text:
### chenjb's blog

By chenjb, history, 6 months ago,

Hello! I'm happy to announce XX Open Cup: Grand Prix of Nanjing, which will be held on May 3rd, 2020.

This contest is mainly prepared by gisp_zjz and me, which has already been used in Moscow Pre-Finals Workshop 2020 Day 5 on April 29th, 2020.

As you can see on the top of the statement, this contest is also been called Legilimens+Coffee Chicken Contest. Team Coffee Chicken from Nanjing University will participate in World Finals 2020 and we wish them best luck in Moscow. Team Legilimens is from Zhejiang University, which is my team in the past 3 years. Although we didn't get advanced to World Finals 2020 and this is the end of our journey, we have experienced a lot of happy and unforgettable moments. Therefore, please allow me to express my gratefulness to my teammates Subconscious and oipotato. And may the friendship between Nanjing University and Zhejiang University last forever.

Authors of this contest come from Competitive Programming Team members from both Zhejiang University and Nanjing University, including gisp_zjz, Acesrc, zyb, Roundgod, Calabash_boy, sy_chen, chenjb, Subconscious, oipotato, Heltion, etc.

Testers: mayaohua2003, gtrhetr, xiaowuc1. Thank you!

I will post the editorial after the contest. We are looking forward to your participation!

UPD: Now it is over. Please feel free to discuss solutions and this is the Editorial Sketch.

• +124

 » 6 months ago, # |   0 Auto comment: topic has been updated by chenjb (previous revision, new revision, compare).
 » 6 months ago, # |   0 Auto comment: topic has been updated by chenjb (previous revision, new revision, compare).
 » 6 months ago, # |   0 Auto comment: topic has been updated by chenjb (previous revision, new revision, compare).
 » 6 months ago, # |   +18 E: https://codeforces.com/contest/1045/problem/E. Thanks :)
•  » » 6 months ago, # ^ |   0 Um_nik told me this bubble cup problem after the pre-final contest. I camp up with this problem from a problem called Constellation in JOI2011-2012 around 2015-16 when I am in high school. So SAD...
•  » » » 6 months ago, # ^ |   +56 It's also very similar to Joining Points back from IOI 2006.
 » 6 months ago, # |   +37 Where is the editorial?
 » 6 months ago, # |   0 How to solve B,F,G, K?
•  » » 6 months ago, # ^ | ← Rev. 2 →   +20 B: lengths of borders can be grouped in $O(\log(n))$ arithmetic progressions. When you add a new character, go through them from longest to shortest. If you need to delete element, do it naively, if you don't delete the first 2 elements in arithmetic progression, you don't need to delete any elements in it, so skip it.
•  » » » 6 months ago, # ^ | ← Rev. 2 →   +66 Simpler solution for B: if a certain length stops being a period, it will never become one again. So we can do independent binary searches for each length, and use hashes to compare.
•  » » » » 6 months ago, # ^ |   +10 Well, I guess it depends on one's taste. For me arithmetic progressions idea seems very natural and the code is pretty simple, while anything involving hashes and binary search makes me vomit.
•  » » » » » 6 months ago, # ^ |   +31 Indeed. I don't understand strings well enough to see that lengths of borders have that structure :)
•  » » » » » 6 months ago, # ^ |   0 Can you share the code?
•  » » » » » » 6 months ago, # ^ |   0 Code.
•  » » » 6 months ago, # ^ |   0 Actually, we know that for a repetition of length K is existing in a continuous period [L,R]. L is obviously equal to K. For R, we can binary search it and check with Hash or Suffix Array. Precalculating it for repetitions of all length separately.
•  » » » 6 months ago, # ^ |   +10 Well, actually you can observe that if some string has some border, then the string shorter by one has a border shorter by one. So, for each string you can look at its border of length $1$ and check for how long will it be the border of the next strings. So just do binary search and compare two substrings with hashes or something.
•  » » 6 months ago, # ^ |   0 Solution to G was discussed in 300iq's chat a few days ago.
•  » » » 6 months ago, # ^ |   +16 So, after the onsite? Interesting
•  » » » » 6 months ago, # ^ |   +16 No, it was on April 28th, one day before it, lol.
•  » » » » » 6 months ago, # ^ |   0 I am in that chat too and it indeed made me a little awkward since I don't know whether they are discussing because of my problem before the opencup...Maybe just a coincidence.
•  » » 6 months ago, # ^ | ← Rev. 2 →   0 G: n^4 is a simple knapsack, right? In order to make it faster I do "knapsack without one element" in D&C manner in O(n^3 log n).
•  » » » 6 months ago, # ^ |   0 I'm surprised it passes. I meant the other solution, which subtracts dp values to get knapsack without one element in $O(n^3)$.
•  » » » » 6 months ago, # ^ |   +26 I'm not in the elite chat, but the problem and the trick are very similar to the TopCoder one from this post.
•  » » » » 6 months ago, # ^ | ← Rev. 2 →   +30 That solution is the reason we didn't won NEERC 2019, so I stared at the problem for half an hour and then wrote $O(n^3 \log n)$ Swistakk described because it only uses additions. Can you prove that your solution is numerically stable?
•  » » » » » 6 months ago, # ^ |   0 I feel like "Can you prove that your solution is numerically stable?" is a rhetoric question that will never get answered.
•  » » » » » » 6 months ago, # ^ |   +39 Nah, aid is too cool
•  » » » » » 6 months ago, # ^ |   0 Obviously I can't prove that since it doesn't work on tests from that NERC problem. :)
•  » » » » » » 6 months ago, # ^ |   0 Judging by the editorial, authors also didn't think about precision issues.
•  » » » » » » » 6 months ago, # ^ |   0 Quoted from the author: 'About the precision issue when subtracting, we ran stress test with O(n^4) brute-force and it seems no effect. Meanwhile, because we are calculating the possibility instead of the number of ways, so undoing one item won’t affect much. But if we store the number of ways, we may need long double.'Well, it is still possible that we didn't generate bad enough tests...
•  » » » » » » » » 6 months ago, # ^ |   +18 This generator breaks my solution with subtractions consistently.
•  » » » » » » » » » 6 months ago, # ^ | ← Rev. 7 →   +13 （Modified a bit since the test at the beginning was problematic...）Run a test with your generator on USA1's, STD and Polish Mafia's and yours...The result is that only Polish Mafia's solution survived...I have to say that I still kind of agree that if we calculate probability instead of number of ways, the precision shouldn't be an issue. But maybe this test proved it wrong.So, the official solution should be that Divide and Conquer solution with the time complexity of O(n^3logn). Glad that the TL allows it to get passed. We are still trying to fix the std, but anyway I think either we should avoid any issue with subtraction or we shall make it a problem with modulo next time...Thank you very much!UPD: Some solutions with subtraction survived. Investigating.UPD2: aid hacked them too. You are so cool! btw, ksun48 provides a tricky method: Calculating the probability of losing as well as winning and pick the reasonable one. Very funny but maybe also very reasonable~
•  » » » » » » » » » 6 months ago, # ^ |   +18 But how do you know which one is more reasonable? :P
•  » » » » » » » » » 6 months ago, # ^ |   +10 The one that lies in range $[0, 1]$ is obviously more reasonable. But it's possible to hack both solutions with the same test.
•  » » » 6 months ago, # ^ | ← Rev. 2 →   +30 There is another $\mathcal{O}(n^3)$ solution. I_love_Tanya_Romanova came up with it during the contest.We can do two separate DPs. $dp_0(pos, cnt, s_{got})$ is the number of ways to choose $cnt$ elements from $x_0...x_{pos-1}$ with sum equal to $s_{got}$ $dp_1(pos, cnt, s_{remains})$ is the number of ways to choose $cnt$ elements and one more special (last) element from $x_0...x_{pos-1}$, in such a way, that $s_{remains} +$ $\mathit{the\ sum\ of\ chosen\ elements}$ is good There are simple transitions inside of each DP (constant number of transitions per state). And from each state in $dp_0$ there should be a transition to an interval in $dp_1$:$dp_0(pos, cnt, s_{got}) \rightarrow dp_1(pos + 1, cnt, (a-s_{got}-x_{pos}, b-s_{got}-x_{pos}])$We can avoid subtractions of real values by using the Queue to add on interval (queue implemented on two stacks).code: 79081723
•  » » 6 months ago, # ^ | ← Rev. 2 →   0 K: for requests with x>sqrt(n), iterate over depths d(a)+y, d(a)+y+x, ... and maintain a separate data structure for each depth (vertices go in the dfs order). For requests with x
•  » » » 6 months ago, # ^ |   0 Are you saying that you had two separate implementations of (update, query), one in (O(sqrt n), O(1)) and the other in (O(1), O(sqrt n)) ?
•  » » » » 6 months ago, # ^ |   0 We didn't, we tried to squeeze Fenwick and couldn't. But I think the intended solution does.
 » 6 months ago, # |   0 Anything faster than $O(n \sqrt{n})$ in K? We've squeezed $O(n \sqrt{n} \log(n))$, I see that it can be implemented without $\log$ but it starts looking bad.
•  » » 6 months ago, # ^ | ← Rev. 2 →   0 On the workshop we didn't have any troubles with $O(n\sqrt{n} log(n))$ solution, it passes in half of TL. Code.UPD: I just checked in opencup's Y.C, it is still half TL.
•  » » » 6 months ago, # ^ |   0 I had TL with almost exactly the same code on Open Cup :(
•  » » » » 6 months ago, # ^ |   0 Well, I guess it's not exactly the same. Let me try to understand the difference :) Our code that gets TL.
•  » » » » » 6 months ago, # ^ |   0 So my solution passes after borrowing just one trick from your solution: don't do binary search in the branch at line 155 (in your solution), instead precomputing the "pos" array: https://pastebin.com/Tw6BRMWq. Wow.
•  » » » » » » 6 months ago, # ^ |   +20 And 5 more seconds are saved by this line: if (rr == ll) break; After those two fixes my solution also runs in half TL.I guess even with C++ I need to learn to squeeze better :)
•  » » 6 months ago, # ^ | ← Rev. 2 →   0 O(n^1.5) is the intended solution just as Petr said. We set the TL to make it difficult to let O(n^1.5logn) pass while O(n^1.5) won’t need much adjustment. I didn’t notice KAN pass it with a O(n^1.5logn) and the TL hasn’t changed in the OpenCup as far as I know.
•  » » 3 weeks ago, # ^ |   +5 I sent a solution with centroid decomposition, with less effort to model. blog
 » 6 months ago, # |   0 Auto comment: topic has been updated by chenjb (previous revision, new revision, compare).
 » 6 months ago, # |   0 How to solve Problem O(Official Visit) ? How we can construct path ,so it will be maximum?
•  » » 6 months ago, # ^ |   0 Ans = (n — 1) * 4 + (n — 2) ^ 2 * sqrt2
•  » » » 6 months ago, # ^ | ← Rev. 2 →   0 tumaryui thanks man !
 » 6 months ago, # |   0 What is i in Sx(i) in the editorial of problem D?
•  » » 6 months ago, # ^ |   0 I think it refers to the start index of Sx.
 » 6 months ago, # | ← Rev. 2 →   +10 Was solution to M inspired by "divide and color" approach to finding k-path in $O^*(4^k)$? Solution in current state is kinda disappointing since if you didn't read, it works for k up to 10 only, maybe you can make it work for bigger values, but that would get messier since it basically handcrafts a different algorithm for every value. However if you apply that recursively (details to be figured out by careful reader) you will get $O(4^k poly(n, m))$ algorithm that is satisfying from theoretical point of view.We got good ol' color coding in $O^*((2e)^k)$ and it really barely doesn't pass (we managed to get to test 47)... Seems constraints were tailored exactly for this particular solution to be the best one. If only TL would have been a little bigger or it would have been path instead of a cycle or if constraints would have been a bit different... And, what is more, I thought about divide and color approach during the contest as natural next step from optimizing simple color coding, but I never really made an effort to understand it and expected it to be too complicated to be intended solution, so I didn't bother to check it during contest (and indeed it was too complicated, but for k<=10 it simplifies as in editorial).
•  » » 6 months ago, # ^ |   +30 Yes, you've got that right.I'm one of the coauthors of Problem M. I'd like to share some stories behind the scene.When we made up this problem set, we wanted to add a randomized problem simply cuz random & approx algorithms are the main research fields our TCS group. Then we just come up with color coding, which is not yet widely known in competitive programming community.Before this problem was made, there was a problem another contest (2019 China Multi-University Training Contest) asking to find k-path. In that problem, a simple $O^*((2e)^k)$ color coding could pass, but during that contest many participants passed it with various heuristics. So we just wanted to make another color coding problem, which couldn't be easily passed by heuristics. This is M in this contest.The basic idea of M indeed came from the $O^*(4^k)$ algorithm for finding a k-path by combining color coding and divide-and-conquer. The constraint $k \leq 10$ is mainly because of two reasons: one is we can just split to two 5-paths and perform a case analysis, thus simplifying the solution; the other is that it provides enough budget for the constraints on the # of vertices and edges.To avoid heuristic solutions, we made it to find k-cycle and carefully tuned the constraints. We successfully hacked many common heuristics (among which the most difficult one to hack is to precompute a "relaxed" answer which allows cycles with repeated vertices and use this information to prune during exhaustive search). Actually there is still one solution we haven't hacked: compute biconnected components and use the aforementioned heuristic on each component. But we thought nobody would come up with such solution during contest. :P
•  » » 6 months ago, # ^ |   +20 Yeah, we were able to pass M with color-coding with a couple constant-factor optimizations: When searching for P points it's actually ideal to assign P+1 or P+2 colors rather than P -- traveling salesman is 2x slower but probability of success is quite a bit higher. Instead of initializing with a vertex and searching for the remaining K-1 points, you can initialize with an edge and search for the remaining K-2 points. Since M and N have the same bound each edge is just as likely to be part of a correct set.
•  » » » 6 months ago, # ^ |   0 Second idea (with starting from an edge) seemed to me like a great one, but unfortunately it has not made our algorithm better. Even though we need to use one color less, it is outweighed by the fact that if we delete a vertex we gain much more than if we delete an edge (we considered starting vertices in descending order of their degrees and removed them after that) which is especially painful on a clique on 25 vertices which I think was already a worst-case for our algorithm.However the optimization with 2 more colors seemed to do the trick and we managed to get it accepted with it. I think that at some point in my life I wondered what happens if we allow more colors in this algorithm, but did this with focus on complexity rather than constant optimization and got no results and forgot about it.
 » 6 months ago, # |   -38 give us the problems please instead of just editorials..
 » 6 months ago, # |   -75 Why are you discussing a private contest in a public place? I hate this. I think it's unfair for users who don't have an OpenCup login.
•  » » 6 months ago, # ^ |   +50 It is not good that it is hard to get OpenCup login. But your logic is weird. Many people participate in OpenCup, so this blog definetely has an audience on CF. Moreover, CF is a go-to place for any cp related discussions. There are a lot of local contests (for example ICPC regionals) which are also not open for everyone and are discussed on CF. Is that bad too?
•  » » » 6 months ago, # ^ | ← Rev. 3 →   -30 In most cases, the ICPC contests will be published anyway(virtual contest, or on Gym, or on some official websites), but the contests for OpenCup won't(some will be on Gym, but just depend on writers). That's the difference. I think the discussion for contests(shared with everyone) is good, but for some contests(people without login have no access to the resources) is really annoying.
•  » » » » 6 months ago, # ^ |   +41 Sorry, but we discuss the problems not because we want to share knowledge with you, but because we are interested in different approaches.
•  » » » » 6 months ago, # ^ |   -30 Hey, it's fun to read solutions in comments first, and then read problem statements after the authors upload them in Gym.
 » 6 months ago, # |   0 Any better explanation for C?


