Wikipedia:Reference desk/Archives/Mathematics/2008 March 6

= March 6 =

Cake Cutting
A circle has N points distributed around its circumference, with a line segment connecting each pair of points. No three segments are concurrent in the interior of the circle. Consider each segment a slice. How many pieces of cake does this produce, as a function of N? Black Carrot (talk) 01:35, 6 March 2008 (UTC)
 * A quess $$(n-1)^n+n.$$ I will explain my reasoning if this happens to be correct. (Igny (talk) 03:05, 6 March 2008 (UTC))
 * I drew it out. Doubt this is right but, $$f(N)=N\,\!$$.  Nah, couldn't be.  —Preceding unsigned comment added by 12.213.80.54 (talk) 03:11, 6 March 2008 (UTC)
 * If I understand your idea correctly, for N odd the number of such segments is the number of diagonals of an n-sided polygon plus n, or $$\frac{1}{2} (N-1) N$$. For N even, at least $$\frac{N}{2}$$ diagonals must be discarded (since they meet at the center of the circle (that is, of course, if the N points are evenly distributed around the circumference). So my guess for this case is $$\frac{1}{2} (N-2) N$$. Pallida  Mors  03:56, 6 March 2008 (UTC)
 * Alex has a big head for life regions in the traditional sense (and under all your caveats), you can find the formula there here. Pallida  Mors  04:30, 6 March 2008 (UTC)


 * Well, you start with 1. Every line you cut adds a piece, so you get 1 + NC2. But then you need to consider that choosing any 4 points gives you a crossover, which adds an extra slice since your line cuts through 2 pieces, hence $$1 + ^NC_2 + ^NC_4$$. Gotta love combinatorics. -mattbuck (Talk) 09:47, 6 March 2008 (UTC)
 * This happens to be one of my favorite philosophical mathematics example ever. With the sequence beginning 1, 2, 4, 8, 16, ..., it is easy to be fooled into thinking the answer is $$2^{n-1}$$.
 * Also see OEIS A006533 for the (much harder) case of the points being symmetrical. -- Meni Rosenfeld (talk) 20:26, 6 March 2008 (UTC)


 * See Dividing a circle into areas. --Lemontea (talk) 03:32, 13 March 2008 (UTC)

Thanks for the help. Black Carrot (talk) 01:40, 7 March 2008 (UTC)

Eigenvectors
Hello. Let A be an nxn square matrix such that A^2 is the identity. Show that any nx1 vector can be expressed as a sum of at most two eigenvectors of A. Can someone give me any hints please.--Shahab (talk) 07:16, 6 March 2008 (UTC)
 * First show that A is diagonalisable (so that any vector is a sum of eigenvectors of A). The think about how many different eigenvalues A can have. 131.111.28.33 (talk) 10:48, 6 March 2008 (UTC)
 * Pardon me. I don't know how to show A to be diagonalizable. I believe A has 2 eigenvalues though, 1 and -1.--Shahab (talk) 04:27, 7 March 2008 (UTC)
 * OK. I got it--Shahab (talk) 04:51, 7 March 2008 (UTC)

Number of faces of the projection of a convex polyhedron
Let $$\pi$$ be the intersection of $$n$$ half-spaces in a $$\mathbb{R}^d$$. We project $$\pi$$ along $$p$$ dimensions, and obtain another convex polyhedron, this time the intersection of $$n'$$ half-spaces (we choose $$n'$$ to be minimal). I'm looking for bounds on $$n'$$.

Fourier–Motzkin elimination provides an upper bound: $$n' \leq n^{2^p}/2^p$$. Yet, I've never seen any proof that this bound is tight for $$p > 1$$.

Any idea? David.Monniaux (talk) 09:26, 6 March 2008 (UTC)


 * Amenta and Ziegler show that the projection may have as many as &Omega;(nfloor(d/2)) vertices. By McMullen's upper bound theorem (why don't we have an article on this??) on the number of vertices a polytope in the lower dimensional space may have in terms of its number of facets, it can therefore also have &Omega;(nfloor(d/2)/floor((d-p)/2)) facets, although maybe a more careful examination of their construction would give a stronger lower bound. In the other direction, your upper bound seems very pessimistic when p is larger than log d: each feature of the projection corresponds to a feature of the original polytope, so (again by the upper bound theorem), we can get an O((nfloor(d/2))) upper bound on the number of facets of the projected polytope, independent of p. And for small p, each facet of the projection comes from the projection of a face of codimension p+1 in the original polytope, so there should be an O(np+1) bound without the powers of two. —David Eppstein (talk) 23:39, 6 March 2008 (UTC)

This bit of Quaternion source code -- are my assumptions correct?
Here is the source code in question that I came across:

Now, the [x,y,z] you pass to this function, makes me think it's a point in space. But Quaternions are merely expressions of rotation, yes?

