April 29

I'm attempting to find if the distance from a point to a cubic curve exceeds a certain threshold, and I've gotten it down to a sextic equation in t. I need to determine if there is a root (or roots) between t = 0 and t = 1. I do not need to know the value of this, only that it exists. Is there a fast method to do this? As for why I'm doing this, if that gives any clarity, it's here: http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Computing#Detecting_click_on_cubic_curve. Thanks for any help! KyuubiSeal (talk) 00:40, 29 April 2011 (UTC)[reply]

Yes, you can use Sturm chains for this, which gives a method for any polynomial equation (with real coefficients). Our article about this is called Sturm's theorem. – b_jonas 09:12, 29 April 2011 (UTC)[reply]

Btw, here is the working link for the question OP refers to: Wikipedia:Reference_desk/Computing#Detecting_click_on_cubic_curve. – soon to be Wikipedia:Reference desk/Archives/Computing/2011 April 29#Detecting_click_on_cubic_curve.. – b_jonas 09:15, 29 April 2011 (UTC)[reply]

I read the Sturm's theorem article, but I don't understand how I would apply it to this situation? KyuubiSeal (talk) 02:47, 30 April 2011 (UTC)[reply]

I would solve the equation numerically and look at the six roots. Use the

Durand-Kerner method or the p. rootfinder of the J (programming language). I think that "saying something about the roots without actually solving the equation" is a pre-computer sport. Bo Jacoby (talk) 07:27, 30 April 2011 (UTC).[reply

]

All the coefficients of the equation depend on a couple variables, so I can't solve the roots beforehand. :\ KyuubiSeal (talk) 13:33, 30 April 2011 (UTC)[reply]

Sturm's theorem gives an algorithm to find number of positive roots of the equation. You compute this, and you then compute the number of real roots greater than 1. If the latter number is less, then there's a root between 0 and 1. If they're equal, there's none. – b_jonas 18:36, 30 April 2011 (UTC)[reply]

What condition on the dimensions of matrices A and B preclude the existence of ${\displaystyle (AB)^{-1}}$? Widener (talk) 06:38, 29 April 2011 (UTC)[reply]

When you say "the existence of ${\displaystyle (AB)^{-1}}$" I take it you mean that AB would be invertible. If so, then an invertible matrix is usually assumed to be a square matrix. So there are immediately simple conditions on the dimensions of A and B which, if not satisfied, would prevent AB being invertible. Gandalf61 (talk) 08:47, 29 April 2011 (UTC)[reply]

I meant assuming AB is square. Perhaps I should have said what conditions on the dimensions of A and B ensure det(AB)=0 ? Widener (talk) 11:03, 29 April 2011 (UTC)[reply]

Okay, I am confused by that. There are obvious conditions on the dimensions of A and B that ensure that AB is square. If A and B meet these conditions, I don't see what further conditions you can place on their dimensions that will ensure that det(AB) is 0 or is not 0. Once you have ensured that AB is square then det(AB) depends on the elements of A and B, but not on their dimensions. Gandalf61 (talk) 11:56, 29 April 2011 (UTC)[reply]

The rank of AB is less than or equal to the ranks of both A and B. So if m<n, A is an n by m matrix, and B is an m by n matrix, then AB has rank at most m even though it's an n by n matrix. It would therefore have no inverse.--RDBury (talk) 12:16, 29 April 2011 (UTC)[reply]

Ah, yes, I missed that. Gandalf61 (talk) 12:42, 29 April 2011 (UTC)[reply]

A matrix is used to represent a linear map between two vector spaces. Let α be a linear map between R^k and Rⁿ, in symbols α : R^k → Rⁿ. Assume we have the standard bases in R^k and Rⁿ. The linear map α will have a matrix given by A, which will be an n × k matrix. Let β be a linear map between Rⁿ and R^k, in symbols β : Rⁿ → R^k. Assume we have the standard bases in Rⁿ and R^k. The linear map β will have a matrix given by B, which will be an k × n matrix. Let v be a vector in Rⁿ, and consider Bv. This will be a vector in R^k because β : Rⁿ → R^k. Since Bv is in R^k, it follows that A(Bv) = (AB)v will be in Rⁿ since α : R^k → Rⁿ. It should be clear then that AB takes the vector v, in Rⁿ, and give another vector in Rⁿ, namely (AB)v; so AB is the matrix of a linear map from Rⁿ to Rⁿ. In fact:

${\displaystyle {\mathbf {R} }^{n}{\stackrel {\beta }{\longrightarrow }}{\mathbf {R} }^{k}{\stackrel {\alpha }{\longrightarrow }}{\mathbf {R} }^{n}\,.}$

In other words, the

injective. If β is not injective then it can't have an inverse: you can't say which guest sat in which chair (maybe several shared the same one). Because α goes between R^k and Rⁿ you get the same equality. So, as RDBury said: you need k ≤ n for any hope of an inverse. If k > n then you have no chance. — Fly by Night (talk) 03:06, 30 April 2011 (UTC)[reply

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Thanks, everyone. Widener (talk) 03:52, 30 April 2011 (UTC)[reply]

Could anyone tell me how to show whether these two connected double-covers of the wedge union of 2 circles (S^1 v S^1) are homeomorphic or not homeomorphic? That is, the top one is 3 circles joined at 2 points (corresponding to 2 of the connected double-covers of S^1 v S^1 and the bottom is 2 "double-loops", joined at 2 points, one on each loop. I apologise for the image quality and the fact it appears to be displaying twice, I haven't used images in wikipedia before - I'm not much of an artist but since there are only 3 connected double-covers of the wedge union, one of which is the mirror image of the other and one which is distinct, hopefully you get my meaning. Nothing particularly obvious came to mind: both for example have 2 points which look locally like 2 lines crossing and so on. Could anyone suggest a topological invariant not shared by these 2 shapes or suggest a way to show they're homeomorphic please? Many thanks. Simba31415 (talk) 18:03, 29 April 2011 (UTC)[reply]

Just to describe these: the second picture is the double cover obtained by taking two copies of the double cover of the circle, and gluing them together at two points. The first picture takes the double cover of the first circle, and then glues two copies of the second circle two it. These have different connectivities: the first can be separated by deleting one or the other of the two vertex points. The second remains connected if any point is deleted. Sławomir Biały (talk) 18:30, 29 April 2011 (UTC)[reply]

And because of bijectivity I take it that connectivity is preserved under homeomorphism? Simba31415 (talk) 18:39, 29 April 2011 (UTC)[reply]

It's a topological invariant, yes. Sławomir Biały (talk) 18:41, 29 April 2011 (UTC)[reply]

Great, thankyou! What was I doing wrong with the picture out of interest? It's no longer doubled up. Simba31415 (talk) 18:50, 29 April 2011 (UTC)[reply]

I'm not sure what's wrong with the picture, other than it not being a good picture :-) Sławomir Biały (talk) 19:10, 29 April 2011 (UTC)[reply]

Haha, can't argue with that! Simba31415 (talk) 19:34, 29 April 2011 (UTC)[reply]

Well, in case you still want to know what was wrong with the picture, here's my guess. In this edit, you added the image to the page without creating a new heading – you weren't logged in, so it could have been someone else, but that's unlikely. This was probably an accidental edit. Then a bot signed it for you. Then in this edit you asked the actual question properly, with a heading, and including the image. You didn't notice that the image was already there above the heading, probably because you only ever clicked on the "new section" and the "edit section" links, which don't show the wiki source of the section above. Kainaw then fixed the mistake in this edit by removing the first image. – b_jonas 09:11, 1 May 2011 (UTC) [reply]

In the UK lottery, 6 balls are drawn from 49. What is the maximum possible number of draws such that no two drawn sets of six balls share three (or more) balls in common? How does one go about answering the general question (with arbitrary numbers in place of the 6, 49 and 3 of this specific question)? —Preceding unsigned comment added by 86.184.110.152 (talk) 20:22, 29 April 2011 (UTC)[reply]