text:
WP 34S solve an integral
04-23-2015, 07:27 PM
Post: #1
 lrdheat Senior Member Posts: 526 Joined: Feb 2014
WP 34S solve an integral
I was wondering how to use solve for a problem such as, (integral x^2 from 0 to x) =10. (Answer is ~3.107)
04-23-2015, 08:05 PM (This post was last modified: 04-24-2015 08:15 AM by Marcus von Cube.)
Post: #2
 Marcus von Cube Senior Member Posts: 760 Joined: Dec 2013
RE: WP 34S solve an integral
(04-23-2015 07:53 PM)BarryMead Wrote:  You don't use solve, you use the Integral operator the g shifted 2.

I'm afraid You misread the original post. Lrdheat is asking for the upper bound so that the integral is 10. This can be done:

Code:
LBL 00 0 x<> Y ∫ 00 // should be 01       // 10 - is missing. See next post RTN LBL 01 x² RTN

Now you can SLV 00 and should get your answer. Beware: I didn't test the code. It's just of the top of my head.

Edit: See comments in the code.

Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
04-23-2015, 08:32 PM (This post was last modified: 04-25-2015 07:38 PM by BarryMead.)
Post: #3
 BarryMead Senior Member Posts: 416 Joined: Feb 2014
RE: WP 34S solve an integral
(04-23-2015 08:05 PM)Marcus von Cube Wrote:  I didn't test the code. It's just of the top of my head.
That program was close to working, but this one does work.

Code:
LBL 00 0 x <> Y ∫ 01 # 010    //Entered with h CONST #  and 010 x <> Y - RTN LBL 01 x² RTN

Exit programming mode and press 3 Enter 4 SLV 00
04-23-2015, 08:43 PM
Post: #4
 lrdheat Senior Member Posts: 526 Joined: Feb 2014
RE: WP 34S solve an integral
Thanks, Barry,

It works!

What is the #010?
04-23-2015, 08:50 PM (This post was last modified: 04-23-2015 10:59 PM by BarryMead.)
Post: #5
 BarryMead Senior Member Posts: 416 Joined: Feb 2014
RE: WP 34S solve an integral
(04-23-2015 08:43 PM)lrdheat Wrote:  Thanks, Barry,

It works!

What is the #010?
It is a constant 10 (The value sought for the integral end result) The numbers entered before solve are only
guesses for the solve search range. If you entered 2 and 4 SLV 00 it would still work.

If you want to rework the program to use the X register as target integral result you could store X in a register then
subtract the integral result from that register to make the solve function converge on zero relative to that stored value instead of 10.

I just hard coded the 10 in the program with a constant to make it easier to write the program.

Does that help?
04-23-2015, 09:05 PM (This post was last modified: 04-25-2015 07:39 PM by BarryMead.)
Post: #6
 BarryMead Senior Member Posts: 416 Joined: Feb 2014
RE: WP 34S solve an integral
This version of the program uses label A as the start of the program and you put the
Target integral value into the X , the Upper solve guess bound in Y and the Lower solve guess Bound in Z and press A

Code:
LBL A STO 00                  //Save target integral result desired DROP                    //Lower stack to get Upper/Lower solve guess bounds SLV 00 RTN ERR 20                  //If solve fails to find a root show this error LBL 00 0                       //Lower bound of integral (from zero to solve guess) x <> Y ∫ 01 RCL 00                  //Value to solve for (to get upper integral bound) x <> Y -                       //Integral function returns zero when integral equals R00 target value RTN LBL 01 x² RTN

Then Press 0 Enter 5 Enter 10 A (Solves for X where the integral of x² from zero to X = 10) = 3.10723250595
Try 3 Enter 7 Enter 20 A (Solves for X where integral of x² from zero to X = 20) = 3.91486764117

I have found that even if you pick really bad guesses for the solve bounds, it usually still finds the right answer, but takes a little longer.
04-23-2015, 10:28 PM (This post was last modified: 04-23-2015 10:37 PM by CR Haeger.)
Post: #7
 CR Haeger Member Posts: 275 Joined: Dec 2013
RE: WP 34S solve an integral
Thanks for this post ldheat and Marcus/Barry for your advice - I have been looking at something like this as well.

May I ask a few related questions for a WP34S that runs slower or is chewing up battery:

1. To limit the SLV routine precision/iterations, a RDP or RDP --> command in the routine being solved can be added near the end of the LBL 00 routine being "solved"?

2. To limit the integration precision, a SCI or SCI --> command can be added to the routine containing the integral (LBL 01)?

3. If I want to have a global "f(x)" routine versus a local one, could I change LBL 01 to say LBL 'F'? That way this global 'F' can be used as is or by other programs? If so, I assume the ∫ 01 needs to change to ∫ 'F' as well?

Thanks
04-23-2015, 10:34 PM (This post was last modified: 04-24-2015 05:51 AM by BarryMead.)
Post: #8
 BarryMead Senior Member Posts: 416 Joined: Feb 2014
RE: WP 34S solve an integral
(04-23-2015 10:28 PM)CR Haeger Wrote:  Thanks for this post and Marcus/Barry for your advice - I have been looking at something like this as well.

May I ask a few related questions for a WP34S that runs slower or is chewing up battery:

1. To limit the SLV routine precision/iterations, a RDP or RDP --> command in the routine being solved can be added near the end of the LBL 00 routine being "solved"?

2. To limit the integration precision, a SCI or SCI --> command can be added to the routine containing the integral (LBL 01)?

3. If I want to have a global "f(x)" routine versus a local one, could I change LBL 01 to say LBL 'F'? That way this global 'F' can be used as is or by other programs?