Is it accurate for me to assume that this code "aims" a Quaternion in a certain direction? ie, what rotation to apply to a hypothetical, imaginary "arrow" with its base point at (0,0,0) (not sure its "default" pointing would be) to make it point at this (x,y,z) point in space? Or does this do something completely different?

edit: be nice if there was a button in the formatting bar to let me paste this code in a way that doesn't get scrambled up by the Wiki engine... even the code tag doesn't quite cut it

--75.165.41.216 (talk) 17:01, 6 March 2008 (UTC)


 * I'm not sure about the quaternions, but I've fixed your formatting - use the  tag. (I guessed your code was php, if it's not, it's close enough that for formatting works!) --Tango (talk) 17:19, 6 March 2008 (UTC)
 * The code is written in C++. I changed the formatting. Morana (talk) 19:37, 6 March 2008 (UTC)


 * Rotations are (or can be) represented specifically by unit quaternions, and q and –q represent the same rotation, so you can specify any rotation with a quaternion of the form $$-\sqrt{1 - |\mathbf{v}|^2} + v_xi + v_yj + v_zk$$, which is what the function above returns when given $$\mathbf{v} = (v_x, v_y, v_z)$$. This should correspond to a rotation by an angle of $$2 \sin^{-1} |\mathbf{v}|$$ around the v axis. It could be clockwise or counterclockwise depending on the sign convention. There's an article on quaternions and spatial rotation which may be helpful. -- BenRG (talk) 20:41, 6 March 2008 (UTC)

generalization of Taylor's theorem for multivariate functions
Hello,

We have for multivariate functions:

$$ f(x) = f(a)+\nabla f(a) \cdot (x-a) + \frac{1}{2}(x-a)^T \mathbb{H}(a) (x-a)+ o((x-a)^{2}) $$

What comes afterwards? That is, how come we express $$o((x-a)^{2})$$ more precisely? It seems we first start with a scalar $$f(a)$$, then a vector $$\nabla f(a)$$, then a matrix $$\mathbb{H}(a)$$, then a third rank tensor?

Also, does what comes afterwards have a special name (like gradient, Hessian, etc.)?

Thanks! —Preceding unsigned comment added by Randomblue (talk • contribs) 21:11, 6 March 2008 (UTC)


 * Yes, you'll have a rank 3 tensor of all the 3rd derivatives. I've never actually used it (2nd order is usually enough), so I don't know if it has a name. --Tango (talk) 22:07, 6 March 2008 (UTC)

Calculating the birth date
Well i've a little arithmetic problem. If somebody is on the 6th June 1944 exactly 9 years & 263 days old, when the person was born? Thanks in advance for your help ... —Preceding unsigned comment added by 84.190.225.150 (talk) 21:19, 6 March 2008 (UTC)


 * I thought there was a question very similar to this recently, but it must have rolled off to the archives. Anyway this calculator looks like it should help.  -- LarryMac  | Talk  21:41, 6 March 2008 (UTC)


 * I found also this calculator but how I should type in the months? I've only the days? —Preceding unsigned comment added by 84.190.225.150 (talk) 22:23, 6 March 2008 (UTC)
 * Can you not just say 0 months and 263 days? --Tango (talk) 23:06, 6 March 2008 (UTC)


 * Oh, sorry, it works, but the result looks wrong: 17. September 1937? From 1937 to 1944 = seven years, not nine ... —Preceding unsigned comment added by 84.190.225.150 (talk) 23:12, 6 March 2008 (UTC)
 * When I try it using that site I get September 17, 1934 - are you sure you typed the numbers in right? --Tango (talk) 23:15, 6 March 2008 (UTC)


