Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

October 1

The answer to this system of equations?

I would appreciate it if anyone could tell me the answer to this system of equations:

x + y = 50

x*y = 60

Thanks. 74.12.20.180 (talk) 04:43, 1 October 2009 (UTC)

Multiply the first equation by x giving xx+yx=50x, and subtract the second equation giving xx=50x−60. Now the variable y has been eliminated from the problem. You know how to solve this quadratic equation, right? Bo Jacoby (talk) 06:24, 1 October 2009 (UTC).

I might have done it like this: since

x + y = 50,

we get

y = 50 − x,

so we can put 50 − x in place of y in the second equation:

x(50 − x) = 60.

Then we have a quadratic equation in one variable. Michael Hardy (talk) 19:22, 4 October 2009 (UTC)

If s/he knows how to solve a quadratic equation, then maybe everyone should stop reading at this point. Michael Hardy (talk) 19:20, 4 October 2009 (UTC)

(Edit Conflict) There are a few aspects of this question which are unclear and thus I wish to clarify them here:

Precision - The domains of the variables x and y are not specified; that is, are you attempting to find integer solutions to the system of equations, or are you attempting to find rational or real solutions to the system of equations? Thus, embedded within your question are infinitely many other questions, each question associated to a particular ring, which one cannot hope to answer here.

Choice of wording - You seem to request "an answer to this system of equations". This remark is not only imprecise (the precise statement would be "the solutions to this system of equations"), but suggests that this is a homework problem. Usually, if this were not a homework problem, the questioner would make some remark regarding his/her attempts at the solution and interest in the question. I encourage you to do the same.

In fact, politely speaking, you should reword your question in a precise manner and present it again, so that it not only demonstrates your understanding of the question, but also the degree of effort you have put into it. This comment should not indicate that I am unwilling to help you, but rather that your expression conveys that you will not learn anything from a straightfoward answer. Nevertheless, I will show you the solution to a similar problem, which, after precisely stating the above problem, should enable you to derive its answer:

Problem - Suppose that the sum and difference of two numbers are given. How does one compute the numbers themselves? (Here, by "number" it is meant "real number")

Solution - Write x = a + b, and y = a - b, where a and b are the two numbers in question. Observe that x + y = a + a, and thus (x + y)/2 = a. Similarly (x - y)/2 = b (why?). This was what was to be demonstrated, since x and y are known.

With the ideas that above the underlying technique in mind, solve the slightly harder problem that you wish to solve. Note the identity (hint) - $x^{2}+2xy+y^{2}=(x+y)^{2}$ for all real numbers x and y (notice how I have incorporated precision into my remark). Also note that there may be more than one way to solve this problem.--PST 06:33, 1 October 2009 (UTC)

Come on, give your head a wobble! You know what the OP was trying to say, so there's no need to be condescending. Your little monologue helped whom? How?! To the OP: First, try drawing a picture. You'll see that there are two solutions. You can get these by substitution. Solve x + y = 50 to give y = 50 − x. Sub this into the second equation to give x² − 50x + 60 = 0. Solve this for x to give x = 25 ± √565. Sub this solution into x + y = 50 to give 25 ± √565 + y = 50; then solve for y. There are two solutions and they are

$(x,y)=\left(25\pm {\sqrt {565}},25\mp {\sqrt {565}}\right).$ ~~ Dr Dec (Talk) ~~ 11:13, 1 October 2009 (UTC)

If you did not understand the purpose of my comment, then do not comment upon it. The OP wanted an answer to a question which he did not state precisely. Although there is no problem with this since he surely would not do this intentionally, the fact that he did not give any indication of his attempts gives reason for one to not answer his question. In fact, it is against Wikipedia guidelines to do so. I do not imply by this that answering the question is an offence since I am sure that thus far everyone has answered his question with good intentions. However, I believe that a lesson should be learnt from this experience, which will be a useful one, should he pursue mathematics or in fact anything else. PST 12:47, 1 October 2009 (UTC)

I found PST's post well written and helpful as usual (whereas Declan's little irony helped whom?) --pma (talk) 13:05, 1 October 2009 (UTC)

If we're taking sides, I'm going with Declan. I think it is very safe to assume the OP is talking about real numbers (could be complex numbers, but the answers are real, so it doesn't matter). If they were talking about anything else it would have been a key part of the question and they would have mentioned it. It is also very clear that "the answer" is a solution set. The OP may not have realised that there will be two answers (for 2 linear simultaneous equations in 2 variables, the solution would be unique, the OP could easily have thought the same applied here), but we don't need clarification on the question to know that we should give both (or rather, how to get both). I wouldn't have actually given the answer, though, since this looks very much like a homework question. I think the Bo Jacoby's answer was the best. --Tango (talk) 14:31, 1 October 2009 (UTC)

I agree that it is possible to interpret the OP's question as it stands and that Bo Jacoby's answer was the most appropriate one. However, in my post I did not solely aim to to answer the question (although I did attempt to give the OP a hint), but rather I expressed my concerns that at some point, precision will be of importance in his/her life. Succintly, I suggested that carefully expressing his/her question would help him/her more than it would help us in answering his question. Perhaps I should have framed my remark in a less confrontational manner, but at its present length, it will at least leave an indelible mark on the OP's mind, which, I hope, will help him/her to realize the meaning of my comment later on. --PST 14:52, 1 October 2009 (UTC)

The OP came on asking how to solve two relatively simple questions. You can therefore assume that s/he may lack the necessary mathematical sophistication to phrase the problem perfectly. My guess is that s/he's been solving simultanious equations and has asked themselves "What if one of the equations was not linear?" I'd be surpised if the OP knows very much about complex number, much less about rings! I'll assume good faith on your part and assume that it wasn't an attempt to show your mathematical and linguistic superiority. That given, you've most probably totally baffled the OP and left him/her with more questions than when s/he started. Ask yourself this: If someone comes on here asking how to solve x + y = 50 and xy = 60 then are they going to have the faintest idea of what you were talking about? Of course not! ~~ Dr Dec (Talk) ~~ 15:57, 1 October 2009 (UTC)

Clearly, I should defend my remark. First of all, I did not intend to display any "mathematical or linguistic superiority"; in fact, why either is demonstrated in my post is beyond my comprehension. Secondly, as I have already made clear, I attempted to illustrate the unclear aspects of the OP's post. In my opinion, this serves to convey the importance of mathematical precision, which, at high school level, few students appreciate. Surely, my post will be in the forefront of the OP's mind when he does hear a mathematician speaking of precision. In fact, it will be important in his life later on, since no matter what you do, clarity of expression is crucial. At the reference desk, we work as a group of people who wish to help. Some people answer all sorts of questions that are posed (independent of what subject matter the question embodies) whereas some choose to discuss the aspects of the question other than the answer. If you would like to think of it in that manner, I am the latter for the simple questions. I know already that many dedicated people here would respond to the OP, and thus, in order to add another dimension to the OP's algebra skills, I explained the importance of precision. Although I do not think that your intentions were malacious, this discussion was started because of your unnecessary remark; not only the content, but the way in which you phrased it. You are certainly an important contributor to the reference desk, but at times your attitude is not. In effect, you must assume good faith and at least attempt to see the importance of others' remarks. --PST 01:28, 2 October 2009 (UTC)

type 1 and 2 errors of statistical hypothesis testing

There seems to be a bit of confusion here regarding the definitions of α, 1-α and 1-(1-α), and similarly, β, 1-β and 1-(1-β). Is the following true:

• α = FP
• 1-α = specificity = significance
• 1-(1-α) = rate of FP
• β = FN
• 1-β = sensitivity = power
• 1-(1-β) = rate of FN

Thanx! DRosenbach ^{(Talk | Contribs)} 12:55, 1 October 2009 (UTC)

α is the probability of a false negative and is called the "level of significance." I believe that β is the probability of a false positive. Wikiant (talk) 14:28, 1 October 2009 (UTC)

1-(1-α) is just α of course. As far as I can recall (from many years ago) α is the probability of a false positive, as stated in the article, and is the significance level of the test, with 1-α being the specificity. Similarly, β (called the sensitivity) is the probability of false negative and 1-β is called the power or sensitivity of the test. The article does seem slightly confusing at first reading, but I think it is correct. Dbfirs 16:37, 1 October 2009 (UTC)

I believe that is incorrect. The verbal definition of α is, "the probability of rejecting the null hypothesis when, in fact, the null hypothesis is true." That's a false negative. Wikiant (talk) 18:33, 1 October 2009 (UTC)

No, it's a false positive. Rejection of the null hypothesis is the positive result. Algebraist 21:25, 1 October 2009 (UTC)

Can no one help me? -- the aforementioned has been culled from the confusing, double-talk of this section. DRosenbach ^{(Talk | Contribs)} 01:01, 2 October 2009 (UTC)

The terminology can be confusing, but the article is correct (I've just fished out a very old text book to check).

A Type I (α) error occurs when we reject the null hypothesis even though the null hypothesis is true. This is called a "false positive" and α is the significance level of the test. 1 - α is the specificity.

A Type II (β) error occurs when we fail to reject the null hypothesis (i.e. accept the null) even though the null hypothesis is false. This is called a "false negative" and β is the sensitivity of the test. 1 - β is called the power or sensitivity of the test. When I have time, I'll have a go at setting out the article more clearly, unless someone else beats me to it (I hope they do!) Dbfirs 12:47, 2 October 2009 (UTC)

Dbfirs -- I'm sort of glad that I have at least one person engaging my question, but I'm also sort of frustrated that you make conflicting seemingly conflicting statements of the article is correct and α is the significance level of the test -- as significance is listed above as being equivalent to not α, but 1-α. Perhaps I'm either misunderstanding your comments or the logic, but to me, x ≠ 1-x. DRosenbach ^{(Talk | Contribs)} 13:27, 2 October 2009 (UTC)

No, the article says that 1 - α is the specificity. α is the significance level of the test. The article does say this, but not expressed very clearly. Dbfirs 17:51, 2 October 2009 (UTC)

If α is the probability of a false positive, then clearly 1-α is the significance of the test. A test at 95% significance does not have a 95% chance of a false positive. Taemyr (talk) 17:56, 2 October 2009 (UTC)

No, that's wrong. The significance level is the rate of false positives, i.e. it is the rate of falsely rejecting the null hypothesis. Michael Hardy (talk) 23:17, 2 October 2009 (UTC)

Taemyr, you're confused. Nobody ever tests at the 95% significance level. 5% is commonplace. Michael Hardy (talk) 23:18, 2 October 2009 (UTC)

( ... sorry about the delay in replying, I 've just returned home ...) People often confuse specificity with significance (it probably doesn't really matter). A significance test is normally done at the 5% significance level (and sometimes 1%). This corresponds to a specificity of 95% (or 99%), but, informally, we often speak of testing at the 95% level of significance when we mean the 5% level of significance i.e. the 95% level of specificity. Hence the confusion. Dbfirs 23:08, 2 October 2009 (UTC)

Maybe the section's been edited since the time it was commented on above. Here's what I find there:

The false positive rate is the proportion of negative instances that were erroneously reported as being positive.

It is equal to 1 minus the specificity of the test. This is equivalent to saying the false positive rate is equal to the significance level.