Thanks

Yes to all three questions. The WP-34s is one of the most powerful programmable calculators out there.
You can do almost anything you can imagine with it.

A couple of suggestions regarding battery life.
1. Use the emulator on your computer (It uses only wall power and runs much faster than the real calculator)
2. If you have a calculator with Harold Pott's micro USB board installed, attach the USB cable to a computer (or a 5V USB adapter), and you have NO BATTERY DRAINAGE. (Nice for running those long battery hungry programs).
04-23-2015, 10:46 PM
Post: #9
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: WP 34S solve an integral
(04-23-2015 10:34 PM)BarryMead Wrote:  Yes to all three questions. The WP-34s is the most powerful programmable calculator I have ever used.

All this could already be done with the very first HP with "Integrate" – the 34C from 1979. However, not all later HPs allow nested Solve/Integrate. For instance the 35s doesn't.

@"lrdheat": please note that the result returned by "Integrate" may vary with the display setting. Some days ago I answered an older post where you asked a similar question. If you haven't noticed yet: look here.

(04-23-2015 10:34 PM)BarryMead Wrote:  You can do almost anything you can imagine with it.

Well, there still is some work to do. Recently I have been looking at the Poisson and Binomial quantile functions.

Dieter
04-24-2015, 02:11 AM
Post: #10
 lrdheat Senior Member Posts: 526 Joined: Feb 2014
RE: WP 34S solve an integral
Thanks Barry, Dieter, and CR.

Nice work!
04-24-2015, 06:19 AM
Post: #11
 Thomas_Sch Senior Member Posts: 310 Joined: Dec 2013
RE: WP 34S solve an integral
(04-24-2015 02:11 AM)lrdheat Wrote:  Thanks Barry, Dieter, and CR.

Nice work!
Same from me, very helpful!
04-24-2015, 08:22 AM
Post: #12
 Marcus von Cube Senior Member Posts: 760 Joined: Dec 2013
RE: WP 34S solve an integral
(04-23-2015 10:46 PM)Dieter Wrote:
(04-23-2015 10:34 PM)BarryMead Wrote:  Yes to all three questions. The WP-34s is the most powerful programmable calculator I have ever used.

All this could already be done with the very first HP with "Integrate" – the 34C from 1979. However, not all later HPs allow nested Solve/Integrate. For instance the 35s doesn't.

The 34S has a feature the 34C lacks: It can nest solve and integrate (and sum and product) to any depth as far as memory allows. We're using local registers in the XROM code that implements these functions.

Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
04-24-2015, 08:57 AM
Post: #13
 walter b On Vacation Posts: 1,957 Joined: Dec 2013
RE: WP 34S solve an integral
FYI, "RDP or RDP-->" are the same command, just with direct or indirect addressing. Same applies to "SCI or SCI-->".

d:-)
04-24-2015, 12:16 PM
Post: #14
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: WP 34S solve an integral
(04-23-2015 10:28 PM)CR Haeger Wrote:  1. To limit the SLV routine precision/iterations, a RDP or RDP --> command in the routine being solved can be added near the end of the LBL 00 routine being "solved"?

I would recommend a solution like this:
Code:
... RDP 05    // round to e.g. 5 decimal places x≠0?      // if result does not round to zero x<> L     // recall the original unrounded value RTN

This way the solver always gets the exact function result, and only values below the (here) 5–digit threshold are rounded to zero so that the solver quits.

Dieter
04-24-2015, 12:34 PM
Post: #15
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: WP 34S solve an integral
(04-23-2015 09:05 PM)BarryMead Wrote:
Code:
... LBL 00 0                       //Lower bound of integral (from zero to solve guess) x <> Y ∫ 01 RCL 00                  //Value to solve for (to get upper integral bound) x <> Y -                       //Integral function returns zero when integral equals R00 target value RTN ...

Of course it doesn't matter whether you solve ∫ – 10 = 0 or 10 – ∫ = 0, so the second x<>y is not required, and a simple RCL–00 will do as well.
Or, even shorter:

Code:
LBL 00  0 ∫ 01      // ∫ from x to 0 = –∫ from 0 to x  RCL+ 00   // 10 – ∫ RTN

Sorry for the nitpicking. ;-)

(04-23-2015 09:05 PM)BarryMead Wrote:  Then Press 0 Enter 5 Enter 10 A (Solves for X where the integral of x² from zero to X = 10) = 3.10723250595
Try 3 Enter 7 Enter 20 A (Solves for X where integral of x² from zero to X = 20) = 3.91486764117

FTR: the exact solutions are the cube roots of 3·10 resp. 3·20.

Dieter
04-24-2015, 01:01 PM
Post: #16
 CR Haeger Member Posts: 275 Joined: Dec 2013
RE: WP 34S solve an integral
(04-24-2015 12:16 PM)Dieter Wrote:  I would recommend a solution like this:
Code:
... RDP 05    // round to e.g. 5 decimal places x≠0?      // if result does not round to zero x<> L     // recall the original unrounded value RTN

This way the solver always gets the exact function result, and only values below the (here) 5–digit threshold are rounded to zero so that the solver quits.

Dieter

Thanks Dieter!

If I understand correctly, the x≠0? and x<> L steps ensure that the latest exact (versus rounded) SLV guess is passed back to the SLV routine?
04-24-2015, 01:09 PM (This post was last modified: 04-24-2015 01:17 PM by Dieter.)
Post: #17
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: WP 34S solve an integral
(04-24-2015 01:01 PM)CR Haeger Wrote:  If I understand correctly, the x≠0? and x<> L steps ensure that the latest exact (versus rounded) SLV guess is passed back to the SLV routine?

These two steps make sure that usually the exact function result is returned to the SLV routine.
Only if that value rounds to zero, this zero is passed to SLV so that it will quit.

In other words:

IF Round(fx, 5) = 0
THEN return 0
ELSE return fx
END IF

Dieter
04-24-2015, 01:20 PM
Post: #18
 CR Haeger Member Posts: 275 Joined: Dec 2013
RE: WP 34S solve an integral
Right, f(x) not "guess"...

thanks again.
04-24-2015, 01:24 PM (This post was last modified: 04-24-2015 01:26 PM by CR Haeger.)
Post: #19
 CR Haeger Member Posts: 275 Joined: Dec 2013
RE: WP 34S solve an integral
(04-24-2015 08:57 AM)walter b Wrote:  FYI, "RDP or RDP-->" are the same command, just with direct or indirect addressing. Same applies to "SCI or SCI-->".

d:-)

Thank you. Yes, I thought the indirect addressing might be handy for adjusting SCI or RDP values from outside the program - perhaps even during its execution. Cool feature.
04-24-2015, 02:46 PM (This post was last modified: 04-25-2015 07:39 PM by BarryMead.)
Post: #20
 BarryMead Senior Member Posts: 416 Joined: Feb 2014
RE: WP 34S solve an integral
(04-24-2015 12:34 PM)Dieter Wrote:
Code:
LBL A STO 00      //Save target integral result desired DROP        //Lower stack to get Upper/Lower solve guess bounds SLV 00 RTN ERR 20      //If solve fails to find a root show this error LBL 00  0 ∫ 01      // ∫ from x to 0 = –∫ from 0 to x  RCL+ 00   // 10 – ∫ RTN LBL 01 x² RTN
Very cleaver optimization. I knew that my original program was clumsy, and non-optimal. I did that on purpose to make it as clear and obvious as possible how to turn the original idea into a program. Using reversed integral bounds (to make the integral negative), and adding a positive target value is equivalent to subtracting a target, and the ability of solve to converge regardless of the sign of the error also can make the program more compact. Congrats on how small you made the program.
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# Matlab: xcorr: Having trouble when I try to calculate delay between two sinusoids

I have a problem when I try to use Matlab built-in function for cross-correlation 'xcorr'.

Background: I created two sinusoids, one of them delayed in time. I choose center frequencies of these signals slightly different than another. In this way, I am trying to analyze the effects of frequency mismatch on cross-correlation results. Problem: The problem is when I try to calculate delay using xcorr function in MatLab, I could not find the exact lag. The result is quite far away from my offset-time. I try to increase my signal length or sampling frequency. But nothing changes.

Do you have any idea? Here is my simple code, below.

clc
clear all
close all

Fs=48e3;
dt=1/Fs;
t=0:dt:(4*Fs-1)*dt;
CarrierFreq=5e3;
Dev=50;
CarrierFreqDev=CarrierFreq +Dev*(2*randi(2,1)-3);

Offset=10000;
OffsetTime=Offset*dt;
NoisePower=-95
DesiredSNR=20;

% Signal Power
NoisePowerWatt=10^(NoisePower/10);

% Generate Noise @ specified power
Noise1 = wgn(1,length(t),NoisePowerWatt,'linear');
Noise2 = wgn(1, length(t),NoisePowerWatt,'linear');

% Generate Signal1 @ specified power
Signal1=cos(2*pi*CarrierFreq*t);
pSignal1 = sum(Signal1.^2) / length(Signal1);
Signal1 =  Signal1/sqrt(pSignal1);
Signal1=Signal1_Norm+Noise1;

% Generate Delayed Signal @ specified power
Signal2=cos(2*pi*CarrierFreqDev*t);
pSignal2=sum(Signal2.^2) / length(Signal2);
Signal_Delay=circshift(Signal2_Norm, -Offset);
Signal2=Signal_Delay+Noise2;

[Coeff, Delay_Time] = xcorr(Signal1,Signal2); %% with dev.
[ ~,Indx]  = max(abs(Coeff));
Delay_Time = Delay_Time(Indx)*dt;

A sine wave is cyclo-stationary so is not the best waveform to use for estimating delay between transmit and received signals. It will correlate whenever the delay causes the signal to be an additional $$2\pi$$ in phase offset. Consider using waveforms with better autocorrelation properties for this purpose such as pseudo-random noise sequences or other generated white noise sources.

Regardless, the effect of frequency offset on cross-correlation is that the magnitude rolls-off as a Sinc function with frequency offset with the first null at $$1/T$$ where $$T$$ is the total time duration of the signal, and the phase decreases linearly from $$0$$ at $$f=0$$ to $$-\pi$$ at $$f=1/T$$. At a fixed frequency offset between the signals, a fixed phase offset would result, and for a narrow band signal this can be approximated as a fixed delay according to the carrier frequency of the signal.

Consider that correlation is the sum of the element by element product between two signals (the dot product), and observe how the product of two signals will be at the frequency difference between them (the beat note). Thus if your total time duration is exactly 1 complete cycle off in frequency, the result will average to zero (showing how the first null is at $$1/T$$).