This sound like a typical homework question. Take a look at the inclusion–exclusion principle article and let us know how you get on. We will help you with your homework if you show us what you have attempted. Otherwise you lose the chance to learn for yourself. — Fly by Night (talk) 23:24, 29 April 2011 (UTC)[reply]

This is not homework. I have not done homework for 40 years. —Preceding unsigned comment added by 86.184.110.152 (talk) 00:32, 30 April 2011 (UTC)[reply]

I have noticed many homework posts from IP address that make a single post, or maybe just a few posts to the reference desk. I notice that a lot of single posters make a question, get an answer, and don't even say thanks. They are either using us to do their homework, or they are hiding their identities. I don't like this idea of being used. We should be directing people towards, and indeed promoting, relevant articles. I think we should review our policy on this reference desk. What do we think of asking people to register an account before they post a question? It's just an idea, I don't have a manifesto. I'd just like to hear my piers' ideas. — Fly by Night (talk) 23:31, 29 April 2011 (UTC)[reply]

I don't mind at all if people don't say thanks. When I feel like people are just idly taking advantage, I stop helping them. Staecker (talk) 00:36, 30 April 2011 (UTC)[reply]

I generally try to say a big thankyou every time someone helps me, but to the best of my recollection only once or twice has it been acknowledged, so I got the impression people weren't that fussed. I myself am generally more of an 'asker' than a contributor, primarily because I'm only a student and feel a lot of the time like there isn't a great deal of insight I can give on many problems - in my experience people don't appear to have noticed the fact that they've been thanked, but perhaps they appreciate it and just don't say anything! At any rate, as long as you only post help on the questions rather than answering them in their entirety, I don't see why it does any harm to continue giving advice on homework. That's just my opinion, anyway :) Spamalert101 (talk) 00:41, 30 April 2011 (UTC)[reply]

I do see your point though re: the above question, that would certainly annoy me in your position. Spamalert101 (talk) 00:47, 30 April 2011 (UTC)[reply]

When I reply to a question I expect to be informed if my answer helped. It is very frustrating when no such feedback is given. It is an added bonus if they actually use the word "thanks" and I think common courtesy requires it.

I don't generally acknowledge being thanked because that seems like a waste of space to me. I can see how cultural difference can cause one to view the lack of "you're welcome" the same way I view the lack of "thank you". -- Meni Rosenfeld (talk) 08:53, 1 May 2011 (UTC)[reply]

Posters failing to acknowledge replies happens frequently in most Internet forums, in my experience. If you get upset at not being thanked then probably you shouldn't be replying. Remember that many people have dynamic IP addresses, so you have no way of telling how many posts they make over time. Talk of people "hiding their identities" seems over-dramatic. Also, you may be incorrectly diagnosing some questions as homework, as you did with mine above. If the reference desk was flooded with uncontrollable spam and crap then I'd agree with the idea of requiring registration, but it doesn't seem that way to me. 86.184.110.152 (talk) 01:06, 30 April 2011 (UTC)[reply]

While that may be true, there's no need to be so terse about a perfectly reasonable assumption. It sounds like a homework question, and given that you're asking for help here it doesn't seem unreasonable that the people helping you expect you to exercise some semblance of independent thought, and if not try to make some visible headway with the question yourself so that people can help you, at least say something like 'I tried but I haven't approached a problem like this in years, so it's unlikely I'll be able to be of any use in solving this myself' so people don't waste their time directing you to useful articles to help you figure it out for yourself. Many people come here for help with homework and it's more than likely a question of that nature appearing here is some kind of learning exercise, particularly as you ask for a general method to solve such problems, but then appear to show no interest in actually learning how to do so. Delaypoems101 (talk) 01:23, 30 April 2011 (UTC)[reply]

