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February 18

Topological relationship between a sphere and a cone?

slice of sphere and imaginary paraboloid. Dmcq (talk) 14:56, 19 February 2012 (UTC)[reply]

Specifically, for a story I'm writing, I'm curious if there's any special mathematical relationship between the surface of the sphere of Earth and the hollow stepped cone of Dante's Hell. Is there some clever topological way one can be folded to produce the other? Please to remember, I'm a lay questioner - keep it simple...

Adambrowne666 (talk) 07:58, 18 February 2012 (UTC)[reply]

Well, for the simplest topological relationship, homeomorphism, sharp edges and spikes don't matter, so you can smooth them out (our article has a picture of a coffee cup being folded and stretched to make a doughnut). You could easily stretch a closed stepped cone to make a sphere (you could even flip it, so the inside of the cone becomes the outside of the sphere, as long as you're talking about mathematical abstractions rather than solid rock). The problem is that the pit is, if I recall correctly, open at the top - certainly it is in pictures like this: File:Stradano Inferno Map Lower.jpg. You could flatten out the steps of the pit, and stretch it around into a kind of ball shape, but you'd still have a hole that you couldn't close. I don't think there's a way to fold that hole away (or to be more rigorous, I don't think there's a way to map every point on a solid sphere onto a punctured ball continuously), since you fundamentally change the properties of the shape (a ball with a hole in it doesn't have a clearly defined interior - a sphere does). Smurrayinchester 11:52, 18 February 2012 (UTC)[reply]

Thanks, Smurray - that's interesting - so you'd end up with a sphere with a hole; makes me think of the 19th Century speculation on the idea of there being an entry to the hollow Earth somewhere around the Arctic. I must admit, I'm being sort of plagiaristic here, referring to the great old Christopher Priest novel The Inverted World, where the protagonists' altered perceptions causes them to see the Earth and Sun etc as pseudospheres. Thanks again Adambrowne666 (talk) 13:14, 18 February 2012 (UTC)[reply]

If you allow an imaginary radius as you go up past the north pole you get a paraboloid going off to infinity there from

r={\sqrt {R^{2}-z^{2}}}

. Dmcq (talk) 15:37, 18 February 2012 (UTC)[reply]

Thanks, Dmcq, that's a sort of magnificent answer, but I can't see it - I see from your profile you're interested in the visualisation of mathematics - maybe you could take pity on me and try and depict it to someone mathematically shortsighted? (BTW, I'm fascinated by your link to EPI - gonna look into that further.) Adambrowne666 (talk) 05:07, 19 February 2012 (UTC)[reply]

How about this then? If Chariots of the Gods can make money there must be something in trying to sell Dante's hell at the poles with this. ;-) Dmcq (talk) 14:56, 19 February 2012 (UTC)[reply]

Ultracool! Thanks heaps. I can certainly use that.

Strange voting system

I'm not sure if I read about this somewhere or made it up. Does the following system have a name? Every eligible voter is also a potential candidate. All votes cast for a particular person X get "paid forward" to whoever X votes for. Anyone wishing to be elected votes for themself and receives all votes cast directly for them as well as any that were paid forward from other voters. Votes are paid forward until they arrive at a self-voter, and otherwise (if there's a loop of votes) they never get counted. Sound familiar to anyone? Staecker (talk) 21:37, 18 February 2012 (UTC)[reply]

Delegated voting, also known as liquid democracy, is close, although it doesn't quite involve the same pooling of votes that goes on here (people nominate themselves as candidates, rather than voting for themselves, though apart from preventing loops that is just a semantic quibble really). Smurrayinchester 21:58, 18 February 2012 (UTC)[reply]

Thanks! That's close enough to probably be what I was thinking of. Staecker (talk) 22:34, 18 February 2012 (UTC)[reply]

3y-2x=5+9y-2x

How to graph this equation? I just curious. — Preceding unsigned comment added by 65.92.151.169 (talk) 23:59, 18 February 2012 (UTC)[reply]

First solve for X or Y (in this case, only Y is possible, and even that will make for a rather dull graph). StuRat (talk) 00:04, 19 February 2012 (UTC)[reply]

Add 2x to both sides to get 3y=5+9y. Subtract 3y from each side and subtract 5 from each side to get -5=6y. Divide both sides by 6 and switch the sides to get y=-5/6. The graph will be a straight horizontal line at -5/6.--Mattmatt1987 (talk) 17:34, 19 February 2012 (UTC)[reply]

I was hoping they would do their own homework, after I told them what they needed to do. StuRat (talk) 22:26, 19 February 2012 (UTC)[reply]

February 19

Geometric algebra question: degrees of freedom to a k-blade

I am well-aware that for a geometric algebra over the reals, the number of (real) degrees of freedom to a k-vector is given by the appropriate binomial coefficient, i.e. ${\tbinom {n}{2}}$ is the number of real numbers required to specify a 2-blade for a geometric algebra over an $n$ -dimensional real space. However, clearly the number of real numbers required to specify a 2-blade is often less than this; given that every pair of vectors can be specified by $2n$ , and any 2-blade can be expressed as the exterior product of 2 vectors, a 2-blade cannot represent more degrees of freedom than this, yet the space of 2-vectors is larger for $n>5$ . So, how many degrees of freedom are required to specify a general k-blade?--Leon (talk) 14:55, 19 February 2012 (UTC)[reply]

For a general k-vector, it's n choose k (see exterior algebra). The subvariety of k-blades is (projectively) the Grassmannian of k-dimensional subspaces of the n-dimensional space and has dimension k(n-k) (projectively) or k(n-k)+1 (nonprojectively). Sławomir Biały (talk) 17:11, 19 February 2012 (UTC)[reply]

The number of degrees of freedom for picking 2 vectors would be n², not 2n, so it's not less than the binomial coeff. This corresponds (sort of) to the dimension of the tensor space being n²; the space of 2-vectors is a quotient of this.--RDBury (talk) 21:22, 19 February 2012 (UTC)[reply]

But the question isn't how many degrees of freedom are involved in picking two vectors. A two blade is the wedge product of two vectors (not the tensor product.) The question is equivalent to asking how many degrees of freedom there are in picking two dimensional subspaces (projectively, add one for a scale). In R³ for instance, this is equal to 2+1: a two-dimensional subspace is determined by its unit normal, plus one for scaling. Similarly a k-blades is the wedge product of k vectors. These are in one-to-one correspondence with a trivial line bundle over a Grassmannian. See exterior algebra for a discussion. Sławomir Biały (talk) 02:06, 20 February 2012 (UTC)[reply]