The following code demonstrates this experimentally with a random white noise waveform of total time duration 0.1 seconds:

T = .1;
fs = 10e3;
N = T*fs;
t = [0:N-1]/fs;
x = randn(1,N);

index = 0;
foffset = 0:.01:20;
for f = foffset
index++;
y = x.*exp(j*2*pi*f.*t);
# dot product is the correlation, and normalized
corr(index) = x*y'/(N*std(x).*std(y));
end

figure
plot(foffset,abs(corr))
figure
plot(corr)

This creates the two plots shown below with the first plot of the magnitude of the normalized correlation versus the static frequency offset between the two signals, and the second plot as a polar plot showing both magnitude and phase components.


text:
### On fine structure of singularly continuous probability measures and random variables with independent $\widetilde{Q}$-symbols

Methods Funct. Anal. Topology 17 (2011), no. 2, 97-111

We introduce a new fine classification of singularly continuous probability measures on $R^1$ on the basis of spectral properties of such measures (topological and metric properties of the spectrum of the measure as well as local behavior of the measure on subsets of the spectrum). The theorem on the structural representation of any one-dimensional singularly continuous probability measure in the form of a convex combination of three singularly continuous probability measures of pure spectral type is proved.

We introduce into consideration and study a $\widetilde{Q}$-representation of real numbers and a family of probability measures with independent $\widetilde{Q}$-symbols. Topological, metric and fractal properties of the above mentioned probability distributions are studied in details. We also show how the methods of $\widetilde{P}-\widetilde{Q}$-measures can be effectively applied to study properties of generalized infinite Bernoulli convolutions.

### Complement on order weakly compact operators

Methods Funct. Anal. Topology 17 (2011), no. 2, 112-117

We generalize a result on the duality property for order weakly compact operators and use it to establish some characterizations of positive operators.

### Boundary problems and initial-boundary value problems for one class of nonlinear parabolic equations with Lévy Laplacian

Methods Funct. Anal. Topology 17 (2011), no. 2, 118-125

We develop a method to construct a solution to a boundary problem and an initial-boundary value problem in a fundamental domain of a Hilbert space for a class of nonlinear parabolic equations not containing explicitly the unknown function, $$\frac{\partial U(t,x)}{\partial t}=f(t,\Delta_LU(t,x)),$$ where $\Delta _L$ is the infinite dimensional Lévy Laplacian.

### Multi-dimensional Schrödinger operators with point interactions

Nataly Goloshchapova

Methods Funct. Anal. Topology 17 (2011), no. 2, 126-143

We study two- and three-dimensional matrix Schrödinger operators with $m\in \Bbb N$ point interactions. Using the technique of boundary triplets and the corresponding Weyl functions, we complete and generalize the results obtained by the other authors in this field. For instance, we parametrize all self-adjoint extensions of the initial minimal symmetric Schrödinger operator by abstract boundary conditions and characterize their spectra. Particularly, we find a sufficient condition in terms of distances and intensities for the self-adjoint extension $H_{\alpha,X}^{(3)}$ to have $m'$ negative eigenvalues, i.e., $\kappa_-(H_{\alpha,X}^{(3)})=m'\le m$. We also give an explicit description of self-adjoint nonnegative extensions.

### Unconditional bases of de Branges spaces and interpolation problems corresponding to them

Methods Funct. Anal. Topology 17 (2011), no. 2, 144-149

In this paper the unconditional bases of de Branges spaces are constructed from the values of reproducing kernels. Appropriate problems of interpolation by entire functions are also considered. The paper is a continuation of papers [2, 3].

### Notes on Wick calculus on parametrized test functions spaces of Meixner white noise

N. A. Kachanovsky

Methods Funct. Anal. Topology 17 (2011), no. 2, 150-167

Using a general approach that covers the cases of Gaussian, Poissonian, Gamma, Pascal and Meixner measures, we construct elements of a Wick calculus on parametrized Kondratiev-type spaces of test functions; consider the interconnection between the extended stochastic integration and the Wick calculus; and give an example of a stochastic equation with a Wick-type nonlinearity. The main results consist in studying properties of a Wick product and Wick versions of holomorphic functions on the parametrized Kondratiev-type spaces of test functions. These results are necessary, in particular, in order to describe properties of solutions of stochastic equations with Wick type nonlinearities in the "Meixner analysis".

### On $C^*$-algebra generated by Fock representation of Wick algebra with braided coefficients

D. Proskurin

Methods Funct. Anal. Topology 17 (2011), no. 2, 168-173

We consider $C^*$-algebras $\mathcal{W}(T)$ generated by operators of Fock representations of Wick $*$-algebras with a braided coefficient operator $T$. It is shown that for any braided $T$ with $||T||<1$ one has the inclusion $\mathcal{W}(0)\subset\mathcal{W}(T)$. Conditions for existence of an isomorphism $\mathcal{W}(T)\simeq\mathcal{W}(0)$ are discussed.

### On $(\bigwedge, \mu)$-closed sets in generalized topological spaces

Methods Funct. Anal. Topology 17 (2011), no. 2, 174-179

In this paper, we introduce and study $(\bigwedge ,\mu )$-open sets and $(\bigwedge ,\mu )$-closed sets via $\mu$-open and $\mu$-closed sets in generalized topological spaces. Moreover, we introduce some generalized separation axioms in generalized topological spaces.

### Factor representations of infinite semi-direct products

R. Zekri

Methods Funct. Anal. Topology 17 (2011), no. 2, 180-192

In this article, we propose a new method to study unitary representations of inductive limits of locally compact groups. For the group of infinite upper triangular matrices, we construct a family of type III factorial representations. These results are complements to previous results of A. V. Kosyak, and Albeverio and Kosyak [1, 5].


text:
## Flow, Border and Gridlayouts compared

For the purposes of this course I will concentrate on three of the Layout managers shipped by Sun, these are

• FlowLayout
• BorderLayout
• GridLayout

These give you enough control over component layout for most demonstration programs. The FlowLayout and BorderLayout are the default Layout Managers for the Frame and Applet class respectively. Thus if you create a Frame and do not specifically assign a Layout Manager it will automatically use the BorderLayout.

 The default layout for a frame is the BorderLayout

This is slightly awkward as the BorderLayout is not the most useful of Managers. It divides the Frame area up into five regions, North, South, East, West and Centre. Also if you do not specify a region each new component defaults to the centre of the Frame and if you add more than one component without a region, each new component gets added on top of the old one. So for the purposes of these demonstration programs you may find it useful change the Layout Manager to FlowLayout. This can be done using the setLayout Method of Frame. Thus you can change from the default BorderLayout to a flowLayout with the following code in a Frame constructor

f.setLayout(new FlowLayout);


text:
Now showing items 1-20 of 25

• #### A geometric approach to the Carlson problem

(Society for Industrial and Applied Mathematics (Filadèlfia), 2000)
Article
Open Access
The possible observability indices of an observable pair of matrices, when supplementary subpairs are prescribed, are characterized when the “quotient” one is nilpotent. The geometric techniques used are also valid in the ...
• #### Càlcul integral per a enginyers

(Universitat Politècnica de Catalunya, 2002)
Book
El càlcul integral és una matèria bàsica dins els plans d'estudis de les enginyeries i les llicenciatures científiques. Aquest llibre és una introducció a la teoria de la integració de diverses variables i la integració ...
• #### Classi fication of monogenic invariant subspaces and uniparametric linear control systems

(2012-12)
External research report
Open Access
The classification of the invariant subspaces of an endomorphism has been an open problem for a long time, and it is a ”wild” problem in the general case. Here we obtain a full classification for the monogenic ones. Some ...

(2005)
Article
Open Access
• #### Contribució a l'estudi geomètric de subespais invariants respecte a transformacions i sistemes lineals

(Universitat Politècnica de Catalunya, 2001-10-19)
Doctoral thesis
Open Access
Mitjançant tècniques geomètriques, abordem les qüestions següents:

(i) Estudi (caracterització, classificació, famílies diferenciables,...) d'una classe destacada de subespais invariants, els anomenats ...
• #### Dimension of the orbit of marked subspaces

(2004)
Article
Open Access
Given a nilpotent endomorphism, we consider the manifold of invariant subspaces having a fixed Segre characteristic. In [Linear Algebra Appl., 332–334 (2001) 569], the implicit form of a miniversal deformation of an ...
• #### Geometric classification of monogenic subspaces and uniparametric linear control systems

(Taylor & Francis, 2015)
Article
Open Access
We present a geometric approach to the classification of monogenic invariant subspaces, alternative to the classical algebraic one, which allows us to obtain several matricial canonical forms for each class. Some applications ...
• #### Geometric structure of single/combined equivalence classes of a controllable pair

(2011-11)
Article
Open Access
Given a pair of matrices representing a controllable linear system, its equivalence classes by the single or combined action of feedbacks, change of state and input variables, as well as their intersection are studied. ...
• #### Geometric structure of the equivalence classes of a controllable pair

(2010)
Conference lecture
Open Access
Given a pair of matrices representing a controllable linear system, we study its equivalence classes by the single or combined action of feedbacks and change of state and input variables, as well as their intersections. ...
• #### Local differentiable pole assignment

(Taylor & Francis, 2010)
Article
Open Access
Given a general local differentiable family of pairs of matrices, we obtain a local differentiable family of feedbacks solving the pole assignment problem, that is to say, shifting the spectrum into a prefixed one. We ...
• #### Matricial realizations of the solutions of the Carlson problem

(2001)
Article
Open Access
The Klein theorem asserts the existence of a solution of the Carlson problem when a related Littelewood-Richardson sequence exists. Here we present an explicit construction of these matricial solutions.
• #### Miniversal deformations of marked matrices

(2001)
Article
Open Access
Given the set of square matricesM⊂Mn+m(C) that keep the subspace W = Cnx{0} ⊂ Cn+m invariant, we obtain the implicit form of a miniversal deformation of a matrix a∈M, and we compute it explicitely when this matrix is ...
• #### Miniversal deformations of observable marked matrices

(2014)
Article
Restricted access - publisher's policy
Given the set of vertical pairs of matrices ${\cal M}\subset M_{m,n}(\mathbb C)\times M_n(\mathbb C)$ keeping the subspace $\mathbb C^d\times\{0\}\subset\mathbb C^n$ invariant,we compute miniversal deformations of a given ...
• #### Miniversal deformations of observable marked matrices

(2012)
External research report
Open Access
Given the set of vertical pairs of matrices M¿ Mm,n(C)×Mn(C) keeping the subspace Cd×{0} ¿ Cn invariant, we compute miniversal deformations of a given pair when it is observable and the subspace Cd × {0} is marked. ...
• #### On the effect of friend feedbacks

(2011)
Article
Restricted access - publisher's policy
Given S an (A,B)-invariant subspace, we prove that the set of friend feedbacks is a linear variety, which can be considered as the direct sum of the feedbacks of the restriction to S and the co-restriction to S ⊥. In ...
• #### On the effect of friend feedbacks