${\text{false positive rate}}={\frac {\text{number of false positives}}{\text{total number of negative instances}}}$

In statistical hypothesis testing, this fraction is given the Greek letter α, and 1 − α is defined as the specificity of the test. Increasing the specificity of the test lowers the probability of type I errors, but raises the probability of type II errors (false negatives that reject the alternative hypothesis when it is true).

That is correct in saying that α is the false-positive rate, i.e. α is the probability of rejecting a null hypothesis, given that the null hypothesis is true, i.e. α is the significance level. Michael Hardy (talk) 23:31, 2 October 2009 (UTC)

I've just realised that I've made a serious error above. The alpha parts are correct, but beta is not the sensitivity. This may be why we were getting confused. I had assumed a symmetry when there isn't one (at least not in the terms used). Dbfirs 19:27, 3 October 2009 (UTC)

Limits on discrete sets

What is $\lim _{n\to 0^{+}}{\frac {1}{{\frac {1}{n}}!}}$? And $\lim _{n\to {\frac {1}{4}}}{\frac {1}{{\frac {1}{n}}!}}$? --88.78.224.81 (talk) 15:23, 1 October 2009 (UTC)

What do you mean by x! for non-integer x? — Emil J. 15:30, 1 October 2009 (UTC)

Good point. But we can generalise the factorial function with the Gamma function. For z ∈ C with Re(z) > 0 we have

$\Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt\;$

It has the nice property that for n ∈ N we have Γ(n) = (n+1)! ~~ Dr Dec (Talk) ~~ 16:02, 1 October 2009 (UTC)

We need to use the Gamma function. So we can define ƒ : C → C given by

$f(z):={\frac {1}{\Gamma (z^{-1}-1)}}\ .$

This generalises the OP's formula because Γ(n−1) = n! for n ∈ N. The second limit is easier than the first, we have

$\lim _{z\to 1/4}f(z)={\frac {1}{2}}\ .$

As for the limit as z → 0, well, I'm still working on that one... ~~ Dr Dec (Talk) ~~ 16:17, 1 October 2009 (UTC)

...et ecce trabis est in oculo tuo! ( ↑ ) --pma (talk) 17:26, 1 October 2009 (UTC) ;-)

Couple of slips there, I think. For a start, Γ(n) = (n − 1)!, not (n+1)!. Gandalf61 (talk) 16:38, 1 October 2009 (UTC)

Ah, yes, quite right. But hey, what's a sign error between friends? ~~ Dr Dec (Talk) ~~ 16:49, 1 October 2009 (UTC)

The values of the limits are trivial to compute, once you manage to make any sense of the original notation by plugging in the Γ's (I know about this function, but I was not sure the OP does, or even is aware that there is a problem). The first limit is 0, as $\lim _{n\to 0^{+}}{\frac {1}{n}}!=\lim _{x\to +\infty }x!=+\infty$. The second limit, restated using Γ, is a limit of a continuous function, hence its value is simply the value of the function at 1/4, i.e., $\lim _{n\to {\frac {1}{4}}}{\frac {1}{{\frac {1}{n}}!}}={\frac {1}{4!}}=1/24$ (I'm not sure where you got the 1/2). — Emil J. 16:43, 1 October 2009 (UTC)

As Gandalf pointed out: I had a sign error. I had Γ(n) = (n+1)! instead of Γ(n) = (n−1)!. And in fact ƒ(z) → 1/2 as z → 1/4. The problem is that I should have had a plus in the denominator instead of a minus. When you replace it with a plus the limit is 1/24 as you say. I was having trouble with the case of z → 0 at the time of writing the post because

$\Gamma \left({\frac {1}{z}}+1\right)=\int _{0}^{\infty }t^{1/z}e^{-t}\,dt\;$

and I wasn't sure whether the limit of the integral on the right was well defined as z → 0 in the integrand. But it does seem that Γ(z⁻¹+1)⁻¹ → 0 as z → 0. ~~ Dr Dec (Talk) ~~ 17:07, 1 October 2009 (UTC)

Of course not. That limit doesn't exist, even for z in the real numbers... --pma (talk) 17:52, 1 October 2009 (UTC)

However, the original limit calls for z → 0+, and that one does exist. — Emil J. 18:01, 1 October 2009 (UTC)

The assumption that we consider the limit z → 0+ is tacit, since we assume that Re(z) > 0. I didn't see much point writing z → 0+, when it's clear that Re(z) > 0, and so z → 0− isn't well defined. ~~ Dr Dec (Talk) ~~ 18:15, 1 October 2009 (UTC)

Well, actually, now I finally noticed the title of this section, and it seems that the OP is after all not interested in computing the limits, but in the question of how to make sense of them, as the functions are not defined on a neighbourhood of the offending point. The answer is that in this particular case, one could interpret x! as a shorthand for Γ(x + 1) as above, which makes the limits well-defined. In general, it is possible to define a limit relative to a set $\lim _{\begin{smallmatrix}x\to a\\x\in A\end{smallmatrix}}f(x)$ provided a is a limit point of A. This would take care of $\lim _{n\to 0^{+}}{\frac {1}{{\frac {1}{n}}!}}$, except that one should use a different notation to make it clear that the limit is only taken relative to $\{1/n\mid 0<n\in \mathbb {N} \}$. The second limit is still not well-defined using this approach, as 1/4 is isolated point of the domain of the function. — Emil J. 17:00, 1 October 2009 (UTC)

P value

Please see here for question. DRosenbach ^{(Talk | Contribs)} 17:45, 1 October 2009 (UTC)

The answer is that your word "still" is wrong. The p-value goes from above 5% to below 5%. Taemyr (talk) 21:18, 1 October 2009 (UTC)

Absolute value has nothing to do with this -- if a p value is .01%, .1% or even 1%, that's acceptable, because it's below 5%. In the example given (copied from the p value article), the p value is 4.41% -- still below 5%, yet is said to allow the null hypothesis to be rejected. Why? DRosenbach ^{(Talk | Contribs)} 00:03, 2 October 2009 (UTC)

The null hypothesis is accepted failed to be rejected when the p-value is greater than 5%. That means the result is within the range of what we'd expect to happen 95% of the time if the null hypothesis were true. If the p-value is less than 5% that's when we suspect something fishy is happening and reject the null hypothesis. The lower the p-value, the more extreme the result, and the less likely it is that the null hypothesis would lead to that result. Rckrone (talk) 05:32, 2 October 2009 (UTC)

You are right, of course, but I find your terminology misleading. When the p-value is greater than 0.05, I find it more suggestive to say "we failed to reject the null hypothesis" or "we cannot reject the null hypothesis"; under no conditions do we "accept" the null hypothesis, because even a high p-value does not lend support for the null hypothesis -- it is merely consistent with the null hypothesis being true. Eric. 131.215.159.109 (talk) 11:21, 2 October 2009 (UTC)

Thanks for that correction. I'll admit I'm no good at statistics. Rckrone (talk) 16:58, 2 October 2009 (UTC)

I make no statement as to the acceptance of a null hypothesis. And if I am correct, then why does the article say we use the p value to reject the null hypothesis if it is below 5%? DRosenbach ^{(Talk | Contribs)} 13:29, 2 October 2009 (UTC)

The p-value tells us the probabilty that random chance could account for the observed outcome given the null hypothesis. So if the p-value is low we can reject the null hypothesis because the observed outcome would have been too unlikely. In short, below 5%, then we can reject the null hypothesis. Above 5% and we can not reject the null hypothesis. Taemyr (talk) 14:56, 2 October 2009 (UTC)

Not to nitpick, but this explanation needs some clarifying. The p-value is the probability that random chance would obtain an outcome at least as inconsistent, if not more so, as the observed outcome is, with respect to the null. The idea is that if one rejects the null with this particular outcome, then any more anomolous outcomes would have drawn the same conclusion. In rigorous foundational terms, the probability of this particular outcome tends to become a measure-theoretical morass. Thus as a decision procedure, determined before any outcome was obtained, one can still speak of not rejecting 95% of the time if the null is true. Baccyak4H (Yak!) 16:49, 2 October 2009 (UTC)

@DRosenbach: Sorry if I wasn't clear, my comment @11:21 was directed towards Rckrone, not towards you. Eric. 131.215.159.109 (talk) 18:48, 2 October 2009 (UTC)

Great -- got it now. Thanx everyone! DRosenbach ^{(Talk | Contribs)} 02:05, 5 October 2009 (UTC)

October 2

Collatz conjecture

It may sound pretentious and I know that most chances are that I'm taking it wrong -but I would be happy to get your evaluation for my "proof" (actually may be disproof) of Collatz conjecture.

strategy: I have learned the compactness theoram in a logic course I participated in during my (uncompleted) BA degree in math and I thought that it may be possible to implement it on Collatz conjecture. I defined infinite collection of formulas whose overall meaning is that there is a Collatz sequence that do not reach to 1. I have shown that every subgroup have suitable solution and that by applicating compactness theoram Collatz conjecture is refuted. Actually, as compactness theoram is valid for first order logic I proofed that, if not wrong, there is a model of first order number theory which contained Collattz sequence that not reach 1. So if Collatz was right then it is only for second order numbers.

a. k(x₁,x₂) ≡ (∃z((x₁=z+z)$\land \!\,$ x₂=z)) V ($\neg \!\,$ ∃z((x₁=z+z))$\land \!\,$ x_2=3*x1+1)

we define statment:

b. φ₀≡x₀≠1

c.for every natural i (1,2,3 etc) we define:

φ₂≡ k(x_2-1, x₂)$\land \!\,$(x₂≠1)

d. P is Peiano axioms group.

f. we define

Q=P $\cup \!\,${x₂}₂=1,2,3...

e. this is infinite group of first order formulas. every sub set is finite S of Q is satisfied by the natural numbers and by:

x₂=2^(n+1-₂)

with n as the maximal index of this statments from those we add in statments b and c in sub set S.

f. that is,according to the compactness thaorem there is a model that hold for P in all of Q formulas and espcially it hold for all Peiano axioms P, and it contain countable sequence of constants of which no one=1 and constitute Collatz sequence.--Gilisa (talk) 10:12, 2 October 2009 (UTC)

That's all very nice, but it has nothing to do with refuting the Collatz conjecture. For example, your model also contains an element a which is larger that every standard natural number 0, 1, 2, …, but you cannot take this to mean that you have refuted the "conjecture" that every natural number has only finitely many predecessors. If you want to show that something holds in the standard model of arithmetic, you can in principle do it by finding its elementary extension where the statement is satisfied (for example, using the compactness theorem), but only if you formulate your statement as a single first-order sentence. It is no longer true for realizability of types (infinite sets of formulas) with free variables, as you have attempted here. Since the Collatz sequence is computable, you can actually formulate the Collatz conjecture as a single sentence ($\Pi _{2}^{0}$ in fact) in the language of arithmetic by the usual Gödel-style coding tricks, but then your compactness argument will not work (precisely because it is a single sentence). In the model you constructed above, the relevant Collatz sequence does actually reach 1 (or can be assumed to, at least), but the number of steps needed is a nonstandard natural number. — Emil J. 11:06, 2 October 2009 (UTC)

In principle, a method such as you developed could only show that the conjecture is not provable in Peano Arithmetic, not that the conjecture is false in the standard model. And even a nonprovability result would be very interesting. But in this case you cannot even get the nonprovability result.