Sławomir, I think this is a pretty useful result in general, since blades serve a significant role in GA. Do you know of a suitable reference, so that it could be added perhaps to Geometric_algebra#Representation_of_subspaces or some similar section? — Quondum ^☏✎ 06:17, 20 February 2012 (UTC)[reply]

I don't know about any references that focus on geometric algebra, but for Grassmannians (including the result about dimension) a good book is: Griffiths, Phillip; Harris, Joseph (1994), Principles of algebraic geometry, Wiley Classics Library, New York: John Wiley & Sons, ISBN 978-0-471-05059-9, MR 1288523. The proof of the dimensionality is actually straightforward. Take k vectors and wedge them together

v_{1}\wedge \cdots \wedge v_{k}

and perform elementary column operations on these (factoring the pivots out) until the top

k\times k

block are elementary basis vectors of

\mathbb {F} ^{k}

. The wedge product is then parametrized by the product of the pivots and the lower

k(n-k)

block. Sławomir Biały (talk) 11:29, 20 February 2012 (UTC)[reply]

Thanks for the detail; I've taken the liberty of copying the reference and your explanation into Blade (geometry) in the interim. Later, we could copy it to the Geometric algebra article. — Quondum ^☏✎ 13:59, 20 February 2012 (UTC)[reply]

It looked to me like there were two questions, first how many degrees of freedom are there in a picking a blade; and second, how can this number be greater than 2n if that's the degrees of freedom in picking the vectors individually. I was answering the second question.--RDBury (talk) 09:17, 20 February 2012 (UTC)[reply]

There was one question, and I (think I) knew that the dimension of a

k

-blade could not have dimension greater than

kn

, for an underlying basis consisting of

n

basis vectors. Anyway, thank you, Slawomir, you have answered my question completely! As for adding the result to the geometric algebra article, I have several textbooks on the matter that don't answer the question (otherwise I wouldn't have had to ask), and given that these are fairly "standard" textbooks on GA I must conclude that it may be difficult to find a reference on GA specifically that is suitable. Is there a name for this result? If so, I can have a glance through Google Scholar to find a citation.--Leon (talk) 12:04, 20 February 2012 (UTC)[reply]

I didn't really understand the last part of the question, and it looked like you were replying to me instead of the original poster. (The reply did seem strange to me.) Sorry for my confusion. Sławomir Biały (talk) 11:29, 20 February 2012 (UTC)[reply]

No problem, confusion often abounds with text based dialog and I was a bit confused myself.--RDBury (talk) 13:41, 20 February 2012 (UTC)[reply]

describe conditions that would allow you to martingale

here is an example of a system that allows you to martingale: let's simplify roulette to be a game of red, black, or green, with red and black having a nearly 50%, and equal, chance each of occurring (e.g. 48%), and green being the house's edge (e.g. 4%). The green pays off only slightly better than red or black (e.g. 4x, 10x, whatever), and red or black pay off at an "even" rate (ten dollars gets you twenty plus change - whatever makes it an even bet at 48% chance of winning - if you win, zero if you lose).

a casino feels that roulette players are put off by excessive strings of reds or blacks, so it tweaks the random number generator to work thusly: rather than pick each red/black/green in succession based on a probability, instead it allots 12 red, 12 black, and 1 green token into randomly into a block of 25, doles out the 25 results in succession, then repeats this random allotment. This has the advantage, that, for example, nobody could bet green 50 times in succession without a payoff, nor would it be common for anyone to observe, joining the table, that out of 20 bets on red, only three or four pay off. This would otherwise be a common occurrence.

naturally, although this "seems" more random to the observers or players, it is in fact less random. this is so because rather than being a random walk, the system in some sense has memory.

it is hard to game, as it is difficult to know where the block boundaries are. To really have some confidence in a bet on red, you would have to join after observing 24 blacks: 12 at the end of the first block, and 12 at the beginning of the next. Then you could be sure the halfway point was the 25-block boundary, and now you know the rest of the block will be reds or greens. You can wait for a green if you want to have 100% payoff on a red.

All right, then, it's a bit difficult to game, but nevertheless imminently possible. The reason it is possible is obvious: given sufficient observation and the nature of the system, there are locations wherein a bet has positive expectation.

Now suppose that some stock on the stock market were literally cyclic (like a sine wave) and you had some guarantee of this. Naturally, if you have a guarantee that it will reach 0 again, any bet below zero will have a positive expectation (you can just hold the stock until it reaches 0). Of course, the time value may be such that this is not a worthwhile investment. So, we must consider stocks guarantee4d to follow sine waves and guaranteed to do so with high frequency. For best results, with high amplitude as well, though of course you can repeat smaller bets by buying low and selling high.

Thus, were any stock to be a sine wave of high frequency, we would be guaranteed to make money by placing bets with positive expectation.

returning to games. I would like a generalization of the sufficient and necessary conditions a system must meet, in as general terms as possible, to allow an observer to join and make bets with positive expectation. I gave two systems above: a stock roughyl following a sine wave in price, and a casino that doles out roulette results by allotting them in chunks instead of randomly one after the other. I would like a generalization that extends to every system that can be gamed, and how one can prove or disprove that in such a system, an observer can place bets with probability 1 of a positive payoff, and probability 0 of a negative payoff: and we are talking a single bet. (So that someone with this in proof who is infinitely risk-averse will still place the bet, provided he or she is a good mathematician and believes the premises or conditions are really a correct model of the system in question). --80.99.254.208 (talk) 15:14, 19 February 2012 (UTC)[reply]

What you're describing is basically the martingale characterization of algorithmic randomness. In this case, the system must not be computably random.--121.74.109.179 (talk) 20:15, 19 February 2012 (UTC)[reply]

Thanks for the link, feel free to make my ramblings leading to it smaller. Could you tell me what it means to be "computably random" and general strategies for deciding this. Could you give me a formalism that I could apply to arbitrary systems to determine whether they are "computably random"... --80.99.254.208 (talk) 20:32, 19 February 2012 (UTC)[reply]