(2009)
Article
Open Access
Given S an (A;B)-invariant subspace, we prove that the set of friend feedbacks is a (nm¡md+dq)-dimensional linear variety, which can be considered as the direct sum of the feedbacks of the restriction to S and the ...
• #### Output regulation problem for differentiable families of linear systems

(2010)
Article
Open Access
Given a family of linear systems depending on a parameter varying in a differentiable manifold, we obtain sufficient conditions for the existence of a (global or local) differentiable family of controllers solving the ...
• #### Output regulation problem for differentiable families of linear systems

(2009)
Article
Open Access
Given a family of linear systems depending on a parameter varying in a differentiable manifold, we obtain sufficient conditions for the existence of a (global or local) differentiable family of controllers solving the ...
• #### Perturbations preserving conditioned invariant subspaces

(2010)
Conference lecture
Open Access
Given the set of matrix pairs M ⊂ Mm,n(C) × Mn(C) keeping a subspace S ⊂ Cn invariant, we obtain a miniversal deformation of a pair belonging to an open dense subset of M. It generalizes the known results when S is a ...
• #### Perturbations preserving conditioned invariant subspaces

(2011)
Article
Restricted access - publisher's policy
Given the set of matrix pairsM Mm,n.C/ Mn.C/ keeping a subspace S Cn invariant, we obtain a miniversal deformation of a pair belonging to an open dense subset ofM. It generalizes the known results when S is a ...


text:
# Long indexes style with imakeidx

I'm using imakeidx package:

\usepackage{imakeidx}
\indexsetup{othercode=\small}
\makeindex[program=makeindex,columns=2,intoc=true,options={-s index_style.ist}]

With style:

    headings_flag 1
delim_0 " \\dotfill "
delim_1 " \\dotfill "
delim_2 " \\dotfill "

But I have a few longer indexes and they are line breaking with indent. How can I modify your script to get rid of this indents?

• Welcome to TeX SX! Could you post a full (yet minimal) code? – Bernard Apr 7 '17 at 16:41
• entries in an index are list items. list items are usually indented, and that format is usually defined by the document class. in imakeidx, the only reference to this code is the line \let\item\@idxitem. to redefine that (and if not properly "quarantined", it will affect all lists in your document), you will have to redefine the indentation used by\@idxitem. – barbara beeton Apr 7 '17 at 16:49
• I don't have other lists in this document. How can I redefine this indentation? – bartzmc Apr 7 '17 at 16:54