In the model you construct, there is a particular element a such that:

• "The Collatz sequence for a does not stop after 1 iteration"
• and "The Collatz sequence for a does not stop after 2 iterations"
• and "The Collatz sequence for a does not stop after 3 iterations"
• and so on. For each standard natural number k, your model satisfies "The Collatz sequence for a does not stop after k iterations"

The problem is that you want to get "The Collatz sequence for a does not stop at all". In the nonstandard model you construct, the sequence could well stop after a nonstandard number of iterations; your sequence of formulas does not rule this out. Indeed, if a happens to be a nonstandard power of 2 then the Collatz sequence for a will stop, but the number of iterations required will be nonstandard. So you do not know that the model you construct fails to satisfy the Collatz conjecture.

This is a general property of $\Pi _{2}^{0}$ sentences in nonstandard models: the witnesses selected by the existential quantifier can be nonstandard, and there is no way to force them to be standard. — Carl (CBM · talk) 11:21, 2 October 2009 (UTC)

It's clear that the conjecture is not provable in Peiano arithmetic (indeed, it's very easy to build model for first order Peiano axioms that include term that is bigger than any natural number and will never reach 1). My opinion is that if an argument is proofable from a set of certain axioms (here Pieano's) so for any model they are valid for, the same argument must be valid as well (no matter how non standard it's). So maybe my model is non standard and the interesting question remain whether the conjecture is valid for the standard model of natural numbers. Maybe the distinction between this two kinds of models (standard vs. non s) can't be described in terms of Peiano axioms. --Gilisa (talk) 13:49, 2 October 2009 (UTC)

It is not clear that the conjecture is not provable in Peano (note spelling) arithmetic, and your argument does not work. Carl has explained this already. All you have shown is that a modified version of the Collatz conjecture is unprovable by PA, and PA can't even state your modified version (in a single formula), so the notion of proving it makes little sense. Algebraist 13:57, 2 October 2009 (UTC)

The problem is that the Collatz conjecture is of the form "for every natural number a, there is a finite sequence τ of natural numbers such that φ(a,τ) holds and τ ends with 1", where φ is the formula that says τ is the Collatz sequence for a. But when this formula is interpreted into a nonstandard model, the "finite" sequences are now indexed by nonstandard numbers. For example, Peano arithmetic is able to prove by induction on k that the number 2^k must have a Collatz sequence that ends in 1. When k is nonstandard, the Collatz sequence will be a nonstandard "finite" sequence. — Carl (CBM · talk) 14:15, 2 October 2009 (UTC)

Oh well, I can see it now...what a shame--Gilisa (talk) 14:25, 2 October 2009 (UTC)

It was actually a very good idea, it's just that it didn't work. It takes a little experience to get used to the way that nonstandard models of arithmetic behave with regard to quantifying over finite sequences. — Carl (CBM · talk) 15:40, 2 October 2009 (UTC)

Theorem

how do cauchy's integral theorem apply? —Preceding unsigned comment added by 41.204.168.3 (talk) 13:29, 2 October 2009 (UTC)

Your question isn't very clear. Have you read Cauchy's integral theorem? Algebraist 13:46, 2 October 2009 (UTC)

You usually apply it by replacing a path that is difficult to integrate over with a path that is easy to integrate over (something piecewise linear, for example). You then show that the assumptions of the theorem hold and, therefore, the integral over the nice path is equal to the answer you were looking for. --Tango (talk) 17:04, 2 October 2009 (UTC)

I had a professor who thought Cauchy's integral theorem was the foundation of modern civilization. It may not seem like much a first but it grows on you.--RDBury (talk) 20:29, 2 October 2009 (UTC)

See Methods of contour integration for various applications of that theorem. Michael Hardy (talk) 20:27, 2 October 2009 (UTC)

Achilles numbers: Never consecutive?

Are there any consecutive Achilles numbers? --88.78.2.158 (talk) 16:35, 2 October 2009 (UTC)

So, you want a solution $(x,y,n,m)$ of the Diophantine equation $nx^{2}-my^{2}=1$, where $n$ and $m$ are cubes. To satisfy the definition of Achilles number, you also need $nx^{2}$ and $my^{2}$ not to be perfect powers. I'd try fixing two cubes, non-square numbers $n,m,$ and then solving the quadratic equation in $x,y.$. Try this [1]. Good luck. --pma (talk) 18:43, 2 October 2009 (UTC)

Are both 5425069447 and 5425069448 Achilles numbers? --88.77.253.57 (talk) 19:26, 2 October 2009 (UTC)

Are there any smaller consecutive Achilles numbers? --88.77.253.57 (talk) 19:32, 2 October 2009 (UTC)

> Are both 5425069447 and 5425069448 Achilles numbers?

Apparently they are:

5425069447 = 7 × 7 × 7 × 41 × 41 × 97 × 97

5425069448 = 2 × 2 × 2 × 26041 × 26041

Michael Hardy (talk) 20:25, 2 October 2009 (UTC)

Are 5425069447 and 5425069448 the smallest consecutive Achilles numbers? --88.77.255.219 (talk) 20:28, 2 October 2009 (UTC)

Looks like it from here - the next pair appears to be 11968683934831 and 11968683934832 Fletch79 (talk) 23:35, 2 October 2009 (UTC)

I have added this fact to our Achilles number article. Gandalf61 (talk) 07:04, 3 October 2009 (UTC)

Question about shoelaces

I was walking along and noticed that both of my shoelaces were untied, and it prompted a study from me. I determined that one of my shoes comes untied average of once every 75 minutes of walking. If I am walking 4 miles an hour (I timed it, and got 3.8, but just use 4), and my shoelaces have an equal likelihood of coming untied, and on average, I notice that a shoe is untied in 1 minute of it occurring and retie it as soon as I notice, how far would I need to walk until I had neither of my shoes tied? Googlemeister (talk) 20:49, 2 October 2009 (UTC)

They might never untie, but if you read shoelaces you'd have a better chance at that. Dmcq (talk) 21:54, 2 October 2009 (UTC)

Can we assume a Poisson distribution with a mean of 5 miles? Dbfirs 23:14, 2 October 2009 (UTC)

For what? Algebraist 23:34, 2 October 2009 (UTC)

Sorry, I've just re-read the question properly. I meant a Poisson distribution with a mean of 0.2 (per mile) to estimate the probability of both laces coming undone, but I've realised that this is not a valid model. Simply using product of probabilities would give 0.04 for both laces, i.e 25 miles for the expected distance. Is this valid? Of course, when I suffer from this problem (common when walking through heather), I just use a double knot. Dbfirs 01:10, 3 October 2009 (UTC)

Well what you've got is a queueing theory problem with one server, you, who has a one minute service time. The question then is on average how long is it till the second shoelace is waiting for you to finish serving the first shoelace? It does seem a bit strange to me though assuming the second shoelace is quietly trying to untie itself even whilst you are stopped doing up the other one. Sorry I see you're still be walking in practically all the service time. Dmcq (talk) 09:12, 3 October 2009 (UTC)

But queueing theory usually assumes the Poisson Distribution which is not valid here because you (presumably) have only two feet? Dbfirs 17:12, 3 October 2009 (UTC)

Might as well have some sort of answer here. For each shoe the men time between it getting untied is about 150 minutes. While say the left shoelace is untied there is a two minute window in which there's problems if the right shoelace gets untied. The chances of the right shoelace getting untied in any particular two minute window is 2/150 = 1/75. SO the mean time between collisions is 150 * 75 minutes. And for an exponential distribution that's how long you'd have to wait on average. So you'd have to walk 750 miles on average. Far longer than I've ever done at a stretch. Dmcq (talk) 22:09, 3 October 2009 (UTC)

But, in practice, this is not what happens (from personal experience of walking through heather). I think the reason is that each lace is gradually loosening through tiny tugs as I walk. Even my 25 miles (above) seems excessive, though perhaps I don't always notice within one minute. Because I tend to tighten the other lace as I re-tie the undone one, there is a greater chance of both laces coming undone near the expected "undo point" of five miles. Dbfirs 10:27, 6 October 2009 (UTC)

October 3

Calculator

The number pattern on mine has the 1, 2, 3 at the bottom and the 7, 8, 9 at the top. Now on my phone it is the opposite 1,2,3 at the top and the 7, 8, 9 at the bottom. Why is this? It obviously leads to mistakes. I am certain there would be less pissed off people answering wrong numbers and almost-graduates who managed to flunk out of school due to this mismatch. What gives? 174.152.243.23 (talk) 00:09, 3 October 2009 (UTC)

The dial phone came first, with had letters corresponding to numbers, and when phones went "digital" with tone dialling buttons they logically put ABC before WXYZ, but meanwhile Texas Instruments had brought out their calculator with higher numbers at the top, and computer keyboards followed with the same arrangement. I have to mentally adjust every time I dial a number on a phone because I learnt to "touch-type" on a calculator. Dbfirs 01:18, 3 October 2009 (UTC) Thanks to Anonymous for pointing out (below) that the calculator arrangement is over a hundred years old. Dbfirs 17:23, 3 October 2009 (UTC)

See also the article on numeric keypad: "The arrangement of digits on numeric keypads is different from that of telephone “Touch-Tone” keypads — this may be confusing for those who use one of these arrangements more often". The wording "may be confusing" is an understatement. It is downright irritating. Bo Jacoby (talk) 03:38, 3 October 2009 (UTC).

I don't think I have problems with this. I do the keypad on the keyboard with all 5 fingers. I dial my phone with 1 finger, my thumb. Maybe that's why. StatisticsMan (talk) 04:31, 3 October 2009 (UTC)

I don't have problems with it either and you may have found the reason - I don't tend to use to use multiple fingers on a keypad or any kind, but I use my middle finger on a calculator/computer keyboard and my thumb on my phone. --Tango (talk) 05:27, 3 October 2009 (UTC)

I might be wrong about Texas Instruments (theirs was the first pocket calculator I saw around 1972, and it was very expensive, certainly well beyond my means!), but did Sanyo produce one in Japan before then? I'm trying to remember whether any of the electro-mechanical calculators that I used around 1970 had a layout of numbers like a modern calculator. My recollection is that they didn't. The ones I used didn't, but some did!

My confusion with layouts arises because I tend to use the same fingers for both layouts, but I have to concentrate harder to dial a number on a phone, and yes, it does help to use just one finger. Dbfirs 07:52, 3 October 2009 (UTC)

You guys are too young! The "calculator-style" numeric keypad goes back to mechanical adding machines. According to this page, this machine was the first to use it and was first produced in about 1912 or 1914. Nobody worried about this when the phone keypad was designed because phones were meant for the general public whereas only a few people, such as computers, would make heavy use of adding machines. And, as noted above, people wouldn't like it if the phone company messed with the alphabet. --Anonymous, 10:28 UTC, October 3, 2009.

