Wikipedia:Reference desk/Archives/Mathematics/2012 February 6

= February 6 =

Max Quad Area
How do I calculate the maximum area of a real quadrilateral given sides a, b, c, and d. The diagonals p and/or q are the variables. While we're at it, how about the minimum area as well. Thanks, hydnjo (talk) 00:05, 6 February 2012 (UTC)


 * The area is given in quadrilateral. Try the formula with the two opposite angles when they add up to two right angles - that gives a cyclic quadrilateral.. Dmcq (talk) 00:27, 6 February 2012 (UTC)


 * Umm... How do I calculate the maximum area of a convex quadrilateral given sides a, b, c, and d. The diagonals p and/or q are the variables. This isn't homework. hydnjo (talk) 03:33, 6 February 2012 (UTC)


 * Use Bretschneider's formula. Looie496 (talk) 04:01, 6 February 2012 (UTC)


 * @Looie496: Can't use as the interior angles are variable along with the diagonals. hydnjo (talk) 19:03, 6 February 2012 (UTC)


 * And as for max and min when you have sides A, B, C, and D and can arrange them in any order, there aren't many ways to arrange the sides, so try them all. I get three ways to arrange them, with the rest being mirror reflections and or rotations of one of those three (and thus with the same area):

ABCD ABDC ACBD ACDB mirror and rotation of        ABDC ADBC mirror and rotation of               ACBD ADCB mirror and rotation of ABCD BACD mirror and rotation of        ABDC BADC mirror and rotation of ABCD BCAD mirror and rotation of               ACBD BCDA           rotation of  ABCD BDCA           rotation of         ABDC BDAC           rotation of                ACBD CABD           rotation of         ABDC CADB mirror and rotation of               ACBD CBAD mirror and rotation of ABCD CBDA           rotation of                ACBD CDAB           rotation of  ABCD CDBA mirror of                     ABDC DABC           rotation of  ABCD DACB           rotation of                ACBD DBAC mirror and rotation of        ABDC DBCA mirror of                            ACBD DCAB           rotation of         ABDC DCBA mirror of              ABCD
 * StuRat (talk) 04:51, 6 February 2012 (UTC)


 * @StuRat: Assume sides a, b, c, and d are successive and of arbitrary length. hydnjo (talk) 19:03, 6 February 2012 (UTC)


 * There is a formula using p and q in the quadrilateral article but you are far better off looking at the angles in the formula pointed at above and ignoring p and q and just consider that the cos is squared so always positive or zero. Dmcq (talk) 09:12, 6 February 2012 (UTC)


 * @Dmcq: I understand that the maximum area is achieved when the quadrilateral with successive sides a, b, c, and d of arbitrary length is arranged to be a cyclic quadrilateral. I just don't know how to make angle ab + angle cd = 180º (or angle bc + angle da = 180º). Also, assume a convex quadrilateral. hydnjo (talk) 19:03, 6 February 2012 (UTC)


 * Well consider stretching out one diagonal as far as one can, then one side or the other of the diagonal forms a straight line, the angles at the end must be less than two right angles as a triangle is formed (or equal in which case an infinite radius circle goes through the points as they form a line). Then try doing the same with the other diagonal, since the other angles are now less than two right angles there must have been a point where each pair was two right angles. Dmcq (talk) 19:39, 6 February 2012 (UTC)


 * OK, got it. First find the area of the cyclic quadrilateral using Brahmagupta's formula and then find the interior angles from: area = 1/2 (ab + cd) sin(angle ab). Thanks for your help. hydnjo (talk) 20:17, 6 February 2012 (UTC)

Category theory homework problem - subobject classifier and showing a morphism is self-inverse
Hello all, I'm working on the following Topos/Category Theory problem: you'll have to forgive me for my inability to LaTeX commutative diagrams, but I have in mind a commutative rectangle here composed of 2 commutative squares side by side, with both commutative squares being pullbacks. I'm really struggling with this one so if you think you can help please do.

Let $$\epsilon$$ be a category with finite limits and a subobject classifer $$\Omega $$, "True" map $$T: 1 \to \Omega$$ (where $$1 $$ is the terminal object) and let $$f: \Omega \to \Omega$$ be a monomorphism in $$\epsilon$$. By considering the pullback square of $$f$$ and $$T$$, which pulls back to an object $$U$$ and morphism $$u: U \to \Omega $$ and the unique morphism $$U \to 1 $$ say, and the pullback square of morphisms $$T$$ and $$u$$ to some object $$V $$ (and the composite of these 2 pullback squares, 'joined' by $$u$$), show that $$f \circ f$$ is the identity morphism: $$1_{\Omega}$$.