What I mean here, is I would like to be able to describe the properties of a system, and apply whatever you tell me to, to decide whether I can ever game it as described. Of course, my assumptions are very important, but so are the tools I'm asking from you in order to be able to follow through on deciding the consequences of those assumptions. Thanks for anything you might have along these lines. --80.99.254.208 (talk) 20:35, 19 February 2012 (UTC)[reply]

Unfortunately, no. The definition of computable randomness is basically "can't be gamed". So I'd have a hard time giving you an alternate way to recognize it. A good rule of thumb, though, is to consider the law of large numbers; is every sequence of outputs possible? In the roulette case, 49 blacks isn't possible, so you know right away that the system can be beaten.--121.74.109.179 (talk) 21:16, 19 February 2012 (UTC)[reply]

To get back to the hypothetical roulette game, actually it would be relatively easy to game. The probably that a green follows another green would drop from 1 in 50 to 1 in 2500 and it's certain that someone would notice. Knowing this, even without knowing there the boundaries are, you'd just have to wait for a green to appear and bet on a color other than green on the next roll to beat the house advantage. This is actually a variation of what card counters do in blackjack, in fact it would be much simpler than card counting.--RDBury (talk) 21:42, 19 February 2012 (UTC)[reply]

followup question

in this hypothetical roulette game, after joining randomly how many turns would you have to wait on average before you were 100% sure of the boundary? 84.2.147.177 (talk) 15:06, 20 February 2012 (UTC)[reply]

A number n is a possible boundary if the preceeding 25 outcomes contains exactly 12 reds, 12 blacks, and 1 green. The probability for this to happen by chance can be computed, see multinomial distribution. It is

{\frac {25!}{12!12!1!}}\left({\frac {12}{25}}\right)^{12}\left({\frac {12}{25}}\right)^{12}\left({\frac {1}{25}}\right)^{1}=0.0605092.

So the first 50 outcomes contains one true boundary and, with 6% probability, one additional false boundary. The true boundary is reproduced after the next 25 outcomes, while the false one is eliminated with 94% probability. Bo Jacoby (talk) 06:53, 21 February 2012 (UTC).[reply]

Sorry if I misunderstand you, but I am not interested in confidence levels, but with 100% confidence. Let me give you an example. If you join and the first two results are greens, you have just gained 100% certitude, in two turns, of where the boundary is. (It's in between the two, the first is at the end of the previous group of 25, and the second is the first or the next group of 25 results). So, this is ONE case where you reach 100% certitude. This case has length 2.

And so when you join, there is always a length at which you reach 100% certitude of the boundaries.

What is the average of all of these lengths? (In other words, if I have just joined, I have an average wait time of x turns before I can become 100% sure of where the boundaries are. In practice, it could turn out that I reach certitude after 2 turns, but before I have seen a single result, what is my expected wait for 100% certitude?) --80.99.254.208 (talk) 10:59, 21 February 2012 (UTC)[reply]

I do understand your question, but I did not provide the complete answer. I merely provided a step. After two results you know the answer (2) with probability 1/25^2=0.0016. So the product 2*0.0016=0.0032 contributes to the average length. After three results you know the answer (3) with probability 24/25^3=0.001536 and the product 3*0.001536=0.004608 contributes to the average length. The calculations becomes increasingly complicated, so it is tempting to make some approximations. The probability that you know the answer when n=25 is very low, and the probability that you know the answer when n=50 is very high, so the average is somewhere in between. Bo Jacoby (talk) 15:19, 21 February 2012 (UTC).[reply]

f(1/x)

If I have the graph for a function f(x), what happens to the graph when I turn the function into f(1/x)?190.24.187.123 (talk) 15:39, 19 February 2012 (UTC)[reply]

First off, I'd think of the fixed points of the transformation. That is, where x = 1/x. Those would occur at x = 1 and x = -1. The graph at exactly those points will be the same. As the transformation on the x coordinate is continuous and differentiable in those regions, the area around the fixed points before will also be around the fixed points after. However, as x -> 1/x flips the ordering ( 0.99 < 1 before, but 1.0101... > 1 after), the graph will likewise be flipped around the points at x = 1 and x = -1. So if the graph before is increasing through x = 1, afterwards it will be decreasing, and vice versa. The other place to look is at discontinuities. The behavior around x=0 will be interesting. Points just near the right hand side of zero will be thrown right, toward positive infinity, and points just to the left of zero will be thrown further left, towards negative infinity (the zero point itself will drop off the graph). As the transformation is self-inverse, we can also conclude the reverse will happen - points near positive and negative infinity will be brought in toward zero. As the transformation on the x coordinate is continuous and differentiable on each side from very near zero to both positive and negative infinity, you can envision the transformation as a flipping and stretching. - In summary, take each of the positive and negative sides of the graph, flip them around ±1, as appropriate, and then stretch or squash them to fit in their new ranges (non-uniformly, so the most stretching/squashing happens near zero/infinity). Given the non-uniformity in the scaling and the fact that the infinities don't stay out at infinity, it's difficult to exactly visualize what the after graph will look like, but that's the general idea. -- 67.40.215.173 (talk) 18:57, 19 February 2012 (UTC)[reply]

February 20

Trig functions

Hey I am in precalculus and need help (not answers) on solving two problems.

My job is to find the solutions of the equation that are in the interval [0, 2π) [couldn't find pi in the special characters.

sin 2t + sin t = 0 -- I know this can simplify to "2 sin t cos t + sin t = 0" but if I was trying to find solutions within 2π, where would I go from here? Am I allowed to factor sin t out?
cos u + cos 2u = 0 -- same problem. I know it simplifies to "cos u + cos^2 u = 0" or "cos u + 1 - 2 sin^2 u = 0", but same with the first problem, can I simplify cos u out?

