and started looking for other avenues to volunteer." If there was so much concern about victims as to make judgment that was made in the first statement, one must never give up the efforts for few people not collaborating. You brought up interesting points and I completely agree with you. I agree that the first statement is judgmental and will not seem good to the adcom. Also, if you are so concerned about the 'cause', then you would certainly not give it up because others are not sincerely working for it. In fact, my belief is that for every one person that is completely devoted to the cause, there are another 50+ people who would start working for the cause with a bang only to move away in a few days or weeks. Those 50+ people may have started the work as a fashion/status symbol, they may have had just a brief urge to do something for the community or humanity or they may have found the work more involving than they first thought. But this does not mean that they don't want good for the cause. And this does not mean that the cause can't get any value out of them. Anybody who has attempted to raise awareness for a good cause would know that you desperately need those 50+ people even if you know that they aren't completely devoted to the cause. If nothing else, they can help you tremendously with viral marketing (By telling friends, posting on their forums / social websites etc). Even their brief urges to do good for the community provide you$ through donations and viral marketing that are simply invaluable for the cause. There is this one girl who had this urge to help our 'school for poor ' one day. She told me that she was sending an email about our school to 6000+ members of her community in a social website. I don't care if that girl did this only to show off to her friends that she was doing something good or if she just had a moment of truth. Bottomline is that she had put our school on the radars of 6000+ people in one go. Her contribution is by no means lesser than that of people like this Wharton applicant who has perhaps spent many many hours volunteering or that of anybody else.
There are millions of people who turn a blind eye to all social causes or make fun of people involved in social causes. In such a world, how can you complain about people who are at least aware of the needs of the community and who are at least making the effort of coming to a social event?
BTW, great post, dosa. This application review thread may help 2009 applicants a lot (Depending on the kind of comments people post here). Kudos to you.
Director
Joined: 18 Dec 2007
Posts: 983
Location: Hong Kong
Concentration: Entrepreneurship, Technology
Schools: Hong Kong University of Science and Technology (HKUST) - Class of 2010
Followers: 12
Kudos [?]: 134 [0], given: 10
### Show Tags
26 Jan 2008, 11:21
I agree with rhyme, but also, i think there are some serious logical problems as well e.g., His career has plateaued, yet he is only 23 and achieved a promotion in the last year? How does that show it`s plateaued?
Not that saying your career has plateaued is a good thing.
One thing i do during interviews when explaing a why mba and why not engineering; i use a comparative description that most people understand. Typecasting and bring in an example of acting.
GMAT Club Legend
Status: Um... what do you want to know?
Joined: 03 Jun 2007
Posts: 5464
Location: SF, CA, USA
Schools: UC Berkeley Haas School of Business MBA 2010
WE 1: Social Gaming
Followers: 72
Kudos [?]: 398 [0], given: 14
### Show Tags
28 Jan 2008, 15:28
darn, I guess I missed it. He removed his application already.
But this discussion is good, so it'll stay linked in the Knowledge vault.
_________________
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http://gmatclub.com/forum/123
http://gmatclub.com/forum/128-t62555
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Re: Application review... [#permalink] 28 Jan 2008, 15:28
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1 Application's on Director's desk for review 9 07 Apr 2008, 12:30
Display posts from previous: Sort by | 3,962 | 16,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-04 | latest | en | 0.925504 |
http://puzzle.queryhome.com/773/how-you-can-get-number-using-following-numbers-and-operators | 1,503,496,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.0/warc/CC-MAIN-20170823132736-20170823152736-00540.warc.gz | 355,300,689 | 31,647 | # How you can get number 28 by using following numbers and operators.
73 views
How you can get number 28 by using following numbers and operators
Numbers: 1, 5, 7 and 8.
Operators: +, -, * and /
You can use ( ) brackets as well.
Rules:-
You can use each number only once.
You can not use 1 and 5 as 15.
You have to use all of the above number mentioned (1,5,7 and 8)
posted May 12, 2014
## 3 Solutions
+1 vote
solution May 12, 2014
We can make 28 as follow as:
(7*5)-8+1 = 35-8+1
Hence It is 28
solution Mar 28, 2016 by
We can find 28 in this format :-
(5*7)+1-8 = 28
solution Mar 28, 2016
Similar Puzzles
–1 vote
By using the numbers 8,5 1 and 1 exactly once and with the help of basic operation (+, -, /, *), Can you form an expression with a value of 10?
Note: Parentheses are allowed
If repetition of numbers is not allowed then how many numbers of five digits can be formed by using 1,2,3,4,5.
If 4 is necessarily taken at hundred's place and 2 is not allowed at unit's place | 315 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-34 | latest | en | 0.925301 |
https://mathoverflow.net/questions/143143/interesting-families-of-groups-as-group-extensions | 1,582,844,554,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00523.warc.gz | 409,773,558 | 27,915 | # Interesting families of groups as group extensions
Let me start this question with an example that hopefully makes clear what I am looking for:
A discrete subgroup $G$ of the group of euclidean isometries of $\mathbb{R}^d$ is called a crystallographic group if it contains a lattice $L$ of full rank. In this case, $G/L \cong P$ is a finite crystallographic subgroup of the orthogonal group. So what we get is a short exact sequence $$0 \rightarrow \mathbb{Z}^d \rightarrow G \rightarrow P \rightarrow 1,$$ which can also be expressed as stating that $G$ is isomorphic to an extension of $\mathbb{Z}^d$ by $P$. Remarkably, the converse is also true. Namely that any such extension is isomorphic to a crystallographic group. So the family of crystallographic groups is in $1$-to-$1$ correspondence with group extensions of $\mathbb{Z}^d$ by a crystallographic point group.
Now, for my question. What other interesting examples of families of groups are there that can also be described by group extensions in a similar way as crystallographic groups can?
Let me add a few sentences to maybe prevent misinterpretation of my question. I am certainly aware of the theorem of Jordan-Hölder and its consequence that one in principle could survey all finite groups by using the classification of all finite simple groups and solving the extension problem. But this also means allowing for very complicated sets of groups that can be used as a first part and last part in the above sequence. I am looking for more reasonable examples, where "reasonable" appeals to everyone's intuition.
• Nice question, thank you. I would also add the tag of geometric topology to it. – Sasha Anan'in Sep 25 '13 at 11:56
• Would something like the class of metabelian groups qualify? en.wikipedia.org/wiki/Metabelian_group Also people study "$x$-by-$y$ groups" where $x$ and $y$ are adjectives such as cyclic, abelian, finite, nilpotent... I must confess, I think you need to make your question a little more precise. – Mark Grant Sep 25 '13 at 11:59
• Lots of group extensions appear in Galois embedding problems. But I am not sure if they fit in your criterion for being interesting! – user23860 Sep 25 '13 at 13:25
• This example has a quasi-isometric generalization: any group QI to $\mathbb Z^d$ is ((finite)-by-($\mathbb Z^d$-by-$P$)). So this applies, for example, to any group containing a finite index $\mathbb Z^d$ subgroup. – Lee Mosher Sep 25 '13 at 13:39
• Your statement is not quite true: Direct product of a finite group and a free abelian group is not crystallographic. A better definition is of a "c-type group", as a group which acts properly discontinuously isometrically and cocompactly on a Euclidean space. In this definition, non-effective action with finite kernel are allowed. Then, a f.g. group is virtually abelian iff it is a c-type group. – Misha Sep 25 '13 at 14:40
• Derek, what is your convention for "$G$-by-$H$" notation? $1 \to G \to X \to H \to 1$? or $1 \to H \to X \to G \to 1$? – Lee Mosher Sep 25 '13 at 13:43
• I am using $G$-by-$H$ to mean normal subgroup $G$ with quotient $H$. So polycyclic groups have a normal nilpotent subgroup with virtually abelian quotient group. (I know A-by-B-by-C for classes of groups A,B,C, is not necessarily associative, but I think it makes no difference in this case.) – Derek Holt Sep 25 '13 at 15:27 | 875 | 3,362 | {"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} | 2.8125 | 3 | CC-MAIN-2020-10 | longest | en | 0.939213 |
http://cboard.cprogramming.com/c-programming/97872-pithagore-fast-computation-printable-thread.html | 1,406,753,833,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510271648.4/warc/CC-MAIN-20140728011751-00096-ip-10-146-231-18.ec2.internal.warc.gz | 42,868,193 | 3,809 | # Pithagore-fast computation
• 01-14-2008
invinciblevn
Pithagore-fast computation
Dear all.
I'm doing the computation rectangle's edge using Pithagore theory.
All 3 rectangle's edge are integer.
Is there any algorithm for fast computation that don't use sqrt() function?
Thanks.
• 01-14-2008
vart
c is less than a+b but greater than max(a,b)
you can use some search algorithm in this range with test condition c*c == a*a +b*b
to find c
not guaranteed to be faster, but it uses only integer math
• 01-14-2008
iMalc
Post the function in your current code that uses it and we'll show you an optimised version.
• 01-14-2008
invinciblevn
just now I use this function:
c=(int) sqrt(a*a+b*b);
• 01-14-2008
vart
Quote:
Originally Posted by invinciblevn
just now I use this function:
c=(int) sqrt(a*a+b*b);
that can be 1 less than the actual value
• 01-14-2008
invinciblevn
Quote:
Originally Posted by vart
that can be 1 less than the actual value
I'm doing compute distant in an image which use pixel as unit.
sqrt() require a lot resource.
• 01-14-2008
matsp
So don't calculate the sqrt, just use the c * c value - for all intents and purposes, it's the same thing, just bigger.
--
Mats
• 01-14-2008
Dino
Pythagorean. Pythagoras invented it.
• 01-14-2008
iMalc
Quote:
Originally Posted by invinciblevn
just now I use this function:
c=(int) sqrt(a*a+b*b);
That's not what I meant. I meant, post the whole function that includes this line of code. Assuming of course that it isn't longer than a couple dozen lines...
It's extremely difficult to optimise 1 line of code, but if we can see the lines of code around it then it becomes quite easy.
• 01-14-2008
invinciblevn
Quote:
Originally Posted by iMalc
That's not what I meant. I meant, post the whole function that includes this line of code. Assuming of course that it isn't longer than a couple dozen lines...
It's extremely difficult to optimise 1 line of code, but if we can see the lines of code around it then it becomes quite easy.
Thank you very much, iMalc.
I'm doing the image rotation. I use this for point distance computation.
This computation is used many times in my program, so that is require a lot resource.
I wonder whether there is any other way of point distant computation that does not use pythagore?
That is my problem.
Thanks.
• 01-15-2008
Your problem must be a common one in gaming forums. (as opposed to general purpose programming forums like this). I'd check the gaming forum, and also game.dev for more detailed answers.
You can find the hypotenuse of a triangle by using it's two sides and the angle between them, but I have no idea if that is any faster, at all. The game programmers will know what's quick.
• 01-15-2008
matsp
Quote: | 737 | 2,722 | {"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-2014-23 | longest | en | 0.804882 |
http://mathhelpforum.com/discrete-math/37892-de-morgan-s-second-law-proof.html | 1,511,517,261,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807344.89/warc/CC-MAIN-20171124085059-20171124105059-00090.warc.gz | 194,041,190 | 10,557 | # Thread: De Morgan's second law Proof
1. ## De Morgan's second law Proof
Prove De Morgan's second law, ~(AB) = ~A v ~B, by applying De Morgan's first law to ~A and ~B.
I think I know how to do it but I just need some clarification to justify my answer... Thanks
2. Okay, then post your solution to see if it's correct.
P.S.: that should be $\sim(A\wedge B).$
3. I don't understand how this can be used to prove anything since $\sim(A\wedge B) \not{\equiv} \sim A ~\wedge \sim B$.
4. similar problem done here | 150 | 516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-47 | longest | en | 0.922259 |
https://edurev.in/chapter/6781_Chapter-14-Theoretical-Distribution | 1,653,033,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00649.warc.gz | 278,860,550 | 52,520 | # Chapter 14 - Theoretical Distribution - Quantitative Aptitude for CA CPT | CA Foundation | Notes, Videos & Tests
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## More chapters similar to Chapter 14 - Theoretical Distribution for CA Foundation
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Chapter 14 - Theoretical Distribution | Quantitative Aptitude for CA CPT | 955 | 4,607 | {"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-2022-21 | latest | en | 0.807855 |
https://tex.stackexchange.com/questions/495628/align-does-not-increment-the-equation-counter | 1,713,663,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00090.warc.gz | 514,521,150 | 35,315 | # Align does not increment the equation counter
The align environment is not advancing the equation counter. Here's the code that produced the output below:
\documentclass{report}
\usepackage{mathtools,amsthm}
\begin{document}
\numberwithin{section}{chapter}
\numberwithin{equation}{section}
\let\realequation\equation
\def\equation{\setcounter{equation}{\arabic{subsection}}%
\refstepcounter{subsection}%
\realequation}
\theoremstyle{definition}
\newtheorem{thm}[subsection]{Theorem}
\newtheorem{lemma}[subsection]{Lemma}
\newtheorem{corol}[subsection]{Corollary}
\newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
\newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
\newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
\newtheorem{exmp}[subsection]{Example} % same for example numbers
\newtheorem*{remark}{Remark}
\newtheorem{enum_remark}[subsection]{Remark}
\newtheorem{exercises}[subsection]{Exercises}
\newtheorem*{opex}{Opening Exercises}
\begin{align} \label{nchoosed}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&= \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}
some text
$$a = b$$
\end{document}
The document is set up so that a single counter runs through each chapter and gets advanced by subsections, theorems, equations, etc.
• the numbering is determined by code you are not showing so it is impossible to guess what is wrong. Please edit your example so that it is a complete small document that shows the problem. Jun 13, 2019 at 15:11
• @DavidCarlisle thanks I've updated the code Jun 13, 2019 at 15:40
• well it wasn't runnable as posted but I guessed some needed packages, it doesn't make the posted image but i suppose now shows the issue Jun 13, 2019 at 15:45
• align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1 Jun 13, 2019 at 15:48
• ahhh, ok, so I need to redefine align so that it also uses the subsection counter? Jun 13, 2019 at 15:50
align advances the equation counter but you have redefined equation to ignore that and use the subsection counter, so naturally they both use 1
This makes subsection and equation counter the same, so if any construct increases one, the value of the other will have changed.
\documentclass{report}
\usepackage{mathtools,amsthm}
\numberwithin{section}{chapter}
\numberwithin{equation}{section}
% make equation counter an alias for subsection
\makeatletter
\let\c@equation\c@subsection
\makeatother
\theoremstyle{definition}
\newtheorem{thm}[subsection]{Theorem}
\newtheorem{lemma}[subsection]{Lemma}
\newtheorem{corol}[subsection]{Corollary}
\newtheorem{defn}[subsection]{Definition} % definition numbers are dependent on theorem numbers
\newtheorem{defns}[subsection]{Definitions} % definition numbers are dependent on theorem numbers
\newtheorem{prop}[subsection]{Proposition} % proposition numbers are dependent on theorem numbers
\newtheorem{exmp}[subsection]{Example} % same for example numbers
\newtheorem*{remark}{Remark}
\newtheorem{enum_remark}[subsection]{Remark}
\newtheorem{exercises}[subsection]{Exercises}
\newtheorem*{opex}{Opening Exercises}
\begin{document}
first align
\begin{align} \label{nchoosed}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&= \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}
some text
$$a = b$$
second align
\begin{align} \label{nchoosedz}
\begin{split}
\prescript{}{n}C_d = \binom{n}{d}
&= \frac{n \cdot (n-1) \cdot (n-2) \cdots (n - d + 1)}{d!} \\
&= \frac{n!}{d! (n-d)!}
\end{split}
\end{align}
some text
$$a = b$$
\end{document}
• Thanks so much! that's the fix. Jun 13, 2019 at 16:02 | 1,128 | 3,869 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 3, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.734326 |
http://mathhelpforum.com/advanced-statistics/218091-compare-expectation-two-functions-same-r-v.html | 1,511,476,801,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806979.99/warc/CC-MAIN-20171123214752-20171123234752-00657.warc.gz | 197,093,801 | 10,578 | # Thread: Compare the expectation of two functions of same r.v.
1. ## Compare the expectation of two functions of same r.v.
Hello everybody,
does anyone have an idea how to prove the following inequality
$E(\log(1+c_1(e^{Z+b}-1)))\geq E(\log(1+c_2(e^{Z+b}-1)))$?
where $Z$ is normally distributed with a non-negative mean and $c_1,c_2$ are constants with $c_1,c_2 \in [0,1],c_1\geq c_2$
$b>0$ is also a constant.
2. ## Re: Compare the expectation of two functions of same r.v.
Hey Juju.
Try showing that the function is monotonic increasing and using that, show that if CDF_X(p) > CDF_Y(p) for all p in [0,1] then E[X] > E[Y].
3. ## Re: Compare the expectation of two functions of same r.v.
Thanks!
Unfortuanately, I do not completely understand your hint. The function $log(1+c(e^{Y-b}+1)))$ is increasing and convex in $Y.$
The expectation is given by
$\int log(1+c(e^{y-b}+1)))\frac{1}{\sqrt{2 \pi \sigma^2}}e^{-0.5\frac{(x-\mu)^2}{\sigma^2}} dx$
Or do you mean that I should consider the cdf of the function of r.v. $log(1+c(e^{Y-b}+1)))$?
$P(log(1+c_1(e^{Y-b}+1))) \leq y)=P(e^Y\leq \frac{e^y-1-c_1}{c_1})$
But now it does not hold
$\frac{e^y-1-c_1}{c_1}\leq \frac{e^y-1-c_2}{c_2}\; \forall y$
4. ## Re: Compare the expectation of two functions of same r.v.
Yes, consider the CDF of the log function of the random variable. That was my initial intent and idea for you, | 479 | 1,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.762444 |
https://byjus.com/question-answer/a-g-rightarrow-2b-g-c-g-initially-at-t-0-gas-a-was-present-6/ | 1,680,344,010,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00153.warc.gz | 193,214,737 | 28,621 | Question
# A(g)→2B(g)+C(g) Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is 13. After some time t=t1, the total pressure is half of the final total pressure at t=tx (a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at t=tx, the final total pressure is 35 bar. Ratio of rate constant at t = 0 to t=t1 to t=tx is:
A
2 : 3 : 4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1 : 1 : 1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1 : 3 : 5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1 : 3 : 5
No worries! We‘ve got your back. Try BYJU‘S free classes today! | 249 | 763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.911279 |
https://www.hsslive.co.in/2021/12/1-square-feet-in-square-inch-odisha.html | 1,713,823,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00490.warc.gz | 722,743,293 | 40,865 | # HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## 1 Square Feet in Square Inch in Odisha
In this article, you will learn how to convert 1 Square Feet in Square Inch in Odisha. Square Feet is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to Square Feet for measurement of land. Here in this page you will learn how to calculate 1 Square Feet in Square Inch unit.
## How to Convert from 1 Square Feet to Square Inch?
You can easily convert Square Feet to Square Inch or the reverse with a simple method. You can multiply the figure in Square Feet by 143.999938 to determine the Square Inch value. This is all you need to do for undertaking the conversion procedure of Square Feet to Square Inch in Odisha.
## Relationship between Square Feet and Square Inch
It is not difficult to work out the actual relationship between Square Feet to Square Inch in most cases. You should know the proper calculations existing between Square Feet to Square Inch before you venture to conversion procedures. 1 Square Feet to Square Inch calculations will be 143.999938 Square Inch.
Once you know these calculations, it will not be hard for you to calculate Square Feet in an Square Inch. One Square Feet can be worked out to various other units as well.
## Formula for Converting Square Feet to Square Inch
The formula to convert Square Feet to Square Inch is the following-
1 Square Feet = 143.999938 Square Inch
Conversely, you can also use this formula-
Square Feet = Square Inch * 143.999938
This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.
### Other Plot Areas in Square Inch in Odisha
• 1 Square Feet in Square Inch in Odisha = 143.999938
• 2 Square Feet in Square Inch in Odisha = 287.999876
• 3 Square Feet in Square Inch in Odisha = 431.999814
• 4 Square Feet in Square Inch in Odisha = 575.999752
• 5 Square Feet in Square Inch in Odisha = 719.99969
• 6 Square Feet in Square Inch in Odisha = 863.999628
• 7 Square Feet in Square Inch in Odisha = 1007.999566
• 8 Square Feet in Square Inch in Odisha = 1151.999504
• 9 Square Feet in Square Inch in Odisha = 1295.999442
• 10 Square Feet in Square Inch in Odisha = 1439.99938
• 15 Square Feet in Square Inch in Odisha = 2159.99907
• 20 Square Feet in Square Inch in Odisha = 2879.99876
• 25 Square Feet in Square Inch in Odisha = 3599.99845
• 30 Square Feet in Square Inch in Odisha = 4319.99814
• 35 Square Feet in Square Inch in Odisha = 5039.99783
• 40 Square Feet in Square Inch in Odisha = 5759.99752
• 45 Square Feet in Square Inch in Odisha = 6479.99721
• 50 Square Feet in Square Inch in Odisha = 7199.9969
• 55 Square Feet in Square Inch in Odisha = 7919.99659
• 60 Square Feet in Square Inch in Odisha = 8639.99628
• 65 Square Feet in Square Inch in Odisha = 9359.99597
• 70 Square Feet in Square Inch in Odisha = 10079.99566
• 75 Square Feet in Square Inch in Odisha = 10799.99535
• 80 Square Feet in Square Inch in Odisha = 11519.99504
• 85 Square Feet in Square Inch in Odisha = 12239.99473
• 90 Square Feet in Square Inch in Odisha = 12959.99442
• 95 Square Feet in Square Inch in Odisha = 13679.99411
• 100 Square Feet in Square Inch in Odisha = 14399.9938
## FAQs About 1 Square Feet in Square Inch in Odisha
#### Q: Ek Square Feet main Kitne Square Inch in Odisha Hote hain?
Ek Square Feet main 143.999938 Square Inch in Odisha hote hain.
#### 1 Square Feet main kitna Square Inch in Odisha Hota Hain?
1 Square Feet main 143.999938 Square Inch in Odisha hote hain
Share: | 996 | 3,651 | {"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-2024-18 | latest | en | 0.888677 |
https://cboard.cprogramming.com/c-programming/80383-3plusd-array.html | 1,495,682,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607963.70/warc/CC-MAIN-20170525025250-20170525045250-00109.warc.gz | 744,724,001 | 16,415 | 1. ## 3+d array?
Is it possible to have a 3+ dimensional array?
I think a 3d array would work when working with 3d, but 4d+? Wouldnt that defy the laws of physics or something? What would it look like?
2. Code:
```int three_dimensional_array[x][y][z];
int four_dimensional_array[x][y][z][a];
int eight_dimensional_array[x][y][z][a][b][c][d][e];```
You'll run out of memory pretty quickly if you have lots of large arrays, though.
3. wait, how can you simulate a 4d+ array? Make a 4dimensional world?!
Isnt 4d time, 5d space, 6d amtter, 7d i have no clue... but im gonna google it.
4. What are you talking about? You're not creating physical dimensions here. Just a series of arrays of arrays of arrays ...
Quzah.
5. You are creating a huge mess. The most dimensions I'd ever use is 2, and I normally use 1D arrays or linear arrays.
6. I know, but arrays can be just like dimensions in real life. Isnt there a way to simulate it somehow? How 1d is like a string (a line) while 2d is like a square (down adn across) and 3d is like a 3d cube(detph/z axis). So what would 4d be?
Oh and this is just for informational purposes... Its not like im really gonan be using anything more then 3d tops.
7. Originally Posted by SG57
wait, how can you simulate a 4d+ array? Make a 4dimensional world?!
Isnt 4d time, 5d space, 6d amtter, 7d i have no clue... but im gonna google it.
The fact that humans perceive only 3 physical dimensions is coincidental to the fact that arrays can be multi-dimensional. Each dimension of an array represents whatever you are needing to represent. If you want to use a 4 dimension array to keep track of x,y and z coordinates for each time interval a then you could use a 4 dimension array.
Code:
`world[x][y][z][a]`
You could also use a larger dimension array something like
Code:
`data[row][column][sheet][workbook][folder][company][city][state][country][etc...]`
where the row and column are what is stored in the row and column of a spreadsheet that is in a particular workbook that is located in a particular folder that resides within a particular company that is based out of a certain city in a certain state in a particular country... This could be accomplished much more efficiently using other methods of course, but is an example of a 9+ dimension array.
Originally Posted by SG57
I know, but arrays can be just like dimensions in real life. Isnt there a way to simulate it somehow? How 1d is like a string (a line) while 2d is like a square (down adn across) and 3d is like a 3d cube(detph/z axis). So what would 4d be?
What do you want 4d to be? There are widely differing opinions of what the 4th dimension is, whether it is actually time, or it is some measurement humans are unable to perceive naturally, or whatever. If you are the program writer, you decide what 4d is. I like to think of the 4th dimension as time because it is conceptually easy for me to understand, but I have heard compelling arguments otherwise.
Either way, you should define your world (3d or 4d or 12d?) before you start trying to figure out how to represent it in a data structure.
8. you ready for a mess of uneccessary code?