Hard to figure how this wound up in the Math page, but whatever. It's important to note that the original arrangement of the numbers on adding machines and dial phones were both done for practical reasons. Numbers in a calculator are entered left-to-right, and 0, 1, etc., tend to be more common in the high-order digits, hence they are closest to the user. Dial phones, which were clearly constructed with right-handers in mind, are pulled clockwise to generate the pulses, so naturally the 1 was closer to the finger-stop. 0 was put last, as it was used to suggest "O" for "Operator", though not to be confused with the M-N-O over the 6. The letters were originally connected with exchanges, long before exchanges became all digits and 800 numbers co-opted the letters to make clever mnemonics. And as noted above, because of the association of letters with numbers, the touchtone phone put 1-2-3 at the top instead of the bottom. However, both phones and calculators still have 0 at the bottom. There are also calculators with extra keys for 00 and 000, for further entry convenience. Touchtone phones had the * and the # keys from the beginning, although it was awhile before they were used very much. →Baseball Bugs ^{What's up, Doc?} carrots 11:45, 3 October 2009 (UTC)

Calculator is a math device so the math page could easily be the first place someone would think to ask such a question. StatisticsMan (talk) 19:39, 3 October 2009 (UTC)

The original use of "0"=Operator, surely does not explain the position of 0 on the dial. The dial emits pulses down the line to the exchange, according to the distance the dial travels in returning to home base. Dialling "1"=1 pulse, "2"=2 pulses, etc and "0" produces 10 pulses. [Obviously there is no such thing as zero pulses.] The digit "0" therefore belongs after "9" on the telephone but before "1" on a calculator (and also on the numeric keypad on a PC). Sussexonian (talk) 20:55, 4 October 2009 (UTC)

Calculator (solution)

What solutions are available for the above? It seems that all are in agreement that the above key arrangement is unacceptable.

1. I still have my Ma Bell touch tone desk phone ( I think I have a wall model boxed up in the basement as well). I plan on opening it up and reconnecting the DTMF connections for the 1st and 3rd rows of buttons. I can then use white out and get the number pad back in the correct order.
2. This won't work for phones with newer/smaller/unmodifiable circuitry. I am thinking that the phone companies could reconfigure their switches (as to how they process the DTMF). Possibly stickers could be passed out to relable all of the existing phones.

This might seem like a lot of trouble, but I am sure that there would eventually be a payoff from the resulting increase in efficiency. 70.4.71.74 (talk) 11:01, 3 October 2009 (UTC)

That's an interesting idea. But do you really use the phone and the calculator both heavily? Usually someone using a calculator (or the numeric keypad on a computer keyboard) is doing a large amount of data entry - dozens of numbers per minute, let's say. How often do you make a phone call? →Baseball Bugs ^{What's up, Doc?} carrots 11:50, 3 October 2009 (UTC)

I already gave a solution above, and Tango backed it up. Do each of the two in different ways. I use my full right hand to use the computer number pad with starting position thumb through pinky on 0, 4, 5, 6, and enter, respectively. I have also practiced typing this way to gain speed a long while ago. If I dial a cell phone, I hold it in my hand and use my thumb of that hand only to dial. If I have a landline phone that is bigger, I might do that or I might hold it in my left hand and dial with my pointer finger of my right hand. It's a completely different method so I don't get confused. Or, perhaps the solution is because I have practiced both. I don't know why, but for me this isn't a problem at all. StatisticsMan (talk) 19:32, 3 October 2009 (UTC)

A guy could obviously rewire his own home phone to match a calculator, but that might give him even more trouble when he goes out into the world and all the phones are at odds with his. And how would you rewire a cellphone, for example? Data entry people have a method, as you do, for typing the number pad quickly, a skill which also works for calculators (and vice versa). I'm trying to figure out why on earth someone using the telephone would ever need to enter numbers at such a fast and voluminous rate as with data entry. Maybe telemarketing? But they might well have VOIP, in which case they could use the keyboard number pad anyway. It makes more sense to simply learn one skill for one operation and another skill for another operation. It reminds me of some who fear driving because they might hit the accelerator instead of the brake. Well, you simply train your mind that when a braking situation arises, or a traffic slowdown, or anything unusual, you remove your foot from the accelerator. Practice makes perfect, or at least makes "permanent". →Baseball Bugs ^{What's up, Doc?} carrots 20:14, 3 October 2009 (UTC)

I think telemarketers don't even dial. I think they have a list of phone numbers in a computer and the computer dials. However, I am not sure on this. And, I agree that it's probably not necessary to dial at extreme speeds. StatisticsMan (talk) 04:52, 4 October 2009 (UTC)

Rectangle under pinhole projection

I have an image of a rectangle under pinhole projection of unknown parameters. Can I solve the rectangle's aspect ratio? --194.197.235.240 (talk) 16:15, 3 October 2009 (UTC)

In liner optics it's identical to this of the original rectangle.--Gilisa (talk) 19:47, 3 October 2009 (UTC)

Afraid not, the cross ratio you get from joining opposite corners and projecting lines will be the same as for a square. You can project a rectangle onto a square. Dmcq (talk) 20:13, 3 October 2009 (UTC)

If the line on which the the pinhole camera and the center of the square are located on is vertical to the rectangle plane then we can follow the simple solution of linear optics. --Gilisa (talk) 07:06, 4 October 2009 (UTC)

If the original rectangle and the projection are on parallel planes, they have to be similar. If the planes aren't parallel then opposite sides of the projected quadrilateral won't be parallel and you can calculate how far off from parallel the two planes are, so you have all the information you need. I don't see how a non-square rectangle could ever be projected to a square. Rckrone (talk) 22:29, 3 October 2009 (UTC)

Correct me if I'm wrong, but I think projecting a non-square rectangle to a square is really easy if the two planes are not required to be parallel. Michael Hardy (talk) 22:33, 3 October 2009 (UTC)

I'm still not seeing it. I think you're right it would be possible to project both to the same quadrilateral (but the quadrilateral won't be a rectangle), so I am wrong about that. However, if in addition to your projected shape you also know where the pinhole is, I think you have all the information you need. Rckrone (talk) 23:18, 3 October 2009 (UTC)

I seem to recall a theorem of elementary geometry stating that any rectangle at all can be realized as the projection of a square from a point. If so, the shape of the rectangle would not give enough information to determine the aspect ratio: it could be a square, but other rectangles would also have irregular quadrilaterals as projections from a point. Michael Hardy (talk) 22:32, 3 October 2009 (UTC)

I'm reversing what I said originally to agree a bit with Rckrone. You can't get to a square directly with a projection except with an infinitely long projection. If you know the distance from the pinhole to the projection then yes measuring that ought to give you the original aspect ratio. If you don't know the relationship between the pinhole and the screen then no you won't be able to calculate the original aspect ratio. Dmcq (talk) 10:42, 4 October 2009 (UTC)

Thanks --194.197.235.240 (talk) 20:27, 5 October 2009 (UTC)

October 4

Math

3^x-2^x+5^x=30 —Preceding unsigned comment added by THKhamosh (talk • contribs) 04:42, 4 October 2009 (UTC)

Just to rewrite this: 3^x − 2^x + 5^x = 30. ~~ Dr Dec (Talk) ~~ 11:11, 4 October 2009 (UTC)

x = 2. I don't know how to solve such a problem but it's easy to guess easy solutions and x = 2 works. StatisticsMan (talk) 04:50, 4 October 2009 (UTC)

...and no other real solution exists, for the LHS is strictly increasing in x. --84.221.208.114 (talk) 09:20, 4 October 2009 (UTC)

There appear to be a whole bunch of complex solutions. Newton-Raphson gives 2.310309-7.566221i, 2.35323-19.4722i, 1.90159+11.73987i, etc etc depending on starting value. HTH, Robinh (talk) 10:09, 4 October 2009 (UTC)

• This is probably best solved graphically, if x is real and you don't want to be guessing. I don't know of an algebraic method, and neither do the people I asked on IRC. —Anonymous Dissident^Talk 12:06, 4 October 2009 (UTC)

Newton-Raphson method would only yield approximation (2.0000...)and such is not needed in this case, you probably better follow the advise of StatisticsMan-it's not too elegant or sophisticated, but here it's very obvious that placing 2 instead of the Xs would give you the solution.--Gilisa (talk) 12:37, 4 October 2009 (UTC)

Hi guys. I used Newton-Raphson with complex arithmetic. Seems that the real part of the answer is always close to 2.0 (at least, I failed to find any roots with real part far away from 2.0). Comments anyone? Robinh (talk) 20:05, 4 October 2009 (UTC)

That makes sense. The real part corresponds to moduli of the terms and the imaginary part to the arguments. You can choose the imaginary part to get pretty much whatever arguments you want for each of the three terms since ln2, ln3 and ln5 don't have rational ratios. The real part is is bounded by the real solution to 5^x + 2^x + 3^x = 30 and the real solution to 5^x - 2^x - 3^x = 30. Those are both around 2. In fact the real parts of all the solutions should be dense in that interval. Rckrone (talk) 21:22, 4 October 2009 (UTC)

A good way to find exact solutions of difficult equations is to apply Newton-Raphson, plug the result in Plouffe's inverter, and validate. Of course, in this case, once we get to 2.000000000, we don't need Plouffe's. -- Meni Rosenfeld (talk) 20:34, 5 October 2009 (UTC)

The transcendental function 3^x-2^x+5^x-30 can be expanded as a taylor series, truncated to say 9 terms, and the roots of the resulting polynomial computed by standard methods. Using the J (programming language) you write p.(_30 + (3&^) + (5&^) - (2&^)) t. i. 9 and get the answer

┌───────┬────────────────────────────────────────────────────────────────────────────┐
│0.00117│_3.96 _2.93j2.62 _2.93j_2.62 _0.452j3.79 _0.452j_3.79 1.83j3.24 1.83j_3.24 2│
└───────┴────────────────────────────────────────────────────────────────────────────┘

The last root in the list is x=2. The other roots in the list are no good because the degree of the polynomial is too small to provide a good approximations out there. Bo Jacoby (talk) 23:33, 4 October 2009 (UTC).

In this specific case I think that better way than using numerical analysis would be to think on P^x space as a vectorial space and to dismantl and compare the coordinates of the measured vector.--Gilisa (talk) 07:36, 5 October 2009 (UTC)

Interesting! Could you elaborate on that? Bo Jacoby (talk) 14:23, 5 October 2009 (UTC).

trigonometry

what is sin cos and tan in a triangle —Preceding unsigned comment added by Ashishsihora (talk • contribs) 08:19, 4 October 2009 (UTC)

See Trigonometry. They are the ratios of sides of a triangle. -- SGBailey (talk) 08:46, 4 October 2009 (UTC)

Best way to calculate if a line crosses another line, or a polygon.

Dealing with X,Y coordinates, I have two lines within a large rectangle, with the x,y coordinates of all four ends of the lines being known. Neither of the lines extend from side to side of the rectangle. What would be the best way of calculating if they cross each other or not? This is for a computer program. And a more advanced question, how can I calculate if a line crosses a polygon or not? Thanks. 89.242.93.56 (talk) 11:36, 4 October 2009 (UTC)

• How are you defining the lines and the polygons? If you have equations then solve the equations simultaneously. Consider lines in the real xy-plane. If you have the lines given by the equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 where a_i, b_j, c_k ∈ R then there are three cases. Assume that a₁ and b₁ are not both zero and that a₂ and b₂ are not both zero. (This means the solutions are actually lines!)