How would I solve these problems? Thanks for your help!--Prowress (talk) 16:35, 20 February 2012 (UTC)[reply]

You can factor something, but not simply remove a common factor. So:

2 sin t cos t + sin t = 0 ⇒ sin t (2 cos t + 1) = 0 ⇒ sin t = 0 or 2 cos t + 1 = 0

The first equation of the last pair tells you t = nπ are solutions, n ∈ ℤ. The second equation of the pair gives futher solutions. Simply discarding a factor would hide half the solutions. — Quondum ^☏✎ 16:53, 20 February 2012 (UTC)[reply]

Double-check your first step in the second problem. --COVIZAPIBETEFOKY (talk) 17:13, 20 February 2012 (UTC)[reply]

So for #1, I got sin t = 0 and cos t = -1/2 but I cannot think of any configuration for both of them. --Prowress (talk) 23:08, 20 February 2012 (UTC)[reply]

Sorry, I meant by "configuration" that I cannot find any radians of pi that would fit both the answers.--Prowress (talk) 23:10, 20 February 2012 (UTC)[reply]

But you don't need a solution to both of them. You have two things which you're multiplying together, and they're supposed to make 0. So it's enough that one of them be 0.--130.195.2.100 (talk) 23:26, 20 February 2012 (UTC)[reply]

Meaning that for #1 you get sin t = 0 or cos t = -1/2 rather than sin t = 0 and cos t = -1/2 . Bo Jacoby (talk) 05:59, 21 February 2012 (UTC).[reply]

February 21

phi(n) question

On page 224 of Handbook of Number Theory II, by Sandor and Crstici, it says that C. A Nicol proved that there exist infinitely many numbers n such that phi(n) <= phi(n-k) for all 1 <= k <= n-1. (It references problem E2590 in AMM, vol 83, p 656, with the solution in vol 85, p. 654, but I don't have access to those.) But the statement doesn't make sense to me - when k=n-1, you have phi(n) <= phi(1) = 1, and only phi(1) and phi(2)=1. Is there an error? Bubba73 ^{You talkin' to me?} 03:46, 21 February 2012 (UTC)[reply]

I don't know whether this link is stable but it displays problem E2590 which asks to show that there are infinitely many numbers n such that phi(n) <= phi(k) + phi(n-k) for 1 <= k <= n-1. It sounds like Sandor and Crstici forgot phi(k). PrimeHunter (talk) 04:17, 21 February 2012 (UTC)[reply]

Resolved

Thanks, that makes sense because the next line says that there are infinitely many n such that phi(n) >= phi(k) + phi(n-k) for that range of k. Bubba73 ^{You talkin' to me?} 04:41, 21 February 2012 (UTC)[reply]

Statistics and entailment

On page 173 of Cognitive Strategy Research by McCormick, Miller and Pressley, there is an F-statistic table that evaluates causal inferences, but I've never come across this use of the F-statistic. Part of the table is given here:

Entailment	BETA	F
Importance -> learning	0.13	F(2, 503) = ...
Importance -> attention -> learning	0.12	F(3, 502) = ...

In the experiment (a study of attention-focusing strategies), students had to read a technical passage, then sit a brief test. Those parts of the text that were examined on the test were deemed important by the experimenters (the "importance" part of the table). Students were measured on their ability to focus their attention on these important parts of the text (the "attention" part of the table), as well as their actual performance (the "learning" part). So intuitively, I hope that is clear: the causal chain involving attention (line 2) describes the link between "importance of the text" and "learning of the text", when it is mediated by conscious attention on the part of the reader. The chain without attention (line 1) is blind to this intermediate step.

So I can get the idea but not how the table works. The F-statistic is being used for some kind of test where the numerator steals a degree of freedom from the denominator, whenever an extra link is added to the causal chain. Can anyone explain? IBE (talk) 19:43, 21 February 2012 (UTC)[reply]

F-test is the article you want. Thats got some typical examples of its use.--Salix (talk): 21:22, 21 February 2012 (UTC)[reply]

That doesn't mention anything about entailment, nor does it tell me what Beta might represent. I know how to use an F-test for ANOVA and regression, but I've never seen anything about entailment before. Is it meaningful statistically, or is it just based on assuming causation from correlation? IBE (talk) 15:39, 22 February 2012 (UTC)[reply]

Yes, I realised that its not of much help. I realised you probably knew about F-test as soon as I posted my comment. Wondering if its anything to do with Regression analysis. I've looked at a few papers on Causal inference [1] which seem to involve a lot of regression. The beta could come from the linear regression equation

Y=\beta X+\varepsilon

. But this is a shot in the dark.--Salix (talk): 17:04, 22 February 2012 (UTC)[reply]

I think it's probably better than a shot in the dark - you are probably correct. I just thought Beta might be specific to the application. IBE (talk) 18:51, 22 February 2012 (UTC)[reply]

Could be some form of mediation analysis, but hard to be sure without looking at the source (which I don't have). As explained in that article, in recent years considerable doubt has been cast on simple ways of testing for mediation. Qwfp (talk) 18:09, 22 February 2012 (UTC)[reply]

I think this might be it, but I'll go over it more carefully when I have a moment. More info welcome from anyone, but thanks to you both for the help. IBE (talk) 18:51, 22 February 2012 (UTC)[reply]

February 22

Radians and calculus

Is there an intuitive/easy way to see why measuring angles in radians leads to nicer formulas in calculus? 74.15.139.132 (talk) 01:30, 22 February 2012 (UTC)[reply]

If you plot a curve of sine as a function of degrees with equal scaling on the x & y axes, then it is a long, long, stretched-out graph, with the maximum slope going up the curve of only π/180 ≈ 0.0175, while if you plot it as a function of radians, then it is a more "natural" looking graph, with a maximum slope going up the curve of 1. In fact, the slope of sin(x) at each point is exactly cos(x). Thus the first derivative of the sine is the cosine, or in formula d/dx sin(x) = cos(x). If we were to define a function sin_d(x) which was the sine as a function of degrees, then the derivative would be d/dx sin_d(x) = π/180 * cos_d(x). That is messy enough, but in radians we know that the fourth derivative of the sine function is the sine function itself, where in degrees, the fourth derivative of sin_d(x) is (π/180)^4 * sin_d(x). What a mess!. Radians give these functions the right horizontal scale, somewhat in the same sense that the e, the natural log base, is just the right choice for the exponential function so that it's derivative is itself. -- 203.82.91.152 (talk) 04:25, 22 February 2012 (UTC)[reply]

I know that, I'm wondering if there's an intuitive way of seeing why this is true for radians, defined as 1/(2*pi) of a circle, w/o explicit computation. 74.15.139.132 (talk) 18:35, 22 February 2012 (UTC)[reply]