Code:
``` int i = 3;
int* pi = &i;
int** p2i = π
int*** p3i = &p2i;
int**** p4i = &p3i;
int***** p5i = &p4i;
int****** p6i = &p5i;
int******* p7i = &p6i;
int******** p8i = &p7i;
int********* p9i = &p8i;
int********** p10i = &p9i;
int*********** p11i = &p10i;
int************ p12i = &p11i;
int************* p13i = &p12i;
int************** p14i = &p13i;
int*************** p15i = &p14i;
int**************** p16i = &p15i;
int***************** p17i = &p16i;
int****************** p18i = &p17i;
int******************* p19i = &p18i;
int******************** p20i = &p19i;
int********************* p21i = &p20i;
int********************** p22i = &p21i;
int*********************** p23i = &p22i;
int************************ p24i = &p23i;
int************************* p25i = &p24i;
int************************** p26i = &p25i;
int*************************** p27i = &p26i;
int**************************** p28i = &p27i;
int***************************** p29i = &p28i;
int****************************** p30i = &p29i;
int******************************* p31i = &p30i;
int******************************** p32i = &p31i;
int********************************* p33i = &p32i;
cout << *********************************p33i << endl;```
This is an attempt to find out if my compiler (MSVC 2003) put a limit on the levels of indirection I can have. Looking at the asm, she accounted for all 33. I don't want to go any further.
9. My compiler (MinGW) supports over 3000 levels of indirection before it runs out of stack space (it must be recursive).
10. The impossibility of 4 dimensions in real life does not in any way affect the plausibility of having a multidimensional array in C. In any case, even a 2D array is stored one-dimensionally in your computer's RAM. In fact, most processors don't support 2+D arrays; they are only provided in C as a convenience.
11. A line of text. - 1 dimension.
A page of lines. - 2 dimensions.
A folder of papers. - 3 dimensions.
A box of folders. - 4 dimensions.
A truck of boxes. - 5 dimensions.
A warehouse of trucks. - 6 dimensions.
A field of warehouses. - 7 dimensions.
We could go on, but I'll assume you've grasped the concept.
Quzah.
12. This thread is a clear example of the complete abuse of arrays and their usefulness.
This should answer all your questions. Notice it's from webster.cs.ucr - the same site you can get Randall Hyde's Art of Assembly Language. Definitely not a coincedence.
http://webster.cs.ucr.edu/AoA/Window.../Arraysa2.html
Now you can see why using more than 1 dimension will severely impact the performance of the array, especially if does NOT have dimensions that are a power of 2. Even using bit shifting for multi-dimensional arrays with dimensions that are a power of 2 will still be slower than using a 1 dimensional array.
So how do you solve the problem? In C you use structs, and in C++ you use classes.
Code:
```struct Person
{
int iAge;
int iHeightInches;
int iShoeSize;
int iYearBorn;
int iMonthBorn;
int iDayBorn
};
Person Crowd[1000];```
Much better than:
Code:
`int Crowd[1000][6];`
To fill the array using structs watch:
Code:
```for (int i=0;i<1000;i++)
{
Crowd[i].iAge=0;
Crowd[i].iHeightInches=0;
Crowd[i].iShoeSize=0;
Crowd[i].iYearBorn=0;
Crowd[i].iMonthBorn=0;
Crowd[i].iDayBorn=0;
}```
1000 iterations, 6 variables set per iteration.
As opposed to the two dimensional method:
Code:
```for (int i=0;i<1000;i++)
{
for (int j=0;j<6;j++)
{
Crowd[i][j]=0;
}
}```
6000 iterations.
You could unroll the inner loop and still get 1000 iterations, but you still have the fact that in the array, none of the elements have any apparent meaning. They are just memory locations. With the structure you can tell exactly what each variable is representing.
The only reason I would ever allocate a huge linear array is if I wanted to do my own memory management for something like a game. Then I would make functions and sort of a small memory management API to manage that huge chunk of memory. This has the benefit of, hopefully, being faster, easier to use, etc.
Hope this helps. | 1,970 | 7,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-22 | longest | en | 0.900878 |
https://antescofo-doc.ircam.fr/Library/HowTo/PlotLocalTempo/plotlocaltempo/ | 1,721,110,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514737.55/warc/CC-MAIN-20240716050314-20240716080314-00218.warc.gz | 85,605,915 | 10,833 | # Visualizing the tempo changes and the beat progression in a group¶
This how-to details a possible approach to visualize the local position progression and the local tempo change in an arbitrary group. We illustrate the approach on a specifioc example but the reader may adapt it immediately to its proper needs.
We suppose that some external process drive a tempovar $pos_to_play. To make the example self-contened, we emulate this process by a loop that shrink in time: @tempovar$pos_to_play(85,1) := 100
$p := 1 Loop DRIVE$p s
{
$ref_tempo := 60./$p
$old_p :=$p
if ($p > 0.5) {$p *= 0.9 }
$pos_to_play := 1 } during [20#] The loop DRIVE is used as a kind of metronome implementing an accelerando. The period is given by $p and starts at 1 second, i.e. a tempo of 60 BPM. At each iteration, the period is shortened until it drops below $0.5$, i.e. a tempo of 120.
The tempovar is initialized with a tempo of $80$. Therefore, the tempo computed by the tempovar will start to decrease (slowly) and then increase until is alignment to the tempo fixed by the limit valiuue of $p. We suppose that the tempovar is used to control the timing of a loop spanned by a whenever: $graincurvesduration := 50
$t := []$tt := []
$pos := []$rpos := []
$rcpos := []$tpo := []
$ftpo := []$xx := []
$top_x := []$top_t := []
$top_delta := [] whenever ($pos_to_play == $pos_to_play) @exclusive { @local$x := 0
_ := $tt.push_back($NOW)
_ := $ftpo.push_back(60./$p)
Loop L $graincurvesduration ms @sync$pos_to_play
@Abort {
$top_x.push_back(60. *$x)
$top_t.push_back($NOW)
}
{
$x +=$graincurvesduration/1000.
_ := $t.push_back($NOW)
_ := $pos.push_back($pos_to_play.position)
_ := $rpos.push_back($pos_to_play.rnow)
_ := $rcpos.push_back($pos_to_play.rcnow)
_ := $tpo.push_back($pos_to_play.tempo)
_ := $xx.push_back(60.*$x)
}
}
The nternal loop L is used to emulate an activity whose timing is driven by the tempovar. The period is fixed in absolute time and the variable $x is incremented at each iteration. The others variables are tabs recording various informations during the run. The expression ::antescofo$pos_to_play.position returns the current position (in beat) of the tempovar. This expression could have been replaced by @local_beat_position() which is more general. Function @local_beat_position returns the position in beat of the current temporal scope. So it can be used for any kind of synchronization reference. Similarly for ::antescofo $pos_to_play.tempo and @local_tempo. The information recorded can be ploted using @gnuplot: 11 s$gnuplot_path := "/usr/local/bin/gnuplot"
; $gnuplot_linestyle := "lines" _ := @gnuplot( "tempo",$t, $tpo, "forced tempo",$tt, $ftpo, "position",$t, $pos, "rnow",$t, $rpos, "curveX",$t, $xx, "maxX",$top_t, $top_x ) and the execution of the full program builds the following plot: The timing and synchronization behavior can be read as follow: • The violet staircase curve tempo represents the tempo computed by the tempovar. It decrease and the it converges to the tempo forced by the DRIVE loop. The curve is a staircase because the tempo computed by a tempovar (using the E. Large approach) is updated only when the tempovar is updated. • The tempo emulated by the DRIVE loop is pictured in green (curve labeled forced tempo). It increases starting from $60$ until it exceeds $120$. • The value of $x starts at $0$ eachtime the whenever is activated. It increase linearly with the iteration of the loop L. It is ploted in yellow and labeled curveX. The body of this loop records also the local tempo and the local position.
• When the whenever is activated, the previous activation is killed (as a result of the @exclusive attribute). The abord handler of the internal loop is used to record the maximal value reached by $x. This value is ploted on the blue circled curve labeled _maxX_q. Because the period is shrinking, the maximal value reached by $x decreases until the period stabilizes. | 1,051 | 3,940 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.829436 |
https://www.causal.app/excel/counting-groupings-below-a-threshold | 1,653,454,427,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00242.warc.gz | 748,597,078 | 4,041 | Excel Guides
## Counting Groupings Below a Threshold in Excel
To count the number of groups below a threshold in Excel, you can use the COUNTIF function. For example, if you have data in cells A1:A5 and you want to count the number of groups that are less than 3, you can use the following formula:
=COUNTIF(A1:A5,"<3")
This will return a value of 2, since there are two groups that are less than 3.
You can also use the COUNTIF function to count the number of groups that are greater than or equal to a certain threshold. For example, if you want to count the number of groups that are greater than or equal to 3, you can use the following formula:
=COUNTIF(A1:A5,">=3")
This will return a value of 3, since there are three groups that are greater than or equal to 3.
### Excel
Get started with Causal today.
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http://www.educator.com/physics/physics-b/jishi/circular-motion-part-2.php | 1,508,739,850,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00653.warc.gz | 414,150,091 | 42,020 | Start learning today, and be successful in your academic & professional career. Start Today!
• ## Related Books & Services
1 answerLast reply by: Vasilios SahinidisSat Aug 13, 2011 8:15 AMPost by malika eradi on February 21, 2011why Fn is down on the top?
### Circular Motion, Part 2
• If a jet moves in a vertical path, the centripetal force on the pilot is determined by a combination of the pilot’s weight and the normal force exerted by the seat.
• For an object attached to one end of a string and moving in a vertical circle, a combination of the weight of the object and tension in the string provide the necessary centripetal force.
### Circular Motion, Part 2
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Normal Force by a Pilot Seat 0:14
• Example : Pilot Rotating in a Circle r and Speed s
• Pilot at Vertical Position in a Circle of Radius R
• Net Force on Pilot Towards Center (At Bottom)
• Net Force on Pilot Towards Center (At Top)
• Object Attached to a String in Vertical Motion 10:46
• Example: Object in a Circle Attached to String
• Case 1: Object with speed v and Object is at Bottom
• Case 2: Object at Top in Vertical Motion
• Object at Angle ø (General Position)
• 2 Radial Forces (Inward & Outward)
• Tension of String
• Extra Example 1: Pail of Water in Vertical Circle
• Extra Example 2: Roller Coaster Vertical Circle
• Extra Example 3: Bead in Frictionless Loop | 392 | 1,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-43 | latest | en | 0.837229 |
https://brainmass.com/math/graphs-and-functions/graphs-three-equations-common-point-19121 | 1,685,432,406,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00121.warc.gz | 187,210,504 | 75,706 | Explore BrainMass
# Draw the Graphs for Three Equations and Find a Common Point
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Using first two equations determine 3 possible points that the 2 conic sections that you get from equations meet. Add a third equation figure to out the one point that all three conic sections meet.
First equation is 9(x-squared)+25(y-squared)-72x=81
Second equation is 9(x-squared)-15(y-squared)=9
Graph these two conical sections and let us know how you get the graphs with such things as vertex and foci. Show three common points of these two conical sections. Then add the third equation:12x+6y=-12 and determine one point that all three conical sections share. Please go in to the basics of how to make these graphs from the equations.
© BrainMass Inc. brainmass.com May 24, 2023, 1:22 pm ad1c9bdddf
https://brainmass.com/math/graphs-and-functions/graphs-three-equations-common-point-19121
#### Solution Preview
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From equation (1): , we get , then we have . Thus it is an ellipse with ...
#### Solution Summary
Three graphs are drawn and their common point is found. Details are given as to how to how the draw the graphs and how to find vertices, foci and directrices. The drawing of the graph with all three equations is shown.
\$2.49 | 358 | 1,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-23 | longest | en | 0.926659 |
https://homework.cpm.org/category/CC/textbook/cca2/chapter/7/lesson/7.1.4/problem/7-55 | 1,611,083,967,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00424.warc.gz | 395,469,257 | 16,298 | ### Home > CCA2 > Chapter 7 > Lesson 7.1.4 > Problem7-55
7-55.
The measure of $∠ROS$ in $ΔROS$ below is $60^\circ$.
1. The curved arrow represents the rotation of $\overline { O R }$, beginning from the positive $x$-axis. Through how many degrees has $\overline { O R }$ rotated?
Remember that a whole circle is $360^\circ$.
2. If $OR = 1$, what are the exact length of $OS$ and $SR$?
Remember the $30^\circ-60^\circ-90^\circ$ triangle.
3. What are the exact coordinates of point $R$?
$\left(\text{cos}(300^{^{\circ}}), \text{sin}(300^{^{\circ}})\right)\text{ or }\left(\frac{1}{2}, - \frac{\sqrt{3}}{2}\right)$ | 209 | 619 | {"found_math": true, "script_math_tex": 13, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.658744 |
https://oeis.org/A056566 | 1,657,047,672,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00060.warc.gz | 466,319,526 | 4,321 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A056566 Fibonomial coefficients. 3
1, 34, 1870, 83215, 3994320, 186135312, 8771626578, 411591708660, 19344810307020, 908637119420910, 42689423937884208, 2005443612183077232, 94214069697350815795, 4426039514623184676790, 207929935924379904006970 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..100 FORMULA a(n) = A010048(n+8, 8) = Fibonomial(n+8, 8). G.f.: 1/p(9, n) with p(9, n)= 1 - 34*x - 714*x^2 + 4641*x^3 + 12376*x^4 - 12376*x^5 - 4641*x^6 + 714*x^7 + 34*x^8 - x^9 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2)* (1 + 18*x + x^2)*(1 - 47*x + x^2) (n=9 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references). Recursion: a(n) = 47*a(n-1) - a(n-2) + ((-1)^n)*A001658(n), n >= 2, a(0)=1, a(1)=34. MAPLE with(combinat): a:=n-> 1/65520*fibonacci(n) *fibonacci(n+1) *fibonacci(n+2) *fibonacci(n+3) *fibonacci(n+4)*fibonacci(n+5)*fibonacci(n+6)*fibonacci(n+7): seq(a(n), n=1..17); # Zerinvary Lajos, Oct 07 2007 MATHEMATICA a[n_] := (1/65520) Times @@ Fibonacci[n + Range[8]]; Array[a, 20, 0] (* Giovanni Resta, May 08 2016 *) PROG (PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(20, n, b(n-1, 8)) \\ Joerg Arndt, May 08 2016 CROSSREFS Cf. A010048, A000045, A001654-8, A056565, A001906 (signed), A004187, A049660 (signed), A049668. Sequence in context: A296673 A086881 A212023 * A335090 A242177 A187591 Adjacent sequences: A056563 A056564 A056565 * A056567 A056568 A056569 KEYWORD nonn,easy AUTHOR Wolfdieter Lang, Jul 10 2000 STATUS approved
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Last modified July 5 14:54 EDT 2022. Contains 355099 sequences. (Running on oeis4.) | 798 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-27 | latest | en | 0.434069 |
https://math.stackexchange.com/questions/1483707/singular-covariance-matrix-understanding-the-beginning-of-a-proof | 1,696,082,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00037.warc.gz | 426,621,834 | 34,985 | # Singular covariance matrix, understanding the beginning of a proof
Let $\left\{X_t,t\in T\right\}$ be a stationary process such that $\text{Var}(X_t)<\infty$ for each $t\in T$. The autocovariance function $\gamma_X(\cdot)=\gamma(\cdot)$ of $\left\{X_t\right\}$ is defined to be $$\gamma(h)=\text{Cov}(X_{h+t},X_t)~\forall h,t\in\mathbb{Z}.$$ Moreover, assume that $EX_t=0$ for each $t\in T$.
There is the following statement:
If $\gamma(0)>0$ and $\gamma(h)\to 0$ as $h\to\infty$, then the covariance matrix $\Gamma_n=[\gamma(i-j)]_{i,j=1,\ldots,n}$ of the column vector $(X_1,\ldots,X_n)'$ is non-singular for every $n$.
The proof of this starts as follows:
Suppose that $\Gamma_n$ is singular for some $n$. Then since $EX_t=0$ there exists an integer $r\geq 1$ and real constants $a_1,\ldots,a_r$ such that $\Gamma_r$ is non-singular and $$X_{r+1}=\sum_{j=1}^r a_j X_j.~~~~~(*)$$
I do not completely see this argument. In particular, there are two inaccuricies to my opinion.
Here's how I do understand it:
Suppose $\Gamma_n$ is singular. This means, its determinant is zero. So there is at least one column (one row) that is a linear combinations of the other columns (rows). If we delete this column (row), we get $\Gamma_{n-1}$, if its determinant is again zero, we repeat this. Eventually, we get $\Gamma_r$, for some $r\geq 1$, such that all columns (rows) are independent, meaning that $\Gamma_r$ has positive determinant, i.e. is non-singular.
So far so good.
It remains to argue, how to get $(*)$.
In the proof, there is said to look at the following statement that one probably should use for my problem:
If $X=(X_1,\ldots,X_n)'$ is a random vector with covariance matrix $\Sigma$, then $\Sigma$ is singular if and only if there exists a non-zero vector $b=(b_1,\ldots,b_n)'\in\mathbb{R}^n$ such that $\text{Var}(b'X)=0$.
So, I apply this as follows:
As done above, let $\Gamma_r$ be non-singular. Since $r$ was the smallest $r\geq 1$ such that all columns are linear independent, the covariance matrix $\Gamma_{r+1}$ of $X=(X_1,X_2,\ldots,X_{r+1})'$ is singular. By the cited statement, there is some $b=(b_1,b_2,\ldots,b_{r+1})'\in\mathbb{R}^{r+1}$ such that $$\text{Var}(b'X)=0.$$ But this means that $$b'X=\sum_{i=1}^{r+1}b_iX_i=E(b'X)=0\text{ almost surely },$$ implying that $$b_{r+1}X_{r+1}=-\sum_{i=1}^r b_iX_i.$$ Imho, we can suppose that $b_{r+1}\neq 0$ since if $b_{r+1}=0$, we have that $0=\text{Var}(b_1X_1+b_2X_2+\ldots + b_rX_r+b_{r+1}X_{r+1})=\text{Var}(b_1X_1+b_2X_2+\ldots + b_rX_r)$, meaning by the cited statement that $\Gamma_r$ is singular. But $\Gamma_r$ is non-singular. Hence $$X_{r+1}=\sum_{i=1}^r a_iX_i,~~~a_i:=-\frac{b_i}{b_{r+1}}$$
or, more precisely, this identity holds almost surely. I don't know why the "almost surely" is omitted in the proof.
Is this okay?
$$\Gamma_n$$ is singular $$\Rightarrow \exists b_n s.t. \Gamma_nb_n=0 \Rightarrow b_n'\Gamma_nb_n = 0 \iff Var(b_n'X)=0$$ where $$X = (X_1,X_2,...,X_n)'$$
Since $$E(X)=0$$, $$Var(b_n'X) = E((b_n'X)'(b_n'X))=0 \Rightarrow b_n'X = 0$$ a.s. | 1,067 | 3,060 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-40 | latest | en | 0.844128 |
https://mechanics.stackexchange.com/questions/27426/how-does-the-ignition-system-control-the-energy-of-the-spark | 1,713,624,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00437.warc.gz | 341,892,551 | 36,173 | # How does the ignition system control the Energy of the spark?
An automotive ignition system needs to provide a specific amount of energy to properly ignite the air/fuel mixture in the combustion chamber. Energy = voltage * amperage * duration.
A specific energy must be produced to overcome the quenching effects present in the environment of the combustion chamber and produce a viable flame kernel which will develop into a flame front that advances at a sufficient speed to provide efficient and timely combustion.
In the 2000 edition of "Auto Fundamentals", chapter 8, Ignition Systems, page 122 it notes that in older ignition systems a resistor is used to lower the input voltage
to around 9.5 volts during normal engine operation
Further on it says:
At high speeds, when a hotter spark is needed, the coil receives full battery voltage.
It then notes that:
The majority of modern electronic ignition systems use full battery voltage at all times.
This all brings up a number of questions. When the book talks about needing a `hotter spark` at high speeds, it could be talking about either a spark with higher voltage, higher amperage or longer duration. It's my understanding that the duration of the spark is usually fixed at about 1/1000th of a second, so they must be talking about either voltage or amperage.
Now I know the voltage needed to jump the gap can in general vary depending on both the amount of compression and the composition of the A/F mixture with both leaner mixtures and higher compression requiring more voltage, all this based on the application of Paschen's law to determine the breakdown voltage. However, compression is pretty much a fixed value, and under normal operation, even at high speeds, the A/F ratio doesn't vary that much ( excluding wide open throttle or heavy load which are very rich and would require less voltage if I understand correctly ).
So when they talk about requiring a `hotter spark`, I can only think that they mean a more energetic spark with higher amperage, which brings me to my question.
How does the ignition system in a modern vehicle built in the last 20 years control the output wattage of the spark ( voltage * amperage ) and what might effect the wattage negatively other than the ignition coil, wires and plugs?
Here is an example ignition system diagram from this book on Fuel Efficiency: | 483 | 2,375 | {"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} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.939194 |
https://ncertmcq.com/mcq-questions-for-class-11-physics-chapter-6/ | 1,716,399,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00434.warc.gz | 356,404,246 | 11,313 | Check the below Online Education NCERT MCQ Questions for Class 11 Physics Chapter 6 Work, Energy and Power with Answers Pdf free download. MCQ Questions for Class 11 Physics with Answers were prepared based on the latest exam pattern. We have provided Work, Energy and Power Class 11 Physics MCQs Questions with Answers to help students understand the concept very well.
## Online Education for Work, Energy and Power Class 11 MCQs Questions with Answers
Work Energy And Power Class 11 MCQ Question 1.
A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platfrom from where the took the jump the total work done by the man at the instant he lands is
(a) 300 J
(b) 150 J
(c) 600 J
(d) zero
Class 11 Physics Chapter 6 MCQ Question 2.
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
(a) 7.2 J
(b) 3.6 J
(c) 120 J
(d) 1200 J
MCQ Questions For Class 11 Physics Chapter 6 Question 3.
A body of mass 20 kg is initially at a height of 3 m above the ground . It is lifted to a height of 2 m from that position. Its increase in potential energy is
(a) 100 J
(b) 392 J
(c) 60 J
(d) -100 J
MCQ On Work Power And Energy Class 11 Question 4.
A body of mass 10 kg is travelling with uniform speed of 5 m/s. Its kinetic energy is
(a) 25 J
(b) 125 J
(c) 1250 J
(d) 1000 J
Work Power And Energy Class 11 MCQ Question 5.
A ball is dropped from a height of 1 m. If the coeffcient of restitution between the surface and ball is 0.6, the ball rebounds to a height of
(a) 0.6 m
(b) 0.4 m
(c) 1 m
(d) 0.36 m
Work Energy And Power Class 11 MCQs Pdf Question 6.
A quantity of work of 1000 J is done in 2 seconds. The power utilised is
(a) 998 W
(b) 1002 W
(c) 2000 W
(d) 500 W
Class 11 Work Energy And Power MCQs Question 7.
When the linear momentum of a particle is increased by 1% its kinetic energy increases by x%. When the kinetic energy of the particle is increased by 300%, its linear momentum increases by y%. The ratio of y to x is
(a) 300
(b) 150
(c) 100
(d) 50
MCQ Of Work Energy And Power Class 11 Question 8.
A solid cylinder of length 1 m and diameter of cross section y 100 cms is first placed with its axis vertical its then slowly inclined till its axis is horizontal. The loss in its potential energy if the mass of the cylinder is 10 kg is
(a) 5g J
(b) 10g J
(c) Zero
(d) g J
Class 11 Physics Work Energy And Power MCQ Question 9.
A body of mass 10 kg moving at a height of 2 m, with uniform speed of 2 m/s. Its total energy is
(a) 316 J
(b) 216 J
(c) 116 J
(d) 392 J
Ch 6 Physics Class 11 MCQ Question 10.
Two masses 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
(a) 4 : 1
(b) 1 : 2
(c) 0 : 1
(d) 1 : 6
Class 11 Physics Ch 6 MCQ Question 11.
A body of mass 10 kg is initially at a height of 20 m above the ground. It falls to a height of 5 m above the ground. Its potential energy in the new position is
(a) 490 J
(b) 50 J
(c) 100 J
(d) 300 J
Class 11 Physics Chapter 6 MCQ With Answers Question 12.
A man of 60 kg weight is standing at rest on a platform. He jumps up vertically a distance of 1 m and the platform at the same instant moves horizontally forward with the result that the man lands 1 meter behind the point on the platfrom from where the took the jump the total work done by the man at the instant he lands is
(a) 300 J
(b) 150 J
(c) 600 J
(d) zero
Chapter 6 Physics Class 11 MCQs Question 13.
A marble moving with some velocity collides perfectly elastically head-on with another marble at rest having mass 1.5 times the mass of the colliding marble. The percentage of kinetic energy by the colliding marble after the collision is
(a) 4
(b) 25
(c) 44
(d) 67
Work Energy Power Class 11 MCQ Question 14.
A body of mass 100 kg falls from a height of 10 m. Its increase in kinetic energy is
(a) 9800 J
(b) 1000 J
(c) 5000 J
(d) 3000 J
Work Power Energy MCQ Class 11 Question 15.