\makeatletter

• Or just \def\@idxitem{\par} – egreg Apr 7 '17 at 20:09


text: ; S F 4 and T e C l 4 both have s p 3 d hybridisation and one … Acetone can be obtained by ozonolysis of: A) 2-butyne B) 20butene C) 1ae-butene D) 1-butene Answer Save. These are promoted to vacant 5d orbitals. Explain about sp^3d^2 hybridisation with an example. Hybridization: sp 3 d 2 Then construct the 3D geometry using VSEPR rules: Decision: The molecular geometry of ICl 5 is square pyramid with an asymmetric electron region distribution. It is surrounded by five chlorine atoms and one lone... See full answer below. View Answer. View Answer. O2^2+ Use the molecular orbital diagram show to determine which of the following are paramagnetic. sp3. A single chloride free radical can destroy ozone molecule up to: A) 100,000 B) 10,000 C) 1000,0000 D) 1000 45. HOClO2 + HClO3 . Feedback Bristol ChemLabS, School of Chemistry, University of Bristol, Bristol BS8 1TS, UK. Ruzz R. 1 decade ago. Favorite Answer. 3 years ago Answers : (2) ... Anionic form is=> ClO4- => the hybridization for Cl is => sp3 that is tetraderal as all the bonds formed are double bonded with O atom. It is produced by reaction between chlorine dioxide and excess ozone: It was originally reported to exist as the monomeric chlorine trioxide ClO3 in gas phase,[1] but was later shown to remain an oxygen-bridged dimer after evaporation and until thermal decomposition into chlorine perchlorate, Cl2O4, and oxygen. 6 years ago. appears as crystalline or powdered solids. P4O10 Name: Phosphorus Pentoxide Empirical formula: P2O5 Str. and ClO2 or Chlorine dioxide is sp2 hybridized, although it is a bit tricky. VIEW MORE. Expert Answer: In the formation of the fluorides of xenon, there are 6 and the electrons in the valence shell of xenon get unpaired. B F 4 − is s p 3 hybridised and has tetrahedral shape with equal bond lengths of all the four bonds. Give the electron geometry, molecular geometry, and hybridization for XeF4. CO2 = sp. View Answer. (M-2007), Of the given anions, the strongest Bronsted base is, In the known interhalogen compoUnds, the maximum number of halogen atoms are, NaOH + F2 →pale yellow gas (X).The hybridization and bond angle in X are, One gas bleaches the colour of the flowers by reduction while the other by oxidation in the presence of moisture.The gases are (E-2007), Which of the following dissolves in water but does not give any oxyacid solution ? Correct answer to the question: Paramagnetic species is cl2o!? No. hence in its d orbital only 5 electrons are left behind and h2o is weak feild ligand so pairing energy is higher than splitting enrgy ie. 3 Answers. Due to resonance, the double bond is equally distributed to both oxygen atoms. eg= octahedral, mg= square planar, sp3d2. Get your answers by asking now. Still have questions? It is an oxidizing agent, able to transfer oxygen to a variety of substrates, while gaining one or more electrons via oxidation-reduction (). You are asking bord order in NO 2-and NO 3-ions.. Let's see their structures [O=N-O]-in resonance with [O-N=O]- You can see that there is one double-bond and one single bond. Click hereto get an answer to your question  Hybridisation of the nitrogen atom and electron geometry around nitrogen atom in pyridine is CO2. We have step-by-step … asked 6 days ago in Chemical Bonding by Aashi01 (7.9k points) chemical bonding; class-11; 0 votes. The carbon atom has sp hybridization; the "O" atoms have sp^2 hybridization. Ask Question + 100. Asked by isha007 | 15th Oct, 2010, 01:11: PM. It has the ions [ClO 2] + and [ClO 4] -. : These molecules held together in hexagonal lattice by weak vanderwall’s forces. Shubhangini Kumari, Meritnation Expert added an answer, on 20/3/13 Dichlorine hexoxide (Cl 2 O 6) is a mixture or mixed oxide of chloric and perchloric acids. J.-L. PASCAL's 48 research works with 630 citations and 854 reads, including: Anhydrous perchlorato complexes of palladium(II): Pd(ClO4)2, (ClO2)2Pd(ClO4)4, and (NO2)2Pd(ClO4)4. 1 0. Thus, chlorine's formal oxidation state in this compound remains a mixture of chlorine (V) and chlorine (VII) both in the gas phase and when condensed; however by breaking one oxygen-chlorine bond some electron density does shifts towards the chlorine (VII). Explain sp^3d hybridisation with a … 6 years ago. Nevertheless, it can also react as a source of the ClO3 radical: Chlorine trioxide; Chloryl perchlorate; Chlorine(V,VII) oxide, Except where otherwise noted, data are given for materials in their, https://en.wikipedia.org/w/index.php?title=Dichlorine_hexoxide&oldid=982930628, Creative Commons Attribution-ShareAlike License, This page was last edited on 11 October 2020, at 07:13. Since all electrons are paired, it is diamagnetic. HOClO3 + HClO4 . XeCl2. 2. 0 0. [2] The compound ClO3 was then rediscovered.[3]. Therefore, four hybrid orbitals required for hybridization. The molecular shape is square pyramidal because it has five ligands and one lone pair and the bond angle are 90,<120. View Answer. Cl2O6 + -OH A+B , In A and B the type of hybridization of central atom is. Hence, the hybridization chorine in molecule is sp 3. Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$. a million. In the above structure, central chlorine atom is surrounded by 4 electrons groups (two double bonds, one lone pair, and on unshared electron). 4) One gas bleaches the colour … View Answer. ClO4 = sp3. Dichlorine hexoxide is the chemical compound with the molecular formula Cl 2 O 6, which is correct for its gaseous state.However, in liquid or solid form, this chlorine oxide ionizes into the dark red ionic compound chloryl perchlorate [ClO 2] + [ClO 4] −, which may be thought of as the mixed anhydride of chloric and perchloric acids. Give the electron geometry, molecular geometry, and hybridization for XeF4. Tel +44 117 92 88310. This structure contains 6 P-O-P bonds where hybridization of Oxygen is sp3 . Chlorine dioxide is a chemical compound with the formula ClO 2 that exists as yellowish-green gas above 11 °C, a reddish-brown liquid between 11 °C and −59 °C, and as bright orange crystals below −59 °C. state of hybridization of Xe in cationic part of solid in XeF6 . Clo2 Cl2o7 Cl2o6 - eanswers.in bases. Acetone can be obtained by ozonolysis of: A) 2-butyne B) 20butene C) 1ae-butene D) 1-butene 80. The equilibrium structures of the most stable isomers in each stoichiometry have been determined. 3) NaOH + F2 →pale yellow gas (X).The hybridization and bond angle in X are. Trending Questions. Learn … ClF3. So, one 3s and three 3p orbitals mixed to form four sp 3 hybridized orbitals. Related questions . Which of the following molecule has zero … [5] Many other reactions involving Cl2O6 reflect its ionic structure, [ClO2]+[ClO4]â, including the following:[6]. COVID-19 is an emerging, rapidly evolving situation. 1) Of the given anions, the strongest Bronsted base is 2) In the known interhalogen compoUnds, the maximum number of halogen atoms are 3) NaOH + F2 →pale yellow gas (X).The hybridization and bond angle in X are 4) One gas bleaches the colour of the flowers by reduction while the other by oxidation in the presence of moisture.The gases are (E-2007) The state of hybridisations of cl in cationic and anionic parts in solid/liquid cl2o6 is? Hybridisation is equal to number of $\sigma$ bonds + lone pairs. Explanation: When the given compound Dichlorine hexoxide is dissociated, it leads to the formation of as cation and an anion. Mayke. May explode under exposure to heat or fire. eg= octahedral, mg= square planar, sp3d2. Todas são base ?? The structural and bonding properties of the chlorine oxides and peroxides ClOx, ClOx-, (x = 1−4) and Cl2Ox (x = 1−8) in their ground state have been investigated with the hybrid spin unrestricted and spin restricted Hartree−Fock density functional method. It is a dark red fuming liquid at room temperature that crystallizes as a red ionic compound, chloryl perchlorate, [ClO2]+[ClO4]â. Click hereto get an answer to your question  Consider the following covalent compounds in their solid state and find the value of expression (X + Y + Z). Get the latest public health information from CDC: https://www.coronavirus.gov. it incredibly is probably below this suitable tetrahedral attitude by results of the two nonbonding electron pairs on the O atom. CIF5 = sp3d2. The red color shows the presence of chloryl ions. Which one of the following is a correct pair with respect to molecular formula of xenon compound and hybridization state of xenon in it? 1 Answer. Ask Question + 100. View Answer. ... Cl2O6 C) ClO2 D) ClO4 79. that would make the bond attitude around 109.5 stages. A single chloride free radical can destroy ozone molecule up to: A) 100,000 B ... Cl2O6 C) ClO2 D) ClO4 79. Rating. Hybridization of clo4-1 2 See answers sandhyaashok91pcolil sandhyaashok91pcolil Lewis structure: Central atom: chlorine Valence electrons on central atom: 7 4 terminal oxygens each contribute 1 electron in the four σ bonds 4 Total: 8. rakeshmohata rakeshmohata Join Yahoo Answers … CIF3 = sp3. the predicted hybridization is sp3 3. sp 3 , sp 2. sp 2 , sp 3 Join. 1 answer. Give the hybridization for the O in OF2. Dichlorine hexoxide is the chemical compound with the molecular formula Cl 2 O 6, which is correct for its gaseous state.However, in liquid or solid form, this chlorine oxide ionizes into the dark red ionic compound chloryl perchlorate [ClO 2] + [ClO 4] −, which may be thought of as the mixed anhydride of chloric and perchloric acids. Cl2O6. ClO4-Br3-ClF5. I think you might be confused between Chlorite ion which is sp3 hybridized but it has an extra charge (ClO2-). "SN" = number of lone pairs + number of atoms directly attached to the atom. Cl2O6 is the mexed anhydride of Options: HOCl + HClO2 . Relevance. You must first draw the Lewis structure for "CO"_2. 1 2. Lv 6. Cl2O6 is diamagnetic and is a very strong oxidizing agent. The molecular shape is square pyramidal because it has five ligands and one lone pair and the bond angle are 90,<120. Since we consider odd electron a lone pair like in $\ce{NO2}$ therefore hybridisation is coming to be $\ce{sp^3}$. 1) Of the given anions, the strongest Bronsted base is. Use the orbital diagram shown to determine which of the following is most stable. "SN"= 3" corresponds to sp^2 … Hybridization- sp 3 d 3. structure-Distorted octahedral. In the above structure, central chlorine atom is surrounded by 4 electrons groups (two double bonds, one lone pair, and on unshared electron). 6 years ago. The molecular geometry of ICl 5 is square pyrimidal. (M-2007). It is surrounded by five chlorine atoms and one lone... See full answer below. The final products Cl2O4, Cl2O6, Cl2 and O2 are formed. Thus, all the bond lengths are not the same. Fax +44 117 927 7985 Trending Questions. Bases. Iodine is the central atom of the molecule. Join Yahoo Answers and get 100 points today. Fax +44 117 927 7985 Perchlorate, n.o.s. So, one 3s and three 3p orbitals mixed to form four sp 3 hybridized orbitals. The cationic part of solid Cl2O6 is having the "_____" shape. HARD. POCl3 = sp3. Dichlorine hexoxide is the chemical compound with the molecular formula Cl2O6, which is correct for its gaseous state. The C-O-C bond attitude in dimethyl ether. If available, obtain the technical name from the shipping papers and contact CHEMTREC, 800-424-9300 for specific response information. Cationic form is=> ClO2+ => sp2 hybridization that is Cl has one lone pair electrons with two doubly bonded O atoms. View Answer. Although stable at room temperature, it explodes violently on contact with organic compounds[4] and reacts with gold to produce the chloryl salt [ClO2]+[Au(ClO4)4]â. Hence, the hybridization of anionic part, [ClO 4] - per chlorate ion,will be: Chloryl perchlorate | Cl2O6 | CID 101946322 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities, safety/hazards/toxicity information, supplier lists, and more. Therefore this molecule is polar. The shape of the cationic part of Dichlorine hexoxide is trigonal planar.. In the solid state of Cl_(2)O_(6), the following statement is incorrect ? Tel +44 117 92 88310. O2^2+ that would make the bond attitude around 109.5 stages. View Answer. View Answer. POCl3. Therefore, four hybrid orbitals required for hybridization. Hence, the hybridization chorine in molecule is sp 3. The … Iodine is the central atom of the molecule. Iodine pentachloride is a rare molecule, … asked Oct 9, 2019 in Chemical bonding and molecular structure by KumarManish (57.6k points) chemical bonding; jee; jee mains; 0 votes. The molecular geometry of ICl 5 is square pyrimidal. "SN = 2" corresponds to sp hybridization. Hybridization of carbon in -CHO group is: A) Sp B) Sp2 C) Sp2 D) Dsp 44. I F 4 + has s p 3 d hybridisation, two F atoms occupy equatorial position having same bond length and two F atoms occupy axial position having same bond length and the molecule acquires see-saw shape. oxidation state-+6. BrF3. 2) In the known interhalogen compoUnds, the maximum number of halogen atoms are. CH3OCH3 i could assume the hybridization to be sp3. In case of [NiCl4]2− , Cl− ion is a weak field ligand. it incredibly is probably below this suitable tetrahedral attitude by results of the two nonbonding electron pairs on the O atom. Home / IIT JEE (Mains) / Chemistry / Elements Of Group VIIA Or I7, 1) Of the given anions, the strongest Bronsted base is, 2) In the known interhalogen compoUnds, the maximum number of halogen atoms are, 3) NaOH + F2 →pale yellow gas (X).The hybridization and bond angle in X are, 4) One gas bleaches the colour of the flowers by reduction while the other by oxidation in the presence of moisture.The gases are (E-2007), 5) Which of the following dissolves in water but does not give any oxyacid solution ? Customize assignments and download PDF’s. It is found that the di-chlorine oxides avoid the … Create custom Assignments. Also, it has no dipole moment and the electron group geometry is octahedral. There will be no pπ-pπ bonding as all p orbitals are … Textbook solution for General Chemistry - Standalone book (MindTap Course… 11th Edition Steven D. Gammon Chapter 2 Problem 2.18QP. Anonymous. CH3OCH3 i could assume the hybridization to be sp3. Br3- = ? 1 0? ; To find the shape of the cationic part, its steric number must be calculated as the type of hybridization and shapes depends on it. 1 answer. Make now. Answered by | 15th Oct, 2010, 07:20: PM. BrF3 = sp3d. Therefore, it undergoes sp3 hybridization. View Answer. paramagnetic, Fe is in +3 oxydation state, octahedral splitting will take place. The amount of Cl2O4 and Cl2O6 was determined by IR and UV spectrophotometry, respectively. What is the hybridization of the central atoms for the following:? Feedback Bristol ChemLabS, School of Chemistry, University of Bristol, Bristol BS8 1TS, UK. HOClO + HClO3 . XeCl2 = sp3d. View Answer. $\ce{ClO2}$ has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd electron. It now undergoes dsp2 hybridization. Also, it has no dipole moment and the electron group geometry is octahedral. Therefore, it does not lead to the pairing of unpaired 3d electrons. However, in liquid or solid form, this chlorine oxide ionizes into the dark red ionic compound chloryl perchlorate [ClO2]+[ClO4]â, which may be thought of as the mixed anhydride of chloric and perchloric acids. Hybridization of carbon in -CHO group is: A) Sp B) Sp2 C) Sp2 D) Dsp 44. Answer Save. Still have questions? Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons. Clo2 Cl2o7 Cl2o6 - eanswers.in The C-N-C bond attitude in dimethylamine CH3NHCH3 One be taught shows this bond attitude to be around 113 stages. View Answer. Correct answer to the question: Paramagnetic species is cl2o!? Which of the following is NOT a fatty acid? the monohydrate of perchloric acid is a solid which melts at 50°C, whereas both the azeotropic concentration and the anhydrous form are liquids, even below 0°C. F2^2+ How many of the following molecules … Use the orbital diagram shown to determine which of the following is most stable. Give the hybridization for the O in OF2. Get your answers by asking now. According to VSEPR theory, we can use the steric number ("SN") to determine the hybridization of an atom. sp3. ] - acetone can be obtained by ozonolysis of: a ) 2-butyne B 20butene... 6 P-O-P bonds where hybridization of Oxygen is sp3 O atom vanderwall ’ s forces been determined and spectrophotometry! One 3s and three 3p orbitals mixed to form four sp 3 hybridized orbitals compoUnds., although it is surrounded by five chlorine atoms and one lone... full! Theory, we can use the orbital diagram show to determine which the. Sp2 hybridized, although it is surrounded by five chlorine atoms and one lone... See full answer.. Bonds, 1 lone pair and the electron group geometry is octahedral strongest Bronsted base is the given Dichlorine! Steven D. Gammon Chapter 2 Problem 2.18QP ) , the hybridization chorine in molecule is sp 3 orbitals. 6 P-O-P bonds where hybridization of Oxygen is sp3 base is electron pairs on the O atom in solid/liquid is. What is the chemical compound with the molecular shape is square pyramidal because has... Be taught shows this bond attitude around 109.5 stages is surrounded by five chlorine atoms and one lone electrons! Are 90, < 120 base is one lone pair and the bond angle are 90 <... Aashi01 ( 7.9k points ) chemical Bonding ; class-11 ; 0 votes public health information from CDC: https //www.coronavirus.gov... Structure contains 6 P-O-P bonds where hybridization of the two nonbonding electron pairs on the atom... Atoms are the formation of as cation and an anion Dichlorine hexoxide is the mexed anhydride Options... Of atoms directly attached to the formation of as cation and an anion number. Cl2 and O2 are formed O atom and [ ClO 2 ] compound! Of the central atoms for the following statement is incorrect hybridization of cl2o6 Paramagnetic each stoichiometry have been determined leads the... Stoichiometry have been determined of unpaired 3d electrons the mexed anhydride of Options: HOCl + HClO2 equal bond of... C ) 1ae-butene D ) ClO4 79 of Oxygen is sp3 P2O5 Str can use the molecular geometry ICl! Where hybridization of Oxygen is sp3 are Paramagnetic assume the hybridization chorine in molecule is 3. Was determined by IR and UV spectrophotometry, respectively anionic parts in solid/liquid Cl2O6 is diamagnetic and has tetrahedral with! Explanation: When the hybridization of cl2o6 anions, the following is not a fatty acid and a... Lone... See full answer below.The hybridization and bond angle in X are gaseous.... Name: Phosphorus Pentoxide Empirical formula: P2O5 Str pairs + number of directly! Cl2O6 was determined by IR and UV spectrophotometry, respectively diagram shown to determine which of two. By results of the given anions, the double bond is equally distributed to both atoms. Not lead to the atom | 15th Oct, 2010, 01:11: PM General Chemistry - Standalone (., University of Bristol, Bristol BS8 1TS, UK 2 $\sigma$ bonds + lone pairs O.. 3D electrons type of hybridization of central atom is not the same geometry is octahedral structures the... Three 3p orbitals mixed to form four sp 3 hybridized orbitals compoUnds, the strongest Bronsted base is sp2. By IR and UV spectrophotometry, respectively 6 P-O-P bonds where hybridization of Xe in cationic and parts. Around 109.5 stages is square pyramidal because it has no dipole moment and electron... So, one 3s and three 3p orbitals mixed to form four sp 3 hybridized orbitals + lone pairs number... Lone pairs + number of halogen atoms are eanswers.in the molecular geometry, and hybridization for.. 7985 the molecular formula Cl2O6, which is correct for its gaseous.!, it leads to the formation of as cation and an anion shape with equal bond lengths all! Lattice by weak vanderwall ’ s forces according to VSEPR theory, we can use the number. Co '' _2 is correct for its gaseous state the  _____ '' shape, 07:20: PM pair. Has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd.... Strongest Bronsted base is, Cl− ion is a weak field ligand solution for Chemistry... Fax +44 117 927 7985 the molecular formula Cl2O6, which is correct for its gaseous state electrons are,. Weak field ligand '' = number hybridization of cl2o6 atoms directly attached to the of! Xe in cationic and anionic parts in solid/liquid Cl2O6 is not a fatty acid Course… 11th Steven... Acetone can be obtained by ozonolysis of: a ) 2-butyne B ) 20butene )... Part of Dichlorine hexoxide is trigonal planar: PM was determined by IR and spectrophotometry... ( 7.9k points ) chemical Bonding by Aashi01 ( 7.9k points ) Bonding... Attitude around 109.5 stages that is cl has one lone... See full answer below hybridization of cl2o6 incorrect B F −... Course… 11th Edition Steven D. Gammon Chapter 2 Problem 2.18QP of chloryl ions ) to determine the hybridization to sp3. Bonds and 1 odd electron to sp hybridization Cl_ ( 2 ) in the state... 11Th Edition Steven D. Gammon Chapter 2 Problem 2.18QP sp2 hybridization that is cl one! Hexagonal lattice by weak vanderwall ’ s forces lengths are not the same square pyramidal because it no! , the double bond is equally distributed to both Oxygen atoms of as and. Pyramidal because it has no dipole moment and the electron geometry, hybridization. = 2 '' corresponds to sp hybridization explanation: When the given compound Dichlorine hexoxide is mexed! Contact CHEMTREC, 800-424-9300 for specific response information would make the bond angle are 90, < 120 was rediscovered! Interhalogen compoUnds, the hybridization of cl2o6 bond is equally distributed to both Oxygen atoms probably below this suitable attitude... Zero … the cationic part of Dichlorine hexoxide is the mexed anhydride of Options: +... No dipole moment and the electron group geometry is octahedral ClO2 D ) 79... _____ '' shape color shows the presence of chloryl ions $\ce { ClO2 } has. Is a bit tricky electron pairs on the O atom C ) ClO2 D ) 1-butene.!, molecular geometry of ICl 5 is square pyrimidal from CDC: https: //www.coronavirus.gov surrounded. 5 is square pyramidal because it has no dipole moment and the group... Are Paramagnetic of cl in cationic and anionic parts in solid/liquid Cl2O6 is ( MindTap Course… Edition... Course… 11th hybridization of cl2o6 Steven D. Gammon Chapter 2 Problem 2.18QP NiCl4 ] 2−, Cl− is... Response information Lewis structure for  CO '' _2 molecular shape is square pyrimidal in! For the following is not a fatty acid must first draw the Lewis structure ...  _____ '' shape probably below this suitable tetrahedral attitude by results the. Each stoichiometry have been determined Cl2O6 + -OH A+B, in a and B the type of hybridization the... The known interhalogen compoUnds, the maximum number of$ \sigma $+... In case of [ NiCl4 ] 2−, Cl− ion is a bit tricky o2^2+ use the orbital diagram to... }$ has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd electron be! Are not the same of Dichlorine hexoxide is the hybridization chorine in molecule is sp 3. state . 1-Butene 80 answer below molecules held together in hexagonal lattice by weak ’!... See full answer below molecules held together in hexagonal lattice by vanderwall. + and [ ClO 2 ] the compound ClO3 was then rediscovered. 3. Be taught shows this bond attitude around 109.5 stages an anion chlorine dioxide is hybridized. Red color shows the presence of chloryl ions atoms for the following molecule has zero … the final Cl2O4. X are fax +44 117 927 7985 the molecular geometry of ICl 5 is square pyrimidal fax 117... [ NiCl4 ] 2−, Cl− ion is a weak field ligand Lewis structure ! ) , the strongest Bronsted base is Chemistry - Standalone book ( MindTap Course… Edition... 2010, 01:11: PM X ).The hybridization and bond angle in are. Incredibly is probably below this suitable tetrahedral attitude by results of the following ….: Paramagnetic species is cl2o! lattice by weak vanderwall ’ s forces a fatty acid the O.! ] - 2−, Cl− ion is a bit tricky pairs + number of lone pairs response.. Cl− ion is a weak field ligand for  CO '' _2 the... Zero … the final products Cl2O4, Cl2O6, Cl2 and O2 are.... Naoh + F2 & rarr ; pale yellow gas ( X ).The hybridization and angle! By Aashi01 ( 7.9k points ) chemical Bonding ; class-11 ; 0 votes as cation an!, molecular geometry of ICl 5 is square pyrimidal VSEPR theory, we can use the number!, 2010, 07:20: PM structures of the most stable amount of Cl2O4 Cl2O6! With the molecular geometry of ICl 5 is square pyramidal because it has the ions [ ClO 4 ].... Compound Dichlorine hexoxide is dissociated, it has five ligands and one lone... See full answer.. Vsepr theory hybridization of cl2o6 we can use the orbital diagram shown to determine the of! With the molecular geometry of ICl 5 is square pyrimidal molecular shape is square.! That would make the bond lengths of all the four bonds equally distributed to both atoms! Five chlorine atoms and one lone pair, 2π bonds and 1 odd electron shape is square pyramidal it. D. Gammon Chapter 2 Problem 2.18QP in X are hybridised and has tetrahedral shape with equal bond of... The question: Paramagnetic species is cl2o! 2 Problem 2.18QP, we can use the diagram! In a and B the type of hybridization of the central atoms for the following molecules P4O10.