OK. Our trig functions are easily represented on a unit circle because that sets the hypotenuse to 1 which make the ratios which define the trig functions trivial. And an angle is not just the central wedge shape that you measure with a protractor, but it can also be thought of as the length of the arc of the unit circle that is cut out by the wedge, and that is the definition of radians. A 1 radian angle cuts out an arc of the unit circle with length 1. Why is this important to calculus? The coordinates of a point on the unit circle, θ up (CCW) from the x-axis, are (x, y) = (cos θ, sin θ), for whatever measure of angle you are using (assuming that your sine and cosine function are intended to take that measure of angle). When you choose radians you get the advantage that for the initial, very small movement up from θ = 0, you find ∆θ ≈ ∆y, so that lim ∆θ -> 0 of ∆y/∆θ = 1. Since y = sin θ, we just showed that d/dθ sin θ = 1 for θ = 0, and in general we find that d/dθ sin θ = cos θ without some arbitrary and annoying scaling factor, such as π/180. So the intuitive feature of radians you are seeking is linked to their property of representing distance around the unit circle. -- ToE 02:36, 23 February 2012 (UTC)[reply]

One nice thing is looking at the area element

rdrd\theta

when integrating in polar coordinates. This its quite easy to see working in radians as the length of an arc is the radius times the angle.--Salix (talk): 21:45, 22 February 2012 (UTC)[reply]

As alluded to above, if we only used degrees, the derivative of sin at 0 would be π/180. In radians, it's exactly 1 instead, which is nice. For this to happen, we need the limit as θ tends to 0 of (sin θ)/θ to be 1. (This is because if you look at the derivative of sin at 0 through its definition as a limit, this is precisely the limit you obtain.) So what it boils down to is we want to know why sin θ and θ are roughly equal when θ is very small and measured in radians. Now have a look at this illustration. sin θ is the length of the dotted vertical segment. θ is the length of the circular sector to the right of the vertical dotted segment (but only if θ is measured in radians). Now imagine that θ becomes very small, so that the point on the circle slides towards (1,0). You can see, at least intuitively, that these two lengths are going to be very close in relation to one another (that is, their ratio is going to become close to 1). All of this only works if the angle is measured in radians. 96.46.204.126 (talk) 22:11, 22 February 2012 (UTC)[reply]

In addition, the series expansion for sin and cos are in radians. In all of these cases, using radians (the natural unit of angles) eliminates constants (which are akin to unit conversions). And because of the series, sin(x) is approximately x when x is small, and x is in radians. And the inverse trig functions naturally give their result in radians. Bubba73 ^{You talkin' to me?} 03:03, 23 February 2012 (UTC)[reply]

Matrix Eigenvalues

I need to find the eigenvalues and eigenvectors of the matrix

\mathbf {A} ={\begin{bmatrix}\mathbf {O} &\mathbf {I} &\mathbf {O} &\mathbf {O} \\-\mathbf {B} &\mathbf {O} &\mathbf {S} &\mathbf {O} \\\mathbf {O} &\mathbf {O} &\mathbf {O} &\mathbf {I} \\{\mathbf {DB} }&\mathbf {O} &-{\mathbf {D(C-S)} }&\mathbf {O} \\\end{bmatrix}}

where O is the zero matrix, I is identity matrix, B is a circulant tridiagonal matrix with elements (-1,2,-1) and C and D are diagonal matrices with constant diagonal terms (in other words, a scalar times the identity matrix.) S is a circulant matrix (or a diagonal matrix, if that helps.) I am hoping that the presence of large number of O's , I's and simple matrices would lead to a closed form solution for the eigenvalues and the eigenvectors. Any help will be sincerely appreciated. deeptrivia (talk) 03:35, 22 February 2012 (UTC)[reply]

you can at least evaluate the determinant more easily: Swap rows and columns around in groups. You can easily make the two I's appear on the top left and lower right by swapping rows 1 and 2, and rows 3 and 4. Then you have

\pm \left|B\right|\times \left|D(C-S)\right|\pm \left|S\right|\left|DB\right|

. The signs are given by the parity of the size of the two I's. HTH, Robinh (talk) 20:38, 22 February 2012 (UTC)[reply]

New method for long division

Please excuse my math ignorance, but I have no way of asking this question without seeming like a complete dunce when it comes to math (true): My son is starting long division in grade school. I tried helping him with his homework, but I can't make heads or tails of what is going on here. When I was in school (back in the 80's) we did long division by dividing into the first set of numbers, subtracting out, dropping down the next set of numbers, and continuing the process until completed with a remainder. It still works, but it definitely not the way they are teaching it now. And I am stumped. I know you can't help "do my son's homework for him," so I won't include the actual problem, but is there someone out there who is privy to this "new way" of teaching long division to elementary aged kids that can point me to some helpful resources? The problems are basic long division, no variables or square roots or anything like that yet. Quinn ^{░ RAIN} 05:34, 22 February 2012 (UTC)[reply]

PS:It has something to do with doubling the divisor, if that helps. Quinn ^{░ RAIN} 05:39, 22 February 2012 (UTC)[reply]

I'm not familiar with a different way to do division, but the doubling you mention sounds like it might be a square-root algorithm. Could it be this: http://www.homeschoolmath.net/teaching/square-root-algorithm.php (scroll down a bit)?--121.74.109.179 (talk) 05:51, 22 February 2012 (UTC)[reply]

Yes it seems that Long Division Teaching Aid, "Double Division" is basically what is going on here. Can anyone "dumbdown" the method for me so I can grasp what they're wanting? We have to show-the-work and I'm an English major, so I "need" to understand the concept before I can wrap my brain around it to try to explain to my son. 06:32, 22 February 2012 (UTC)

Or maybe an example with explanation? I don't have to understand how A=B, I just need to know how A-gets-to-B (if that' makes sense) so I have a broad understanding of what I'm trying to do here. After that, I can trial and error the problems, and hopefully come up with a way to assist my son with his homework. Quinn ^{░ RAIN} 06:37, 22 February 2012 (UTC)[reply]

It's not really all that different from long division. They explicitly fill in the zeros to the right, where long division leave potentially confusing blanks. They also precompute multiples of the divisor, but only the 1x, 2x, 4x, and 8x multiples. Doing this involves the three doublings you mentioned. Doubling is cheap, but the price they pay is that where you would get a 7 in the answer via long division, they get a 4, a 2, and a 1 in three different steps, and have to sum them for a final answer. If this is used commonly enough, and we can find references in educational literature, it would certainly deserve mention in long division, if not its own article. -- ToE 06:41, 22 February 2012 (UTC)[reply]

I just noticed that the link I gave above has a nice applet on the right which will step you through example problems of your choice, with explanation for each step taken. Why don't you give that a try and ask here if something doesn't make sense? I was about to write up an example along the lines of those in Long division#Method, but that applet will probably do a better job of explaining things. Cheers! -- ToE 06:50, 22 February 2012 (UTC)[reply]