I can see how it could look abrupt. It wasn't really meant that way. Equally, responders need to be cautious about making inappropriate and patronising remarks such as "otherwise you lose the chance to learn for yourself" when they know nothing about the person to whom they are replying or the circumstances of the question. 86.184.110.152 (talk) 02:07, 30 April 2011 (UTC)[reply]

I mainly work at Wikipedia:Help desk and not the reference desks but there are some similarities. I always watch for follow-ups and it's nice to be thanked but the main concern is knowing whether my reply was read and useful. "Problem solved" is as good as "Thank you" to me (getting both is even better). Definitely appreciated, but I rarely acknowledge it. I have answered a few math questions at Yahoo! Answers but the poster usually gave no indication whether any of the replies had been read (and they even have a rating system for replies). That was unsatisfactory to me and has discouraged me from spending more volunteer time there. PrimeHunter (talk) 01:27, 30 April 2011 (UTC)[reply]

On that note, how do you add 'problem solved'? I never figured it out! Delaypoems101 (talk) 01:46, 30 April 2011 (UTC)[reply]

The precise text was an example. You could simply write it manually but you can also add {{resolved}} at the start of the section. See Template:Resolved. PrimeHunter (talk) 03:47, 30 April 2011 (UTC)[reply]

For me, it's more than just being thanked or not. The lack of thanks was a secondary issue. My main point, although maybe not well explained, is that people ask questions anonymously. They get the answer and they run. (And often they don't say thank you.) That means that we don't get to know the people we help, we aren't able to identify the those making a good use of this resource, as a posed to those just taking advantage of it. If users had to register then we would get to know them and their needs. We could often build working relationships with the sincere posters, but also be able to spot the time wasters, the ingrates and the homework cheats. But as with all policy suggestions: it is a suggestion. It's a "what if?" I'm not saying it's a good idea. I am just putting it to the floor to ask the reference desk what it thinks. If we decide it's a bad idea then it'll get binned and we'll move on.— Fly by Night (talk) 01:40, 30 April 2011 (UTC)[reply]

I've long been in favor of requiring people to have accounts in order to edit articles, but I don't see the need just for asking ref-desk questions, except to the extent that it would suppress the small number of trolls who sometimes plague us. Looie496 (talk) 02:30, 30 April 2011 (UTC)[reply]

As someone who has been asking questions on this desk for a long time (and learning a lot in the process for which I am grateful to the regulars here) I think FlybyNight has a point, for it helps if the regulars have an idea about the questioners level. However, to limit anonymous users to ask for references only is a bit too far though. Instead I suggest that anonymous users are advised to include some background information (by a note on top of the page) about their mathematical standard if they are posting questions, and if they just want a reference they neednt do that. This accompalishes the same purpose.-

talk) 05:44, 30 April 2011 (UTC)[reply

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Hello Fly by Night. To me this is not a problem. I simply answer if I like to answer and if I know the answer. Of course I enjoy when being thanked, but I do not expect it. Any contribution or question tells a lot about the author, including the way he (or she) introduces himself (or herself), the attitude shown, the wording used, the name chosen in the account, and so on. The nice thing about wikipedia is the freedom. Let's all behave nicely, and there will be no reason for imposing rules upon one another. (Rules may be considered necessary, but they are rarely sufficient anyhow). Bo Jacoby (talk) 07:11, 30 April 2011 (UTC).[reply]

For myself, I just reply whenever the fancy takes me. It doesn't really bother me if it's homework or just an "honest" question. Being thanked is always pleasant. Sławomir Biały (talk) 12:42, 30 April 2011 (UTC)[reply]

Same here, this is a lot more fun if you don't feel the obligation to answer every question you know the answer to. I don't take the "no homework" thing too seriously as it's impossible to enforce rigorously, though I suppose we need it to keep the more egregious abuse at bay. I post answers much more frequently than questions, though still not that often, but I think I learn at least as much from the forum as I contribute, so I don't think being "used" is really an issue.--RDBury (talk) 15:40, 30 April 2011 (UTC)[reply]