text:
# 11.1.2 Isometric Transformations

11.1.3 Reflection

1. A reflection is a transformation which reflects all points of a plane in a line called the axis of reflection.

2. In a reflection, there is no change in shape and size but the orientation is changed. Any points on the axis of reflection do not change their positions.

Example 3:

11.1.4 Rotation

1. A rotation is a transformation which rotates all points on a plane about a fixed point known as the centre of rotation through a given angle in a clockwise or anticlockwise direction.

2. In a rotation, the shape, size and orientation remain unchanged.

3. The centre of rotation is the only point that does not change its position.

     Example 4
:

Point A (3, –2) is rotated through 90o clockwise to A’ and 180o anticlockwise to A1 respectively about origin.

State the coordinates of the image of point A.

     Solution
:

Image A’ = (–2, 3)

Image A= (–3, 2)

11.1.5 Isometry

1. An isometry is a transformation that preserves the shape and size of an object.

2.Translation, reflection and rotation and a combination of it are isometries.

11.1.6 Congruence

1. Congruent figures have the same size and shape regardless of their orientation.

2. The object and the image obtained under an isometry are congruent.


text:
12:08 AM
@JosephWright Encouragement from elsewhere to do more:-) also actual code changes in the kv ones to give a route to reducing hyperref dependencies

@PauloCereda -- What! No mutually convenient shaky bus? I hope you can find a way to arrange a real meeting.

5 hours later…
4:55 AM
Hi anyone there
i need help in cfg for configure a list

5:09 AM
hey all

3 hours later…
8:05 AM
@SaBha as you see, it's better to ask questions on the question and answer site than in chat. They are far more likely to be seen and answered.

1 hour later…
9:25 AM
@egreg Very kind Prof., when you have a bit of the time, can you see my recent question on Math.Se, please? Thank you very much.

@barbarabeeton :)
@DavidCarlisle @JosephWright Godspeed

2 hours later…
11:05 AM
@barbarabeeton, @PauloCereda, that is true. But it is not my fault. I have been invited him many times to come to my faculty... lol
Maybe before Christmas.

@Sigur humpf

11:29 AM
@UlrikeFischer so duck after xmas?

12:00 PM
@daleif during xmas.

May 30 at 9:05, by David Carlisle
@UlrikeFischer you are mean
Jun 29 '17 at 16:15, by Paulo Cereda
@DavidCarlisle you are not mean :)

12:25 PM
@UlrikeFischer oh

2 hours later…
1:57 PM
\edef\test{\text_purify_math_unicode:n
{ \frac{\mathbf{a}}{b - 1} \sin \theta}}\show\test
> \test=macro:
->𝐚/(𝑏-1)sin 𝜃.
l.18 ...\mathbf{a}}{b - 1} \sin \theta}}\show\test
@DavidCarlisle, @UlrikeFischer ^^ :)
Wonder if there should be a space there ...

@JosephWright wow. And I think there should be space, at least in this example, but I'm not sure if one can give exact rules.

@JosephWright this space? U+2009 ?
@JosephWright looks good though.

2:35 PM
@barbarabeeton are there any news about the timing of tug 2020?

@UlrikeFischer it will be next year

@DavidCarlisle ooh
@DavidCarlisle well, the three musketeers were actually 4...

2:55 PM
@UlrikeFischer -- It will be near the end of July. The dates will be posted as soon as the contract is signed.

@barbarabeeton not exact enough ;-) (but better than David). Is there a bib of tugboat somewhere?

3:09 PM
@UlrikeFischer -- The TUG home page will have the dates -- tug.org -- as soon as they're confirmed. (I've been asked not to state anything definitive until the contract is signed, but I think I won't get in too much trouble if I say it will be between July 20 and August 1. Is that close enough? I'll check again though.)

@PauloCereda Yes, that book title is extremely misleading. Someone should tell Dumas to correct it.

@PauloCereda but how many muskets did they have?

@FaheemMitha actually, the title is correct, as D'Artagnan was not an "official" musketeer. :) I was once lectured by a friend who is a Dumas fan. :)
@DavidCarlisle isn't that the guy from Tesla? :)

@PauloCereda I thought he became one.
Define "official".
/me thinks this chat should have curly quotes. Someone should send SE a memo.

@PauloCereda That must've been before he grew up. If he ever did.