I don't think they should change methods like this unless the new method is much better than the old one. After all, people need to be able to check each other's work, which we can't do if we all use different systems. I've sometimes experienced this when getting change back, when instead of counting up the change, they count down from the amount tendered until they get to the amount of the bill. I find that very confusing, and insist on recounting it myself. So, this sure doesn't save any time. StuRat (talk) 07:11, 22 February 2012 (UTC) [reply]

The method sounds related to Ancient Egyptian multiplication or Egyptian multiplication and division. Bubba73 ^{You talkin' to me?} 17:35, 22 February 2012 (UTC)[reply]

Well I've got the hang of it now (the applet, which I didn't notice at first, was the key, so thanks for that), but I have to say in the long run it is not a time saver vs. the "old" version, but oh well. Concur with StuRat above. Thanks all! Quinn ^{░ RAIN} 01:16, 23 February 2012 (UTC)[reply]

Great, I will mark this resolved. StuRat (talk) 04:41, 24 February 2012 (UTC)[reply]

Resolved

Quinn, do you know if your son's school is teaching this method instead of long division, or as a precursor to long division? -- ToE 15:20, 24 February 2012 (UTC)[reply]

Quadratic Eigenvalue Problem with Circulant Matrices

I have solve the following quadratic eigenvalue problem:

det(s^{2}\mathbf {I} +s\mathbf {B} +\mathbf {C} )=0

where B and C are circulant matrices. Since the eigenvalues of circulant matrices are known in closed form, I'm hoping there would be a closed form solution in terms of eigenvalues of B and C. deeptrivia (talk) 11:17, 22 February 2012 (UTC)[reply]

This is just a stab in the dark, so check it thoroughly. Since all three of the summands are circulant, isn't the matrix inside your determinant circulant itself? After that, it looks like its eigenvalues are the sum of the eigenvalues of the three matrices. That sounds to me like a connection you're looking for. I'm doing this hastily, so apologies if I made a mistake and I mislead you. Rschwieb (talk) 18:59, 23 February 2012 (UTC)[reply]

I removed "Matrix Eigenvalues" as the title, since that is also a title of a previous section, and section names must be unique, both for proper human indexing and wiki indexing. StuRat (talk) 19:16, 24 February 2012 (UTC) [reply]

Finding quadratic formula

long ago I found a wikipedia article about how to "get" the quadratic formula from $ax^{2}+bx+c=0$ ?? But I cannot find it now? someone may help? Thanks! 190.158.184.192 (talk) 17:53, 22 February 2012 (UTC) (PS: The process of getting the formula itself.)[reply]

Try Quadratic equation#Derivations of the quadratic formula. Qwfp (talk) 18:12, 22 February 2012 (UTC)[reply]

Curl

In maxwell's equations the curl of electric and magnetic fields are used, but curl only operates on functions of 3 variables and E and B generally are 4 variables because they depend on time as well. Can someone explain? Money is tight (talk) 18:20, 22 February 2012 (UTC)[reply]

The curl acts only on the spatial variables. Sławomir Biały (talk) 22:59, 22 February 2012 (UTC)[reply]

The E and B fields are generally functions of four variables (coordinates x, y, z and t), but the curl only involves partial derivatives with respect to three of these variables. Each partial derivative treats the other three variables as constant. The partial derivative with respect to the fourth variable t generally occurs in the same equations alongside the curl (or div). This 3+1 separation is an artifact of the choice of opertors. In a treatment such as in geometric algebra that puts all four the these coordinates on an equal footing replaces the curl, div and time derivative with one that combines all four partial derivatives into one operator, and in the process simplifies the statement of Maxwell's equations. — Quondum ^☏✎ 04:52, 23 February 2012 (UTC)[reply]

February 23

Math Problem

5≤ -s/12 -3 — Preceding unsigned comment added by 68.80.157.180 (talk) 02:17, 23 February 2012 (UTC)[reply]

Did you have a question? RudolfRed (talk) 02:48, 23 February 2012 (UTC)[reply]

I added a title. I think we can assume you want us to help solve the inequality for s. It works pretty much like solving an equality, with one big exception: "When you multiply or divide both sides of an inequality by a negative number, you also flip the sign of the inequality". Thus, you will flip ≤ to ≥. (There are other differences in solving inequalities, such as dealing with square roots, but those don't come into play here.) If you show us your work, we would be glad to check it for you. StuRat (talk) 03:09, 23 February 2012 (UTC)[reply]

Try www.wolframalpha.com : [2] Bo Jacoby (talk) 08:36, 23 February 2012 (UTC).[reply]

Trig function help

Hello, how would I accomplish this trig problem?

cos (arcsin (5/13) - arctan (3/4))

thanks.--Prowress (talk) 03:43, 23 February 2012 (UTC)[reply]

Hint: determine the missing lengths of the triangles involved, then use the identity $\cos {(A-B)}=\cos {A}\cos {B}+\sin {A}\sin {B}$ . --Kinu ^t/_c 04:19, 23 February 2012 (UTC)[reply]

Difference between a series and a progression

Could you please explain the difference between a Series and a progression with examples? Kasiraoj (talk) 08:09, 23 February 2012 (UTC)[reply]

You could start by working out the difference between a sequence and a series, and then seeing how the listed examples of a progression fit into the picture. — Quondum ^☏✎ 09:04, 23 February 2012 (UTC)[reply]

Uniqueness of injective/flat resolutions

The article resolution (algebra) article mentions projective resolutions are unique up to chain homotopy. I would guess the same can be said for injective and flat resolutions, but I'm not familiar enough with the subject matter to verify. Is this the case? Thanks! Rschwieb (talk) 18:27, 23 February 2012 (UTC)[reply]

Heart-shaped function

I found it quite interesting to find out that the function:

$(\cos(100x){\sqrt {\cos(x)}}+{\sqrt {|x|}}-0.7)(4-x^{2})^{0.01}$

gives a good approximation of a heart shape when viewed graphically (e.g. here).