An isolated particle of mass m is moving in a horizontal plane (x-y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4 . An instant later, the smaller fragment is at y = + 15 cm. The larger fragment at this instant is at
(a) y = -5 cm
(b) y = +20 cm
(c) y = +5 cm
(d) y = -20 cm
Answer: (a) y = -5 cm
Question 16.
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?
(a) 7.2 J
(b) 3.6 J
(c) 120 J
(d) 1200 J
Question 17.
A body of mass 20 kg is initially at a height of 3 m above the ground. It is lifted to a height of 2 m from that position. Its increase in potential energy is
(a) 100 J
(b) 392 J
(c) 60 J
(d) -100 J
Question 18.
A body of mass 10 kg is moved parallel to the ground, through a distance of 2 m. The work done against gravitational force is
(a) 196 J
(b) -196 J
(c) 20 J
(d) zero
Question 19.
A ball is dropped from a height of 1 m. If the coeffcient of restitution between the surface and ball is 0.6, the ball rebounds to a height of
(a) 0.6 m
(b) 0.4 m
(c) 1 m
(d) 0.36 m
Question 20.
An electric heater of rating 1000 W is used for 5 hrs per day for 20 days. The electrical energy utilised is
(a) 150 kWh
(b) 200 kWh
(c) 100 kWh
(d) 300 kWh | 1,698 | 5,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-22 | latest | en | 0.880082 |
https://metanumbers.com/203897 | 1,723,455,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00744.warc.gz | 312,984,739 | 7,446 | # 203897 (number)
203897 is an odd six-digits prime number following 203896 and preceding 203898. In scientific notation, it is written as 2.03897 × 105. The sum of its digits is 29. It has a total of one prime factor and 2 positive divisors. There are 203,896 positive integers (up to 203897) that are relatively prime to 203897.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 6
• Sum of Digits 29
• Digital Root 2
## Name
Name two hundred three thousand eight hundred ninety-seven
## Notation
Scientific notation 2.03897 × 105 203.897 × 103
## Prime Factorization of 203897
Prime Factorization 203897
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 203897 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -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 12.2254 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 203897 is 203897. Since it has only one prime factor, 203897 is a prime number.
## Divisors of 203897
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 203898 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 101949 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 451.55 Returns the nth root of the product of n divisors H 1.99999 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 203897 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 203897) is 203898, the average is 101949.
## Other Arithmetic Functions (n = 203897)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 203896 Total number of positive integers not greater than n that are coprime to n λ 203896 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 18254 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 203,896 positive integers (less than 203897) that are coprime with 203897. And there are approximately 18,254 prime numbers less than or equal to 203897.
## Divisibility of 203897
m n mod m
2 1
3 2
4 1
5 2
6 5
7 1
8 1
9 2
203897 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion 203897
Base System Value
2 Binary 110001110001111001
3 Ternary 101100200202
4 Quaternary 301301321
5 Quinary 23011042
6 Senary 4211545
8 Octal 616171
10 Decimal 203897
12 Duodecimal 99bb5
20 Vigesimal 159eh
36 Base36 4dbt
## Basic calculations (n = 203897)
### Multiplication
n×y
n×2 407794 611691 815588 1019485
### Division
n÷y
n÷2 101948 67965.7 50974.2 40779.4
### Exponentiation
ny
n2 41573986609 8476811147615273 1728396362565311318881 352414833137979281985879257
### Nth Root
y√n
2√n 451.55 58.8577 21.2497 11.5314
## 203897 as geometric shapes
### Circle
Diameter 407794 1.28112e+06 1.30609e+11
### Sphere
Volume 3.55076e+16 5.22434e+11 1.28112e+06
### Square
Length = n
Perimeter 815588 4.1574e+10 288354
### Cube
Length = n
Surface area 2.49444e+11 8.47681e+15 353160
### Equilateral Triangle
Length = n
Perimeter 611691 1.80021e+10 176580
### Triangular Pyramid
Length = n
Surface area 7.20083e+10 9.99002e+14 166481
## Cryptographic Hash Functions
md5 a23b649beeb717af54f71a6670c72c34 2b6ce7dd0dae5e482d22fc322cae04cceaaa539b 126314a6b3924969bb11c614dd224cf746e236b4582f2b2d3940feedcb73e750 24ca6f84e16ff55eee52ce0f9615974945641f28402e1ac8f4bad1cfe9ef3622c37a94fc12a7ddea325b1f87c1e25e3e74bc859700dd439329684bfefb751f60 71e9fba62a65f0f31f7fb9384a1b990db3991cc1 | 1,368 | 4,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-33 | latest | en | 0.813921 |
https://testbook.com/question-answer/what-was-the-day-of-the-week-on-22-february-2012--610a8d42b27021a91bf82d41 | 1,639,043,032,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00205.warc.gz | 600,905,370 | 31,787 | # What was the day of the week on 22 February 2012?
1. Thursday
2. Wednesday
3. Friday
4. Tuesday
Option 2 : Wednesday
## Detailed Solution
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Code 6 2 3 6 1 4 6 2 5 0 3 5
Date: 22 February 2012
Steps to find the day of the week:
Last two digit of the year 12 No. of the leap year (divisor) 12 ÷ 4 Divisor = 3 Code of the month 2 Date 22
Total = (12 + 3 + 2 + 22) = 39
And 39 ÷ 7, we get 4 as the remainder.
Code for Centuries:
1600 → 0, 1700 → 5, 1800 → 3, 1900 → 1, 2000 → 0, 2100 → 5, 2200 → 3, 2300 → 1.
So, for 2012 → 0 odd days will be added.
Total odd days = (4 + 0) = 4
Code Day 1 Sunday 2 Monday 3 Tuesday 4 Wednesday 5 Thursday 6 Friday 0/7 Saturday
So, ‘4’ is the code for Wednesday.
Hence, ‘Wednesday’ is the correct answer.
Mistake PointsIn trick solution, the day code starts with 1, and 1 is considered as Sunday.
Year - One year has 365 days or 366 days on this basis the year is divided into two parts:
• Normal year - It has 365 days.
• Leap year - It has 366 days and completely divisible by 4.
Century leap year - The year which is exactly divisible by 400. Example - 1200, 1600, 2000 etc.
Counting of Odd days
Ordinary Year: 365 days (52 weeks + 1 days)
= 1 Odd days
Leap Year: 366 days (52 weeks + 2 days)
= 2 Odd days | 460 | 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} | 4.46875 | 4 | CC-MAIN-2021-49 | latest | en | 0.904429 |
https://mapleprimes.com/users/mathiaszip/questions | 1,582,269,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145443.63/warc/CC-MAIN-20200221045555-20200221075555-00022.warc.gz | 443,511,338 | 412,441 | 2 years, 85 days
## Complex solutions...
Maple 18
Hi
I d like to limit my solution to a real (non complex) solution
There should be a simple solution to my calculation but maple can t process the solutions in one of my "solve" commands.
How can do I tell maple to limit itself to one
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simple non complex solutions ?
## nested loop not working...
Maple 18
Hi
I m trying to do a nested loop but for some reason maple tells me my for loop is unterminated
I d love some feedback on my code and on how to make it work
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Thanks in advance
## Help ...Maple will not plot...
Maple
Hi
II ve managed to build an expression with one variable
I me trying to plot this expression on a defined range but maple doesn t not let me.
How do I manage to plot this expression ?
If you have any advise on how to improve my code I m open to comments
Thanks a lot in advance
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## HELP !! Maple doesn t want to give me th...
Maple
Hi
I have to create a function "I(epsilonc)" by integrating a function "sigma(y)" from 0 to the variable "epsilonc"
but maple doesn t give me the actual formula of the integrated function, it only gives me equation 4 which is utterly useless to me.
How do I get Maple to actually calculate and give me the integrated function ??
Any help will be greatly appreciated
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## Help !! maple won t plot and won t integ...
Maple
Hi I have a function i d live to plot and integrate but maple tells me there is a probleme with the range when i want to plot and will not give me a numerical value of the integral : | 886 | 2,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.739969 |
https://emba.gnu.org/emacs/emacs/-/commit/db37d2572b49123aba6cdc611cef0d6e6411615f | 1,620,517,501,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00548.warc.gz | 254,606,685 | 22,003 | Commit db37d257 by Daniel E. Doherty Committed by Jay Belanger
### (Tutorial): Use ^{\prime} to indicate primes.
parent cd03c74a
2010-05-01 Daniel E. Doherty (tiny change) * calc.texi (Tutorial): Use "^{\prime}" to indicate primes. 2010-05-01 Michael Albinus * tramp.texi (Inline methods, Default Method): Mention ... ...
... ... @@ -6234,7 +6234,7 @@ new_x = x - f(x)/f'(x) @end ifnottex @tex \beforedisplay $$x_{\rm new} = x - {f(x) \over f'(x)}$$ $$x_{\rm new} = x - {f(x) \over f^{\prime}(x)}$$ \afterdisplay @end tex ... ... @@ -7901,7 +7901,7 @@ term. So we can take it out to get an equivalent formula with @tex \turnoffactive \beforedisplay $$9 a + 3 b + c - 511 n'$$ $$9 a + 3 b + c - 511 n^{\prime}$$ \afterdisplay @end tex
Markdown is supported
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You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first! | 296 | 873 | {"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-2021-21 | longest | en | 0.507714 |
http://mathhelpforum.com/differential-geometry/130784-identify-rigid-motion-t-o-f-o-t-1-a.html | 1,529,543,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00581.warc.gz | 205,385,885 | 10,968 | # Thread: Identify the rigid motion T o f o T^(-1)
1. ## Identify the rigid motion T o f o T^(-1)
If f is any rigid motion and T is any translation, identify the rigid motion T o f o T^(-1).
Can I split the f into three reflections since there is a theorem that says "Every rigid motion is the composition of at most three relfections and every rigid motion is a translation, a rotation, or a glide relfection"?
My intuition tells me that the final answer would be f by itself since T o T^(-1) will be cancelled out ... I just guess.
2. Originally Posted by rainyice
If f is any rigid motion and T is any translation, identify the rigid motion T o f o T^(-1).
Can I split the f into three reflections since there is a theorem that says "Every rigid motion is the composition of at most three relfections and every rigid motion is a translation, a rotation, or a glide relfection"?
My intuition tells me that the final answer would be f by itself since T o T^(-1) will be cancelled out ... I just guess.
In what form can every translation be written?
3. I have the same exact question actually. I know that every translation is the composition of two reflections with the two axes of reflection being parallel to each other...
4. Originally Posted by Pinkk
I have the same exact question actually. I know that every translation is the composition of two reflections with the two axes of reflection being parallel to each other...
If you don't mind me asking, what class is this for?? Some of your questions suggest you're in an introductory analysis course...but....what's with all the isometry/rigid motion questions? I mean, a couple are a nice supplement to a discussion of mappings but the majority of your questions are these.
5. Introduction to Modern Geometry. First portion of the class is examining Euclid's Postulates and Hilbert's Axioms and classifying/identifying/composing/etc rigid motions.
6. Originally Posted by Pinkk
Introduction to Modern Geometry. First portion of the class is examining Euclid's Postulates and Hilbert's Axioms and classifying/identifying/composing/etc rigid motions.
That's really cool!
7. I'm still completely stuck on this.
8. Originally Posted by Pinkk
I'm still completely stuck on this.
Do you agree with my intuition that the outcome will be f by itself? I viewed this like: go up one block, then go right one block, and then go down one block. The displacement is one block to the right.
9. That's only if $\displaystyle f$ is another translation.
10. Originally Posted by Drexel28
If you don't mind me asking, what class is this for?? Some of your questions suggest you're in an introductory analysis course...but....what's with all the isometry/rigid motion questions? I mean, a couple are a nice supplement to a discussion of mappings but the majority of your questions are these.
do you have any good suggustion to do this problem? | 649 | 2,896 | {"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} | 3.375 | 3 | CC-MAIN-2018-26 | latest | en | 0.930295 |
https://www.airmilescalculator.com/distance/vns-to-imk/ | 1,606,869,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00125.warc.gz | 498,609,301 | 47,516 | # Distance between Varanasi (VNS) and Simikot (IMK)
Flight distance from Varanasi to Simikot (Lal Bahadur Shastri Airport – Simikot Airport) is 318 miles / 511 kilometers / 276 nautical miles. Estimated flight time is 1 hour 6 minutes.
Driving distance from Varanasi (VNS) to Simikot (IMK) is 621 miles / 999 kilometers and travel time by car is about 15 hours 13 minutes.
## Map of flight path and driving directions from Varanasi to Simikot.
Shortest flight path between Lal Bahadur Shastri Airport (VNS) and Simikot Airport (IMK).
## How far is Simikot from Varanasi?
There are several ways to calculate distances between Varanasi and Simikot. Here are two common methods:
Vincenty's formula (applied above)
• 317.605 miles
• 511.135 kilometers
• 275.991 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 318.626 miles
• 512.779 kilometers
• 276.879 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
City: Varanasi
Country: India
IATA Code: VNS
ICAO Code: VIBN
Coordinates: 25°27′8″N, 82°51′33″E
B Simikot Airport
City: Simikot
Country: Nepal
IATA Code: IMK
ICAO Code: VNST
Coordinates: 29°58′15″N, 81°49′8″E
## Time difference and current local times
The time difference between Varanasi and Simikot is 15 minutes. Simikot is 15 minutes ahead of Varanasi.
IST
+0545
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 72 kg (158 pounds).
## Frequent Flyer Miles Calculator
Varanasi (VNS) → Simikot (IMK).
Distance:
318
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
318
Round trip? | 517 | 1,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.804654 |
https://www.justanswer.com/math-homework/1pvrs-wondering-scott-help-question-take.html | 1,606,825,238,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00169.warc.gz | 699,600,190 | 50,089 | Math Homework
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Show MoreShow Less | 2,313 | 10,203 | {"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} | 3.53125 | 4 | CC-MAIN-2020-50 | latest | en | 0.888446 |
http://docplayer.net/24461599-Lecture-11-graphical-models-for-inference.html | 1,548,136,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583829665.84/warc/CC-MAIN-20190122054634-20190122080634-00348.warc.gz | 66,143,870 | 27,815 | # Lecture 11: Graphical Models for Inference
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1 Lecture 11: Graphical Models for Inference So far we have seen two graphical models that are used for inference - the Bayesian network and the Join tree. These two both represent the same joint probability distribution, but in different ways. Both express local properties which are used to make the inference computations tractable. The Bayesian net has the desirable property that it expresses a causal relationship between the variables which can be used for reasoning about their behaviour. We will now look at some other graphical models that have been used for inference, and show that they are each specific examples of a more general method called belief propagation. Tanner Graphs The Tanner graph was originally used to depict constraints imposed for error recovery using parity checking. It is a bi-partite graph, meaning it has two types of nodes, and each node only has neighbours of the opposite type. A simple example is shown in Figure 1. Here the circles represent bits that are being transmitted down a noisy channel, and the squares represent parity constraints. Each square checks for even parity and will evaluate to 1 if the sum of the three bits connected to it is even. We can think of bits 1,2 and 3 representing the data and bits 4, 5 and 6 as parity bits whose values will help us to recover from an error. Given that an error occurs, we can formulate the recovery procedure as a probabilistic inference problem as follows. Suppose that the probability of a bit being flipped during transmission is P f, and we write X i to be the value (0 or 1) that was transmitted for bit b i and Y i as the value that is received for bit b i. We have that: P (X i Y i ) = 1 P f if X i = Y i P f otherwise and we can work out the probability that a set of bits (X 1, X 2,..X N ) is correct by taking the product of the conditional probabilities: N P (X 1, X 2,..X N ) = P (X j Y j ) Our inference problem is to determine the most likely transmitted string. The above equation simply tells us that the most likely transmitted string is the received string, since the joint probability will always be a maximum for the case where i : X i = Y i. To use it for error correction we need to impose the parity constraints, which is done simply by multiplying by the parity check functions which we will denote Ψ. So for the example graph of Figure 1 we get the joint probability distribution: P (X 1, X 2,..X 6 ) = Ψ(X 1, X 5, X 3 )Ψ(X 1, X 2, X 4 )Ψ(X 2, X 3, X 6 ) N P (X j Y j ) Any bit string that fails the parity check will now have a probability of zero, and, if this is the case, we can select the most probable bit string from the set of possible valid bit strings in order to recover from the error. Notice that the way the parity functions express a relationship (or structure) between the bits is analogous to the way in which the conditional probability matrices P (A B) express relationships between variables in a Bayesian network. Factor Graphs A factor graph is a graphical model representing the factorisation of a function. It is in fact the most general graphical model,and can express all the probabilistic inference procedures we have seen so far. In general if we write a factorisation as: g(x 1, X 2, X 3,...X n ) = f j (S j ) The corresponding factor graph G is a triple (X, F, E) where: X is a set of variables (X 1, X 2, X 3,...X n ) F is a set of factors (f 1, f 2,...f m ) E is a set of edges joining factors and variables DOC493: Intelligent Data Analysis and Probabilistic Inference Lecture 11 1
2 Figure 1: A Tanner graph for parity checking Figure 2: The equivalent factor graph of Figure 1 The factor graph for the six bit parity checker shown in figure 1 is shown in figure 2. Each factor is represented by a square node, and each variable by a round node. We see that the joint probability of the variables is simply the product of all the factors. We have seen that a Bayesian network can be looked on as factorisation of the joint probability distribution which is made on the basis of known (or assumed) conditional dependencies between the variables. In general we write: n P (X 1, X 2, X 3,...X n ) = P (X i P arents(x i )) i=1 So for the metastatic cancer network (Figure 3) we can write: P (A, B, C, D, E) = P (A)P (B A)P (C A)P (D B&C)P (E C) and we see that the factor graph variables are the same as the network variables and the factors are simply the prior and conditional probability matrices. The Bayesian network and its equivalent factor graph are shown below. Figure 3: Bayesian Net for Metastatic Cancer Figure 4: The equivalent factor graph We have also seen how the join tree method allows us to find a different factorisation of the same probability distribution based on finding a particular set of cliques and for each clique defining a potential function. This representation is called a Markov random field. We can write a general Markov random field factorisation as: P (X 1, X 2, X 3,...X n ) = 1 Z Ψ j (V j ) where V j is a subset of variables (belonging to clique j in the case of a join tree) and Z is a normalisation factor Z = X Ψ j (V j ) The sum is take over all the possible states of the complete set of variables X. DOC493: Intelligent Data Analysis and Probabilistic Inference Lecture 11 2
3 Figure 5: Join Tree for Metastatic Cancer Figure 6: Factor Graph of the Join Tree (Figure 5) For the metastatic cancer network (Figure 5) we can write (following the notation in the lecture on join trees): P (A, B, C, D, E) = Ψ(C 1 )Ψ(C 2 )Ψ(C 3 ) = Ψ(A, B, C)Ψ(B, C, D)Ψ(C, E) The normalisation factor Z is 1 in this case since we initialised the potential functions with exactly the prior and conditional probability matrices, and ensured that any message that divided one potential function multiplied its parent. We see that the join tree is a different factorisation of the joint probability distribution that is also represented by a factor graph (Figure 6). However, in this case the factor graph representation is not particularly useful. Pairwise Markov Random Fields Pairwise Markov random fields are commonly used as a graphical structure for making inferences. Their simplicity has meant that they are often used in theoretical studies as a fundamental building block from which other inference algorithms can be derived. In a pairwise field the variables are always joined in cliques of size two, which means that factors join at most two variables. The classic example of the use of pairwise Markov random fields is in the area of image processing. In Figure 7 the image pixels are depicted by filled circles, and are labelled Y i. Note that the two dimensional image pixels are indexed by a single index i. The inferred model of the image is depicted by the unfilled circles and the pixels are labelled X i. The definition of the inferred model depends on the application. For example, if we are interested in image segmentation, then the X i variables may be discrete and used to label segments of the image. In a brain image the labels might be white matter, grey matter, cerebro-spinal fluid and lesion, and the inference problem is to decide which label each pixel should have. The Y i pixels in this case are observed (or evidence) variables, equivalent to the observed bits in the Tanner graph, and the X i are hidden variables whose values are unknown and cannot be measured, equivalent to the true bit string that the Tanner graph is designed to recover. Figure 7: A pairwise Markov random field model of an image Figure 8: The equivalent factor graph of the pairwise markov random field The Markov assumption in the model expresses the local property that a pixel is very likely to be similar to its neighbours. In the simplest case, where an image is corrupted by random noise in the measurement process, this DOC493: Intelligent Data Analysis and Probabilistic Inference Lecture 11 3
4 assumption can be used, analogously to the parity checking example, to restore the true underlying image from the measured pixels. In segmentation applications the underlying assumption is that the segments that we want to label are large regions with some common property, in which case the Markov assumption is reasonable. For images with rapidly changing textures it does not hold, and preprocessing becomes necessary. Overall we can think of the pairwise Markov random field as fusing both the measured and the structural information to make a decision about the image. An image inference problem may be expressed by writing a joint probability distribution over the variables. To do this we need to define two compatibility functions: Φ(X i, Y i ) and Ψ(X i, X j ). The first relates the observed and hidden variables, so we could think of it as a conditional probability P (X i Y i ), for example the probability of a region label given the pixel value. The second expresses the relationship between the pixels, so in the case of pixels that are not adjacent it will be one (expressing no information), and for pixels connected in the grid it can be again thought of as a conditional probability, for example the probability of one region label given the adjacent region label. To get the joint probability we simply take the product (as usual) to get: P (X, Y ) = 1 Ψ(X i, X j ) Z i,j i Φ(X i, Y i ) where X = (X 1, X 2,..X n ) is the set of hidden variables and Y = (Y 1, Y 2,..Y n ) is the set of observed variables. Note that this is a scalar equation equivalent to the joint probability equations we used for Bayesian networks. It is another factorisation of a joint probability distribution, and its factor graph is shown in Figure 8. As before the value of Z is chosen to normalise the joint probability distribution, and is found by summing the products over the entire set of values that the variables can take. Cleary this will create a massive computational demand in the case of an image, and for that reason iterative estimates of the optimal X i values are the only possible solution option. Here we see the advantage of the Markov assumption, since the variable X i depends only on its four neighbours and on its corresponding evidence node. Belief Propagation We have already discussed belief propagation in Bayesian networks extensively in terms of the passing of λ and π messages between connected nodes, and we have observed that in a singly connceted network the propagation is very fast in reaching the correct posterior probability distributions of the unknown variables. The process can be defined for other graphical models, and to illustrate this we will look at the pairwise Markov random field. We will first define a message from variable X i to variable X j which will have a dimension equal to the states of X j and will express the evidence that X i holds for the states of X j. This message is exactly the same as the λ message that we used extensively in Pearl s propagation algorithm. The belief in a variable is the same as the posterior probability over the states of a variable in a Bayesian network. For any one state of variable X i we can write it using the scalar equation: b(x i (s)) = 1 Z Φ(X i(s), Y i ) kɛn(i) m k (X i (s)) where X i (s) means state s of variable X i, m k (X i (s)) is the message (or evidence) from variable X k for state s of variable X i, N(i) is the set of variables adjacent to variable X i, and Z is the normalisation constant that makes the belief a probability distribution over the states of the variable. The message sent from one variable to another depends on the (unnormalised) belief that the sending variable has received from all variables except the recipient (equivalent to the π messages in Pearl s propagation algorithm). We can write this in scalar form as: b \j (X i (s)) = Φ(X i (s), Y i ) m k (X i (s)) kɛn(i)\j where the notation \j means excluding variable j. If we use bold face (b \j (X i )) to indicate the vector of beliefs for the states of a variable we can then define the message, in vector form, from variable X i to X j as: m i (X j ) = b \j (X i ) Ψ(X i, X j ) DOC493: Intelligent Data Analysis and Probabilistic Inference Lecture 11 4
5 Belief propagation is a generalisation of Pearl s propagation algorithm in Bayesian networks. The notion of causality has been dropped, so we have only one type of message which is computed equivalently to the π message. The compatibility matrix Ψ(X i, X j ) can be thought of as a joint probability matrix. In vision applications the X i variables all have the same states, and so the matrix is symmetric and messages can be passed in either direction using the same matrix equation. Loopy Propagation When we discussed Pearl s propagation algorithm we noted that it worked for singly connected networks, and for networks with sufficient instantiated nodes to block any loop. Moreover in this case the propagation was very fast. For networks where the singly connected criterion was not met we could solve the problem by converting the network into a join tree, which is a singly connected Markov random field. Now, looking at the pairwise Markov random field we may wonder how the propagation can work in a vision application since there are multiple loops. The answer is that it is not guarenteed to work, but will do so in many cases. It has been shown that belief propagation with loops is equivalent to an optimisation problem in statistical Physics. There are three possible outcomes of loopy propagation. Firstly that the process converges to the optimal result, second that it converges but to a secondary maximum and thirdly that is does not converge but cycles. Computer vision algorithms based on pairwise Markov random fields are strongly constrained by the evidence variables Y i (pixel intensities) which are constants while the message passing takes place. They therefore tend to converge in a stable manner. Bayesian networks are not necessarily so strongly constrained. Therefore their behaviour in loopy propagation is data dependent and harder to determine. Summary Graphical models in probabilistic inference express local properties that allow us to calculate posterior probability distributions in a reasonable time frame. For any set of variables the probability distribution over an unknown can always be calculated by multiplication and marginalisation of the joint probability distribution. Thus if we want to compute a distribution of one variable, say X n out of a set (X 1, X 2,..X n ) we can do so by first multiplying the joint probability distribution by the values of any measured variable, and then marginalising using the equation: P (X n ) =.. P (X 1, X 2, X 3...X N ) X 1 X 2 X n 1 However, for any problem other than very small ones this is computationally intractible as the size of the state space explodes. Graphical models allow us to decompose large problems into smaller ones by invoking local properties. In the case of Bayesian networks these local properties are the conditional independencies among variables which allows us to express the joint probabilility distribution as a product of several conditional probability distributions. Similarly the join tree algorithm discovers a different factorisation of the joint probability distribution that always results in a singly connected network. However this is at the expense of having fewer factors and therefore greater computational demands to find an individual variable s posterior distribution. The pairwise Markov random field for making inferences about images exploits the local coherence implicit in images. As each of these methods is a different form of factorisation, the factor graph can represent them all, and is (arguably) a simpler and clearer model than the Bayesian network, though its simplifications come at the cost of loosing an explicit representation of cause. Graphical models are used extensively in other fields such as bio-informatics where it is possible to describe the behaviour of, for example, large numbers of genes involved in cell regulation as a collection of random variables with sparse dependencies. Another place where graphical models are used is neural circuitry, and inference using aggreagates of spiking neurones is currently the subject of extensive research. DOC493: Intelligent Data Analysis and Probabilistic Inference Lecture 11 5
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MapReduce Approach to Collective Classification for Networks Wojciech Indyk 1, Tomasz Kajdanowicz 1, Przemyslaw Kazienko 1, and Slawomir Plamowski 1 Wroclaw University of Technology, Wroclaw, Poland Faculty | 9,569 | 42,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-04 | latest | en | 0.950167 |
https://www.sofievandam.nl/05-02/30695/critical-rotating-speed-of-a-mill.html | 1,596,457,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735810.18/warc/CC-MAIN-20200803111838-20200803141838-00198.warc.gz | 859,069,289 | 8,014 | # critical rotating speed of a mill
### What Is The Optimum Rotation Speed For A Ball Mill
Formula for critical speed of a rotating mill. what is the optimum rotation speed for a ball mill critical speed The mill critical speed is the rotational speed at which and mill rotation speed on energy Speed . More; how fast do ball mills rotate rrcserin. Get Price; How To Calculate Optimum Milling Speed …
### Ball Mill Critical Speed & Working Principle - YouTube
Jun 19, 2015· If 75 percent of critical speed is considered desirable for efficient grinding in a 2.4 meter (8 foot) diameter mill, then the same will be true for a 5.0 meter (15½ foot) diameter mill.