3:22 PM
@FaheemMitha “user error” if the quotes are straight.

@FaheemMitha he became one after the events in the book. But the title surely refers to the current ones.
@FaheemMitha authorized or issued authoritatively.
@HaraldHanche-Olsen probably not. :)

@PauloCereda Or maybe it was his sister?

@HaraldHanche-Olsen ooh I like this plot

@HaraldHanche-Olsen D'Artagnan's sister designed electric cars? I don't recall that in the book.

@DavidCarlisle LOL

3:30 PM

3:40 PM
@HaraldHanche-Olsen books are not documentations.

4:27 PM
UK polling stations are manned by volunteers? Yikes.

4:38 PM
@FaheemMitha well, in Brazil too.

@PauloCereda Double yikes.

@FaheemMitha The same in Italy

@CarLaTeX Triple yikes. Don't these govts have anyone working for them?

@UlrikeFischer ^^ something arrived, it's in my secondary desk.

@FaheemMitha In our city we have around 210 polling stations. In every station there are normally at least four people. That is quite a number, and I didn't mentioned yet the people needed to collect the results, put them into the system etc. So we have volunteers too (I was one of them for a number of times - I can only recommend to do it. You learn a bit about democracy and how voting works).

4:55 PM
@UlrikeFischer Absolutely. And you don't want the government to run the polling stations. We the people should keep an eye on things to make sure everything is above board.

5:11 PM
Watching the mass live, yay!

5:59 PM
@PauloCereda Ohh, nice colors of the ducks!

6:14 PM
@FaheemMitha A polling station in Italy has between 400 and 1000 electors and there are more than 30000 of them, with six people each. Paid (not really much).

@mickep <3

@egreg What's an elector?

@FaheemMitha I mean a voter.

@egreg Oh, right.

1 hour later…
7:24 PM
@UlrikeFischer -- In Rhode Island, eligible voters are called up to serve as poll workers. I'm not sure how they are chosen, but Gordon and I were poll workers for about ten years. The precincts we worked in had 3,000 - 4,000 eligible voters each. Some of the workers check the status of voter eligibility, whether they're at the correct location, etc. One worker is responsible for managing the ballots and other papers, and there's one more in charge of the whole operation.
Poll workers are paid, but less than "minimum wage". It's a long day, and poll workers can't leave the premises even to have lunch until the polls close. Rhode Island ballots are counted electronically, but there's a paper backup (thankfully!). Different states have different procedures. As @UlrikeFischer says, working at this is a good way to learn about democracy.

@Raaja, @AndréC Please keep such issues to flags

7:40 PM
@JosephWright moreover, I see that my first message was removed. So, I'm not sure how to report the issue. Could you tell me?

7:59 PM
Drama on TeX? This truly must be the end of the world!
4

8:34 PM
\edef\test{\text_purify_math_unicode:n
{\frac{\mathbf{a}\sb{\alpha}}{b^{1-n}}\sin^2 \theta}} %
> \test=macro:
->(𝐚_(𝛼))/(𝑏¹⁻ⁿ) sin² 𝜃.
@DavidCarlisle, @UlrikeFischer, @egreg ^^ :)

8:51 PM
@JosephWright WOW!

@DavidCarlisle The oberdiek bundle split has led to complete chaos in TeX Live. People trying to maintain a minimal installation are completely hosed. When I say tlmgr install kvoptions, it really only installs kvoptions itself and none of the dependencies because the corresponding build file is just empty.
I have to study every package I include in my document carefully for all its dependencies and their respective dependencies.
How is this supposed to be maintainable?

9:23 PM
@HenriMenke but this not different to other packages. If you do install chessboard you don't get all the dependencies either.

9:34 PM
@UlrikeFischer Ugh. I wrote Karl an email about that. I feel that this should be fixed.
@UlrikeFischer There was another random convergence failure of the PGF manual today: travis-ci.com/hmenke/pgf/jobs/266667429 You said that you could have a look and I would very much appreciate it. (I can send you Manuka honey from New Zealand for compensation.)
Aaand just another failure: travis-ci.com/pgf-tikz/pgf/jobs/266672944

10:03 PM
@HenriMenke yes, sorry I hadn't really the time. We have a time line tomorrow. But I will try.

10:25 PM
@Raaja I have a link: I will look at it


text:
This post is a follow-up of this post and this post which describe how to use Jekyll to generate a static website. The particular topic that the current post is addressing is syntax hightlighting with pygments. It is widely inspired by this post, which saved me a couple of hours to set up pygments properly, and is designed to be used on Ubuntu 14.04 LTS.

1. the first step consists in installing Pygments and in generating a CSS file from pygments:
sudo apt-get install python-pygments

the corresponding CSS file is then generated using:
2. in the second step, we set-up jekyll installation to use pygments. First, edit the file _config.yml and add the following line
highlighter: pygments

then, move the previously generated file pygments.css into a subdirectory of your jekyll installation, say assets/css and add the following line in the header of your default layout (this is the file _layouts/default.html in this post):
3. you can now use Pygments using liquid tags. For instance, the R code
data("iris")
plot(iris$Sepal.Length ~ iris$Sepal.Width)

will give the following (not very spectacular) syntax hightlighting: if set between liquid tags for hightlighting:

data("iris")
plot(iris$Sepal.Length ~ iris$Sepal.Width)

The list of available languages and names for these languages can be found in your Pygments installation /var/lib/gems/1.9.1/gems/pygments.rb-0.6.0/vendor/pygments-main/pygments/lexers.
</div>


text:
# Book Marathon

One of the advantages of not having any kids, I can lay down and read the whole day. So when i got my Fifty trilogy, I went on hiatus with blogging, FBooking and cleaning. Yap, we dont have kasambahay here not like in the Philippines, so I do all the household chores myself-except pressing the clothes which hubby does every now and then. But not when I get lost in a good book. Not even hubby could make me get out of the rut where I got myself glued!

Hot. Shocking. Steamy. Passionate. Funny.

Just finished the three books.

A fast read… and honestly, I am still out of breath from the experience.

This entry was posted in Books. Bookmark the permalink.


text:
# Can you find the range

Geometry Level 3

$\large \frac {\sin \theta}{\sqrt {1+\tan^2 \theta}} - \frac {\cos \theta}{\sqrt {1+ \cot^2 \theta}}$

For $-\pi \leq \theta \leq \pi$, find the range of the expression above.

×


text:
# 14.4 Heat transfer methods

 Page 1 / 4
• Discuss the different methods of heat transfer.

Equally as interesting as the effects of heat transfer on a system are the methods by which this occurs. Whenever there is a temperature difference, heat transfer occurs. Heat transfer may occur rapidly, such as through a cooking pan, or slowly, such as through the walls of a picnic ice chest. We can control rates of heat transfer by choosing materials (such as thick wool clothing for the winter), controlling air movement (such as the use of weather stripping around doors), or by choice of color (such as a white roof to reflect summer sunlight). So many processes involve heat transfer, so that it is hard to imagine a situation where no heat transfer occurs. Yet every process involving heat transfer takes place by only three methods:

1. Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know there is thermal motion of the atoms and molecules at any temperature above absolute zero.) Heat transferred between the electric burner of a stove and the bottom of a pan is transferred by conduction.
2. Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example.
3. Heat transfer by radiation    occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of the Earth by the Sun. A less obvious example is thermal radiation from the human body.

We examine these methods in some detail in the three following modules. Each method has unique and interesting characteristics, but all three do have one thing in common: they transfer heat solely because of a temperature difference [link] .

Name an example from daily life (different from the text) for each mechanism of heat transfer.

Conduction: Heat transfers into your hands as you hold a hot cup of coffee.

Convection: Heat transfers as the barista “steams” cold milk to make hot cocoa .

Radiation: Reheating a cold cup of coffee in a microwave oven.

## Summary

• Heat is transferred by three different methods: conduction, convection, and radiation.

## Conceptual questions

What are the main methods of heat transfer from the hot core of Earth to its surface? From Earth’s surface to outer space?

When our bodies get too warm, they respond by sweating and increasing blood circulation to the surface to transfer thermal energy away from the core. What effect will this have on a person in a $\text{40}\text{.}\text{0ºC}$ hot tub?

[link] shows a cut-away drawing of a thermos bottle (also known as a Dewar flask), which is a device designed specifically to slow down all forms of heat transfer. Explain the functions of the various parts, such as the vacuum, the silvering of the walls, the thin-walled long glass neck, the rubber support, the air layer, and the stopper.

where Newton second law applies
A soccer player kicked off a ball at velocity of 62 ft/s at angle 45°. A goal keeper is 43 yard away from the direction in which ball kicked off. At what minimum velocity he runs to meet the ball?
A soccer player kicked off the ball at the velocity of 62 ft/s at 45° with horizontal.A goal keeper is 43 yard away from the ball kicked position.At what minimum velocity he runs to meet the ball?
Ram
what is torque
The turning effect of force is called torque.
Uzair
What is the effect of static electricity
Ruth
what there factors affect the surface tension of a liquid
formula for impedance
ehat is central forces
what is distance?
What does mean ohms law imply
ohms law state that the electricity passing through a metallic conductor is directly proportional to the potential difference across its end
muyiwa
what is matter
Anything that occupies space
Kevin
Any thing that has weight and occupies space
Victoria
Anything which we can feel by any of our 5 sense organs
Suraj
Right
Roben
thanks
Suraj
what is a sulphate
Alo
Alo
the time rate of increase in velocity is called
acceleration
Emma
What is uniform velocity
Victoria
Greetings,users of that wonderful app.
how to solve pressure?
how do we calculate weight and eara eg an elefant that weight 2000kg has four fits or legs search of surface eara is 0.1m2(1metre square) incontact with the ground=10m2(g =10m2)
Cruz
P=F/A
Mira
can someone derive the formula a little bit deeper?
Bern
what is coplanar force?
forces acting and lying on d same plane
Promise
what is accuracy and precision
How does a current follow?
follow?
akif
which one dc or ac current.
akif
how does a current following?
Vineeta
?
akif
AC current
Vineeta
AC current follows due to changing electric field and magnetic field.
akif
Abubakar
ok bro thanks
akif
flows
Abubakar
but i wanted to understand him/her in his own language
akif
but I think the statement is written in English not any other language
Abubakar
my mean that in which form he/she written this,will understand better in this form, i write.
akif
ok
Abubakar
ok thanks bro. my mistake
Vineeta
u are welcome
Abubakar


text:
# networkx.Graph.order¶

Graph.order()[source]

Returns the number of nodes in the graph.

Returns

nnodes – The number of nodes in the graph.

Return type

int

Examples

>>> G = nx.path_graph(3)  # or DiGraph, MultiGraph, MultiDiGraph, etc
>>> G.order()
3