1. If a₁b₂ − a₂b₁ ≠ 0 then x = (b₁c₂ − b₂c₁) / (a₁b₂ − a₂b₁) and y = (a₁c₂ − a₂c₁) / (a₁b₂ − a₂b₁).
2. If a₁b₂ − a₂b₁ = 0 then a₂ = λa₁ and b₂ = λb₁ for some non-zero λ ∈ R. If c₂ ≠ λc₁ then the lines are parallel; there is no solution.
3. If a₁b₂ − a₂b₁ = 0 and c₂ = λc₁ then the lines are the same line and all points are solutions.

If you have a polygon then the sides of the polygon are line segments and so will have equations. Just check that the intersection occur in these line segments, and not outside. So one side of the polygon might be the segment of the line x − y = 0 where 0 ≤ x ≤ 1 and so 0 ≤ y ≤ 1 too. Just check that the solution, if there is one, has 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. ~~ Dr Dec (Talk) ~~ 12:03, 4 October 2009 (UTC)

• If you're doing this in practice you have to be careful with parallel lines and horizontal and vertical lines, in fact avoid divides. 'Graphics Gems' is a nice introduction to computer code for things like this. Or just do a google of 'line segment intersection' and look around for reasonable code. Dmcq (talk) 12:13, 4 October 2009 (UTC)

Copying the code introduce you to the holy mantra of programmming Reduce reuse recycle ;-) Dmcq (talk) 12:24, 4 October 2009 (UTC)

• This might not be the easiest way, but you can define the orientation of three points in a specific order to positive if when you go from point to point you go counterclockwise and negative if you go clockwise. This is easy to do on a computer since the orientation can be computed as

$\operatorname {sgn} \left({\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}}\right)$

So if your line segments are P₁ = (x₁, y₁) to P₂ = (x₂, y₂) and P₃ = (x₃, y₃) to P₄ = (x₄, y₄) then they cross iff

1. (P₁P₂P₃) and (P₁P₂P₄) have opposite orientations
2. (P₁P₃P₄) and (P₂P₃P₄) have opposite orientations.

You need to find four determinants with this method; if any of these are 0 then one of the endpoints lies on the other line, though not necessarily on the line segment, and it depends on what you mean by "cross" in that case.--RDBury (talk) 12:39, 4 October 2009 (UTC)

I do not have equations for the lines, just the x,y coordinates for the end points. I am not a mathematician, and neither myself or the computer language I am using understand matrices. I've just now thought that I could use a two-stage process to check for crossing lines - I could assume the lines are of unrestricted length, a) calculate where they cross, and then b) check if this crossing point is within the parts of the line that actually exist. Is there a simple equation to do a) please? This would I think work for polygons as well. 89.242.93.56 (talk) 12:47, 4 October 2009 (UTC)

• If a line segment has endpoints (p₁,q₁) and (p₂,q₂) then it is a segement of the line (q₁ − q₂)x − (p₁ − p₂)y + p₁q₂ − p₂q₁ = 0. Now re-apply my method from above. ~~ Dr Dec (Talk) ~~ 13:13, 4 October 2009 (UTC)

The determinant above is just x₁(y₂-y₃)-x₂(y₁-y₃)+x₃(y₁-y₂). Like I said it's easy to put it in a computer. Really though, it you want to know how to do geometry on a computer you pretty much have to learn matrices.--RDBury (talk) 13:27, 4 October 2009 (UTC)

Here's a C++ implementation of RDBury's solution. This is a better approach than the one below using slope-intercept form because it is more numerically stable and it avoids special cases. Even better is Dmcq's suggestion to use code from Graphics Gems, which can be found here. I didn't see code there for this particular problem, though (somewhat surprisingly).

            typedef SOME_NUMERIC_TYPE Coord;
            struct Point { Coord x, y; };
            struct LineSegment { Point p, q; };
            
            Coord TripleProduct(Point p, Point q, Point r) {
                return (q.x - p.x) * (r.y - p.y) - (q.y - p.y) * (r.x - p.x);
            }
            
            bool DoLineSegmentsIntersect(LineSegment a, LineSegment b) {
                return (TripleProduct(a.p, a.q, b.p) >= 0) != (TripleProduct(a.p, a.q, b.q) >= 0)
                    && (TripleProduct(b.p, b.q, a.p) >= 0) != (TripleProduct(b.p, b.q, a.q) >= 0);
            }

That code assumes that your line segments are closed (include the endpoints). To make them open or half-open, change >= to > just to the right of the points that you want to exclude. For example, to exclude b.p, change ... b.p) >= 0 ... by removing the = sign. -- BenRG (talk) 21:49, 4 October 2009 (UTC)

BenRG, please make me a bit smarter and explain me how exactly the ability to consider special cases is a disadvantage?--Gilisa (talk) 22:04, 4 October 2009 (UTC)

The solution above works in the special cases, they just aren't special—all cases are treated in the same way. The problem with slope-intercept form is that there's a danger of catastrophic precision loss in borderline cases. For example if the segment goes from (100, 100) to (100, 200) then you're fine (you can treat it as a special case) but if the segment goes from (100, 100) to (100.00001, 200) then the slope-intercept form is y = -999999900 + 10000000 x. When intermediate quantities change that dramatically with tiny changes to the input it's usually a problem because of roundoff error. With the solution above the numbers used in the calculation only change slightly when one of the input coordinates changes slightly. -- BenRG (talk) 22:38, 4 October 2009 (UTC)

So you see, as you wrote yourself my solution works aside for special cases, but they just aren't special ;-). I still think that for him/her it would be easier solution to implement as he/she don't know how to workout or to read matrix (unless you already wrote the code, and that's kind). Also, I chcked it, my solution work always. Cheers--Gilisa (talk) 06:51, 5 October 2009 (UTC)

There you go :

line A: from pixel P1 (x1,y1) to pixel P2 (x2,y2) line B: from pixel P3 (x3,y3) to pixel P4 (x4,y4)

line A can be described by the linear equation: Y = a1 * X + b1 given that pixels P1 and P2 are on this line, we will get the following equations:

1) y1 = a1 * x1 + b1 => b1 = y1 - a1 * x1 2) y2 = a1 * x2 + b1 => b1 = y2 - a1 * x2

two equations with two unknown parameters (a1, b1)

=> y1 - a1 * x1 = y2 - a1 * x2 => a1 * (x1 - x2) = y1 - y2 => (3) a1 = (y1-y2)/(x1-x2)

from eq. (1) we will get:

b1 = y1 - x1*(y1-y2)/(x1-x2)

=> b1 = [y1*(x1-x2)-x1*(y1-y2)]/(x1-x2)

=> (4) b1 = (x1*y2-x2*y1)/(x1-x2)

in a similar way we will get the next equations for line B:

(5) a2 = (y3-y4)/(x3-x4) (6) b2 = (x3*y4-x4*y3)/(x3-x4)

step 2) now, we will find the cross point (Xc,Yc):

a1* Xc + b1 = a2 * Xc + b2

=> Xc = (b2-b1)/(a1-a2) => Yc = a1*Xc + b1

all that left now is to find if the cross point apears on one of the lines , for example line A:

lets assume that X2 >= X1:

if X1<=Xc<=X2 the lines cross each other, otherwise they don't.

(if X1>=X2 we should check if X2<=Xc<=X1)

This is not the full solution, there are several cases that should be checked first, this is the full solution:

1. if x1=x2 and x3=x4:

          in that case, if max(y1,y2)>=max(y3,y4)  and min(y1,y2)<=max(y3,y4) the lines cross
          otherwise if max(y3,y4)>=max(y1,y2)  and min(y3,y4)<=max(y1,y2) the lines cross
          otherwise they don't.

2. otherwise, if x1=x2 (and x3 != x4)

           in that case: first we have to find a2, b2 as described above.
           then: Xc = X1, Yc = a2*X1 + b2
           if Yc<=max(y1,y2) and Yc>=(min(y1,y2) the lines cross
          otherwise the don't.

3. otherwise, if x3=x4 (and x1 != x2)

           in that case: first we have to find a1, b1 as described above in the regular solution.
           then: Xc = X3, Yc = a2*X3 + b2
           if Yc<=max(y3,y4) and Yc>=(min(y3,y4) the lines cross
          otherwise the don't.

4. otherwise:

   first we have to calculate a1, b1, a2, b2 as described above in the regular solution:
   4.1)   if a1=a2 (parallel lines)
                   if b1 != b2 the lines don't cross
                   otherwise (b1=b2)
                       if min(x1,x2)<=min(x3,x4) and max(x1,x2)>=min(x3,x4) the lines cross
                       otherwise if min(x3,x4)<=min(x1,x2) and max(x3,x4)>=min(x1,x2) the lines cross
                       otherwise they don't
   4.2) otherwise (a1!=a2), continue with the regular solution (from step 2).

definitions:

min(X,Y):

   if (X<Y) min(X,Y) = X
   otherwise min(X,Y) = Y

max(X,Y):

   if (X>Y) max(X,Y) = X
   otherwise max(X,Y) = Y

--Gilisa (talk) 14:07, 4 October 2009 (UTC)

What programming language are you using? There might well be a build in function for this (which will be easier to use than writing it yourself and will also be optimised so will run faster and with less memory than your own implementation is likely to). --Tango (talk) 20:36, 4 October 2009 (UTC)

C, what built in function you know for this? If you know better way feel free to write it, I'm sure it would help 89.242.93.56 alot--Gilisa (talk) 20:50, 4 October 2009 (UTC)

Sorry, my question was meant for the OP. There probably isn't a built in function in C, it's a pretty minimalist language. --Tango (talk) 23:33, 4 October 2009 (UTC)

And I was thinking that in this last reply you meant to OOP for some reason I couldn't realy get to, and only now understand that you meant to Open Protocal. --Gilisa (talk) 08:29, 5 October 2009 (UTC)

That's highly unlikely, he probably meant OP = original poster (i.e., 89.242.93.56). — Emil J. 12:11, 5 October 2009 (UTC)

Well, this is the kind of difficulties I have to face with as a non native English editor. This time I was refering to the link the OP (now I understand what the acronyms stand for) posted here, even it didn't include Open Protocol just few equations-as I couldn't find any other association in this context I get it hard to see how my post connecting to anything here now (probably confused it with Open Source)..Well, I'm hasty indeed.--Gilisa (talk) 12:18, 5 October 2009 (UTC)

Thanks. Sorry, I do not understand any of the above. But this webpage here http://geometryatlas.com/entries/41 seems to give a simple formula for the x.y coordinates. Is it correct please? I would check that the formulas would not be divided by zero - but would that indicate the lines were parallel, or what? (Edit - they would be parallel so can be disregarded). Would lines at right angles a problem? 78.146.166.105 (talk) 22:29, 4 October 2009 (UTC)

It looks correct but it's not a good choice for software implementation (see my replies above). I recommend using the code I wrote. -- BenRG (talk) 22:38, 4 October 2009 (UTC)

It is a very good choice, because it is simple to program and I can understand it. Speed is not an issue here. 78.149.191.96 (talk) 17:45, 5 October 2009 (UTC)

BenRG When I think again about what you wrote om my solution, it have a little sense. First, the screen resolution is limited to produce a case such as (100,100) and (100.0001, 200). And more, if we use variables which have enough bits (float, doubles) then my solution will always work.--Gilisa (talk) 08:50, 5 October 2009 (UTC)

There's code for 'Intersection of Line Segments' by Prasad, Mukesh in Graphics Gems II called xlines.c Dmcq (talk) 11:55, 5 October 2009 (UTC)

And I've just noticed it has another in Graphics Gems III 'Faster Line Segment Intersection' by Antonio Franklin. I must have a good study of those gems again. Perhaps I better get some more money too looking at the price of vol V and the way I buy books up. 12:06, 5 October 2009 (UTC)

Thinking about it; maybe the computing reference desk would have been a better place to post this computing based question. They deal with "Computing, information technology, electronics, software and hardware." ~~ Dr Dec (Talk) ~~ 17:58, 5 October 2009 (UTC)

The original question was about maths but the answers given have been computer related, and of no use to the OP regretably thanks for the effort / good intentions anyway. 78.144.240.57 (talk) 20:21, 5 October 2009 (UTC)

The OP said "This is for a computer program", I'm surprised if computer related answers are of no use. Dmcq (talk) 09:01, 6 October 2009 (UTC)

Powerful numbers

I don't understand the wikipedia article on powerful numbers.