### Critical Rotational Speed For Ball Milling - antoonmahieu.be
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is a function of a ball … Chat Online
### critical speed ball mill - barberasmit.nl
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is a function of a ball .
### difference between critical speed and rotational speed in ...
The critical speed of the mill, &c, is defined as the speed at which a single ball will just remain ... The speed of rotation of the mill influences the power draft through two effects: the ... This difference is ascribed to the presence of a pool of slurry that is present on the bottom ... steel balls in a ball mill, or large lumps of ore in an.
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Mill Speed - Critical Speed - Paul O Abbe- ball mill rotating speed ,No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation Too low a speed and little energy is imparted on the productBall Mills - Mine EngineerComThe rotation is usually between 4 to 20 ...
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critical speed of ball mill from wiki - smartspector. critical speed of ball mill from wiki. critical speed of ball mill from wikiBall millWikipedia, the free encyclopedia A ball mill is a type of grinder used to. live chat TECHNICAL NOTES 8 GRINDING R. P. King - Mineral Tech. Live Chat » Rod mills, how they function - Mine Engineer.Com
### formula for critical speed of rotation of a ball mill
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both. Ball Mills Mine Engineer.Com. This formula calculates the critical speed of any ball mill. Most ball mills operate most efficiently between 65% and 75% of their critical speed.
### critical speed for a ball mill - museum-neftenbach.ch
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both. Ball Mill Critical Speed Mill (Grinding) Cement. Scribd is the world's largest social reading and publishing site.
### what is critical speed of ball mill - intuitievepraktijk.nl
critical rotating speed of a mill customer case. Ball Mill Critical Speed 911 Metallurgist Mar 17 2017 The critical speed of a rotating mill is the RPM at which a grinding medium will begin to centrifuge namely will start rotating with the mill and Contact Us locked charge in your grinding mill Outotec
### ball mill rotating critical speed - designkadoo.nl
critical ball mill rotation speed - Mineral Processing . 25/01/2018 Mining Ball Mill For Hematite,Iron Ore,Copper Ore,Dolomite. mining ball mill grinds material by rotating a cylinder with steel grinding balls, . the mill ...
### critical rotational speed for ball mill - prirodnizena.cz
critical rotational speed nc of ball mill. Design Method of Ball Mill by Discrete Element Method837 . 2 Schematic diagram of the rocking ball mill The mill was filled with 15 mm nylon coated iron balls, and the rotational speed N of the mill was varied in a range from 40% using the critical rotational speed …
### Critical Rotational Speed For Ball Milling
Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is a function of a ball .
### critical rotating speed of a mill - proficiency-fp7.eu
Ball Mills - Mine Engineer.Com provides mining. A Ball Mill grinds material by rotating a cylinder with steel grinding balls, ... Thepoint where the mill becomes a centrifuge is called the "Critical Speed…
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Ball Mill Critical Speed Mill , of a rotating mill is the RPM at which a grinding medium You'll find a good online tool for ball mill critical speed calculation . Live Chat; critical speed of cement mill - citiclubin.
### Critical rotation speed for ball-milling - ScienceDirect
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is …
### Mill Speed - Critical Speed - Paul O. Abbe
Mill Speed - Critical Speed. Mill Speed . No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation. Too low a speed and little energy is imparted on the product.
### formula for critical speed of a rotating mill
formula for critical speed of a rotating mill. Grinding. Critical Speed (r.p.m.) is commonly considered as the hypothetical mill speed ... a tumbling motion is imparted to the charge by rotation of the grinding mill barrel. Chat Now. How to calculate the critical speed of rotating drum .
### TECHNICAL NOTES 8 GRINDING R. P. King - Mineral Tech
The center of gravity first starts to shift to the left as the speed of rotation increases but as critical speed is reached the center of gravity moves towards the center of the mill as more and more of the material is held against the shell throughout the cycle. Since critical speed is larger at smaller radii the centrifuging layer gets
### What it is the optimun speed for a ball mill ...
Oct 19, 2006· What it is the optimun speed for a ball mill ? - posted in Pyrotechnics: I have done a ball mill , recenly finished , but the motor has too rpms , is too fast for use in a ball mill (the pvc cylinder that i use , left of the shafts). With the motor i will use a 40 mm pulley , because i have a 50 mm driven pulley, in one of my two shafts.
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Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by. Get Price. Quick and Easy Black Powder Ball Mill - Skylighter.
### SAGMILLING.COM .:. Mill Critical Speed Determination
Mill Critical Speed Determination. The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where balls will not fall away from the mill's shell.
### difference between critical speed and rotational speed in ...
Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by experiments and by ... we found that the relation between critical rotation speed and ball-containing fraction .... rotation of the balls, we could not observe a significant difference between the...
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Ball Mill Critical Speed - Mineral Processing MetallurgyA Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces
### what is the critical speed for a rotating mill shaft
what is the critical speed for a rotating mill shaft. View this project. Ball Mills - Mine Engineer.Com. information on ball mills ... This formula calculates the critical speed of any ball mill. ... ball mill critical speed calculation - 19 Jan 2014 A Ball Mill grinds material by rotating a …
### Critical Speeds of Rotating Shafts and Mass Review ...
Bearings Application Knowledge Menu Bearings and Bushings Products and Supply. Critical Speeds of Rotating Shafts or Mass: The center of a rotating mass is always offset from the center of rotation. This is due to the reality of imperfect geometry and uneven mass distribution.
### critical rotating speed of a mill
The critical speed of a SAG mill is the speed in RPM's at which centrifugal force causes the material being ground to be held against the inside of the shell. Contact Supplier; Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete ...
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critical rotating speed of a mill bussy. critical rotational speed nc of ball mill. Critical rotation speed for ball-milling ScienceDirect. where D is the inner diameter of a jar and r is the radius of balls. critical rotational speed for ball milling.
### critical rotational speed nc of ball mill - duedonna.nl
critical speed of the ball mill . Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM) The results carried out by both methods showed good agreement It has been commonly accepted that the critical rotation speed is a function of a ball,
### formula for critical speed of a ball mill - socapie.eu
The critical speed of the mill, &c, is defined as the speed at which a single ball will just remain against the wall for a full cycle. .... steel balls in a ball mill, or large lumps of ore in an ... autogenous mills calculated using the formulas of Austin. Get Price
### Ball Mills - Mine Engineer.Com
The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed. Ball Mills are generally used to grind material 1/4 inch and finer, down to the particle size of 20 to 75 microns.
### critical rotational speed ball mill - barberasmit.nl
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is a function of a ball .
### Critical Rotational Speed Ball Mill Benin - entiteb2.be
Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM). The results carried out by both methods showed good agreement. It has been commonly accepted that the critical rotation speed is a function of a ball . | 2,419 | 11,571 | {"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} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.870887 |
https://www.systutorials.com/docs/linux/man/3-DGBEQU/ | 1,679,983,963,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00436.warc.gz | 1,107,862,901 | 8,981 | # DGBEQU (3) - Linux Man Pages
dgbequ.f -
## SYNOPSIS
### Functions/Subroutines
subroutine dgbequ (M, N, KL, KU, AB, LDAB, R, C, ROWCND, COLCND, AMAX, INFO)
DGBEQU
## Function/Subroutine Documentation
### subroutine dgbequ (integerM, integerN, integerKL, integerKU, double precision, dimension( ldab, * )AB, integerLDAB, double precision, dimension( * )R, double precision, dimension( * )C, double precisionROWCND, double precisionCOLCND, double precisionAMAX, integerINFO)
DGBEQU
Purpose:
``` DGBEQU computes row and column scalings intended to equilibrate an
M-by-N band matrix A and reduce its condition number. R returns the
row scale factors and C the column scale factors, chosen to try to
make the largest element in each row and column of the matrix B with
elements B(i,j)=R(i)*A(i,j)*C(j) have absolute value 1.
R(i) and C(j) are restricted to be between SMLNUM = smallest safe
number and BIGNUM = largest safe number. Use of these scaling
factors is not guaranteed to reduce the condition number of A but
works well in practice.
```
Parameters:
M
``` M is INTEGER
The number of rows of the matrix A. M >= 0.
```
N
``` N is INTEGER
The number of columns of the matrix A. N >= 0.
```
KL
``` KL is INTEGER
The number of subdiagonals within the band of A. KL >= 0.
```
KU
``` KU is INTEGER
The number of superdiagonals within the band of A. KU >= 0.
```
AB
``` AB is DOUBLE PRECISION array, dimension (LDAB,N)
The band matrix A, stored in rows 1 to KL+KU+1. The j-th
column of A is stored in the j-th column of the array AB as
follows:
AB(ku+1+i-j,j) = A(i,j) for max(1,j-ku)<=i<=min(m,j+kl).
```
LDAB
``` LDAB is INTEGER
The leading dimension of the array AB. LDAB >= KL+KU+1.
```
R
``` R is DOUBLE PRECISION array, dimension (M)
If INFO = 0, or INFO > M, R contains the row scale factors
for A.
```
C
``` C is DOUBLE PRECISION array, dimension (N)
If INFO = 0, C contains the column scale factors for A.
```
ROWCND
``` ROWCND is DOUBLE PRECISION
If INFO = 0 or INFO > M, ROWCND contains the ratio of the
smallest R(i) to the largest R(i). If ROWCND >= 0.1 and
AMAX is neither too large nor too small, it is not worth
scaling by R.
```
COLCND
``` COLCND is DOUBLE PRECISION
If INFO = 0, COLCND contains the ratio of the smallest
C(i) to the largest C(i). If COLCND >= 0.1, it is not
worth scaling by C.
```
AMAX
``` AMAX is DOUBLE PRECISION
Absolute value of largest matrix element. If AMAX is very
close to overflow or very close to underflow, the matrix
should be scaled.
```
INFO
``` INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
> 0: if INFO = i, and i is
<= M: the i-th row of A is exactly zero
> M: the (i-M)-th column of A is exactly zero
```
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
November 2011
Definition at line 153 of file dgbequ.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code. | 932 | 3,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-14 | longest | en | 0.767151 |
https://dmoj.ca/problem/hkccc15j1 | 1,601,583,480,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00612.warc.gz | 334,011,639 | 7,974 | ## CCCHK '15 J1 - Rock-paper-scissors
View as PDF
Points: 3
Time limit: 1.0s
Memory limit: 256M
Problem type
Allowed languages
Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig
Alice and Bob are playing rock-paper-scissors game. The game rules are simple: rock crushes scissors; paper covers rock; and scissors cut paper. If both players throw the same shape, the game is tied. Your task is to determine the number of games won by Alice and Bob, respectively.
#### Input Specification
• The first line contains one integer that represents the number of games.
• The second line is Alice's shape sequence. The shape sequence contains shapes and they are separated by a space. The th shape in the sequence represents the shape thrown by Alice in the th game. There are only three shape values: rock, paper, and scissors.
• The third line is Bob's shape sequence.
#### Output Specification
Two integers separated by a space, representing the number of games won by Alice and the number of games won by Bob.
#### Sample Input 1
3
rock rock scissors
paper rock rock
#### Output for Sample Input 1
0 2
#### Sample Input 2
4
paper rock rock scissors
rock scissors rock rock
#### Output for Sample Input 2
2 1
• commented on June 11, 2020, 3:21 p.m.
how to do it
• commented on June 11, 2020, 4:53 p.m.
If you need help on a question please join the Slack
• commented on Jan. 9, 2020, 9:10 p.m.
Python
• commented on Feb. 26, 2018, 12:10 p.m. edited
This comment is hidden due to too much negative feedback. Click here to view it.
• commented on Dec. 25, 2017, 1:26 p.m.
I'm like,"Let's submit my code for Java 9!" Gets test case 9 TLE. Spends 4 months writing a different code. Submits the same code in Java 8. Gets test case 9 right. ;-;
• commented on Nov. 26, 2016, 11:30 a.m.
What if it's a tie?? I guess that that doesn't count for anybody, right?
• commented on Nov. 26, 2016, 5:20 p.m.
No one wins.
• commented on Nov. 27, 2016, 11:57 a.m.
Or everyone wins !!! ;)
• commented on Feb. 19, 2018, 8:47 p.m. edited
This comment is hidden due to too much negative feedback. Click here to view it. | 685 | 2,367 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-40 | longest | en | 0.888243 |
https://www.disidu.com/q/157 | 1,726,261,943,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00744.warc.gz | 693,428,415 | 7,862 | # 问题157: It is given that ABCD is a parallelogram, AM=MD and area of △CDN=12 . Find the area of the quadrilateral ABNM(x) .
It is given that $A B C D$ is a parallelogram, $A M=M D$ and area of $\triangle C D N=12$. Find the area of the quadrilateral $A B N M(x)$.
\begin{aligned} & \because \angle M N D=\angle C N B, \angle M D N=\angle C B N \therefore \triangle M N D \backsim \triangle C N B \\ & \because M \text { 是中点 } \therefore M D: A D=M D: C B=1: 2 \\ & \therefore M N: C N=M D: C B=1: 2 \\ & \because \triangle M N D \text { 与 } \triangle C N D \text { 等高且底边 } M N: C N=1: 2 \\ & \therefore S_{\triangle M N D}: S_{\triangle C N D}=1: 2 \\ & \therefore S_{\triangle M N D}=\frac{1}{2} S_{\triangle C N D}=\frac{1}{2} \cdot 12=6 \\ & \because S_{\triangle B N C}=\left(\frac{B C}{D M}\right)^2 \cdot S_{\triangle M N D}=2^2 \cdot 6=24 \\ & \therefore S_{\triangle C B D}=S_{\triangle C B N}+S_{\triangle C D N}=24+12=36 \\ & \therefore S_{\triangle A B D}=S_{\triangle C B D}=36 \\ & \therefore S_{A B N M}=S_{\triangle A B D}-S_{\triangle M N D}=36-6=30 \end{aligned} | 429 | 1,081 | {"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} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.478781 |
https://unige.it/en/off.f/2017/ins/15310 | 1,643,416,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00428.warc.gz | 629,928,066 | 7,484 | # COMPLEX ANALYSIS
iten
Code
84039
2017/2018
CREDITS
7 credits during the 3nd year of 8760 Mathematics (L-35) GENOVA
SCIENTIFIC DISCIPLINARY SECTOR
MAT/05
TEACHING LOCATION
GENOVA (Mathematics)
semester
1° Semester
## OVERVIEW
Language: Italian (unless otherwise requested)
This course consists of 60 hours of lectures, including 12 hours of exercises.
The course consists essentially of two parts. The first, whch is the main part, is a standard introduction to complex analysis, from first definitions to the residue theorem. The second part contains a selection of topics, including the theory of Euler's Gamma function.
## AIMS AND CONTENT
LEARNING OUTCOMES
The couse aims at giving to students the bases of complex analysis. In particular, at the end of the course the students will understand the basic concepts of complex analysis and will be able to use them to solve some problems.
TEACHING METHODS
The course consists of traditional lectures and exercises sessions. Moreover, exercises will be given to students for independent training.
SYLLABUS/CONTENT
Power series: formal power series; convergent power series; analytic functions. Complex differentiation: holomorphic functions; Cauchy-Riemann equations; conformal transformations; elementary functions. Complex integration: integration along paths; primitives; Cauchy's theorem. Consequences of Cauchy's theorem: Cauchy's integral formula; holomorphic functions are analytic; further consequences (theorems by Morera and Liouville, mean value, maximum modulus, Weierstrass' convergence theorems). Singularities and residues: Laurent series; behavior near singularities; residue theorem and applications. Miscellanea: zeros and poles of meromorphic functions; Euler's Gamma function; analytic continuation.
V.Villani - Funzioni di una variabile complessa - ES Genova 1971.
H.Cartan - Elementary theory of analytic functions of one or several complex variables - Dover 1995.
R.Remmert - Theory of complex functions - Springer 1989.
S.Lang - Complex analysis - Springer 1999.
C.Presilla - Elementi di analisi complessa, 2a edizione - Springer 2014 (for exercises).
M.R.Spiegel - Variabili complesse - ETAS Libri 1974 (for exercises).
## TEACHERS AND EXAM BOARD
Office hours: On appointment.
## LESSONS
TEACHING METHODS
The course consists of traditional lectures and exercises sessions. Moreover, exercises will be given to students for independent training.
LESSONS START
The class will start according to the academic calendar.
Class schedule
All class schedules are posted on the EasyAcademy portal.
## EXAMS
EXAM DESCRIPTION
ASSESSMENT METHODS
Evaluation of written and oral parts; during the oral examination, students may present a topic, chosen from an ad hoc list, prepared in advance.
Exam schedule
Date Time Location Type Notes
11/01/2018 09:00 GENOVA Scritto
11/01/2018 14:30 GENOVA Orale
08/02/2018 09:00 GENOVA Scritto
08/02/2018 14:30 GENOVA Orale
12/06/2018 14:30 GENOVA Scritto
13/06/2018 14:30 GENOVA Orale
10/07/2018 14:30 GENOVA Scritto
11/07/2018 14:30 GENOVA Orale
11/09/2018 14:30 GENOVA Scritto
12/09/2018 14:30 GENOVA Orale | 808 | 3,142 | {"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} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.843717 |
https://jp.mathworks.com/matlabcentral/cody/groups/44/problems/1082-lychrel-number-test-inspired-by-project-euler-problem-55/solutions/2088606 | 1,585,507,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00218.warc.gz | 540,276,035 | 16,093 | Cody
# Problem 1082. Lychrel Number Test (Inspired by Project Euler Problem 55)
Solution 2088606
Submitted on 11 Jan 2020 by Asif Newaz
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(islychrel(3763));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint1 (line 1) In solutionTest (line 3)]
2 Pass
assert(islychrel(5943));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint2 (line 1) In solutionTest (line 5)]
3 Pass
assert(islychrel(4709));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint3 (line 1) In solutionTest (line 7)]
4 Pass
assert(~islychrel(3664));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint4 (line 1) In solutionTest (line 9)]
5 Pass
assert(~islychrel(3692));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint5 (line 1) In solutionTest (line 11)]
6 Pass
assert(islychrel(196));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint6 (line 1) In solutionTest (line 13)]
7 Pass
assert(islychrel(8619));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint7 (line 1) In solutionTest (line 15)]
8 Pass
assert(islychrel(9898));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint8 (line 1) In solutionTest (line 17)]
9 Pass
assert(islychrel(9344));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint9 (line 1) In solutionTest (line 19)]
10 Pass
assert(islychrel(9884));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint10 (line 1) In solutionTest (line 21)]
11 Pass
assert(islychrel(4852));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint11 (line 1) In solutionTest (line 23)]
12 Pass
assert(islychrel(7491));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint12 (line 1) In solutionTest (line 25)]
13 Pass
assert(~islychrel(5832));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint13 (line 1) In solutionTest (line 27)]
14 Pass
assert(~islychrel(7400));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint14 (line 1) In solutionTest (line 29)]
15 Pass
assert(~islychrel(2349));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint15 (line 1) In solutionTest (line 31)]
16 Pass
assert(~islychrel(7349));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint16 (line 1) In solutionTest (line 33)]
17 Pass
assert(~islychrel(9706));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint17 (line 1) In solutionTest (line 35)]
18 Pass
assert(~islychrel(8669));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint18 (line 1) In solutionTest (line 37)]
19 Pass
assert(~islychrel(863));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint19 (line 1) In solutionTest (line 39)]
20 Pass
assert(~islychrel(5979));
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [> In islychrel (line 2) In ScoringEngineTestPoint20 (line 1) In solutionTest (line 41)] | 1,527 | 5,542 | {"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-2020-16 | latest | en | 0.769264 |
http://www.us-lotteries.com/Florida/Cash_3/cash_3-straights-analysis.asp | 1,386,332,717,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163051588/warc/CC-MAIN-20131204131731-00084-ip-10-33-133-15.ec2.internal.warc.gz | 594,572,468 | 6,805 | Frequency and last appearance of the Florida Lottery Cash 3. Lists most and least frequently drawn numbers, or numbers that have not been drawn for a long time.
# Florida Lottery Cash 3 Straight Numbers (Exact Order) Analysis
How many times is a Florida Cash 3 straight number drawn? What are the hot and cold Florida Cash 3 numbers? Which Florida Cash 3 numbers have never been drawn? Which Cash 3 numbers were drawn only a long, long time ago? -- These are some of the questons that this page tries to answer. This page is re-written after every draw. Last update was on Thursday, December 05, 2013, with 084 as the latest winning number. The Florida Cash 3 results analyzed are from 4/29/1988 and they constitute 11375 draws. On this page...
## Florida Cash 3 Frequency of Straight Numbers
The number of times that each one of the 1000 Cash 3 numbers are drawn is presented here. Since 11375 draws of the Florida Cash 3 game are analyzed, we expect each straight number to be drawn about 11 times. In fact, we should not expect all the numbers to be drawn that many times inasmuch as the science of statistics tells us that such a frequency should demonstrate a normal distribution. This means that the majority of numbers must be drawn 10, 11, or 12 times; less and less on both the lower and higher frequency sides. The maximum number of times that a Florida Cash 3 number is drawn is 25, and the minimum is 2.
Below, you will be able to list the Florida Cash 3 winning numbers corresponding to a certain frequency. That is, when you select a number from the drop-down list of frequencies that range from 0 to the maximum frequency of 25, all the Cash 3 numbers that have been drawn that many times as a straight will be displayed in the box below. For example, if you select a frequency value of 5, all the Cash 3 straight numbers that have been drawn 5 times will be listed.
So, the the most frequently drawn Florida Cash 3 numbers can be listed by selecting 25 from the menu. For the least frequently drawn Cash 3 numbers select 2. Note that the list of numbers corresponding to a frequency of 0 are those numbers that have never been drawn as a straight in the Cash 3 game. These numbers are listed further below also (if there are any).
Results will be displayed here
TOP
## Florida Cash 3 Straight Numbers Never Drawn
Here are Cash 3 straight numbers that have not been drawn yet in the Florida.
-none-
TOP
## Long-Time-No-See Florida Cash 3 Straight Numbers
Here, you'll find the list of Florida Cash 3 straight numbers that have not been drawn for a long time now.
In general, we have arbitrarily chosen the numbers that have not been drawn in the last 3650 draws (10 years for 1-draw-a-day games and 5 years for 2-draws-a-day game). If the number of draws until today are significantly less than 3650, then the late numbers listed are those not drawn in the last 3/4 draws of the total draws.