How would this fairly complicated equation have been derived? Trial and error, or chance? Is there software for generating a function that will deliver a certain shape? --Iae (talk) 20:01, 23 February 2012 (UTC)[reply]

I don't know, but a Limaçon can be heart-shaped. Bubba73 ^{You talkin' to me?} 20:20, 23 February 2012 (UTC)[reply]

You might also be interested in [3]. There was probably some trial and error involved in finding the function but there is some cleverness involved as well. For example the cos 100x produces the up and down squiggles that fill in the shape. The sqrt(cos x) gives the rounded bits on either side and the sqrt(|x|) produces the V shape in the middle. You might try playing with the formula to see what effects small changes have, for example to find out what happens if you change the .7 to a .8 or a .6.--RDBury (talk) 22:45, 23 February 2012 (UTC)[reply]

MATH GESSUING

I am a two digit number over 50.When you put me in groups of 7,2 are left over.The sum of my digits are 11. What number am I ? — Preceding unsigned comment added by 24.163.1.170 (talk) 22:01, 23 February 2012 (UTC)[reply]

You are not a number! You are a free man. --Trovatore (talk) 22:17, 23 February 2012 (UTC)[reply]

The answer is 65. The title of this question is misleading, though. --COVIZAPIBETEFOKY (talk) 22:50, 23 February 2012 (UTC)[reply]

I'm still trying to figure out why the "over 50" caveat is given. Number is of the form 7n+2 from the one criteria and a subset of those of the form 9n+2 from the other so the answers are a subset of the numbers of the form 63n+2, for n=0 the answer is 2 (which doesn't fit the second criteria exactly, n=1 gives 65 and n=2 gives 128 which is too big. Without the requirement that it be a two digit number, the numbers start with (marked * sum to another 9n+2 other than 11) are 2*,65, 128, 191, 254, 317, 380, 443, 506, 569*, 632, 695*, 758*, 821, 884*, 947*, 1010*, 1073, etc. I don't know if there is a clean way to find out what the last number of the form 63n+2 whose digits sum to 11, but we know it can't be more than 11111111111.Naraht (talk) 03:52, 24 February 2012 (UTC)[reply]

What about 10000001111111111, 10000000000001111111111, 10000000000000000001111111111, ... ?--RDBury (talk) 12:01, 24 February 2012 (UTC)[reply]

86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

Area enclosed

Hello. Suppose I have an arbitrary function defined by a parametric which on some interval of the parameter is closed and simple. How would I find the area enclosed by the curve? (without doing something ugly like splitting it into two regular functions and integrating each, then subtracting of course... *shudder*...) Thanks. 24.92.85.35 (talk) 23:50, 23 February 2012 (UTC)[reply]

See Green's theorem#Area_Calculation – b_jonas 10:08, 24 February 2012 (UTC) (Update: deleted the formula. It was wrong. – b_jonas 10:09, 24 February 2012 (UTC))[reply]

This means that if you have the point of the curve parametrized as (x(t), y(t)) where x runs from a to b, and this is a closed curve so x(a) = x(b) and y(a) = y(b), then you can get the area as the absolute value of

\int _{a}^{b}y{\frac {\partial x(t)}{\partial t}}\,dt

– b_jonas 10:13, 24 February 2012 (UTC)[reply]

t runs from a to b.86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

February 24

Solids of revolution by cylindrical shells

Resolved

– I got

20π/3

which matches the computer answer, and it's due in an hour. Feel free to check my work. --Ks1stm (talk) ^{[alternative account of Ks0stm]} 17:24, 24 February 2012 (UTC)[reply]

My calculus textbook says "the volume of the solid obtained by rotating about the y-axis the region under the curve $y = f(x)$ from $a$ to $b$ is $V = \int b a 2 π x f(x) dx$ where $0 \leq a < b$ ." That's fine and all except one of my homework problems is to find the volume of the solid obtained by rotating the region between $y = x 2$ and $y = 2 - x 2$ about $x = 1$ , and this region would produce an $a$ less than 0. Will it still get me the correct answer if I attempt to find the volume by doing $V = \int 1 -1 2 π (1- x) (2-2 x 2) dx$ or will it be thrown off by $a$ being less than 0? Ks0stm ^{(T•C•G•E)} 00:55, 24 February 2012 (UTC)[reply]

Just impose a new coordinate system such that x=1 in the old one is x'=0 (y axis) in the new one and you won't have that problem. 24.92.85.35 (talk) 03:13, 24 February 2012 (UTC)[reply]

If I'm not mistaken, the definition is worded in a manner to ensure that the radii are all positive (and only included in the integration once, of course). In this case, because r=(1-x) and your limits are x=-1..1, these conditions still hold. --Kinu ^t/_c 03:26, 24 February 2012 (UTC)[reply]

Length of the curve of any polynomial

How is the length of any polynomial between two points calculated? --Melab±1 ☎ 04:24, 24 February 2012 (UTC)[reply]

See Arc_length#Finding_arc_lengths_by_integrating. StuRat (talk) 04:38, 24 February 2012 (UTC)[reply]

Would all polynomials substituted into the equation be integrable though? --Melab±1 ☎ 17:50, 24 February 2012 (UTC)[reply]

No, this problem is one of the motivations behind Elliptic integrals.--Salix (talk): 18:23, 24 February 2012 (UTC)[reply]

Clarification: all the integrands at arc length are integrable functions, in the sense that the definite integrals exist as finite real numbers (they also satisfy the conditions given by Riemann_integral#Integrability whenever f is a poynomial). However, there is no guarantee that the indefinite integral has an analytic expression, which leads us to Salix's comment above. (I only bring this up because, as I understand it, Melab is using 'integrable' to mean 'has an analytic antiderivative', which is not standard usage. (also feel free to chastise/correct me if I've missed something here;)) SemanticMantis (talk) 21:29, 24 February 2012 (UTC)[reply]

Question: limits of partially defined functions on the reals

I was alerted to a quirk about limits which I do not recall hearing before. Suppose f(x)=1 on the irrationals, and is undefined on the rationals. Topologically, when considered as a function from the irrationals into the reals, it has limit 1 as x approaches 0. It is appealing then to say that this function has a limit as a partially defined function on R. However the current version of the wiki article on function limits requires that f be defined at least on a (connected) open interval with endpoint 0. This stricter requirement would exclude the function described above from having a limit, due to the dense set of removable discontinuities.

Question(s): Is there anything foundationally wrong with taking the wiki definition and "excusing" countably many removable discontinuities? (I.e. we would say "if x is less than delta from c AND f(x) is defined, then f(x) is less than epsilon from L") Does anybody know if any authors address this?