Part 1 says (I think) m is powerful if its prime factors all come as at least squares. eg if a,b,c are all prime them we have a^2, a^3, a^77, a^2*b^2, a^3*b^4*c^77 as examples of powerful numbers. The list given appears to confirm this understanding.

Part 2 'rabbits on' about a^2*b^3, but 4 is powerful and that is only 2^2.

What is all this stuff about cubes for? Makes no sense (to me). -- SGBailey (talk) 22:47, 4 October 2009 (UTC)

The a²b³ part only requires a and b to be positive integers, not necessarily primes, so 4 = 2² × 1³ is fine. —JAO • T • C 22:54, 4 October 2009 (UTC)

Indeed. If n is a powerful number then any prime factors of n with odd exponents will have an exponent of 3 or greater. Make b the product of these prime factors with odd exponents (there may be none, in which case set b equal to 1). Then all the prime factors of n/b³ have even exponents and so n/b³ is a perfect square, which we can call a². For example, to put p³q⁴r⁷⁷ into the a²b³ form, we have b = pr, so

$p^{3}q^{4}r^{77}=(q^{2}r^{37})^{2}(pr)^{3}$ Gandalf61 (talk) 23:05, 4 October 2009 (UTC)

October 5

Odds ratio vs. relative risk

To begin, all who have contributed have been a great help.

I've asked a few days ago about the difference between odds and probability -- this is sort of the same idea, but with an example. One could either look here or use another example, but my question is: what is the difference in reporting relative risk (probability) vs. odds? In the example I linked to above, the relative risk is 4.5 times as likely while the odds ratio is 36 times the odds. What is that supposed to mean to someone? I mean, to most people (I assume, because it seems this way to me), if you would ask me to choose between the two, I'd say the latter is WAY more likely, just because I think most people intuitively disregard the words and focus on the number -- 36 being much bigger than 4.5, and people wouldn't even realize that they are equivalent. Is it like metric vs. non-metric...cause that's just units (I don't think they are similar). I'm just confused over this entire thing. DRosenbach ^{(Talk | Contribs)} 02:12, 5 October 2009 (UTC)

Can you copy here your original question?--Gilisa (talk) 07:14, 5 October 2009 (UTC)

The statment 4.5 times as likely are largely meaningless. Compare 1% to 4.5% and 22% to 99%. Taemyr (talk) 07:18, 5 October 2009 (UTC)

The odds ratio takes the sample size into account, the relative risk doesn't. In your example, if the sample size were 200 (instead of 100) then the odds ratio would been 7.364 (3 d.p.) So the odds ratio changes as the sample size changes. But the relative risk would still be 90/20 = 4.5. If your sample size is s and the number of men who drank wine was m_y and those that didn't was m_n, where m_y + m_n = s, and similarly with women. We always have the relative risk given by m_y / w_y, but the odds ratio would be m_yw_n / m_nw_y. The relative risk is not too useful then. This is highlighted by Taemyr's post. ~~ Dr Dec (Talk) ~~ 11:20, 5 October 2009 (UTC)

When the rare disease assumption does not hold, the odds ratio can overestimate the relative risk.

Thanks guys -- but what about this? I think it's basically my point -- I'm sure mathematics people can figure out many ways to represents relativity between two achieved results, but if the concept of the different methods has no bearing to not only the average person, but to even a large proportion of well educated people, isn't it sort of like generating a logical fallacy? DRosenbach ^{(Talk | Contribs)} 12:16, 5 October 2009 (UTC)

The odds ratio is larger than the relative risk if, and only if, m_n < w_n. The difference is zero if, and only if, m_n = w_n. The relative risk is larger than the odds ratio if, and only if, m_n > w_n I'm sure that the use of the measure depends on the input values. Could you leave a link to the quote that "When the rare disease assumption does not hold, the odds ratio can overestimate the relative risk." So that we might read it in context? ~~ Dr Dec (Talk) ~~ 17:10, 5 October 2009 (UTC)

Sorry 'bout that -- I made the quote into a link. I'm appreciative of your effort, but none of what you just said makes any sense to me -- I'm a periodontist studying statistical research papers for validity of methods, and balk at the sight of letters intermingling with numbers as they do in your explanation.

Essentially, after getting a handle on the distinction between the procedure for determining odds vs. that for determining probability, I am having difficulty understanding what I'm supposed to do with those numbers. What in the world does "X times the odds" mean? I'm grappling to wrap my brain around such a concept -- I find probability completely intuitive, yet odds completely out there without anything with which to figure out what it means. DRosenbach ^{(Talk | Contribs)} 19:24, 5 October 2009 (UTC)

Numbers? I didn't use any other numbers than those that you yourself used. My 7.364 (3 d.p.) was a reformulation of your 36. So I'm sorry that you "balk at the sight of letters intermingling with numbers", but maybe you should have tested your gag reflex with your original post? ~~Dr Dec (Talk)~~ 19:42, 5 October 2009 (UTC)

Moreover, your link is littered with number and letters living in harmony. Quite why you "balk" is beyond me. ~~Dr Dec (Talk)~~ 19:44, 5 October 2009 (UTC)

Furthermore, if you are a "periodontist studying statistical research papers for validity of methods" then why did you make your original post? I resolved it with some high-school algebra. ~~Dr Dec (Talk)~~ 19:49, 5 October 2009 (UTC)

The math is merely a tool to reach some figure -- I don't think I understand what 36 units of "times the odds" means versus 4.5 units of "times the chances". Just today, though, someone told me that odds refers to taking the result (such as being drunk) and figuring out what the odds are that that person fits into the more likely category (male vs. female) -- would that be accurate? DRosenbach ^{(Talk | Contribs)} 21:22, 5 October 2009 (UTC)

Having odds of m-to-n against means that in a sample of m + n the number of people with a certain property is n and the number of people without that property is m. The key point is that it depends on both the total sample size and the number of people with a certain property. Odds are a sample-size-invariant measure. ~~Dr Dec (Talk)~~ 22:28, 5 October 2009 (UTC)

dx

What I wanted to ask is, when we integrate a function with respect to x, say, we put dx at the end of the integral. Such that we would write ʃ f(x) dx. Now I understand this is to show us which variable we are integrate with respect to, but when we do the U substitiution it also seems as if the dx is treated as being some type of factor. For example :

ʃ ( 3x²/x³) dx let U = x³ du = 3x²dx, so we turn it into ʃU du, which equals ln |U|+C which equals ln|x³|+C

How does this work ? I realise that the derivative can be treated like a fraction, and can cancel out in that way especially when applying the Chain Rule, but why then is the dx treated as if it is some kind of factor as if we are saying that theIntegral of f(x) dx is the same as saying the Integral of f(x) times dx ? Thanks , The Russian. 202.36.179.66 (talk) 02:59, 5 October 2009 (UTC)

Think of dx as an infinitely small increment of x. An integral is a sum of infinitely many infinitely small quantities. And if, for example, f(x) is in meters per second and dx is in seconds, then f(x) dx is in meters (see dimensional analysis). With derivatives, dy and dx are corresponding infinitely small increments, so that if dy/dx = 3 then y is changing 3 times as fast as x is changing. Michael Hardy (talk) 03:24, 5 October 2009 (UTC)

...and please don't use capital U and lower-case u to represent the SAME variable. That isn't done. Michael Hardy (talk) 03:25, 5 October 2009 (UTC)

While that is a good way to think about it, it isn't how it is usually made rigorous. Technically speaking, the "dx" is purely abstract notation and has no meaning in itself. The notation is a kind of mnemonic - it takes the place of the $\Delta x$ ("change in x") in the sum (that you get by chopping the area under the curve into bits, approximating each by a trapezium and adding together their areas - $\Delta x$ is the width of each bit) that you turn into an integral by taking the limit (you let $\Delta x$ tend to zero, which means the number of bits tends to infinity, each with zero area - when you add up that infinitely many zeros you get the answer you want, we call that integrating). --Tango (talk) 04:42, 5 October 2009 (UTC)

When one first learns any particular technique in calculus, one should ask himself/herself about its purpose. The purpose of the technique is perhaps its best intuitive representation. In the case of integration by substituition, the purpose is quite simple - convert the integrand to a much simpler expression, whose integral is computable by standard techniques. This is of course imprecise, but reveals the general idea behind the technique. However, may I ask whether you have ever wondered why this is so, or how this is accomplished?

Suppose we have an arbitrary function (f) defined on some interval [a, b] (here we consider definite integration). Geometrically, this integral is precisely "the area bounded between the graph of f and the x-axis." Suppose we were to consider the graph of f a few units to the right - would we still obtain the same area? A little thought will reveal that the answer is no. However, let us consider a concrete example of f, and let us consider its integral over [0, 1]. Before doing so, let us consider the notion of "the length of an interval."

The length of [0, 1] is 1 unit, in accord with our natural perception of length. Let us consider the image of [0, 1] under the function, f(x) = x²; does this image have a different length? Well, f(0) = 0, f(1) = 1, f is strictly increasing, and thus the image is simply [0, 1] and thus has length 1. Let us consider a different interval with the same function - [2, 4]. What is the image of this interval under f(x) = x²? Well, f(2) = 4, f(4) = 16, f is strictly increasing and thus the image is simply [4, 16] and thus has length 12. How may we write this in terms of the length of [2, 4] (of length 2)? Well 2 x 6 = 12, so we must find a general method which allows us to obtain 6. The method is actually quite simple - firstly, compute the derivative of f - this is the function g(x) = 2x. The midpoint of [2, 4] is 3, and g(3) = 6, and thus we have obtained 6 by rather simple means (the mean of 2 and 4 is 3 - note the pun!).