The late Florida Cash 3 numbers listed below are those that have not been drawn in the last 3650 draws. The numbers are arranged in chronological order (left to right and down) of their last appearance.
575 - Thu Aug 5, 1993231 - Fri Nov 20, 1998040 - Mon Jan 29, 2001
823 - Wed Oct 24, 2001633 - Mon Sep 13, 2004926 - Wed Nov 24, 2004
958 - Fri Apr 28, 2006921 - Thu May 11, 2006069 - Mon Aug 21, 2006
726 - Wed Sep 27, 2006755 - Sat Nov 11, 2006728 - Sat Dec 23, 2006
708 - Tue Jan 30, 2007594 - Tue Feb 13, 2007603 - Sun Jun 3, 2007
247 - Mon Jun 4, 2007022 - Wed Aug 29, 2007162 - Mon Sep 24, 2007
876 - Sun Oct 7, 2007470 - Fri Jan 25, 2008969 - Wed Apr 9, 2008
041 - Tue May 27, 2008751 - Mon Jun 2, 2008500 - Sat Jul 12, 2008
612 - Tue Aug 5, 2008703 - Tue Aug 12, 2008666 - Fri Sep 5, 2008
094 - Wed Nov 12, 2008972 - Tue Nov 18, 2008647 - Sat Dec 6, 2008
TOP
## List Florida Cash 3 Straight Numbers not drawn for a specified number of draws
Alternatively, you may list all the list of Florida Cash 3 numbers that haven't been drawn in the last specified number of draws. For instance, to see the numbers that haven't fell in the last 2000 draws, just enter 2000 in the box and then click on the 'Show' button. If the total number of draws is less than the number you entered, then the list will be all those numbers that have never been drawn. Note that the list is arranged just in numerical order (not chronological), and that there are a total of 11375 draws so far.
Results will be displayed here
TOP
Florida lottery Games Data and Analysis Menu
Cash 3 - Cash 3 Game - Winning Numbers - Number Search - Yearly Calendar - Boxes(Combos) - Triples - Digits Analysis - Straights Analysis - Boxes Analysis - Pairs - Sums - Combinations Generator - Game Info Play 4 - Play 4 Game - Winning Numbers - Number Search - Yearly Calendar - Boxes(Combos) - Quadruplets - Digits Analysis - Straights Analysis - Boxes Analysis - Pairs - Sums - Combinations Generator - Game Info Fantasy 5 - Fantasy 5 Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Triads Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Mega Money - Mega Money Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Number-Ball Pair Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Lotto - Lotto Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Triads Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Mega Millions - Mega Millions Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Number-Ball Pair Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info Powerball - Powerball Game - Winning Numbers - Number Search - Yearly Calendar - Numbers Analysis - Pairs Analysis - Number-Ball Pair Analysis - Check Combinations - Full Wheel - Abbreviated Wheel - Game Info
- Florida Lottery Recent Results and Tools - Compact page for quick check - Florida Lottery General Information - Florida Lottery Games - US-Lotteries Home Page
• WE DO NOT SELL ANY LOTTERY, WE DO NOT BUY LOTTERY NUMBERS ON BEHALF OF ANYONE; this is only an information and guide web site. | 1,652 | 6,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2013-48 | latest | en | 0.939803 |
https://crypto.stackexchange.com/questions/52777/how-can-shors-algorithm-be-applied-to-ecc | 1,725,918,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00826.warc.gz | 181,875,388 | 41,769 | # How can Shor's Algorithm be applied to ECC?
I have not found a specific answer to this question on here.
Shor's algorithm can be used to factorize a large (semi)prime $N$ by reducing the task to period-finding of a function $f(x)=x^a$ mod $N$.
This is done by creating an equal superposition over all pairs of $a_i$ and $f(x)=x^{a_i}$ for a random $x$, then measuring $f(x)$ causing the superposition to collapse into all $a_i$ for which $f(x)$ is our measures value. Using "Fourier sampling" (I have not fully understood this part) we can then obtain the period of $\ f$ and with .5 probability this yields a non-trivial square root of $N$ which leads to a prime factor.
(Plase correct me if my understanding of the above is flawed)
Now how can this algorithm be applied to Elliptic Curve schemes like ECDSA? I struggle to find an explanation for how the discrete log problem for groups over elliptic curves could be solved using Shor's.
EDIT: I would just as well appreciate a reference to other papers except Shor's, that explain the case of Shor's algorithm on DLPs.
• Do you understand how the discrete logarithm problem is solved over $\mathbb Z^*_p$ using Shor's algorithm? Commented Nov 2, 2017 at 20:20
• I read Shor's paper but did not understand the DLP part. From what I learned, if our $\mathbb{Z}_p^\ast$ DLP is find $r$ so that $g^r \equiv x$ mod $p$, we create superpositions over all $\ket{a}$ and $\ket{b}$ and calculate $g^ax^{-b}$ mod $p$. What is this good for to find the order of the group we are looking for? Commented Nov 2, 2017 at 20:46
• You may find this reference arxiv.org/abs/quant-ph/0301141 , interesting (found via the always useful quantum algorithm zoo math.nist.gov/quantum/zoo) Commented Nov 3, 2017 at 10:02
• @Frédéric Grosshans thanks for the reference, I read this paper but could you maybe give a few sentences describing what is done, like I did for integer factorization in the initial post? I can read the formulas in the paper, but I'm uncertain what they actually do. Commented Nov 4, 2017 at 11:50 | 555 | 2,056 | {"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": 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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.910746 |
https://www.jiskha.com/display.cgi?id=1303174285 | 1,516,789,468,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893629.85/warc/CC-MAIN-20180124090112-20180124110112-00253.warc.gz | 911,837,399 | 3,744 | # Physics
posted by .
Suppose that you want to take a photograph of yourself as you look at your image in a flat mirror 1.6 away.
For what distance should the camera lens be focused?
• Physics -
light is moving 3.2 meters from it's source.
## Similar Questions
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2. ### Physics
Why can't you take a photograph of your image in a plane mirror and focus the camera on both your image and on the mirror frame?
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An image of a candle is produced by a convex lens 14 cm away. If the focal length of the lens is 7 cm, how far from the lens is the candle placed?
5. ### calculus problem
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6. ### physics
Camera lenses are described in terms of their focal length. A 50 mm lens has a focal length of 50 mm. A camera is focused on an object 81 m away using a 88 mm lens. Locate the position of the image. Answer in units of mm
7. ### Physics
Standing 2.5 m in front of a plane mirror with your camera, you decide to take a picture of yourself without turning the camera around. To what distance should the camera be focused to get a sharp image?
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Camera lenses are described in terms of their focal length. A 50.0 mm lens has a focal length of 50.0 mm. (a) A camera with a 50 mm lens is focused on an object 4.0 m away. Locate the image. (b) A 980 mm lens is focused on an object …
10. ### Physics
Camera lenses are described in terms of their focal length. A 50.0 mm lens has a focal length of 50.0 mm (a) A camera with a 50 mm lens is focused on an object 4.0 m away. Locate the image. (b) A 980 mm lens is focused on an object …
More Similar Questions | 586 | 2,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-05 | latest | en | 0.946837 |
https://apboardsolutions.com/ap-board-7th-class-maths-solutions-chapter-5-review-exercise/ | 1,721,811,126,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00246.warc.gz | 88,543,886 | 16,857 | # AP Board 7th Class Maths Solutions Chapter 5 Triangles Review Exercise
SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Review Exercise Questions and Answers.
## AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Review Exercise
Question 1.
Mark any three non collinear points A, B and C in your note book, join them to make a triangle and name it.
Question 2.
Observe the given triangle and answer the following :
(i) Write the interior points of the triangle.
A, C, D and J.
(ii) Write the points marked on the triangle.
B, E, G, P, Q and R.
(iii) Write the exterior points of the triangle.
F, H and I.
Question 3.
Observe the given triangle and answer the following:
(i) The opposite side to vertex L is __________
$$\overline{\mathrm{KM}}$$
(ii) The opposite side to ZK is __________
$$\overline{\mathrm{LM}}$$
(iii) The opposite angle to $$\overline{\mathrm{KL}}$$ is __________
∠M.
(iv) The opposite vertex to $$\overline{\mathrm{LM}}$$ is __________
K.
Question 4.
Classify the following angles into acute, obtuse and right angles :
20°, 50°, 102°, 47°, 125°, 65°, 36°, 90°, 95° and 110°. | 326 | 1,134 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.780141 |
https://math1089.in/tag/maths-trivia/ | 1,696,352,045,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00205.warc.gz | 430,453,507 | 45,146 | ## 40 Interesting Facts About Mathematics
We could use up two Eternities in learning all that is to be learned about our own world and the thousands of nations that have arisen and flourished and vanished from it. Mathematics alone would occupy me for eight million years.
Mark Twain
Glad you came by. I wanted to let you know I appreciate your spending time here at the blog very much. I do appreciate your taking time out of your busy schedule to check out Math1089!
Mathematics is a very interesting subject. Understanding different interesting facts about mathematics helps the students to develop their skills and motivates them to learn more about the subject. In this blog post, we are listing down a few amazing facts about mathematics.
One can observe that all odd numbers, which are those ending in 1, 3, 5, 7, or 9, contain the letter e. Therefore, it can be concluded that the presence of the letter e is a common characteristic of all odd numbers.
There is no zero (0) in Roman numerals. The entire number system is generated by the seven different letters, namely I, V, X, L, C, D and M. Using these, we can write numbers up to 3999.
In the entire Hindu Arabic number system, there is only one number which can be spelled with the same number of letters as itself. That number is FOUR.
In the decimal number system, there are 10 digits from 0 to 9. It is also known as the Hindu Arabic numeral system, invented more than 1000 years ago.
Forty is the only number that is spelt with letters arranged in alphabetical order.
One is the only number that is spelt with letters arranged in descending order.
What comes after million, billion, and trillion? You can term the numbers after these terms as quadrillion, quintillion, sextillion, septillion, octillion, nonillion, and decillion.
The majority of the months have 31 days in a calendar. Only four months have 30 days i.e., April, June, September and November. February typically only has 28 days, with the exception of leap years (there 29 days).
The opposite sides of a die always add up to 7. For example, 6 and 1 will always be on opposite sides which add up to 7.
2 is the only natural number satisfying 2 + 2 = 2 × 2.
The number 2 is the smallest even prime number.
In the Indian number system, when we write numbers from 0 to 1000, the letter A only appears first in 1000 (one thousand).
The idea of zero was invented by an Indian mathematician and astronomer Brahmagupta in around 600 A. D. He contributed a lot to mathematics and astronomy, and is known for explaining how to find the cube and cube-root of an integer.
The nickname of Denver city is the mile-high city because its official elevation is exactly one mile (or 5280 feet or 1609.344 meters) above sea level.
The hour and minute hand of a clock coincide 22 times in a day.
Take any number and multiply it by three. Now find the sum of the digits in the answer. It is always divisible by 3. For example, consider 17. Multiply it by 3 to obtain 17 × 3 = 51 and the sum of digits = 5 + 1 = 6. Clearly, 6 is divisible by 3.
A perfect number refers to a positive integer which is equal to the sum of its positive divisors. Following this rule, 6 is the smallest perfect number. The next perfect number is 28.
From 0 to 1000, the only number that has the letter a in it is one thousand.
The word love in tennis means a score of zero.
Number 9 is also known as the magic number. This is because if we find the digital sum of digits (sum of digits till we get a single digit) of any multiple of 9, we will get 9. For example, consider 9 × 99 = 891. Now the digital sum is 8 + 9 + 1 = 18 and again 1 + 8 = 9.
Googol means 1 followed by 100 zeroes. A googolplex equivalent to ten raised to the power of a googol. A googolplexian equivalent to ten raised to the power of a googolplex.
Teenagers texting in Thailand will send the digits 555 to indicate that something is funny. In the Thai language, 5 is pronounced as ha which when translated becomes ha-ha-ha.
In a sample of 23 people, there are 50% chances that two of them will share the same birthday. This is known as the birthday problem.
Seven is the most significant number across religions and cultures. For example, seven colours in a rainbow, seven days in a week, seven notes on the musical skill, etc.
The signs Plus (+) and Minus (-) were discovered as early as in 1489 A.D.
Two and five are the only prime numbers in the entire number system which ends with two or five.
If you add up all the numbers between 1 to 100 consecutively then the total that you will reach is 5050.
Leonardo Fibonacci from Pisa, in the 13th century, discovered the Fibonacci sequence. This sequence is created as the sum of the two preceding numbers, with starting members as 1, 1. Complete sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, . . .
The minimum number of cuts that are required to cut the figure into eight pieces are three i.e., horizontally, vertically and cutting by the base.
Maps have four colours. According to the four-color theorem, any map can be coloured using only four colours so that regions sharing a boundary do not share the same colour.
−40° C is equal to −40° F.
For a given rate of interest say r, the time T required to double an investment is 72 / r. This is known as the Rule of 72.
26 is the only natural number sandwiched between a perfect square number (25) and a perfect cube number (27).
We all have heard about the Pythagoras theorem. But there is also something which is known as Pythagoras constant. The square root of 2 i.e. 1.41 was the first irrational number to be discovered which is known as the Pythagoras constant.
A standard football is made up of Pentagons and Hexagons. There are 12 pentagons and 20 hexagons.
Eleven plus Two is an anagram of Twelve plus One which is pretty fitting as the answer to both equations is 13. Also, there are 13 letters in both eleven plus two and twelve plus one.
A shirt will never come free if successive discounts of 50%, 30% and 20% are given by the shop keeper.
In any circle, ratio of the circumference and diameter is always constant. This constant quantity is known as π. It is a special number which is irrational. There is a designated day, called Pi day, that we celebrate on March 14.
There are other special numbers also apart from Pythagoras constant and pi. Some of these numbers are Euler’s constant, Golden ratio, etc.
The ratio of the distance between the fingertip and the elbow to the distance between the elbow and the wrist approximately equals the golden ratio ф (phi) = 1·618.
Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call Math1089 – Mathematics for All!“. | 1,616 | 6,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-40 | latest | en | 0.953893 |
https://www.encyclopedia.com/science/encyclopedias-almanacs-transcripts-and-maps/analemma-0 | 1,660,471,521,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00600.warc.gz | 682,534,208 | 16,538 | views updated
# Analemma
The Earths orbit around the Sun is not a perfect circle, but an ellipse. This leads to a number of interesting observational effects, one of which is the ana-lemma, the apparent path traced by the Sun in the sky when observed at the same time of day over the course of a year. The path resembles a lopsided figure eight, which is sometimes printed on a globe, usually somewhere on the Pacific Ocean where there is lots of room to show it.
Suppose you were to go outside and measure the Suns position in the sky every day, precisely at noon, over the course of a year. You would see the Sun appear to move higher in the sky as summer approached, and then move lower as winter approached. This occurs because the tilt of Earths axis causes the Suns apparent celestial latitude, or declination, to change over the course of the year.
However, you would also notice that at some times of the year, the Sun would appear slightly farther west in the sky at noon than it did at other times, as if it were somehow gaining time on your watch. This results from the ellipticity of Earths orbit. According to Keplers second law of motion, planets moving in an elliptical orbit will move faster when they are closer to the Sun than when they are farther away. Therefore, Earths speed in its orbit is constantly changing, decelerating as it moves from perihelion (its closest point to the Sun) to aphelion (its farthest point from the Sun), and accelerating as it then falls inward toward perihelion again.
It would be difficult for watchmakersat least, it would have been difficult for pre-digital watchmakersto make a clock that kept solar time. The clock would have to tick at different rates each day to account for Earths changing velocity about the Sun. Instead, watches keep what is called mean solar time, which is the average value of the advance of solar time over the course of the year. As a result, the Sun gets ahead of, and behind, mean solar time by up to 16 minutes at different times of the year. In other words, if you measured the position of the Sun at noon mean solar time at one time of year, the Sun might not reach that position until 12:16 P.M. at another time of year.
Now all the elements are in place to explain the figure eight. The tilt of Earths orbital axis causes the Sun to appear higher and lower in the sky at different times of year; this forms the vertical axis of the eight. The ellipticity of Earths orbit causes the actual solar time to first get ahead of, and then fall behind, mean solar time. This makes the Sun appear to slide back and forth across the vertical axis of the eight as measured by a fixed clock time each day, forming the rest of the figure.
Clearly, the shape of the analemma depends upon a particular planets orbital inclination and ellipticity. The Sun would appear to trace a unique analemma for any of the planets in the solar system; the analemmas thus formed are fat, thin, or even teardrop-shaped variants on the basic figure eight.
Jeffrey Hall | 658 | 3,033 | {"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-2022-33 | latest | en | 0.972199 |
http://en.wikipedia.org/wiki/Talk:Adele_ring | 1,410,964,832,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657123617.30/warc/CC-MAIN-20140914011203-00233-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 98,967,612 | 11,671 | WikiProject Mathematics (Rated Start-class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
Start Class
Mid Importance
Field: Number theory
I don't understand the definition. What is the topology on adeles to make it a locally compact group? AFAICS the usual product topology is not locally compact in this case. EJ 14:51, 20 Aug 2004 (UTC)
Correct. But the restricted product topology is. The adeles as a whole are the union of more 'reasonable' subsets looking like product of p-adic integers over almost all p (compact), cartesian product with a finite product of p-adic numbers and the real numbers (locally compact). Every given adele is in such a 'reasonable' set.
Charles Matthews 16:24, 20 Aug 2004 (UTC)
Thanks for a reply. Notice that the restricted product page doesn't describe the topology either, which is why I asked in the first place. Just to make sure I understand you correctly: the res. prod. topology coincides with prod. topology on each 'reasonable' subset, thus a subset of the restricted product is open iff it is relatively open in each 'reasonable' subset; in other words, the res. prod. topology is inductively generated by prod. topologies on 'reasonable' subset. Is this correct?
EJ 11:16, 21 Aug 2004 (UTC)
Yes, that sounds right. Thanks for pointing out the deficiency of the RP page - should be fixed up. Charles Matthews 11:47, 21 Aug 2004 (UTC)
## Idèle vs idele
I have reverted this change which changes idèle to idele throughout, and adds the alternative adèle, with an edit summary that fails to mention those changes. This is the sort of change that tends to produce strong reactions, and so should be discussed here per WP:BRD. For what it's worth, Neukirch (1999) uses the form idèle and adèle. Deltahedron (talk) 18:10, 17 April 2013 (UTC)
It was not an arbitrary change. You can see that there was no standard before my edit. I just standardized the spelling that has already been established both on Wikipedia (see also Adelic algebraic group) and in the mathematical literature (confer 1; 2, 3). Neukirch (1999) is obviously not the only English-language source on the subject. --Omnipaedista (talk) 02:03, 18 April 2013 (UTC)
It was a change made without discussion here and without a suitable edit summary. If the spelling has been standardised here on Wikipedia, please link to the discussion where that consensus was established. The spelling has clearly not been standardised in the mathematical literature, and if you think it has please cite an authoritative source. Google book searches do not constitute an authoritative source. Deltahedron (talk) 06:29, 18 April 2013 (UTC)
I should have indeed provided a suitable edit summary. As far as Wikipedia is concerned the "no accent" form has been "established" pragmatically. The original version of 'Adele ring' and the long-standing versions of the articles 'Adelic algebraic group', 'Idele class character' feature no accent. The addition of an accent here was made by a single user who him/herself was not even consistent about it. As far the mathematical literature is concerned I deem that there is no authoritative source available establishing a priori which is the standard, so we could for the moment rely on Google Book searches and ArXiv searches. In any case even if the Wikipedia community eventually chooses for some reason to employ the accented form of the word 'idele', it would be wise to do so consistently throughout Wikipedia, and also to employ the accented form of the word 'adele' to avoid confusion. --Omnipaedista (talk) 11:23, 18 April 2013 (UTC)
## Source of the term "adele"
Oscar Goldman claimed to be the originator of the contraction. He was taking notes for a series of lectures by Chevalley, who referred to them as "additive ideles" at the time. According to Goldman, when Chevalley noticed the contraction Goldman was using, he liked it and adopted it. Goldman always insisted on the pronunciation "add-ell." — Preceding unsigned comment added by 108.28.113.226 (talk) 19:05, 31 May 2013 (UTC)
This comment would be very much more valuable if you could provide a source for it. Colin McLarty (talk) 04:12, 9 June 2013 (UTC)
I personally heard Goldman's pronunciation of "add-ell." The story about taking notes from Chevalley is a personal communication from Goldman's student Robert Rubin. — Preceding unsigned comment added by 108.28.113.226 (talk) 14:27, 12 June 2013 (UTC) | 1,123 | 4,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2014-41 | latest | en | 0.925658 |
http://homeschoolden.com/2013/09/26/free-ultimate-math-game-board-free-addition-cards | 1,571,623,874,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00090.warc.gz | 95,235,679 | 28,642 | # Free Ultimate Math Game Board (+free addition cards)
This week I brought out a math game board to go over some math facts with the kids. The great thing about this board is that the kids can practice any math skill they’re working on.
A player can start on any green start space. If she answers a math fact card correctly, she rolls the die and moves clockwise around the gameboard. If she lands on a space such as double move or draw again, she gets to take another turn (and another) until she lands on a space (such as stay put or a start space) when she can’t go any farther. The first person to land on one of the win spaces wins the game!
I played this yesterday with DD. She’s still working on getting her multiplication facts down cold — and we played four games — going through probably 30 problems without one complaint!
I made some matching cards for ED who is working on doubling (1+1 through 9+9) and addition with +1 or +2. I also have a blank page where you could fill in your own math problems.
Math Cards — Doubling +1 +2
————————————————————————————–
You might also be interested in:
Telling Time Kitty Clocks
Free Daily Calendar Page — In this post I shared some of the activities ED did with the lapbook on the days of the week, months of the year and the daily calendar page I made for her.
Our hands-on activities – when we did our measurement unit on gallons, quarts, pints, etc. (plus the free measurement unit pages below):
Valentine’s Bump Game (for doubling 1+1, 2+2, etc.)
Doubling with Pete the Cat:
Addition Board Games (sums that equal 13 or 14): The Great Pet Races
Here are some fun outdoor math games that we played when the weather warmed up last year. They are good for *any* math practice!:
See you again soon here or Homeschool Den Facebook page. Don’t forget to Subscribe to our Homeschool Den Newsletter! ~Liesl
### 22 Responses
1. September 26, 2013
[…] Homeschool Den is offering the free Ultimate Math Game Board + Free Addition Cards! She also explains how to play the game. It sounds like lots of […]
2. September 26, 2013
[…] Homeschool Den is offering the free Ultimate Math Game Board + Free Addition Cards! She also explains how to play the game. It sounds like lots of […]
3. September 26, 2013
[…] Free Ultimate Math Game Board (+free addition cards) […]
4. September 26, 2013
[…] Free Ultimate Math Game Board (+free addition cards) […]
5. September 26, 2013
[…] Free Ultimate Math Game Board (+free addition cards) […]
6. September 26, 2013
[…] Free Ultimate Math Game Board (+free addition cards) […]
7. September 27, 2013
[…] The Homeschool Den is offering a free Ultimate Math Game Board + Free Addition Cards! […]
8. September 27, 2013
[…] The Homeschool Den is offering a free Ultimate Math Game Board + Free Addition Cards! […]
9. September 30, 2013
[…] هنــــا مجموعة من المطبوعات لمادة الرياضيات مع كروت […]
10. September 30, 2013
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11. October 5, 2013
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12. October 5, 2013
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13. December 9, 2013
[…] the other (uncut) sheet. You can also use these numbers to go around a game board (Here’s a game board I made for the kids that you can download for free). You can pull out festive erasers, flip over a card and have you kids count that same amount. As […]
14. December 9, 2013
[…] the other (uncut) sheet. You can also use these numbers to go around a game board (Here’s a game board I made for the kids that you can download for free). You can pull out festive erasers, flip over a card and have you kids count that same amount. As […]
15. January 30, 2014
16. January 30, 2014
17. June 12, 2014
[…] also made a printable version of this board which I called the “Ultimate Game Board.” You can download it free over at that […]
18. June 12, 2014
[…] also made a printable version of this board which I called the “Ultimate Game Board.” You can download it free over at that […]
19. August 6, 2014
[…] Ultimate Math Board Game: A printable version of the board game above. […]
20. August 6, 2014
[…] Ultimate Math Board Game: A printable version of the board game above. […]
21. December 14, 2014
[…] on the other (uncut) sheet. You can also use these numbers to go around a game board (Here’s a game board I made for the kids that you can download for free). You can pull out festive erasers, flip over a card and have you kids count that same amount. As […]
22. December 14, 2014
[…] on the other (uncut) sheet. You can also use these numbers to go around a game board (Here’s a game board I made for the kids that you can download for free). You can pull out festive erasers, flip over a card and have you kids count that same amount. As […] | 1,219 | 4,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-43 | latest | en | 0.922324 |
http://forums.wolfram.com/student-support/topics/346245 | 1,444,335,508,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737899086.53/warc/CC-MAIN-20151001221819-00248-ip-10-137-6-227.ec2.internal.warc.gz | 118,474,117 | 5,374 | Student Support Forum: 'If same element exists in a list, gives 0.' topicStudent Support Forum > General > Archives > "If same element exists in a list, gives 0."