 * Yeah, you're right, man. Sorry, I'm writing from europe and there it's now twenty past midnight. I should sleep a little bit. Thanks a lot! —Preceding unsigned comment added by 84.190.225.150 (talk) 23:19, 6 March 2008 (UTC)

real valued function continuous only at rational points
does there exist a funtion mapping the real line to itself such that it is continuous at rational points in the domain but discontinuous at irrational ones, and if not why not? —Preceding unsigned comment added by 81.96.160.6 (talk) 21:58, 6 March 2008 (UTC)


 * I would think not, since the rationals are dense in the reals, so any neighbourhood of a rational number will contain irrational numbers, so if a given neighbourhood of a rational in the domain maps to a sufficiently small neighbourhood in the range (such a nbhd must exist, by definition of continuity), that nbhd is also a nbhd of an irrational number, so the function is also continuous at that irrational number. Does that make sense? I can write it out in symbols if you like, it might be a little clearer... --Tango (talk) 22:05, 6 March 2008 (UTC)
 * Actually, that's not denseness. It's true, but not because they're dense - I guess it's because they're measure 0. --Tango (talk) 22:10, 6 March 2008 (UTC)
 * It's easy to construct a function whose set of continuous points has zero measure: e.g. a function discontinuous everywhere (the characteristic function of Q), or everywhere except one point, or any variation of this idea. Bikasuishin (talk) 22:27, 6 March 2008 (UTC)
 * Discontinuous everywhere is a trivial case and can be ignored. Can you give an example of a function continuous at only one point? --Tango (talk) 22:43, 6 March 2008 (UTC)
 * Sure. Take f(x) = x u(x), where u is the characteristic function of the rationals. f is continuous at 0 and nowhere else. Bikasuishin (talk) 22:49, 6 March 2008 (UTC)
 * Yep, that works. My proof is nonsense, isn't it? They'll always be an irrational number in any nbhd of a given rational number, but not always the same one, so the next part of my proof falls apart. Oh well... --Tango (talk) 22:59, 6 March 2008 (UTC)
 * Ok, how's this one: The rationals are dense in the reals, so any given irrational number is contained in an arbitrarily small nbhd of a rational number, which must map to an arbitrarily small nbhd by continuity on the rationals, thus it's continuous everywhere. --Tango (talk) 23:06, 6 March 2008 (UTC)
 * If the proof continues to hold if you exchange the words "rational" and "irrational", it has to be flawed. In this case, if you try to write it down, you'll find that there still is a problem with the way you chose a rational in the neighborhood of your given irrational. Bikasuishin (talk) 00:31, 7 March 2008 (UTC)


 * The set of continuous points of a function on the real line is a G-delta set, which the rationals clearly aren't. Note that it is possible to construct a function on the real line that is continuous exactly at all irrationals, on the other hand (see Riemann function). Bikasuishin (talk) 22:07, 6 March 2008 (UTC)

So, you find yourself on a deserted island...
One thing that's always bugged me about Lost is that they never even make an attempt to find out where they are. I mean, there was a science teacher on the island (remember Arzt?), he should know how to do it! (now it appears that the island exists in some weird time bubble, but they didn't know that from the start)

This got me thinking: how would you figure out where you are? I mean, assuming that you'd have to go old-school, no gps or nothing. I've been thinking about this, and I would like to check with you guys if I'm correct, and possibly get your ideas for how to do it (gonna be a little long, I guess. Sorry 'bout that). I haven't done maths for a while, so don't be too mean to me if I say something really stupid :)

I figure that longitude is reasonably easy (which is ironic, since historically that was way difficult), as long as someone has a functioning (and accurate) wristwatch (I think that's a concession to modernity I'm willing to allow). Just look at the clock, correct for what timezone it was in, and when the sun is exactly to the north, compare the times. I figure that I'd put a long stick in the sand before sun-up and then plot the course of the shadow, that should give you a pretty exact reading for north. Assuming that you'd get a reading accurate to around a minute (that's pretty reasonable, given a long stick, right), you would be at most 4.4 km wrong (the equatorial radius is 6378.1 km, divide that by 24 hours and then 60 minutes), which is pretty darn good for a couple of sticks and a wristwatch, IMO.

If you didn't have a wristwatch, is there any other conceivable way you could measure longitude?