In general, given an arbitrary function h (assume that it is not too complex, in that assume that it is differentiable, or that it is a polynomial), and an interval [a, b], how can we determine the factor by which h "enlarges" (or "shrinks") [a, b]? I will leave this problem to you - note that intuitively, this should be related to "how fast h is changing" (thus the derivative of h), as we observed.

In your example, if we are integrating a function over a particular interval, the function may not always be in a simple form on that particular interval. However, on another interval, it may well be (consider the integral of f(x) = x + 1, over [0, 1], and note that this is the same as the integral of g(x) = x over [1, 2], by "transforming the interval" by f). Thus, perhaps, by defining a transformation from the initial interval to another one, the integral may be simpler to compute. However, the integral is intimately tied to the notion of an "area", and thus we must know the length of the new interval. This may be done by the notion of a derivative, as I have already described. When one refers to "u-substituition", one writes:

u = g(x)

du = g^'(x)dx

In an intuitive sense, the derivative of g multiplied by the length of the old interval, equates the length of the new interval - du (when one considers "small increments", this formula works precisely (in that the error becomes miniscule)). The function u is the transformation onto the new interval. Hope this helps (and remember to answer my question). --PST 08:01, 5 October 2009 (UTC)

In answer to your question, yes, I have thought about why and how over many things in Mathematics, which is why I asked this about dx in the first place, because I have learned never to just accept things or take them at face value - this goes in all subjects, but is especially true for Maths. And it is certain that knowing why or how particular things work, such as the fact that derivatives are in fact functions anyway, which is why we can treat them as such when applying the Chain Rule, will assist anyone in their further exploration of the Atlantis that is Mathematical Truth. IF we just accept things without thinking why or how we may not get very far. The explanations given here do now make a lot of sense, and thank You everyone who contributed. TO me it doesn't make sense to study a discipline, then maybe try to teach it to others, without really considering what it is all about, or what it means. I was initially confused when I first encountered U substitution, but when one begins to think about it, then it starts to make sense. I had another question relevant to this which I will state in another place below here. THANKS. The Russian.202.36.179.66 (talk) 02:01, 6 October 2009 (UTC)

The Integral symbol is actually a stylized S for sum. It started off like Σ f(x) × Δx, the limit of the sum of lots of tiny rectangular bits as the width Δx of the bits tended to zero. Dmcq (talk) 12:19, 5 October 2009 (UTC)

Though this notation inspired the notation used with differential forms and not vice versa; you can look at fdx as a 1-form, in which case it really is f times dx. Additionally, u = x^3 has du = (3x^2)dx, d exterior differentiation. —Preceding unsigned comment added by 71.61.48.205 (talk) 13:28, 5 October 2009 (UTC)

I presumed that the notion of a differential form was too complex (and general) for the OP to comprehend. Some of the essential ideas which undermine these concepts are there in my post, however. --PST 13:33, 5 October 2009 (UTC)

I don't like to presume anything. Also, by reading things I didn't understand and trying to piece them together was how I learned half the things I did; thus, when I could actually learn about advanced topics, they weren't a mystery to me.. Finally, I didn't say anything was wrong with your answer, it is quite good. 71.61.48.205 (talk) 13:39, 5 October 2009 (UTC)

Actually, I take back what I said, your answer is not good at all. Consider g(x) = x^3, g takes [1,3] to [1, 27], length 2 to length 26. However, 2 * (2 ^ 2) = 8 not 26; though g'(2) should be 27 by your ideas. Of course, you might have meant that we can apply the mean value theorem, though your example using the mean of the end points doesn't convey this at all; especially to someone you assumed knew little of mathematics. The above notion of how the size of the intervals we are using in our partition goes to zero makes much more sense (this is the one I assumed was yours, should have been more cautious).

To op, this same notion of treating dy/dx like a fraction comes up in implicit differentiation aswell, as in the case x^2 + y^2 = r^2, we use 2xdx + 2ydy = 0, then solve for dy/dx, obtaining -x/y. Thus, supposing that du/dx = 3x^2, we can view this as du = (3x^2)dx; hence, if we are integrating (3x^2/x^3)dx, then using the identity gives (1/u)du. I'm aware that this isn't the most helpful answer, but you could consider d as a thing that takes functions f(y) to f'(y)dy, dy undefined. As another example of this, y = x^2 -> dy = 2x dx -> dy/dx = 2x. Finaly, then we observe that 3x^2 = d(x^3)/dx, thus (3x^2)dx = d(x^3)/dx * dx = d(x^3); using definite integrals, you have to change the inteval being integrated over, in which case dx is no longer undefined and takes on a geometric meaning, thus we view d(x^3) as being a different increment than dx.

Two things. I'm not big on the whole implicity differentiation idea as above for a number of reasons. Second, what I said above is not precise, nor the best way of viewing things, just a rough idea of how derivatives can have an algebraic nature; leafing through some more advanced things might be better. 71.61.48.205 (talk) 14:39, 5 October 2009 (UTC)

In Calculus I, the "dx" is best thought of as a placeholder only, and expressions such as

if u = x² then du = 2x dx

should be taken as purely formal, nothing more than convenient notation.

In more advanced mathematics, there are at least two ways one can interpret the "dx", but these new interpretations are not necessary to understand the theory of calculus. One of the ways is to think of "dx" as a differential form, and the second is to think of "dx" as representing "dμ" where μ is a certain measure on the real line. Some of this is sketched in the lede section of this nice article on integration of differential forms by Terence Tao. — Carl (CBM · talk) 02:20, 6 October 2009 (UTC)

I disagree. The integral

$\int _{a}^{b}f(x)\,dx$

should be regarded as the sum of infinitely many infinitely small numbers, since dx is infinitely small, and its units are those of ƒ(x) times those of dx.

Of course this is not logically rigorous. That means logical rigor isn't everything. Michael Hardy (talk) 03:37, 6 October 2009 (UTC)

I have no idea which part you claim to disagree with. If it's the first sentence, I don't want to get into a lengthy discussion about how the Riemann integral should be interpreted. Chacun à son goût. My main point was the second paragraph. — Carl (CBM · talk) 10:34, 6 October 2009 (UTC)

You can do calculus in terms in infinitesimals (even rigorously), but it isn't the normal way. See non-standard calculus. --Tango (talk) 14:37, 6 October 2009 (UTC)

You might want to consider this definition of dx. This is easy enough for high school and sufficiently rigorous: Assume f is differentiable at a point x, define dx to be the independent variable that can have any real value, and define dy by the formula dy = f'x)dx. --(1) So dy really represents a value achieved by a function. If dx is non zero then we can divide both sides of (1) by dx to obtain dy/dx = f ' (x) ---(2) Thus we have reached our goal of defining dy and dx so that their ratio is f'(x). Formula (1) is said to express (2) in differential form and dx,dy are called differentials. This is from CALCULUS by Howard Anton.--Shahab (talk) 16:09, 6 October 2009 (UTC)

Constant Morphisms

In category theory a constant morphism is defined as a c:X ->Y so that given any Z and any arrows g,h:Z -> X, cg = ch. That said, I was wondering if anyone was aware of any theorems/theories that make use of this notion, for some reason I find them interesting and all I seem to find about them is their definition. If anyone could point me in the direction of any work done using this concept, I would be extremely grateful. —Preceding unsigned comment added by 71.61.48.205 (talk) 13:02, 5 October 2009 (UTC)

You have not explicitly stated your mathematical knowledge and thus I should not state (or assume) it either. One example of a mathematical discipline which is closely related to category theory is homological algebra. However, there are many other mathematical disciplines which make use of category theory (which I shall leave subsequent users to explain). With regards to the notion of a constant morphism, I should say that it is a mere generalization which plays the same role in category theory as does the constant function in "traditional mathematics". --PST 13:28, 5 October 2009 (UTC)

Thank you for your response, I was mainly interested in the use of constant morphisms though rather than category theory. As for mathematical knowledge, most of my interests are in homotopy theory and logic. I'm curious for two reasons, first,all maps from a terminal into X are constant, second I've found a couple neat ways to use coconstants and related morphisms to define algrbraic-like objects in a category; thus, I wanted to see if anybody else has done anything similair, and if i led anywhere interesting.[Also, there I imagine there are some neat relations between constants into objects in a diagram and constants into the limit of that diagram, etc.] —Preceding unsigned comment added by 71.61.48.205 (talk) 13:35, 5 October 2009 (UTC)

Why won't you publish it here? If you sign it then it's yours.--Gilisa (talk) 15:01, 5 October 2009 (UTC)

Im confused as to what you mean by, "publish it here" I haven't really written an article or anything of that nature, and I don't believe this is the place for such things anyways. I'm here because I'm unfamilair with anyone else making use of constnats for any real purpose, and would like help determining if someone else has; not so much so that I'm not duplicating ideas, but because if someone has, I'm sure they've done much more than myself and I would like to see what. Thank you for the reply though. Also, signing something here doesn't really make it "yours", not that you really own mathematical results anyways. 71.61.48.205 (talk) 16:14, 5 October 2009 (UTC)

Well, guess you are right. Sorry for my silly suggestion :)--Gilisa (talk) 16:26, 5 October 2009 (UTC)

Please look on these (you might saw these already): [2][3] and this [4].

P.S. I don't know enough about this matter but it seem that if you category don't have enough morphisms (e.g., in categories when all of them are reversible) so the definition you provide might be not too successful because when you rely on the arrows you lose a lot of information.--Gilisa (talk) 16:54, 6 October 2009 (UTC)

Proving the existence of a function

I am trying to prove that there exists a function J(x) such that J(J(x))=exp(x). I am also trying to prove that for the same function, J(0)=1/2. I have some reasons to believe that both of these can be proven, but I have no idea how to start a proof of either. If someone could lend some insight that would be great. Thank you Jkasd 20:40, 5 October 2009 (UTC)

See the stub Functional square root. Bo Jacoby (talk) 20:51, 5 October 2009 (UTC).

::Jkasd your wording is not very clear, do you mean that you have to find function so (J(J(x))=exp(x and J(0)=1/2 or that you have to find function that apply (J(J(x))=exp(x and that every such function have J(0)=1/2?

as I understand it, I believe that you can't prove it. You want to take function that match 1 to 1 and then match 1 to 2.7. It's not possible.--Gilisa (talk) 21:53, 5 October 2009 (UTC)

I'm not sure what part of Jkasd's wording you thought was vague; it was pretty clear to me that he/she was trying to prove that any solution of J(J(x)) = exp(x) must have J(0) = 1/2.

How are you concluding that J(1)=1 and J(1)=2.7 both have to be satisfied? --COVIZAPIBETEFOKY (talk) 22:07, 5 October 2009 (UTC)

Jkasd, it appears to me that there exist many solutions to your functional equation. If we let x ≥ 0, we have:

J(J(-x)) = exp(-x)

J(J(J(J(-x)))) = exp(exp(-x))

...

J²ⁿ(-x) = expⁿ(-x)

which defines a chain of values of the (2n)^th power of J on x. It cannot be pushed any further backward, to a solution y of -x=J(J(y)), since that would imply -x=exp(y), and exp does not take on negative values. There do not appear to be any restrictions on J(-x), as far as I can see. So we can make any solution by pairing up values of x≥0 so that every x is in a pair and no x appears twice (let's call each pair x₀ and x₁), and defining:

J(-x₀) = -x₁

J(-x₁) = exp(-x₀)

J(exp(-x₀)) = exp(-x₁)

J(exp(-x₁)) = exp(exp(-x₀))

etc...