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Author Comment/Response yehuda 02/20/13 3:25pm In Response To 'Re: If same element exists in a list, gives 0.' --------- Bill, this is not fully correct, since Union sorts the list, and if the list is not sorted in the first place, you will get a wrong answer Therefore the solution should be f[v_]:=If[Length[Union[v]]==Length[v],1,0]; yehuda [World: Please be aware that Bill posted his erratum before Yehuda's comment become Public. -FM] URL: ,
Subject (listing for 'If same element exists in a list, gives 0.') Author Date Posted If same element exists in a list, gives 0. simple 02/19/13 9:31pm Re: If same element exists in a list, gives 0. Bill Simpson 02/20/13 1:34pm Re: Re: If same element exists in a list, gives... yehuda 02/20/13 3:25pm Re: If same element exists in a list, gives 0. Bill Simpson 02/20/13 4:08pm Re: If same element exists in a list, gives 0. yehuda 02/21/13 3:08pm
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``` if c0,c1,c2,.......cn donets the coefficients ofthe bin. expantion (1+x)^n, then the value of
c1+3c3+5c5+... is
```
6 years ago
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``` Dear manan
from Binomial expansion we know that
(1+x)n =C0 + C1x + C2x2 +C3x3 +C4x4+C5x5+ ......Cnxn
now differentiate both side
n(1+x)n-1 = C1 +2C2x+3C3x2 +4C4x3 + 5C5x4 + ......
now put x=1
n2n-1 = C1 +2C2+3C3 +4C4 + 5C5 + ...... .............(1)
now put x=-1
0 =C1 -2C2+3C3 -4C4 + 5C5 + ...... .............(2)
(1) +(2)
n2n-2 = C1 +3C3 + 5C5 + ...... ............
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
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https://www.transtutors.com/questions/oo-t-mobile-5-06-pm-69-ezto-mheducation-com-e-connect-jose-mejia-mannisto-inc-uses-t-2561879.htm | 1,642,691,850,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00239.warc.gz | 1,101,718,736 | 12,829 | # Oo T-Mobile 5:06 PM 69 % ezto.mheducation.com E connect Jose Mejia Mannisto, Inc., uses the FIFO ...
oo T-Mobile 5:06 PM 69 % ezto.mheducation.com E connect Jose Mejia Mannisto, Inc., uses the FIFO inventory cost flow assumption. In a year of rising costs and prices, the firm reported net income of $206,244 and average assets of$1,457,830, If Mannisto had used the LIFO cost flow assumption in the same year, its cost of goods sold would have been $46,370 more than under FIFO, and its average assets would have been$42,320 less than under FIFO. Required: a. Calculate the firm's ROI under each cost flow assumption (FIFO and LIFO)-(Enter your answers as percentages rounded to 1 decimal place (ie, 12.2%).) b. Suppose that two years later costs and prices were falling. Under FIFO, net income and average assets were $252,954 and$1,787,680, respectively. If LIFO had been used through the years, inventory values would have been $45,400 less than under FIFO, and current year cost of goods sold would have been$23,507 less than under FIFO. Calculate the firm's ROI under each cost flow assumption (FIFO and LIFO). (Enter your answers as percentages rounded to 1 decimal place (e. 12.2%) ReferencBook& Resources Expliain how | 314 | 1,229 | {"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} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.940565 |
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GATE 2017 Previous Year Questions of Strength Of Materials Afternoon Slot Set 2 (in Marathi)
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In this lesson we will discuss GATE 2017 previous Year Questions of Strength Of Materials afternoon slot set 2 (Marathi)
RUTUJA Bothara
Strengthen GATE /PSU/ESE/SSC JE/RRB JE /campus Placements preparation. A passionate mechanical engineer -VIT University
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mam te green book ahe ki greak book ase new papper mdhe ala hot
Dr Preeti Raut
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1. GATE Previous Years Strength of Materials Questions GATE 2017 SOM solved questions set 2 By Rutuja Bothara
2. RUTUJA BOTHARA . Perusing B.Tech in Mechanical Engineering from Vellore Institute of Technology, Vellore. (2015-2019) . Placed in Alstom Transport India Ltd e Cleared JEE Mains 2015 e Cleared VITEEE-2015 with AIR 8718 e Cracked HITSEEE-2015 with AIR 91 . Worked as a Career Counsellor in Teachers Academy (2016- 2018) and BOLO app. https://unacademy.com/user/rurutujajb96-9511
3. WHEN YOU FEEL LIKE QUITTING... THINK ABOUT WHY YOU STARTED
4. Total Weightage 8 marks 4Q 1 mark 2Q 2 marks 4 ADD A FOOTER MM.DD.20XX
5. The state of stress at a p xyz50MPa. The maximum normal stress (in MPa) at that point is A. 49 B. 50 C. 55 D. 60
6. Maximu Noemal sthe is on moz = 2 0 50 n maae = 50 MPa
7. . A cantilever beam of length L and flexural moduluS El is subjected to a point load P at the free end. The elastic strain energy stored in the beam due to bending (neglecting transverse shear) PL3 6EI PL: 3EI P2L3 3EI PL 6EI A. B. C. D.
8. Bending Moment is gven by nergy is given by GET
9. . For a loaded cantilever beam of uniform crosss ection, the bending moment (in N.mm) along the length is M(x) 5X 2+10X, where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x=10 mm is . A. 100 B. 105 C. 110 D. 115
10. sheas force is siven by dae
11. Thetmal Stess is given by r'= 220 MPa
12. Engineeting Strain is True Sttain
13. According to maximum sheat strees theory Fos Fos
14. Thank You For watching!! RATE REVIEW RECOMMEN | 691 | 2,195 | {"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-2020-05 | latest | en | 0.809377 |
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Gij / D 0 with FW Rn ! R a polynomial or a rational function of the entries, hence continuous on some open set, their solutions cut out subgroups that are topologically closed. The closed subgroups are very special. 42))). Let G be a Lie group and let H be a closed subgroup of G. Then H has a unique smooth (in fact analytic) structure that makes it a Lie subgroup of G. 43)). Let G and H be two Lie groups with Lie algebra g and h, respectively, and let FW G ! H be a Lie group homomorphism. g/ 2 h.
G/. 3 Representation Theory Let H1 and H2 be two Hilbert spaces (the corresponding norms and scalar product are simply denoted by k k and h ; i). Suppose that AW H1 ! H1 ; H2 /. Recall that A is an isometry if kAuk D kuk for every u 2 H1 . Since kAuk2 D hAu; Aui D hA Au; ui and kuk2 D hu; ui, the polarization identity implies that A is an isometry if and only if A A D idH1 . Hence, isometries are injective, but they are not necessarily surjective. A bijective isometry is called a unitary map. If A is unitary, such is also A 1 and in this case AA D idH2 .
G/Ái D 0 for every g 2 G, contrary to assumption. Hence M D H and is irreducible. 48. 23) of the full affine group is. The calculations that follow are very basic and important. Rd / by Z fO . / D F f . a /; p F . a; b/f /. a /; F . a; b/f /. / D a 2 R ; b 2 R: We start with and show that it is not irreducible. R/. a /Og. a /. b/ˇ2 db da ; D a G jhF . a; b/f /; F gij2 34 F. De Mari and E. a . a /Og. /. a . /j2 d G Z ÂZ O R C1 0 da ; a à 2 da O jOg. R/ W fO . R/ W fO . 26), a < 0 for every a > 0. | 661 | 2,380 | {"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-2018-13 | longest | en | 0.907283 |
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May 20, 2013
# Homework Help: Math
Posted by Nini on Wednesday, May 16, 2012 at 8:16pm.
29166.66666
round up to the neareast whole number.
• Math - Ms. Sue, Wednesday, May 16, 2012 at 8:18pm
29166.66666 = 29,167
• Math - Nini, Wednesday, May 16, 2012 at 8:20pm
Thank you!
• Math - Ms. Sue, Wednesday, May 16, 2012 at 8:21pm
You're welcome.
• Math - Nini, Wednesday, May 16, 2012 at 8:23pm
Question??? is the rounding up to the whole number is rounding it up to the ten digits (-1)? thanks for your help in advance
• Math - Ms. Sue, Wednesday, May 16, 2012 at 8:30pm
The whole number is the one to the left of the decimal point. That's the one's place.
• Math - Nini, Wednesday, May 16, 2012 at 8:32pm
Thank you so much! now I understand...
• Math - Ms. Sue, Wednesday, May 16, 2012 at 8:46pm
You're very welcome. :-)
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View Full Version : Pipe flow formula
Danno30008
10-17-2004, 04:51 AM
Hey:
I need help!
What is the formula for figuring out gallons per hour of water flowing through a 1/2" and 3/4" pipe.
An example of what I need is:
1/2" pipe at 20PSI... How many gallons per hour flow through pipe, or what is flow rate?
1/2" pipe at 35PSI.....Same question.
3/4" Pipe at 20PSI and 3/4" pipe at 35PSI..
Any help would me most appreciated.
Take care,
Dan
hj
10-17-2004, 05:56 AM
The anwer bill be different for new steel pipe, old steel pipe, PEX, PVC, CPVC, type K copper, type L copper, or type M copper. The are charts that give the answer, but first you have to use the correct one.
jimbo
10-17-2004, 06:05 AM
http://www.tpub.com/content/construction/14259/
See table 7-11 in the above link, for copper pipe. An engineer could give the formula for this, but it's a LuLu! There are many variables. So, as a practical matter this data is usually compiled into tables such as these.
I don't know what project you might be working on, but it will be obvious from the tables that there is no single number for "20PSI ½" pipe". The length of the pipe affects the GPM in a big way due to pressure drop in the pipe as a function of distance. This link has some useful tables directly showing the loss (PSI) as a function of length:
www.*******************/pluminfo.html
Danno30008
10-17-2004, 06:18 AM
Thanks!
Here is the deal.
This is in a new restaurant and the connection goes to a Hatco C36 Booster Heater and then into a Hobart C44A Conveyor dishmachine.
They were suppost to have 3/4" copper into the booster heater and then 3/4" to the dishmachine. They put in 1/2" copper.
Here is my problem, the copper is in the wall and I have no way of knowing how long the run is, I can check with the builder.
My problem is with 1/2" pipe my rinse pressure varries from 35PSI to 20PSI. This is a big problem for me because I add a rinse/drying additive in the line and at 35PSI the glasses look GREAT, at 20PSI they have streaks all over them. This is because at 20PSI my rinse agent is much more concentrated, thus causing the streaks and runs. I need to explain with simple math as a back up the reason for the problem and then see if they want to fix it. It is a brand new high end restaurant in Atlanta and they will not tolerate the sub-par results. So......... I need to have my ducks in a row when I meet with them and explain the problem and offer a solution.
Thanks again for your help and take care,
Dan
daveydo
10-17-2004, 07:41 AM
Hello Danno,
I am not a plumber, I am a Mechanical Engineering Technologist with design experience in fluid mechanics.
I believe you need a practical solution to your problem and without knowing the number and type of fittings, the pipe length and other variables it is not possible to calculate a flowrate.
You can rent or buy a flow meter that will measure the flow at the outlet to the dishwasher. There are available flowmeters made in many ranges. The problem is what range are you in? Further, the test needs to be done at the low pressure figure 20psig. You'll have to be able verify 20psig before doing the test to make it valid.
Here's a simplier solution. If you can find a container such as a beaker or bucket that is measured in the units you wish use. say gallons, and are able to redirect the flow at the dishwasher into it. you can measure the flow rate.
Simply, direct the flow into the bucket for 1 minute. or 1/2 min or some other convient period. Then, measure the water you've collect in that period. for your connection, use the same id as that of the dishwasher line and as short as practically possible. 1/2"pipe, I believe you said? Attaching a long garden hose won't do!
The math is simple. If you have say, 2 gal in a 1 min test then the flow rate is 2 gals per min.
Or 3/4 of a gal in a 1/2 min the you have 1.5 gal per min.
The best part of this test is that it is accurate. It will give you the exact flowrate at the outlet to that dishwasher at that time. Again, you need to verify that the pressure at the time of the test is 20 psig.
You can then write a report simply stating how you did the test at what pressure and the results.
Also, call Hobart and ask for a product or design engineer. Explain your problem. Ask him what minimum supply pressure is required for that specific unit in order for it to rinse streak free. He will most likely be more than happy to help and may be able to give you a minimum flow rate requirement to compare against your test. or maybe another solution. After all, Hobart dishwashers are his business. Engineers always go to the manufacturers when dealing with an equipment problem. You should too.
I believe the proper solution is to change the line and I assume you are looking for ammunition to support this. Hope this has helped.
jimbo
10-17-2004, 08:46 AM
The post from Daveydo gets a little involved, but he did get to the core of the issue: You need to get from your equipment manufacturer the specs on minimum required water supply pressure and GPM. The reason your pressure seems to vary from 20 to 35 is probably that your supply line is inadequate, and when some other demand ( sink, toilet, etc) is placed on the system at the same time, it cannot keep up. The only real solution in this case is to repipe or provide an alternate supply. These things may not be as difficult as you think.
Bob's HandyGuy
10-17-2004, 02:40 PM
Why not experiment with a dilution of the rinse agent that will work at both pressures? The restaurant will save money and you will not have to re-plumb. Sounds win-win to me.
hj
10-17-2004, 07:36 PM
Since it appears that increasing the pipe size will be difficult, will your unit operate at 20 psi, if the additive volume is modulated? If so, install a pressure reducing valve at the DW set at 20 psi, so it maintains a constant pressure. | 1,499 | 5,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2014-15 | latest | en | 0.93161 |
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Find the points at which the function f given by has (i) local maxima (ii) local minima (ii) point of inflexion
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Solution
The given function is, f( x )= ( x−2 ) 4 ( x+1 ) 3 First derivative of f( x )is, f ′ ( x )= d( ( x−2 ) 4 ( x+1 ) 3 ) dx Use product rule to differentiate the function. f ′ ( x )=4 ( x−2 ) 3 ( x+1 ) 3 +3 ( x+1 ) 2 ( x−2 ) 4 = ( x−2 ) 3 ( x+1 ) 2 [ 4( x+1 )+3( x−2 ) ] = ( x−2 ) 3 ( x+1 ) 2 [ 7x−2 ] By substituting the value of f ′ ( x ) zero, we get f ′ ( x )=0 ( x−2 ) 3 ( x+1 ) 2 [ 7x−2 ]=0 Solve for roots of x , we get, x=2,−1, 2 7 At x=−1, Sign of f ′ ( x ) does not change for its nearest value so, this is point of inflexion. At x= 2 7 , Sign of f ′ ( x ) changes from positive to negative value for its nearest points so, it is maxima. At x=2, Sign of f ′ ( x ) changes from negative to positive value for its nearest points so, it is minima. Thus, point of local maxima is x= 2 7 , point of local minima is x=2 and point of inflexion is x=−1.
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Remember those equations that you were told were un-factorable or prime, and thus could not
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You cannot solve this equation unless you use either Completing the Square, or the Quadratic
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the quadratic formula is that, like all formulas, you just plug numbers into it, and then follow
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b b 2 4ac
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a 1 b 10 c 7
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The Academic Support Center at Daytona State College (Math 24 pg 1 of 2)
36
If the number under the radical is
not a perfect square, try to break it
down so that the number under
by Using the Quadratic Formula (continued)
x
b b 2 4ac
2a
Step 1: In order to use the quadratic formula, the equation must be in descending order, and set
to “0”.
Step 2: Label each of the coefficients using “a”, “b”, and “c”.
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a 3 b 5 c 1
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52 43 1
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5 25 12
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x
5 37
6
and
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The Academic Support Center at Daytona State College (Math 24 pg 2 of 2)
5 37
6
``` | 740 | 1,767 | {"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} | 4.84375 | 5 | CC-MAIN-2018-26 | latest | en | 0.81895 |
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A girl had designed a clap switch for a science exhibition that enabled her to switch on or off an alarm just with clapping of hands. While testing her device in a hall she noticed that once the alarm has sounded it followed with another one due to echo of the clap that the sound reflected by the walls. She recorded the two soundings of alarm with her tape-recorder and found out that the time difference in between them is 0.10 seconds. If the distance of the walls be 16 m, what is the maximum possible speed of the sound
A girl had designed a clap switch for a science exhibition that enabled her to switch on or off an alarm just
with clapping of hands. While testing her device in a hall she noticed that once the alarm has sounded it
followed with another one due to echo of the clap that the sound reflected by the walls. She recorded the two
soundings of alarm with her tape-recorder and found out that the time difference in between them is 0.10
seconds. If the distance of the walls be 16 m, what is the maximum possible speed of the sound
```
9 years ago
## Answers : (1)
Sudheesh Singanamalla
114 Points
``` Time of the sound to go hit the wall and then return back is 0.10 seconds.
so time taken by sound to just hit the wall is 0.10/2 = 0.05 seconds.
distance between the walls = 16m
we use the formula : speed = distance / time
which gives us the speed of the sound in air
speed = 16 / 0.05 = 1600/5 = 320 metres per second.
so the speed of sound in air is 320 metres per second.
```
9 years ago
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• Previous Year Exam Questions | 602 | 2,405 | {"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} | 3.546875 | 4 | CC-MAIN-2021-04 | latest | en | 0.941154 |
https://jarviscodinghub.com/product/assignment-4-inductive-proofs-solution/ | 1,723,399,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00731.warc.gz | 249,812,816 | 23,928 | # Assignment 4: Inductive Proofs solution
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## Description
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In this assignment, you will assemble a few automatically verifiable inductive proofs about
pure functional code. All your solutions should be defined within a file named foldfiles.pf.
File names are case sensitive.
You will need to obtain the Haskell source files for the ibis lightweight proof verifier. Once
this is done, you can verify a proof script using a Haskell interpreter by loading the Main
module into the interpreter and evaluating:
ibis “filepath/filename.pf”
where filepath/filename.pf is a relative path from the location of the source files. If you
choose to compile the verifier using the provided Makefile, you can run it from the command
line using the same syntax as above.
Problem 1. (50 pts)
Suppose you have a large collection of files spread across many computers. Each computer
has its own processor and can run its own instance of a Haskell interpreter. You need to
produce an aggregate analysis of all the files as defined by a Haskell function called agg. In
other words, for all the files across all the computers, you must compute
f1 ‘agg‘ f2 ‘agg‘ … ‘agg‘ fN
You do not know anything about the function’s implementation, but you do know that there
is a Haskell value id that acts as both a left and right identity for the function agg, and agg
itself is associative. All this information is summarized in the following proof script.
Introduce agg, id.
Assume \forall f. f ‘agg‘ id = f
Assume \forall f. id ‘agg‘ f = f
Assume \forall f,g,h. f ‘agg‘ (g ‘agg‘ h) = (f ‘agg‘ g) ‘agg‘ h
1
You decide that the best way to accomplish this task is to perform a foldr over all the files.
However, you plan to distribute the workload over all the machines and then combine the
results at the end. You observe that this will be equivalent to folding over all the files on a
single machine only if for any pair of file lists xs and ys, it is the case that
foldr agg id (xs ++ ys) = (foldr agg id xs) ‘agg‘ (foldr agg id ys)
Thus, it is necessary to prove that this equation holds.
(a) A Haskell definition for foldr is provided below.
foldr f b (x:xs) = f x (foldr f b xs)
foldr f b [] = b
Add appropriate assumptions to the proof script that mathematically represent the
equations that constitute the definition of foldr.
(b) A Haskell definition for the list “append” operator is provided below.
(x:xs) ++ ys = x:(xs ++ ys)
[] ++ ys = ys
Add appropriate assumptions to the proof script that mathematically represent the
equations that constitute this definition.
Now that all the assumptions describing the circumstances have been added to the proof
script, it is possible to construct a proof of the desired mathematical statement. Because
the statement is quantified universally over two lists, the proof must proceed by induction.
In order to receive full credit, your solutions must contain no new assumptions and only
automatically verifiable assertions.
(c) First, the statement must be shown to hold in the base case for an empty list. Add
to your proof script and complete the following proof of the base case by replacing
… with an appropriate sequence of true statements, separated by the conjunction
operator.
Assert \forall ys.
/\ (foldr agg id []) ‘agg‘ (foldr agg id ys) = foldr agg id ([] ++ ys)
(d) Next, it remains to prove that the statement holds for any recursive case. It is sufficient
to show that given an inductive hypothesis about some list xs, it is possible to derive
the same statement for a list x:xs. Complete the following proof of the inductive case.
Assert \forall x,xs,ys.
(foldr agg id xs) ‘agg‘ (foldr agg id ys) = foldr agg id (xs ++ ys)
=> …
/\ (foldr agg id (x:xs)) ‘agg‘ (foldr agg id ys) = foldr agg id ((x:xs) ++ ys)
2
Problem 2. (50 pts)
As before, solutions that can receive full credit must contain no new assumptions and only
automatically verifiable assertions.
(a) Suppose you also want to prove that [] is a right identity for ++. That is, for any xs,
xs ++ [] = xs
Add two assertions to your proof script that correspond to a proof of this statement.
The first should be the base case, in which xs is []. The second should be the inductive
case, in which you assume that the statement holds for some xs and derive that it holds
for some x:xs:
Assert \forall x,xs.
xs ++ [] = xs
=> …
/\ (x:xs) ++ [] = x:xs
(b) Suppose we want to show that ++ is associative. This means that for any lists xs, ys,
and zs, it is the case that
(xs ++ ys) ++ zs = xs ++ (ys ++ zs)
Any case in which ys = [] or zs = [] is trivial, since [] is both a left and right identity
of ++. Thus, it is sufficient to prove that this holds for any xs by induction over the
structure of xs. Assemble two assertions that correspond to the base and inductive
cases of a proof of this statement. You must use an approach that is analogous to the
two previous proofs. You may not introduce any additional assumptions.
3 | 1,231 | 5,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.883955 |
https://economicsconcepts.com/types_of_elasticity_of_demand.htm | 1,620,404,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00607.warc.gz | 242,280,802 | 8,243 | Home » Elasticity of Demand » Types of Elasticity of Demand
# Types of Elasticity of Demand:
The quantity of a commodity demanded per unit of time depends upon various factors such as the price of a commodity, the money income of the prices of related goods, the tastes of the people, etc., etc.
Whenever there is a change in any of the variables stated above, it brings about a change in the quantity of the commodity purchased over a specified period of time. The elasticity of demand measures the responsiveness of quantity demanded to a change in any one of the above factors by keeping other factors constant. When the relative responsiveness or sensitiveness of the quantity demanded is measured to changes, in its price, the elasticity is said be price elasticity of demand.
When the change in demand is the result of the given change in income, it is named as income elasticity of demand. Sometimes, a change in the price of one good causes a change in the demand for the other. The elasticity here is called cross electricity of demand. The three main types of elasticity of demand are now discussed in brief.
## (1) Price Elasticity of Demand:
### Definition and Explanation:
The concept of price elasticity of demand is commonly used in economic literature. Price elasticity of demand is the degree of responsiveness of quantity demanded of a good to a change in its price. Precisely, it is defined as:
"The ratio of proportionate change in the quantity demanded of a good caused by a given proportionate change in price".
### Formula:
The formula for measuring price elasticity of demand is:
Price Elasticity of Demand = Percentage in Quantity Demand
Percentage Change in Price
Ed = Δq X P
Δp Q
### Example:
Let us suppose that price of a good falls from \$10 per unit to \$9 per unit in a day. The decline in price causes the quantity of the good demanded to increase from 125 units to 150 units per day. The price elasticity using the simplified formula will be:
Ed = Δq X P
Δp Q
Δq = 150 - 125 = 25
Δp = 10 - 9 = 1
Original Quantity = 125
Original Price = 10
Ed = 25 / 1 x 10 / 125 = 2
The elasticity coefficient is greater than one. Therefore the demand for the good is elastic.
### Types:
The concept of price elasticity of demand can be used to divide the goods in to three groups.
(i) Elastic. When the percent change in quantity of a good is greater than the percent change in its price, the demand is said to be elastic. When elasticity of demand is greater than one, a fall in price increases the total revenue (expenditure) and a rise in price lowers the total revenue (expenditure).
(ii) Unitary Elasticity. When the percentage change in the quantity of a good demanded equals percentage in its price, the price elasticity of demand is said to have unitary elasticity. When elasticity of demand is equal to one or unitary, a rise or fall in price leaves total revenue unchanged.
(iii) Inelastic. When the percent change in quantity of a good demanded is less than the percentage change in its price, the demand is called inelastic. When elasticity of demand is inelastic or less than one, a fall in price decreases total revenue and a rise in its price increases total revenue.
## (2) Income Elasticity of Demand:
### Definition and Explanation:
Income is an important variable affecting the demand for a good. When there is a change in the level of income of a consumer, there is a change in the quantity demanded of a good, other factors remaining the same. The degree of change or responsiveness of quantity demanded of a good to a change in the income of a consumer is called income elasticity of demand. Income elasticity of demand can be defined as:
"The ratio of percentage change in the quantity of a good purchased, per unit of time to a percentage change in the income of a consumer".