Latitude might be a little bit tricker. If you're in the northern hemisphere, the classic way would be to measure the angle between the horizon and the North Star. That would be your latitude, right? I mean, since the star is a direct extension of Earths axis of rotation, and since it lies on optical infinity, it seems to me that you wouldn't need to do anything else. I mean, at the north pole it would be exactly at 90 degrees (above you) and at the equator it would lie at exactly 0 degrees (on the horizon). Am I right about that?

To measure the angle, I would take a longer stick and a shorter stick and put them vertically in the sand so the tops of the sticks line up with the north star, and then I would put a longer stick horizontally, so as to form the base of a right-angled triangle (with the line joining the tops of the two vertical sticks as a hypothenuse). Then, using a lot of sand to scratch out a few terms of the Taylor series for sin and cos to figure out Tan(long vertical stick/horizontal stick). As for keeping the base exactly horizontal, I figure you could fashion a bubble level from a coke-bottle, or something.

However, if, like the losties, you're trapped on the southern hemisphere (at least I assume they are), that wouldn't work. Is there a southern-hemisphere equivalent of the North Star? That you could see with your naked eye, or maybe some binoculars (assuming there was a pair on the plane)?

I figure that you would be able to do it if you know what date it was (which presumably you do) and you measured the angle to the sun, but it would be a lot more complicated. You'd need to know at what latitude the sun is at zenith, and then compensate for that. What's the function for the latitude of the sun's zenith? I guess that if you had enough time, you could figure that out, even if you didn't know it (or you'd just wait until an equinox comes around, and then it'd be much easier). How accurate would these two methods be? How accurate an angle could you reasonably measure, using just some sticks?

So there. That's how I would do it. Sorry for being long-winded. These methods would work, right? Are there any better ways? Any details on how exactly the sun moves across the earth? Any comments would be appriciated. --Oskar 22:21, 6 March 2008 (UTC)


 * You've got that all pretty much correct. As far as I know, there is no way to calculate longitude without some way of determining the time at a known location (a watch you set before you left or a radio tuned in to a radio station with a known location are the two most likely methods, I would think). Determining latitude from the North Star is easy, determining it from the Sun is possible, but I wouldn't like to try (not away from an equinox, anyway). Determining it in the southern hemisphere is possible in the same way, however it wouldn't be as precise since there is no bright star particularly close to the south celestial pole - if you know your astronomy, you can find it pretty precisely from surrounding starts, though. You can make a pretty decent protractor by folding a piece of paper lots of times (start with a 90deg angle, fold it in half to get 45, half again to get 22.5, and so on, the creases will mark angles - I'm not sure how precise you could get it, but +/- 5 deg should be easy enough, which gives you a basic idea of where you are). If you had a calculator, or some trig tables you could do it more accurately. Using Taylor expansions would work if you've got nothing better to do. If you really want to use the Sun, you could wait until summer/winter solstice, time how long the day is and calculate your latitude from that - might be sooner than an equinox, although it's a more difficult calculation. --Tango (talk) 22:55, 6 March 2008 (UTC)


 * I guess you could figure out the approximate location of the southern celestial pole, even if you didn't know it. I mean, you've got nothing but time, and if you observe the sky enough, you might be able to figure out the point around which it rotates, and the points relation to a few neighboring stars. I mean, it's probably easier than doing it using the sun. --Oskar 23:31, 6 March 2008 (UTC)


 * I have no idea about actually working out, but if it's on the Lost island then they have no idea what time zone they're actually in and almost certainly don't have a watch set to that time, as they crashed mid-flight. Having said that, they could work out the difference roughly between the times of sunrise and sunset and the time zone their watches are set to. Plus, as you said, they seem to be in a weird time-bubble thing 81.96.160.6 (talk) 01:15, 7 March 2008 (UTC)
 * With a suitable reference work with details of orbits of various celestial bodies you might be able to work out longitude by observing rising and setting times as the body passes across the sky. This raises a question if you were stranded on a desert island with a copy of wikipedia, would you have enough information to find your longitude? --Salix alba (talk) 01:46, 7 March 2008 (UTC)
 * You don't have to know what time it is on the island.. you have to know what time it is in New York and where the sun is over the island to FIND the time on / position of the island D\=&lt; (talk) 02:19, 7 March 2008 (UTC)
 * Exactly. If your watch is set to GMT and you observe that your watch says 4pm at noon, you know you are in the GMT-4 time zone, which corresponds to a longitude of 60deg west. --Tango (talk) 11:20, 7 March 2008 (UTC)