Your suggestion of J(0) = 1/2 is effectively pairing x₀=-ln(1/2)=ln(2) to x₁=0. This is arbitrary; there are other solutions that work perfectly well without satisfying J(0) = 1/2. --COVIZAPIBETEFOKY (talk) 22:44, 5 October 2009 (UTC)

Of course, I didn't really prove that this works, but it's not too difficult to fill in the missing steps. What needs to be shown is that, in the above process, every real number r is assigned a value J(r) exactly once, so there's no conflict in mapping.

For an example set of pairs x₀ and x₁, let every x₀ be between 0 (inclusive) and 1 (exclusive), and, for each of those, define x₁ = 1/x₀, unless x₀=0, in which case x₁=1. --COVIZAPIBETEFOKY (talk) 22:57, 5 October 2009 (UTC)

There was another thread on exactly this subject recently: Wikipedia:Reference desk/Archives/Mathematics/2009 August 9#Functional square root. -- BenRG (talk) 23:28, 5 October 2009 (UTC)

Thanks for the responses everyone. I didn't realize that this concept had a name, so knowing that helped me be able to research it more fully. So it seems that there are actually a whole class of functions with this property, but I still think that J(0)=1/2 might be a "natural" value for it, mainly because its graph looks nicer than for most other values for J(0). My math professor, who can speak German, is going to go over Hellmuth Kneser's paper about it with me when he has time. Jkasd 03:12, 6 October 2009 (UTC)

Here it's [5] (for those who don't have it)..But that's not a solution that you expect to on a forum like page, at the least not so immediately. BTW, Kneser was dealing with analytical solution for the problem. In this case you must have one solution (i.e., you can't choose the value at 0).--Gilisa (talk) 06:59, 6 October 2009 (UTC)

We can try to have analytical solution if there is no demand for one solution (this is the case here I see now..). You can start the builiding from this direction:

f(x) =

-1/x - 1/ln(½)........................ x in ( -∞ , ln(½) )

exp{ -1 / (x - 1/ln(½)) }.......... x in [ ln(½) , 0 )

continue from here........................................ x in [ 0 , 1 )

Or in other words:

we start with the definition of Injective copy, simple as we can

Ho(x) : (-∞,α] → (α,0]

Ho is pretty arbitrary and it's easy to choose it in the form Ho(x) = a/x + b where we determined a and b by the definition of H0 (x).

we continue with building : H1(x) : (α,0] → (0,e^α]

H(1) is defined by the given functional equation as H1[H(x)] = e^x

from condition f(0) = H1(0) = 1/2 you can extricate your α.

then continue with the building

--Gilisa (talk) 08:57, 6 October 2009 (UTC)

Fourier Series Representation

What is the Fourier series representation for this signal?

Sorry, I could not figure out how to insert a picture into the document so here is a link: Signal

Also, if the power of each component in the Fourier representation of the signal is given by the square of the component's amplitude, what is the medium bandwidth required to pass at least 80% of the power of this signal?

Partial sum for 1 ≤ n ≤ 5

Partial sum for 1 ≤ n ≤ 10

Partial sum for 1 ≤ n ≤ 100

129.186.55.233 (talk) 22:39, 5 October 2009 (UTC)

The Fourier series article supplies the formulas for finding the Fourier coefficients of a periodic function. Have you tried using those? If so, could you be more specific about what problems you're having? Rckrone (talk) 03:08, 6 October 2009 (UTC)

This page [6] should at least get you started if the Fourier series article isn't enough.--RDBury (talk) 12:26, 6 October 2009 (UTC)

• You've got an even function, that is to say ƒ(x) = ƒ(−x) for all x in the domain of definition. In such a case the sine terms are redundant, since the sine function is an odd function. We only need to consider the cosine terms. Well, in the case of an even function ƒ we have

$f(x)={\frac {1}{2}}a_{0}+\sum _{n=1}^{\infty }a_{n}\cos \left({\frac {n\pi x}{L}}\right)$

Where L is the length of periodicity, in this case L = 2. The coefficients are given by

$a_{0}={\frac {2}{L}}\int _{0}^{L}f(x)\ dx\ ,$

$a_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\cos \left({\frac {n\pi x}{L}}\right)dx\ .$

Where ƒ(x) = x for all 0 ≤ x ≤ 1 and ƒ(x) = 2 − x for all 1 ≤ x ≤ 2. It follows that

$a_{n}=\int _{0}^{1}x\cos \left({\frac {n\pi x}{2}}\right)dx+\int _{1}^{2}(2-x)\cos \left({\frac {n\pi x}{2}}\right)dx\ .$

We get the expression for a₀ by deleting the cosine terms in the above integrands. Finally we see that

$f(x)={\frac {1}{2}}+\sum _{n=1}^{\infty }{\frac {4}{\pi ^{2}n^{2}}}\left(2\cos \left({\frac {\pi n}{2}}\right)+(-1)^{1+n}-1\right)\cos \left({\frac {\pi nx}{2}}\right)\$

I've added some plots of the partial sums. I hope this helps. ~~ Dr Dec (Talk) ~~ 17:55, 6 October 2009 (UTC)

Cool! You made a lot of work for the OP, It's always nice to see helpful people--Gilisa (talk) 17:58, 6 October 2009 (UTC)

October 6

Question on topological manifolds

Ok I'm a novice at this so please keep your answers simple. Let R be a square with unit length. Now remove three of its sides, and here's the important part: keep the two end points of the 4th non removed side. Is it a topological manifold with boundary? I've been thinking about what happens to the two end points of the non removed side, and can't seem to stretch any of their neighborhoods into an open ball of the half plane. Can anyone help? Breath of the Dying (talk) 09:40, 6 October 2009 (UTC)

If I understand your question correctly, then yes, it is a topological manifold with boundary. Note that a topological manifold (in the simplest sense), is simply a space (well-behaved geometric figures, for instance), each point of which possesses a neighbourhood homeomorphic to an open ball in some Euclidean space. In one dimension, an open "ball" is simply a line segment without endpoints. Thus, in your question, without the end points, your space is a pure topological manifold. In one dimension, however, a boundary can also be defined, and in this case, the boundary consists of the two endpoints of the line segment. Hence, it is a topological manifold with boundary. Note that the real line can be viewed a subset of the plane (and it is a manifold) but does that mean it can be stretched onto an open ball in the plane? The answer is no because it is a one-dimensional manifold. Similarly, the same logic applies to a line segment - just because you cannot stretch any neighbourhood onto some "open ball in the half plane" does not mean it is not a manifold; it just means that it is not a two-dimensional manifold. --PST 11:03, 6 October 2009 (UTC)

This was how I also interpreted the question originally, but then I thought Breath might be looking at the square as a 2-manifold with boundary and then only removing parts of the boundary while retaining the interior in its entirety, as otherwise it would seem like a very contrived way of simply asking something about a line segment. —JAO • T • C 11:24, 6 October 2009 (UTC)

Yes, the OP seems pretty clearly to be describing the open square (0,1)x(0,1) with the closed side [0,1]*{0} added on. That is not a manifold with boundary, due to bad behaviour at (0,0) and (1,0). Algebraist 12:27, 6 October 2009 (UTC)

(ec) No, it isn't a manifold with boundary. The 4th side and its endpoints are kept. (If they weren't kept it would be a MWB). There aren't any local homeomorphisms at those corners to a halfspace. As our Topological_manifold#Manifolds_with_boundary article says, "If M is a manifold with boundary of dimension n, then Int M is a manifold (without boundary) of dimension n and ∂M is a manifold (without boundary) of dimension n − 1." The boundary is a line segment, which is a manifold with boundary, not a manifold without boundary.John Z (talk) 12:43, 6 October 2009 (UTC)

I agree it's not a manifold with boundary in the subspace topology of R² (which I'm sure is what the OP meant). But I think I can imagine a topology (if you allow a noncountable basis) that makes it a 1-manifold with boundary. Staecker (talk) 14:00, 6 October 2009 (UTC)

It's a set with cardinality continuum, so if we're choosing a topology on it at will, we can make it into any manifold we please. Algebraist 14:16, 6 October 2009 (UTC)

Good point. Staecker (talk) 14:44, 6 October 2009 (UTC)

Geometry

This was a question which was presented to me in an event I attended, which I couldn't solve.. How do you arrange 7 points in a plane such that when any three of them are chosen, atleast two of them are 1cm apart. I'm interested in finding out the solution to this, assuming it exists.... So can someone help me.... Rkr1991 ^{(Wanna chat?)} 17:58, 6 October 2009 (UTC)

Maybe this way:

You choose plane:

1. any point.
2+3. two points that their distance from one is $\scriptstyle {\sqrt {3}}$ and from each other 1 cm.
4+5. two points with 1 cm distance from points 1 and 2.
6+7. two points that their distance from 1 and 3 is 1 cm.

--Gilisa (talk) 18:45, 6 October 2009 (UTC)

I just googled "seven points in the plane". The first hit was this one ~~ Dr Dec (Talk) ~~ 19:27, 6 October 2009 (UTC)

Googling does not count!--Gilisa (talk) 19:33, 6 October 2009 (UTC)

:o) You've got me, I'm sorry! ~~ Dr Dec (Talk) ~~ 19:37, 6 October 2009 (UTC)

Strictly speaking, this is a reference desk, so we are actually supposed to find references for people rather than answering questions from personal knowledge. Therefore, Declan wins! --Tango (talk) 22:51, 6 October 2009 (UTC)

You are forgiven this time! Now I better explain my logic:

• We choose three points without two at distance of 1 cm.
  
• Had we choose 1, we can't choose none of 4, 5, 6 or 7.
  
• Have we didn't choose 1, we can now choose one of these groups of three {2,4,5} {3, 6, 7} to take from two points and the distance between them is 1 cm.
  --Gilisa (talk) 20:02, 6 October 2009 (UTC)

See Hadwiger–Nelson problem for an interesting application of this configuraton. Dmcq (talk) 22:39, 6 October 2009 (UTC)

Since when do centimeters figure into plane geometry? →Baseball Bugs ^{What's up, Doc?} carrots 23:49, 6 October 2009 (UTC)

October 7

Integration by Substitution

I have a problem with integrating a certain function. I realise it can also be done as it is in Anton, Bivens and Davis Calculus, but I am always interested in seeing if a thing can be done in many different ways. The diagram on the link is not very clear. Its base equals √( 4 - x²), its height is equal to x, and the hypotenuse is 2.

[IMG]http://img194.imageshack.us/img194/7444/integration.png[/IMG]

If you have any advice, please do not hesitate. Thanks, The Russian.C.B.Lilly 02:11, 7 October 2009 (UTC) —Preceding unsigned comment added by Christopher1968 (talk • contribs)

You should not have cos(y) under the square root sign in the last integral of your solution. Things come out the way they should if you fix that, I think. 70.90.174.101 (talk) 06:55, 7 October 2009 (UTC)