### Formula:
The formula for measuring the income elasticity of demand is the percentage change in demand for a good divided by the percentage change in income. Putting this in symbol gives.
Ey = Percentage Change in Demand
Percentage Change in Income
Simplified formula:
Ey = Δq X P
Δp Q
### Example:
A simple example will show how income elasticity of demand can be calculated. Let us assume that the income of a person is \$4000 per month and he purchases six CD's per month. Let us assume that the monthly income of the consumer increase to \$6000 and the quantity demanded of CD's per month rises to eight. The elasticity of demand for CD's will be calculated as under:
Δq = 8 - 6 = 2
Δp = \$6000 - \$4000 = \$2000
Original quantity demanded = 6
Original income = \$4000
Ey = Δq / Δp x P / Q = 2 / 200 x 4000 / 6 = 0.66
The income elasticity is 0.66 which is less than one.
### Types:
When the income of a person increases, his demand for goods also changes depending upon whether the good is a normal good or an inferior good. For normal goods, the value of elasticity is greater than zero but less than one. Goods with an income elasticity of less than 1 are called inferior goods. For example, people buy more food as their income rises but the % increase in its demand is less than the % increase in income.
## (3) Cross Elasticity of Demand:
### Definition and Explanation:
The concept of cross elasticity of demand is used for measuring the responsiveness of quantity demanded of a good to changes in the price of related goods. Cross elasticity of demand is defined as:
"The percentage change in the demand of one good as a result of the percentage change in the price of another good".
## Formula:
The formula for measuring, cross, elasticity of demand is:
Exy = % Change in Quantity Demanded of Good X
% Change in Price of Good Y
The numerical value of cross elasticity depends on whether the two goods in question are substitutes, complements or unrelated.
### Types and Example:
(i) Substitute Goods. When two goods are substitute of each other, such as coke and Pepsi, an increase in the price of one good will lead to an increase in demand for the other good. The numerical value of goods is positive.
For example there are two goods. Coke and Pepsi which are close substitutes. If there is increase in the price of Pepsi called good y by 10% and it increases the demand for Coke called good X by 5%, the cross elasticity of demand would be:
Exy = %Δqx / %Δpy = 0.2
Since Exy is positive (E > 0), therefore, Coke and Pepsi are close substitutes.
(ii) Complementary Goods. However, in case of complementary goods such as car and petrol, cricket bat and ball, a rise in the price of one good say cricket bat by 7% will bring a fall in the demand for the balls (say by 6%). The cross elasticity of demand which are complementary to each other is, therefore, 6% / 7% = 0.85 (negative).
(iii) Unrelated Goods. The two goods which a re unrelated to each other, say apples and pens, if the price of apple rises in the market, it is unlikely to result in a change in quantity demanded of pens. The elasticity is zero of unrelated goods.
## Relevant Articles:
» Meaning of Price Elasticity of Demand » Degrees of Elasticity of Demand » Measurement of Price Elasticity of Demand » Types of Elasticity of Demand » Factors Determining Price Elasticity of Demand » Importance of Elasticity of Demand
Principles and Theories of Micro Economics Definition and Explanation of Economics Theory of Consumer Behavior Indifference Curve Analysis of Consumer's Equilibrium Theory of Demand Theory of Supply Elasticity of Demand Elasticity of Supply Equilibrium of Demand and Supply Economic Resources Scale of Production Laws of Returns Production Function Cost Analysis Various Revenue Concepts Price and output Determination Under Perfect Competition Price and Output Determination Under Monopoly Price and Output Determination Under Monopolistic/Imperfect Competition Theory of Factor Pricing OR Theory of Distribution Rent Wages Interest Profits
Principles and Theories of Macro Economics National Income and Its Measurement Principles of Public Finance Public Revenue and Taxation National Debt and Income Determination Fiscal Policy Determinants of the Level of National Income and Employment Determination of National Income Theories of Employment Theory of International Trade Balance of Payments Commercial Policy
Development and Planning Economics Introduction to Development Economics Features of Developing Countries Economic Development and Economic Growth Theories of Under Development Theories of Economic Growth Agriculture and Economic Development Monetary Economics and Public Finance History of Money | 1,812 | 8,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-21 | latest | en | 0.923822 |
http://www.jiskha.com/display.cgi?id=1287031237 | 1,462,459,744,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127496.74/warc/CC-MAIN-20160428161527-00120-ip-10-239-7-51.ec2.internal.warc.gz | 611,393,090 | 3,816 | Thursday
May 5, 2016
# Homework Help: math
Posted by dreww1961 on Thursday, October 14, 2010 at 12:40am.
Let f:A->B, where A and B are nonempty, and let T1 and T2 be subsets of B.
a.Prove that f^-1(T1 U T2)= f^-1(T1) U f^-1(T2).
b.Prove that f^-1(T1 intersects T2) = f^-1(T1) intersects f^-1(T2). I think once I see a I can do b.
c. Prove that f^-1(T1) - f^-1(T2) = f^-1(T1-T2).
d. Prove that if T1 is a proper subset of T2, then f^-1(T1) is a proper subset of f^-1(T2).
I am confused on this one and need some help understanding inverse mapping. Thank you | 207 | 561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2016-18 | longest | en | 0.851562 |
https://quant.stackexchange.com/questions/43883/how-to-determine-the-risk-neutral-measure-in-a-heston-model | 1,721,925,446,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00580.warc.gz | 408,751,435 | 41,014 | # How to determine the risk-neutral measure in a Heston model?
To clarify, I'm quite familiar with the risk-neutral pricing framework, and I know one can efficiently Monte-Carlo a Heston model via the non-central $$\chi^2$$ distribution approach. But so far we're only playing with the real world probabilities, and we can never determine the risk-neutral measure because Heston model is incomplete. So even if we can Monte-Carlo the stock price paths under the real world probabilities, what then? We still cannot decide on the risk-neutral measure.
I also have read somewhere about using variance/vol swaps to make the market complete again, but haven't seen a good explanation (or at least the rough scheme/intuition etc) on how to use var/vol swaps to determine the risk-neutral measure.
Could anybody help? Thanks!
• The wiki page on the Heston model answers your question pretty well. Commented Feb 5, 2019 at 11:24
• @Raskolnikov thanks. So it says the main idea is to use the var/vol swap to "calibrate" the risk-neutral measure. Would you care to provide any more concrete exposition on this method?
– Vim
Commented Feb 5, 2019 at 14:37
• I have no experience with that particular method. The method I used was basically calibrating with vanilla options. Technically, you only need two financial instruments, the underlying asset and an option on the underlying or the underlying and a vol swap. But in practice, since markets are imperfect, people tend to use the complete vanilla option surface (cleaned up for outliers and such). Calibrating in that case amounts to a closest fit (through least squares for instance) to the vanilla option prices. Commented Feb 5, 2019 at 15:13
• If however you only have the price of the asset, then several risk-neutral measures are compatible with it. Then, you can select one based on extra conditions you impose for theoretical reasons. Commented Feb 5, 2019 at 15:13
• @Raskolnikov if my interpretation is correct, in practice, under the Heston model, we still assume the drift of the stock price is $(r-q)dt$ under the risk neutral measure; then what we calibrate are the params of the variance process under the risk neutral measure. But how can we just require the risk neutral price process still has the same drift as in the classical BS model?
– Vim
Commented Feb 6, 2019 at 6:22
First you assume that the Heston model under the risk neutral measure takes the following form:
$$dv = \kappa (\theta - v) dt + \eta \sqrt{v} dW$$
Then you calibrate the model to the available options quoted in the market, i.e. find values for the Heston parameters $$(\kappa, \theta, \eta)$$ such that the options prices generated by the Heston model gives a good fit to the market.
Generating vanilla options prices with the Heston model can be done by MC, but for calibration purposes (and for vanilla options in general) there are semi-analytical methods which will greatly speed up your calibration procedure. Google for example "Heston" + "Fourier" + "Characteristic function".
• Thanks. But my question is how to determine the risk neutral measure?
– Vim
Commented Feb 4, 2019 at 23:35
• I've just explained above how to determine the risk neutral measure.
– user34971
Commented Feb 5, 2019 at 16:11
• Sorry I read it as calibrations done under the real world probabilities. Could you be a bit more specific about where and how you get the risk neutral measure?
– Vim
Commented Feb 5, 2019 at 16:34 | 830 | 3,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.918624 |
https://www.sciences360.com/index.php/alternating-current-ac-and-direct-current-dc-a-simple-explanation-4-25943/ | 1,642,553,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301217.83/warc/CC-MAIN-20220119003144-20220119033144-00513.warc.gz | 1,051,263,120 | 6,068 | Physics
# Alternating Current AC and Direct Current DC a Simple Explanation
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The fundamental difference between Alternating Current (AC) and Direct Current (DC) is this: DC is a steady and relatively unchanging flow of current through a circuit while AC is a dynamic, that is: a continuously changing flow of current through a circuit.
Breaking this down a little further, DC can be likened to water flowing through a pipe from point A to Point B - we call this our circuit. A pump is required to physically move the water and in electrical terms, our pump might be a battery, charged capacitor or DC generator. Our water will continue flowing in one direction at a set rate determined by the power of our pump, which equates to the voltage of our battery, generator or capacitor. The direction of current flow is what determines polarity', which can be either positive or negative to keep things simple, let's just say our water will flow in whatever direction we set our pump motor to turn. Our current flow will stop if we open the circuit or turn off the pump - disconnect our power source. (Assume for this explaination that if we break the loop to open the circuit, we are sealing the break so water remains inside the pipes)
If we apply the same visualisation to AC, picture a lever that pushes a diaphragm fixed inside a water pipe backwards and forwards. The pipe is full of water throughout the entire sealed loop and the water cannot get past the diaphragm, but our diaphragm is fairly flexible as it stretches across the internal diameter of our pipe. As the lever is rapidly pumped from left to right to left and so on, the diaphragm pushes the water in each respective direction at a rate of however fast we are moving the lever. The water inside our pipe is therefore moving first from left to right a certain distance, stops before it reverses direction as the diaphragm flexes back in the opposite direction. This rate of change is what we call frequency.
So in summary, DC is a fixed rate of current flowing through a circuit, and AC is a constantly changing flow of current through a circuit. Both have very specific purposes in making things work, and have very different behaviour characteristics in a given electric circuit.
Tweet | 486 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-05 | longest | en | 0.909518 |
https://questions.examside.com/past-years/medical/question/during-the-electrolysis-of-molten-sodium-chloride-the-time-neet-chemistry-electrochemistry-fle1eny5ekiwo4w0 | 1,638,495,573,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00424.warc.gz | 528,991,308 | 11,649 | 1
### NEET 2016 Phase 2
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
A
55 minutes
B
110 minutes
C
220 minutes
D
330 minutes
## Explanation
At cathode : 2Na+ + 2e $\to$ 2Na
At anode : 2Cl $\to$ Cl2 + 2e
----------------------------------------------
Net reaction: 2Na+ + 2Cl $\to$ 2Na + Cl2
From Faraday’s first law of electrolysis,
w = Z$\times$I$\times$t
= ${E \over {96500}}$$\times$I$\times$t
No. of moles of Cl2 gas × Mol. wt. of Cl2 gas
= ${{Eq.\,wt.\,of\,C{l_2}\,gas \times I \times t} \over {96500}}$
$\Rightarrow$ 0.10 $\times$ 71 = ${{35.5 \times 3 \times t} \over {96500}}$
$\Rightarrow$ t = ${{0.10 \times 71 \times 96500} \over {35.5 \times 3}}$
= 6433.33 sec
= 107.22 min $\simeq$ 110 min
2
### NEET 2016 Phase 2
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 $\times$ 10$-$19C)
A
6 $\times$ 1023
B
6 $\times$ 1020
C
3.75 $\times$ 1020
D
7.48 $\times$ 1023
## Explanation
Q = I × t
Q = 1 × 60 = 60 C
Now, 1.60 × 10–19 C $\equiv$ 1 electron
$\therefore$ 60 C $\equiv$ ${{60} \over {1.6 \times {{10}^{ - 19}}}}$
= 3.75 $\times$ 1020 electrons
3
### NEET 2016 Phase 2
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
A
zinc is lighter than iron
B
zinc has lower melting point than iron
C
zinc has lower negative electrode potential than iron
D
zinc has higher negative electrode potential than iron
## Explanation
The reduction potential values are
Zn2+/Zn = – 0.76 V
Fe2+/Fe = – 0.44 V
Thus, due to higher negative electro potential value of zinc than iron, iron cannot be coated on zinc.
4
### NEET 2016 Phase 2
The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductivity of 5.76 $\times$ 10$-$3 S cm$-$1 at 298 K is
A
2.88 S cm2/mol
B
11.52 S cm2/mol
C
0.086 S cm2/mol
D
28.8 S cm2/mol
## Explanation
${\Lambda _m} = {{\kappa \times 1000} \over {Molality}}$
= ${{5.76 \times {{10}^{ - 3}} \times 1000} \over {0.5}}$
= 11.52 S cm2/mol | 793 | 2,146 | {"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} | 3.40625 | 3 | CC-MAIN-2021-49 | latest | en | 0.682509 |
https://godotengine.org/qa/63322/implementing-a-compass | 1,657,210,767,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00394.warc.gz | 313,896,599 | 6,982 | Hello community, I am once again asking for your implementation aid :)
I am currently trying to implement a compass for VR, it should always face the forward direction. I somewhat reached this using the `look_at()` function. My problem is that when pitching the compass the mesh and the texture showing the cardinal directions will collide in one another resulting in a look like this:
https://imgur.com/a/LWKdcsJ
Anyone got a quick idea on how i could avoid this behaviour? I know it is kind of hard since you will it actually does what it should do, as it always points towards the north (forward) direction.
in Engine
Why are the mesh and the texture two different objects? If you have to keep them separate for some reason, then instead of rotating the mesh, couldn't you rotate a parent common to both the texture and the mesh?
I don't really understand what you want to achieve. If you want to avoid clipping, but still have wind rose flat, you can, for example, make it render on top of that mesh. You could also make that mesh slightly transparent, or just put that wind rose higher.
This help me: using A to B compass angle. If B is north of A the angle is 0 radians:
``````public float getDirection(Vector3 vA, Vector3 vB){
//only plane XZ
vA = new Vector3(vA.x, 0, vA.z);
vB = new Vector3(vB.x, 0, vB.z);
Vector3 dir = vB-vA;
float angle = Vector3.Forward.AngleTo(dir); //angle with "nort"
//clockwise
if (dir.x<0){
angle = (2*Mathf.Pi) - angle;
}
return angle;
}
``````
In python same logic.
by (32 points) | 383 | 1,530 | {"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-2022-27 | latest | en | 0.921605 |
http://oeis.org/A135647 | 1,571,679,782,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987781397.63/warc/CC-MAIN-20191021171509-20191021195009-00351.warc.gz | 147,354,575 | 4,244 | This site is supported by donations to The OEIS Foundation.
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A135647 G.f. = 1/(x^36*p(1/x)) where p(x)=(- 25 - 49 x^9 + x^10)*(- 1 - 2 x^9 + x^10)^3*(- 1 - x^9 + x^10)^6. 0
1, 61, 3070, 150836, 7392650, 362245994, 17750074048, 869753690956, 42617931038803, 2088278621406591, 102325652450274784, 5013956970066973919, 245683891533290673468, 12038510685131268747080, 589887023571432406862284 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Weighted solution of a zero sum game. Let Ma={{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {25, 0, 0, 0, 0, 0, 0, 0, 0, 49}}; a={1,2}; ML={{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {25, 0, 0, 0, 0, 0, 0, 0, 0, 49}}. Such that: 6*Game_value[M1]+3*Game_value[M2]+Game_Value[ML]=0 My first solution was "unweighted". LINKS FORMULA p(x)=(-25 - 49 x^9 + x^10)(-1 - 2 x^9 + x^10)^3(-1 - x^9 + x^10)^6; f(x)=1/(x^36*p(1/x)) a(n) =expansion(f(x)) MATHEMATICA f[x_] = Product[CharacteristicPolynomial[{{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, a}}, x]^(6/a), {a, 1, 2}]*CharacteristicPolynomial[{{0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {25, 0, 0, 0, 0, 0, 0, 0, 0, 49}}, x]; g[x_] = Expand[x^100*f[1/x]]; a = Table[ SeriesCoefficient[Series[1/g[x], {x, 0, 30}], n], {n, 0, 30}] CROSSREFS Sequence in context: A000508 A191092 A234028 * A269025 A207231 A207224 Adjacent sequences: A135644 A135645 A135646 * A135648 A135649 A135650 KEYWORD nonn,uned,obsc AUTHOR Roger L. Bagula, Jan 31 2008 EXTENSIONS The connection with the zero-sum game is not clear to me. Also, how does Ma depend on a? It appears that Ma = ML, so perhaps there are errors in these matrices? - N. J. A. Sloane, May 16 2008 STATUS approved
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Last modified October 21 13:24 EDT 2019. Contains 328299 sequences. (Running on oeis4.) | 1,835 | 3,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-43 | latest | en | 0.449917 |
http://www.sitepoint.com/forums/showthread.php?1164073-How-Can-i-Make-a-Conditon-to-get-two-diffrent-Percentage&p=5556193&viewfull=1 | 1,409,412,449,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500835488.52/warc/CC-MAIN-20140820021355-00231-ip-10-180-136-8.ec2.internal.warc.gz | 569,937,704 | 12,461 | # Thread: How Can i Make a Conditon to get two diffrent Percentage
1. ## How Can i Make a Conditon to get two diffrent Percentage
i have examination result script now i am in trouble plz solve my problem "i have one database and full examination result 9th & 10th Class so 9th total marks 525 and 10th total marks 1050, now when i want to get % of 9th class its divide with 525 and when i want to get 10th class i must divide it to 1050, 9th class roll number start from 1 to 6702 and 7101 to 7152 and 20001 to 103090 rest of roll number are 10th class roll number " So plz tell me how can i make the if condition logic to get accurate % of my result or any other option or solution plz tell me regards
2. Try to rephrase your problem. This makes no sense.
3. Not sure if this helps but this is what I get from your question.
PHP Code:
``` <?php \$testnumbers = array("5240","7000","7142","11000","21050"); //Test Single Number //\$roll_number = "5240"; \$ninth_roll_numbers = array(); foreach(range(1,6702) as \$num){ \$ninth_roll_numbers[] = \$num; } foreach(range(7101,7152) as \$num){ \$ninth_roll_numbers[] = \$num; } foreach(range(20001,103090) as \$num){ \$ninth_roll_numbers[] = \$num; } //run a little test with array \$testnumbers foreach (\$testnumbers as \$roll_number){ \$total_marks = (in_array(\$roll_number,\$ninth_roll_numbers) ? 525 : 1050); echo "Total Marks: \$total_marks<br />"; \$percent = \$roll_number/\$total_marks; echo "Result: \$percent<br /><br />"; } ?> ```
This comes out as
Total Marks: 525
Result: 9.9809523809524
Total Marks: 1050
Result: 6.6666666666667
Total Marks: 525
Result: 13.60380952381
Total Marks: 1050
Result: 10.47619047619
Total Marks: 525
Result: 40.095238095238
Could also do it with a bunch of IF greater than && less than statements but I like working with arrays.
PHP Code:
``` echo "TEST Two<br />"; \$testnumbers = array("5240","7000","7142","11000","21050"); //run a little test with array \$testnumbers foreach (\$testnumbers as \$roll_number){ if ( (\$roll_number>=1 && \$roll_number<=6702) || (\$roll_number>=7101 && \$roll_number<=7152) || (\$roll_number>=20001 && \$roll_number<=103090) ){ \$total_marks = 525; }else{ \$total_marks = 1050; } echo "Total Marks: \$total_marks<br />"; \$percent = \$roll_number/\$total_marks; echo "Result: \$percent<br /><br />"; } ```
Results are the same.
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• | 850 | 2,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2014-35 | latest | en | 0.567502 |
http://www.postgresql-archive.org/numeric-field-overflow-td1924572.html | 1,508,699,706,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00295.warc.gz | 529,292,284 | 11,607 | # numeric field overflow
6 messages
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## numeric field overflow
Good day. I may be misunderstanding the NUMERIC type description in the manual, so can anyone please help me with this? Description says: "The scale of a numeric is the count of decimal digits in the fractional part, to the right of the decimal point. The precision of a numeric is the total count of significant digits in the whole number, that is, the number of digits to both sides of the decimal point. ... Integers can be considered to have a scale of zero. " However, I am not able to insert numbers that have number of digits equal to the precision and the scale equal to 0. F.E.: CREATE TABLE test.test ( rate numeric(5,1) ) INSERT INTO test.test VALUES (10000) Generates an error: ERROR: numeric field overflow SQL state: 22003 Detail: A field with precision 5, scale 1 must round to an absolute value less than 10^4. So, does the precision part of the numeric type really means number of digits to the left of the decimal point, or what ? Thanks. -- Regards, Nurzhan Kirbassov. -- Sent via pgsql-general mailing list ([hidden email]) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general
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## Re: numeric field overflow
On Oct 5, 2009, at 11:20 PM, Nurzhan Kirbassov wrote: > So, does the precision part of the numeric type really means number of > digits to the left of the decimal point, or what ? NUMERIC is behaving as documented. The way to think of it is when you are inserting: INSERT INTO test.test VALUES (10000) into a NUMERIC(5,1), what you are doing is inserting: INSERT INTO test.test VALUES (10000.0) 10000.0 has six significant digits, rather than five, so the insert fails. -- -- Christophe Pettus [hidden email] -- Sent via pgsql-general mailing list ([hidden email]) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general
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## Re: numeric field overflow
In reply to this post by Nurzhan Kirbassov On Tue, Oct 6, 2009 at 12:20 AM, Nurzhan Kirbassov <[hidden email]> wrote: > Good day. > > I may be misunderstanding the NUMERIC type description in the manual, > so can anyone please help me with this? Description says: > > "The scale of a numeric is the count of decimal digits in the > fractional part, to the right of the decimal point. The precision of a > numeric is the total count of significant digits in the whole number, > that is, the number of digits to both sides of the decimal point. ... > Integers can be considered to have a scale of zero. " > > However, I am not able to insert numbers that have number of digits > equal to the precision and the scale equal to 0. > > F.E.: > > CREATE TABLE test.test > ( > rate numeric(5,1) > ) This declares a numeric of 5 digits, with 1 to the right of the decimal point. > > INSERT INTO test.test VALUES (10000) > > Generates an error: Like it should. however this: CREATE TABLE test.test ( rate numeric(5,0)); INSERT INTO test.test VALUES (10000); INSERT 0 1 Works just fine. -- Sent via pgsql-general mailing list ([hidden email]) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general
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## Re: numeric field overflow
In reply to this post by Nurzhan Kirbassov > However, I am not able to insert numbers that have number of digits > equal to the precision and the scale equal to 0. Scale applies to the *column*. You defined your column to have a scale of 1, so you can't just claim that your value has scale 0 and claim an extra digit to the left of the decimal point. -- Scott Ribe [hidden email] http://www.killerbytes.com/(303) 722-0567 voice -- Sent via pgsql-general mailing list ([hidden email]) To make changes to your subscription: http://www.postgresql.org/mailpref/pgsql-general | 957 | 3,903 | {"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-2017-43 | latest | en | 0.770955 |
http://mathhelpforum.com/discrete-math/164548-logic-formulas-proof-print.html | 1,526,979,135,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864648.30/warc/CC-MAIN-20180522073245-20180522093245-00043.warc.gz | 181,204,382 | 3,100 | # Logic: formulas, proof
• Nov 27th 2010, 12:28 PM
somebd
Logic: formulas, proof
Hello!
I have just started learning logic and have to idea how to solve this, can anyone help with ideas please?
Prove that a formula that contains only variables and equivalence and negation as operations is identically true if and only if the number each variable and operation of negation appears in the formula is even.
• Nov 28th 2010, 08:31 AM
emakarov
Interesting problem. An obvious idea is to try proving this by induction on formulas. However, as often happens, one needs a stronger induction statement because in a formula A <-> B, A and B separately don't have to be tautologies or contain even number of occurrences of each variable even if the whole formula does this. So the challenge is to come up with a generalization of the claim that will let induction go through.
I'll denote truth values by T and F. Let is call a formula A(p) containing a variable p stable with respect to p if for all values of other variables, A(T) and A(F) have the same truth values, i.e., A(T) <-> A(F) is a tautology. Similarly, A(p) is called unstable w.r.t. p if for all values of other variables, A(T) and A(F) have the opposite truth values, i.e., A(T) <-> A(F) is a contradiction. Note that if A is not stable w.r.t. p, it does not necessarily follow that it is unstable w.r.t. p.
Lemma 1. Let a formula A(p) be built from variables, truth values, equivalences and negations only. Then A is stable w.r.t. p if the number of occurrences of p is even, and A is unstable w.r.t. p if the number of occurrences of p is odd.
Proved by induction on A.
Corollary. If a formula as above contains even number of occurrences of every variable, then it is either a tautology or a contradiction.
Lemma 2. A formula built from T, equivalences and negations only is true iff it contains an even number of negations.
Proved by induction.
From this the claim follows.