I'm not interested in altering the wiki article, this is just curiosity. Rschwieb (talk) 16:35, 24 February 2012 (UTC)[reply]

You could say that the property holds almost everywhere, if nothing else. Looie496 (talk) 18:18, 24 February 2012 (UTC)[reply]

That doesn't work. Consider the function that is 1 on the irrationals and 0 on the rationals instead of undefined. The limit of this function at 0 doesn't exist. The function you describe is actually continuous (the pre-image of open sets is open), but its domain is the irrational numbers and so we're dealing with the topology of the irrationals. There's no sense in which a function can be undefined at a point in its domain. Rckrone (talk) 18:29, 24 February 2012 (UTC)[reply]

Limit of a function#Functions on metric spaces appears to comfortably accommodate this (including sparse/countable domains); it involves only defining the set (domain) over which the limit is taken (as opposed to allowing a domain on which a function is undefined(!)), and that there must exist elements of the domain within any nonzero distance from the limiting point not counting the limiting point itself. — Quondum ^☏✎ 18:35, 24 February 2012 (UTC)[reply]

(edit conflict) What you're talking about makes perfect sense. It's not usually discussed in elementary calculus texts because it's not necessary for talking about derivatives.

If f is a real-valued function defined on a subset E of R, then it makes sense to speak of the limit of f at a number a whenever a is a limit point of E. This means that every open interval containing a, no matter how small, contains a point of E other than a itself. (In fact, limits will still make sense if you drop the "other than a itself" part, in which case the notion is that of an adherent point.) It is not necessary that the set of points where f is undefined be countable, though this is a sufficient condition to ensure that every point is a limit point.

These kinds of limits will be addressed, at least implicitly, in any mathematical analysis textbook that discusses topology (namely, metric spaces or topological spaces). But this subsumes ordinary limits within a much larger theory and requires a degree of abstraction that goes far beyond what is needed for our purposes. The mathematical analysis textbook by Zorich discusses the more general concept of a limit we've been talking about well before it reaches topology. I can't give you a page number because I don't have the English version.96.46.204.126 (talk) 18:41, 24 February 2012 (UTC)[reply]

Haha, well if it's not in English then don't worry about it. I'm glad to hear my intuition was OK. Historical context: this all arose when I discovered a user "improved" l'Hopital's Rule by showing the g'(x) nonzero hypothesis was superfluous. All the analysis texts on my shelf used that hypothesis, so I was doubtful this person had outsmarted 20 generations of mathematicians. Just recently I found one reference which uses the same logic as that user. (First line of proof: "since lim f'/g' exists, g' does not vanish on a connected interval with endpoint a"). Is this not just a misconception? Maybe it's not so much of a misconception as an overstrong definition of limit. If one demands that in the definition of a limit that the function has to be defined on an interval adjacent to the limit point, then yes, the g'(x) nonzero condition is superfluous. Rschwieb (talk) 19:20, 24 February 2012 (UTC) Update: Checked 10 analysis books on a shelf. One of the authors (Kenneth Rogers) uses the proof I find questionable. S.G. Krantz assumes the two functions to be differentiable on an entire neighborhood including c, which I think probably does imply the g'(x) nonzero hypothesis. The other eight (including Rudin and Stewart) use the g'(x) nonzero hypothesis. Rschwieb (talk) 19:53, 24 February 2012 (UTC)[reply]

No, sorry, I meant that there is an English translation of it, but I don't have the translation, so I can't give you a page number in the English version. In Theorem 5.5.1 of the book you linked to, most mathematicians would include the hypothesis g′(x) ≠ 0 over some interval rather than relying on the technicalities of their particular definitions. It's likely that these technicalities are made clear wherever he defines limits. By the way, I can see no reason that the set of points at which g'(x) = 0 should be countable. 96.46.204.126 (talk) 19:59, 24 February 2012 (UTC)[reply]

I think that if you allow g′(x) to be zero for some x and choose to say nothing about what f′(x) does at those points, then L'Hopital's rule becomes false. For example, let f′ and g′ both be continuous, and let g′ be identically zero on infinitely many intervals approaching a, but nonzero elsewhere. Make f′ identical to g′ except on the intervals where g′ is zero, where you let f′ go crazy (while still tending to zero at a). You'll be able to pick f′ in such a way that L'Hopital's rule fails. (f and g will be the corresponding antiderivatives taking the value 0 at a. For example, f(x) will be the integral of f′ from a to x.) 96.46.204.126 (talk) 20:28, 24 February 2012 (UTC) [Correction: This may only work if you let g′ spend enough time being 0. 96.46.204.126 (talk) 20:37, 24 February 2012 (UTC)][reply]

Products

A product can be expressed in terms of factorials;

\prod _{k=0}^{n-1}(N-k)={\frac {N!}{(N-n)!}}

Can a product of every other number be similarly expressed?

\prod _{k=0}^{n-1}(N-2k)

Thanks, Spinning Spark 19:00, 24 February 2012 (UTC)[reply]

If N is even, then we have something like

\prod _{k=0}^{n-1}(N-2k)={\frac {N!2^{N/2-n}(N/2-n)!}{(N-2n)!2^{N/2}(N/2)!}}.

A similar expression is available if N is odd. Sławomir Biały (talk) 19:33, 24 February 2012 (UTC)[reply]

Beautiful! <3 --COVIZAPIBETEFOKY (talk) 20:09, 24 February 2012 (UTC)[reply]

Thanks for that. That's what I was afraid of, it has to be divided into odd and even expressions. Spinning Spark 21:22, 24 February 2012 (UTC)[reply]

February 25

Video frame-frame homography & camera motion

I want to compute world space camera rotation & translation between two frames of video, but I'd like to check that I have a sensible handle on things. I am using a Kinect, so I have depth data for each pixel - I intend to use the colour data (w/RANSAC) to inform an initial estimate of the transform for ICP to align the full (X,Y,Z,Colour) point clouds. Hopefully none of this is relevant, but just in case.

I have a set of feature point matches between the two frames, from SURF. I'm then going to compute the homography as described at the bottom of this page, where (x_i, y_i) are points in the first frame and (X_i, Y_i) the matched points in the second frame. This gives me the homography.

Questions:

1. Is this understanding sensible?

2. How can I extract camera rotation and translation from the homography?

Thanks, 131.111.255.9 (talk) 02:34, 25 February 2012 (UTC)[reply]