• Nov 29th 2010, 09:25 AM
somebd
Thank you! :) I appreciate it. | 506 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-22 | latest | en | 0.925807 |
https://math.stackexchange.com/questions/4780037/left-invertible-non-square-matrices-and-their-sub-matrices-over-a-commutative-ri | 1,701,765,275,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00407.warc.gz | 462,596,162 | 38,644 | # Left invertible non-square matrices and their sub-matrices over a commutative ring
Let $$A= \begin{pmatrix} 1 & 1 \\ 2 & 1 \\ 1 & 4 \end{pmatrix}$$
The matrix $$A$$ has rank two (=number of independent columns), so it is full, hence has a left inverse which is $$A_L= \begin{pmatrix} -1 & 1 & 0\\ 2 & -1 & 0\\ \end{pmatrix}$$ (It is easy to check that $$A_LA=I_{2 \times 2}$$).
$$A$$ has three invertible square sub-matrices: $$A_{1,2}= \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$$ $$A_{1,3}= \begin{pmatrix} 1 & 1 \\ 1 & 4 \end{pmatrix}$$ $$A_{2,3}= \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix}$$. Each of the three sub-matrices is invertible, but generally it may happen that there are non-invertible sub-matrices, as the following example shows: Let $$B= \begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$B_{1,2}= \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$$ $$B_{1,3}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$B_{2,3}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$. Here $$B_{1,3}$$ is not invertible.
Claim 1: Let $$k$$ be a field of characteristic zero and let $$A$$ be an $$m \times n$$ matrix over $$k$$, $$m > n$$. Assume that $$A$$ is left invertible (hence all $$m$$ columns of $$A$$ are $$k$$-linearly independent). Then there exists an $$n \times n$$ sub-matrix of $$A$$ which is invertible (equivalently, has an invertible determinant = a nonzero determinant, since we work over a field $$k$$).
Question 1: Is Claim 1 true? It seems true and follows from the assumption that $$A$$ has rank $$m$$.
Claim 2: Let $$R$$ be a commutative ring and let $$A$$ be an $$m \times n$$ matrix over $$k$$, $$m > n$$. Assume that $$A$$ is left invertible (hence all $$m$$ columns of $$A$$ are $$R$$-linearly independent, so they generate a free $$R$$-module of rank $$m$$). Then there exists an $$n \times n$$ sub-matrix of $$A$$ which is invertible (equivalently, has an invertible determinant = the determinant is an invertible element of $$R$$, see this).
Question 2: Is Claim 2 true? It seems true if Claim 1 is true or am I missing something? Maybe the result of left invertibility is not equivalent to having full column rank $$m$$? (I think this result stull holds. see this).
Thank you very much! I apologize if my question is trivial.
Edit: Maybe all we can say is that there is square $$n \times n$$ sub-matrix $$S$$ of $$A$$ with non-zero determinant $$d_S \in R$$, but this does not imply that $$S$$ itself is an invertible matrix, since $$d_s$$ may not be invertible in $$R$$. The following is not a counterexample: $$A= \begin{pmatrix} t \\ t \end{pmatrix}$$. $$A$$ has rank one, $$S= \begin{pmatrix} t \end{pmatrix}$$ is not invertible, but $$A$$ itselt is not left invertible!
• The size of the largest non-singular matrix is known as the determinantal rank of the matrix. It is known, in the field case, that it is the same as the row/column rank. I don't know about commutative rings though. Oct 3 at 12:15
• @user376343. thank you! I have changed the $1$ in the second row to $-1$. Oct 3 at 12:16
• @TheoBendit, thank you very much! You can write your comment as a partial answer. I think/hope that the same result holds over commutative rings, but I have not carefully checked all the details yet. Oct 3 at 12:17
• @TheoBendit, maybe there would be a problem if we work over a commutative ring which is not a field...Instead of nonzero minors we should consider invertible minors. I will try to find a counterexample over $R=\mathbb{C}[t]$. Oct 3 at 12:32
Since $$A^T$$ has a right inverse it is surjective and the proper claim is its $$n\times n$$ minors generate the unit ideal. [In a PID like $$\mathbb Z$$ you can also talk about a gcd of $$1$$.] With $$R$$ denoting the commutative ring, define $$B:=A^T$$ and we know $$I_n = BC$$ for some $$C\in R^{m\times n}$$.
$$1 = \det\big(I_n\big) = \det\big(BC\big) = \sum_{S} \det\big(B_{[n],S}\big)\det\big(C_{S,[n]}\big) = \sum_{S} \det\big(B_{[n],S}\big)\cdot \alpha_k$$
by Cauchy-Binet [a polynomial identity that holds over $$\mathbb C$$ so it holds over every commutative ring. Ref e.g. https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula for notation clarifications though the OP has flipped the roles of $$m$$ and $$n$$.] You can of course transpose this at the end to talk about minors of $$A$$.
The following is a counterexample to Claim 2: $$M= \begin{pmatrix} t+1 \\ t \end{pmatrix}$$. $$M$$ is left invertible: $$M_L= \begin{pmatrix} 1 & -1 \end{pmatrix}$$. Each of the two sub-matrices is not invertible: $$M_{1,1}= \begin{pmatrix} t+1 \end{pmatrix}$$, $$M_{2,1}= \begin{pmatrix} t \end{pmatrix}$$. | 1,512 | 4,619 | {"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": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | longest | en | 0.740144 |
https://convert-dates.com/days-from/3458/2024/06/12 | 1,718,932,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00346.warc.gz | 156,227,742 | 4,297 | ## 3458 Days From June 12, 2024
Want to figure out the date that is exactly three thousand four hundred fifty eight days from Jun 12, 2024 without counting?
Your starting date is June 12, 2024 so that means that 3458 days later would be November 30, 2033.
You can check this by using the date difference calculator to measure the number of days from Jun 12, 2024 to Nov 30, 2033.
November 2033
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
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4. 30
November 30, 2033 is a Wednesday. It is the 334th day of the year, and in the 48th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2033 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 11/30/2033, and almost everywhere else in the world it's 30/11/2033.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 3458 weekdays from Jun 12, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Jun 12, 2024, which falls on a Wednesday. Counting forward, the next day would be a Thursday.
To get exactly three thousand four hundred fifty eight weekdays from Jun 12, 2024, you actually need to count 4842 total days (including weekend days). That means that 3458 weekdays from Jun 12, 2024 would be September 14, 2037.
If you're counting business days, don't forget to adjust this date for any holidays.
September 2037
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
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1. 1
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4. 30
September 14, 2037 is a Monday. It is the 257th day of the year, and in the 257th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 30 days in this month. 2037 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 09/14/2037, and almost everywhere else in the world it's 14/09/2037.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 931 | 2,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-26 | latest | en | 0.923655 |
https://web2.0calc.com/questions/problem-help-asap | 1,540,345,813,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00470.warc.gz | 829,945,757 | 5,354 | +0
# problem help asap
0
221
1
how do you solve this equation: 10^.37=x/.0128-x
Guest Apr 11, 2017
#1
0
You can't.
But this equation can be solved
10^.37
--------- =x
.0128-x
It will end up being:
91.57143809843445703125...
Guest Apr 11, 2017 | 97 | 254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-43 | latest | en | 0.709884 |
https://dev.to/ashwinsharmap/what-is-gradient-descent-algorithm-13le | 1,718,722,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00661.warc.gz | 182,905,348 | 19,912 | ## DEV Community
Ashwin Sharma P
Posted on • Originally published at ashwinsharmap.hashnode.dev
# What is Gradient descent algorithm?
The gradient descent algorithm is an approach to find the minimum point or optimal solution for a given dataset. It follows the steepest descent approach. That is it moves in the negative gradient direction to find the local or global minima, starting out from a random point. We use gradient descent to reach the lowest point of the cost function.
In Machine Learning, it is used to update the coefficients of our model. It also ensures that the predicted output value is close to the expected output
For example, in Linear Regression, where we separate the output using a linear equation.
Let us say we have an equation y=mx+c.
Here, m stands for slope and c for the y-intercept. These two values can be optimized and the error in the cost function (the difference between expected and predicted output) can be reduced using the gradient descent algorithm.
### Weight updation
So let us see how weight updation works in gradient descent. Let us consider a graph of cost function vs weight. For improving our model, bringing down the value of cost function is essential. We consider the lowest point in the graph as the winner since the cost function would be minimal at this point.
In the above diagram, we can see that with each iteration, the function tries to bring down the cost value. But that is not the case in real-world datasets. In real-world cases, it moves in a zig-zag manner for most of the datasets. The graph for real-world cases is as shown below.
### What is the maths behind it?
The weight updation takes place by decrementing the cost function in steps of the gradient (derivative) of weight function. The equation used for weight updation is:
ωnew = ωold -η* (δLOLD )
Here ω corresponds the weight and η the learning rate, which determines by what value the descent is made in each iteration.
For every cross mark shown in the graph, we calculate the slope. According to the slope value, we update the weights in the above equation.
### Conclusion
Here is a sequence diagram to brief out the process of updating the weights. This update in weights leads to a reduction in the cost function. | 473 | 2,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.889647 |
http://patricktalkstech.com/standard-error/calculate-standard-error-excel-2003.html | 1,500,777,736,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424239.83/warc/CC-MAIN-20170723022719-20170723042719-00539.warc.gz | 244,205,416 | 5,074 | Home > Standard Error > Calculate Standard Error Excel 2003
# Calculate Standard Error Excel 2003
## Contents
The formula for this, as shown in the formula bar, is:=STDEV( B3:B1002, D3:D1002, F3:F1002 )As shown in cell H3, the standard deviation in the individual heights of the sample group is Please try again later. Now click on the fx symbol again. Choose “Statistical” on the left hand menu, and then “COUNT” on the right hand menu. 7. Dan Rott 261,947 views 5:36 Standard Error Bars on Excel - Duration: 5:01. Check This Out
A menu will appear that says “Paste Function”. Select “Stastical” from the left hand side of the menu, if necessary. Scroll down on the right hand side of the menu and You should still be able to navigate through these materials but selftest questions will not work. While we were able to use a function to directly calculate the mean, the standard error calculation is a little more round about. In this case, the temperature of the metal is the independent variable being manipulated by the researcher and the amount of energy absorbed is the dependent variable being recorded.
## Standard Error Excel Formula
PC Review Home Newsgroups > Microsoft Excel > Microsoft Excel Misc > Home Home Quick Links Search Forums Recent Posts Forums Forums Quick Links Search Forums Recent Posts Articles Articles Quick This information is presented as column A of the spreadsheet (figure 5). The easiest way to do this is to click on the up arrow button as shown in the figure above. There are two common ways you can statistically describe uncertainty in your measurements.
With the standard error calculated for each temperature, error bars can now be created for each mean. Loading... sbbiovideos 41,664 views 7:11 How to find Standard Error of Estimate in Excel - Duration: 1:05. Standard Error Of The Mean Excel Graph Dan Rott 172,176 views 11:19 How To...
First click the line in the graph so it is highlighted. Standard Error Bars Excel Member Login Remember Me Forgot your password? Sign in to add this to Watch Later Add to Loading playlists... The command to find the population standard deviation in Excel is =STDEVP(VALUES) and in this case the command is =STDEVP(A2:A6) which gives an answer of 0.49.
Without going into detail, the mean is a way of summarizing a group of data and stating a best guess at what the true value of the dependent variable value is What Does Standard Error Mean Science Class Online 20,675 views 5:01 Mean and Standard Deviation Graphs - Duration: 11:19. In this case, imagine that the data in column A represent the kilometres per litre found for a sample of 5 new cars tested by the manufacturer. The supplied arguments must provide at least two numeric values to the function.If you are using Excel 2007 a later version of Excel you can enter up to 255 number arguments
## Standard Error Bars Excel
Jim Cone, Jul 13, 2009 #5 Francis Guest assuming that your data are as : Known y's are in range A2: A20 Known x's are in tange B2 : B20 try Luke M, Jul 13, 2009 #3 EricG Guest Depends on your definition of standard error. Standard Error Excel Formula Sign in 22 Loading... How To Calculate Standard Error In Excel 2013 Loading...
I wrote: > "Kyle" <> wrote: >> How do I calculate standard error in excel? > > =STDEV(range) / SQRT(COUNT(range)) As others noted, arguably it depends on what standard error you his comment is here Note: it is critical to highlight the standardard deviation values for all of the temperatures. As you have only made 5 trips you do not have any further information and you are therefore measuring the whole population at this point in time. if your data set records the individual heights of a sample of UK males). How To Calculate Standard Error From Standard Deviation
Advertisement Autoplay When autoplay is enabled, a suggested video will automatically play next. a significance level of 0.05 equates to a confidence level of 95%).standard_dev-The standard deviation of the population.size-The population sample size.To calculate the confidence interval for a population mean, the returned CONFIDENCE About Us PC Review is a computing review website with helpful tech support forums staffed by PC experts. this contact form For an example of calculating the population standard deviation, imagine you wish to know how fuel-efficient a new car that you have just purchased is.
It takes just 2 minutes to sign up (and it's free!). 95 Confidence Interval Excel If you look back at the line graph above, we can now say that the mean impact energy at 20 degrees is indeed higher than the mean impact energy at 0 Close Yeah, keep it Undo Close This video is unavailable.
## In fact, there are a number of measurements at 0 degrees (shown in purple squares) that are very close to measurements taken at 20 degrees (shown in light blue triangles).
This distribution of data values is often represented by showing a single data point, representing the mean value of the data, and error bars to represent the overall distribution of the More About Us... The sample standard deviation will always be greater than the population standard deviation when they are calculated for the same dataset. Sample Variance Excel Register Privacy Policy Terms and Rules Help Popular Sections Tech Support Forums Articles Archives Connect With Us Twitter Log-in Register Contact Us Forum software by XenForo™ ©2010-2016 XenForo Ltd.
Published on Sep 21, 2012In this video I use Excel for Mac 2011 Version 14.2.3 I show you how to calculate mean, standard deviation and standard error in Microsoft Excel. The resulting data (and graph) might look like this: For clarity, the data for each level of the independent variable (temperature) has been plotted on the scatter plot in a different Click on the spreadsheet picture in the pop-up box, and then highlight the list of numbers you averaged. Hit enter and “OK” as before. 8. navigate here About Press Copyright Creators Advertise Developers +YouTube Terms Privacy Policy & Safety Send feedback Try something new!
Loading... Now click on the Custom button as the method for entering the Error amount. EricG, Jul 13, 2009 #4 Jim Cone Guest Look up "standard error" in the Excel help file and see what is says about "STEYX" -- Jim Cone Portland, Oregon USA "Kyle" The same data can be used to demonstrate how to calculate the sample standard deviation in Excel.
Built with Wimba Create. howtechoffice 61,985 views 1:50 How to Calculate ANOVA with Excel (Analysis of Variance) - Duration: 15:12. At -195 degrees, the energy values (shown in blue diamonds) all hover around 0 joules. | 1,461 | 6,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-30 | longest | en | 0.826404 |
http://metamath.tirix.org/geolim2.html | 1,686,030,050,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00419.warc.gz | 28,920,400 | 37,103 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > geolim2 Unicode version
Theorem geolim2 13680
Description: The partial sums in the geometric series A M A (M 1)... converge to . (Contributed by NM, 6-Jun-2006.) (Revised by Mario Carneiro, 26-Apr-2014.)
Hypotheses
Ref Expression
geolim.1
geolim.2
geolim2.3
geolim2.4
Assertion
Ref Expression
geolim2
Distinct variable groups: , , ,M ,
Proof of Theorem geolim2
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2457 . . 3
2 geolim2.3 . . . 4
32nn0zd 10992 . . 3
4 geolim2.4 . . 3
5 geolim.1 . . . . 5
65adantr 465 . . . 4
7 eluznn0 11180 . . . . 5
82, 7sylan 471 . . . 4
96, 8expcld 12310 . . 3
10 oveq2 6304 . . . . . . . 8
11 eqid 2457 . . . . . . . 8
12 ovex 6324 . . . . . . . 8
1310, 11, 12fvmpt 5956 . . . . . . 7
148, 13syl 16 . . . . . 6
1514, 4eqtr4d 2501 . . . . 5
163, 15seqfeq 12132 . . . 4
17 geolim.2 . . . . . . 7
18 oveq2 6304 . . . . . . . . 9
19 ovex 6324 . . . . . . . . 9
2018, 11, 19fvmpt 5956 . . . . . . . 8
2120adantl 466 . . . . . . 7
225, 17, 21geolim 13679 . . . . . 6
23 seqex 12109 . . . . . . 7
24 ovex 6324 . . . . . . 7
2523, 24breldm 5212 . . . . . 6
2622, 25syl 16 . . . . 5
27 nn0uz 11144 . . . . . 6
28 expcl 12184 . . . . . . . 8
295, 28sylan 471 . . . . . . 7
3021, 29eqeltrd 2545 . . . . . 6
3127, 2, 30iserex 13479 . . . . 5
3226, 31mpbid 210 . . . 4
3316, 32eqeltrrd 2546 . . 3
341, 3, 4, 9, 33isumclim2 13573 . 2
3513adantl 466 . . . . . . 7
36 expcl 12184 . . . . . . . 8
375, 36sylan 471 . . . . . . 7
3827, 1, 2, 35, 37, 26isumsplit 13652 . . . . . 6
39 0zd 10901 . . . . . . 7
4027, 39, 35, 37, 22isumclim 13572 . . . . . 6
4138, 40eqtr3d 2500 . . . . 5
42 1re 9616 . . . . . . . . . . 11
4342ltnri 9714 . . . . . . . . . 10
44 fveq2 5871 . . . . . . . . . . . 12
45 abs1 13130 . . . . . . . . . . . 12
4644, 45syl6eq 2514 . . . . . . . . . . 11
4746breq1d 4462 . . . . . . . . . 10
4843, 47mtbiri 303 . . . . . . . . 9
4948necon2ai 2692 . . . . . . . 8
5017, 49syl 16 . . . . . . 7
515, 50, 2geoser 13678 . . . . . 6
5251oveq1d 6311 . . . . 5
5341, 52eqtr3d 2500 . . . 4
5453oveq1d 6311 . . 3
55 1cnd 9633 . . . . 5
56 ax-1cn 9571 . . . . . 6
575, 2expcld 12310 . . . . . 6
58 subcl 9842 . . . . . 6
5956, 57, 58sylancr 663 . . . . 5
60 subcl 9842 . . . . . 6
6156, 5, 60sylancr 663 . . . . 5
6250necomd 2728 . . . . . 6
63 subeq0 9868 . . . . . . . 8
6456, 5, 63sylancr 663 . . . . . . 7
6564necon3bid 2715 . . . . . 6
6662, 65mpbird 232 . . . . 5
6755, 59, 61, 66divsubdird 10384 . . . 4
68 nncan 9871 . . . . . 6
6956, 57, 68sylancr 663 . . . . 5
7069oveq1d 6311 . . . 4
7167, 70eqtr3d 2500 . . 3
7259, 61, 66divcld 10345 . . . 4
731, 3, 14, 9, 32isumcl 13576 . . . 4
7472, 73pncan2d 9956 . . 3
7554, 71, 743eqtr3rd 2507 . 2
7634, 75breqtrd 4476 1
Colors of variables: wff setvar class Syntax hints: ->wi 4 <->wb 184 /\wa 369 =wceq 1395 e.wcel 1818 =/=wne 2652 class class class wbr 4452 e.cmpt 4510 domcdm 5004 cfv 5593 (class class class)co 6296 cc 9511 0cc0 9513 1c1 9514 caddc 9516 clt 9649 cmin 9828 cdiv 10231 cn0 10820 cuz 11110 cfz 11701 seqcseq 12107 cexp 12166 cabs 13067 cli 13307 sum_`csu 13508 This theorem is referenced by: geoisum1 13688 geoisum1c 13689 rpnnen2lem3 13950 rpnnen2lem9 13956 abelthlem7 22833 log2tlbnd 23276 geomcau 30252 stirlinglem10 31865 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1618 ax-4 1631 ax-5 1704 ax-6 1747 ax-7 1790 ax-8 1820 ax-9 1822 ax-10 1837 ax-11 1842 ax-12 1854 ax-13 1999 ax-ext 2435 ax-rep 4563 ax-sep 4573 ax-nul 4581 ax-pow 4630 ax-pr 4691 ax-un 6592 ax-inf2 8079 ax-cnex 9569 ax-resscn 9570 ax-1cn 9571 ax-icn 9572 ax-addcl 9573 ax-addrcl 9574 ax-mulcl 9575 ax-mulrcl 9576 ax-mulcom 9577 ax-addass 9578 ax-mulass 9579 ax-distr 9580 ax-i2m1 9581 ax-1ne0 9582 ax-1rid 9583 ax-rnegex 9584 ax-rrecex 9585 ax-cnre 9586 ax-pre-lttri 9587 ax-pre-lttrn 9588 ax-pre-ltadd 9589 ax-pre-mulgt0 9590 ax-pre-sup 9591 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3or 974 df-3an 975 df-tru 1398 df-fal 1401 df-ex 1613 df-nf 1617 df-sb 1740 df-eu 2286 df-mo 2287 df-clab 2443 df-cleq 2449 df-clel 2452 df-nfc 2607 df-ne 2654 df-nel 2655 df-ral 2812 df-rex 2813 df-reu 2814 df-rmo 2815 df-rab 2816 df-v 3111 df-sbc 3328 df-csb 3435 df-dif 3478 df-un 3480 df-in 3482 df-ss 3489 df-pss 3491 df-nul 3785 df-if 3942 df-pw 4014 df-sn 4030 df-pr 4032 df-tp 4034 df-op 4036 df-uni 4250 df-int 4287 df-iun 4332 df-br 4453 df-opab 4511 df-mpt 4512 df-tr 4546 df-eprel 4796 df-id 4800 df-po 4805 df-so 4806 df-fr 4843 df-se 4844 df-we 4845 df-ord 4886 df-on 4887 df-lim 4888 df-suc 4889 df-xp 5010 df-rel 5011 df-cnv 5012 df-co 5013 df-dm 5014 df-rn 5015 df-res 5016 df-ima 5017 df-iota 5556 df-fun 5595 df-fn 5596 df-f 5597 df-f1 5598 df-fo 5599 df-f1o 5600 df-fv 5601 df-isom 5602 df-riota 6257 df-ov 6299 df-oprab 6300 df-mpt2 6301 df-om 6701 df-1st 6800 df-2nd 6801 df-recs 7061 df-rdg 7095 df-1o 7149 df-oadd 7153 df-er 7330 df-pm 7442 df-en 7537 df-dom 7538 df-sdom 7539 df-fin 7540 df-sup 7921 df-oi 7956 df-card 8341 df-pnf 9651 df-mnf 9652 df-xr 9653 df-ltxr 9654 df-le 9655 df-sub 9830 df-neg 9831 df-div 10232 df-nn 10562 df-2 10619 df-3 10620 df-n0 10821 df-z 10890 df-uz 11111 df-rp 11250 df-fz 11702 df-fzo 11825 df-fl 11929 df-seq 12108 df-exp 12167 df-hash 12406 df-cj 12932 df-re 12933 df-im 12934 df-sqrt 13068 df-abs 13069 df-clim 13311 df-rlim 13312 df-sum 13509
Copyright terms: Public domain W3C validator | 3,202 | 5,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-23 | latest | en | 0.102553 |
https://math.stackexchange.com/questions/2336554/distance-between-a-b%C3%A9zier-curve-and-its-control-points | 1,563,735,644,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527196.68/warc/CC-MAIN-20190721185027-20190721211027-00484.warc.gz | 477,690,308 | 34,723 | # Distance between a Bézier curve and its control points
Let $B_{p_0,p_1,...,p_n}$ be a $n$-th degree concave Bézier curve on $[0,1]$ determined by control points $p_0,p_1,...,p_n$. Let the control points be all non-negative. Such curve admits an explicit definition in terms of Bernstein basis polynomials $b_{j,n}(t): = {n \choose j} t^j (1-t)^{n-j}$, $j \in \{ 0, 1, ..., n\}$, i.e.
$B_{p_0,p_1,...,p_n}(t)=\sum_{j=0}^n p_j b_{j,n}(t), \quad t \in [0,1].$
My question is maybe a trivial one: is there a way to obtain "accurate" upper and lower bounds for the distance between the curve and the $j$-th control point at $t=j/n$, i.e. $p_j -B_{p_0,p_1,...,p_n}(j/n)$, $j=0,1,...,n$? I think the simple ones that I have found are too crude:
$p_j - B_{p_0,p_1,...,p_n}(j/n) \geq p_j( 1 - b_{j,n}(j/n)) - nb_{j,n}(j/n) \max_{0\leq i \leq n} p_i$
$p_j - B_{p_0,p_1,...,p_n}(j/n) \leq p_j( 1 - b_{j,n}(j/n)) - n \left[b_{0,n}(j/n)\mathbb{1}_{\{j\geq n/2\}}+ b_{n,n}(j/n)\mathbb{1}_{\{j < n/2\}} \right]\min_{0\leq i \leq n} p_i$
and better ones can be found. Am I missing something? | 449 | 1,083 | {"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} | 3.3125 | 3 | CC-MAIN-2019-30 | latest | en | 0.614666 |
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