I hope that made some sense, apologies if not. Now I've spent a while playing around with morphisms to try and get somewhere with this but I haven't managed anything so far and I was hoping you might be able to help: I've got an extremely hard problem sheet to hand in and i'm having trouble with a number of the questions so I'd be very grateful for any thoughts you could give whatsoever. Really, so far the only ideas I have are to use the fact that the 'composite' of 2 pullbacks in the way I've described is also a pullback square, and to use the fact that the 'subobject classifier property' applies to all monomorphisms, therefore we could apply the property to $$f$$ to find some unique corresponding morphism $$\chi_f: \Omega \to \Omega $$ to get a third pullback square and then perhaps use that somehow, but otherwise I've really been unable to make progress. Could anyone please help? I'm trying to learn this subject so I would be very grateful for hints rather than answers (but please don't be too vague as I'm finding the problem rather difficult! :)) - many many thanks in advance, Totenines99 (talk) 00:23, 6 February 2012 (UTC)

Does an adjective exist to describe this situation?
Hello,

Is there an adjective to describe a planar entity that can expand infinitely along both dimensions of the plane it's contained within? (For example, a planar square that could hypothetically have infinite width or height would be such an entity.) All I can think of is "unbounded", which isn't sufficiently impressive-sounding for my purposes.

Any suggestions would be appreciated!

Hiram J. Hackenbacker (talk) 14:21, 6 February 2012 (UTC)


 * Unbounded is probably the answer. See articles about 'Unbounded ', especially Unbounded set. --CiaPan (talk) 15:07, 6 February 2012 (UTC)
 * you might also use infinite, although this means an object which actually is infinite, while unbounded denotes one that may be arbitraily large, but not necessarily infinite --CiaPan (talk) 15:10, 6 February 2012 (UTC)
 * I wouldn't suggest "unbounded" for this- the unit square seems to have the property you describe, but it is certainly bounded. But really I don't have any idea what exactly your property really means. Does a triangle in the plane have your property? What's an example of a set in the plane without this property? Staecker (talk) 17:08, 7 February 2012 (UTC)
 * I should have been a bit more specific - I was thinking of a dynamic entity, i.e., one with dimensions that can change (for example, as a function of time) existing on an unbounded plane. A square or a triangle with static dimensions would not possess the property in question, while one that could increase its planar dimensions in an unconstrained way (possibly reaching a point at which it would occupy an infinite area on its resident plane) would. Such an entity is purely theoretical (at least, as far as I know), but I'm curious as to whether there's a way to describe it that's a bit more precise than simply saying that it can expand in an unbounded fashion within the plane on which it lies. I think "unbounded" may end up being the best choice of adjective, but I was hoping there might be a more interesting term from topology or one of the more esoteric fields of mathematics. Hiram J. Hackenbacker (talk) 00:45, 8 February 2012 (UTC)

t-test help please
I'm trying to figure out whether the claim in that white students are more likely to get into Oxford University than black students is justified or not. I created tables where the sample values are either 0 or 1, with sample sizes and success rates as described in the article (for x1, 1089 1s and 2799 0s; for x2, 13 1s and 54 0s). I'm trying to do a one-tailed heteroscedastic unpaired t-test. As my raw data I got:

$$n_1=3888, \bar{x}_1=0.280093, s_1^2=0.280072, n_2=67, \bar{x}_2=0.19403, s_2^2=0.193468$$.

Then I calculated:

$$s_{pooled}=0.054402, t=1.581967, degrees_{freedom}=69$$ using the formulae in.

Am I correct so far? And if so, what do I do now?  It Is Me Here   t / c 15:38, 6 February 2012 (UTC)
 * You want to use Pearson's chi-squared test, not Student's t-test. Qwfp (talk) 16:38, 6 February 2012 (UTC)


 * Sorry, could you be a little more specific?  It Is Me Here   t / c 20:45, 6 February 2012 (UTC)
 * You have four categories (whites that got in, whites that didn't get in, blacks that got in, blacks that didn't get in) with a certain number of people in each and you want to see if those numbers could plausibly come from a particular distribution (namely, two binomial distributions with the same probability of success). That is what the chi-squared test is for. You need to work out the expected number in each category and compare that to the actual number in each category in the way described in the article Qwfp's linked to. --Tango (talk) 21:07, 6 February 2012 (UTC)


 * OK, so I got
 * {| class="wikitable" border="1"


 * + O
 * 13 || 1089
 * 54 || 2799
 * }
 * $$\hat{p}=\frac{13+1089}{13+1089+54+2799}=0.278635 \rightarrow X \sim \mbox{Bin}(n,0.278635)$$
 * {| class="wikitable" border="1"
 * $$\hat{p}=\frac{13+1089}{13+1089+54+2799}=0.278635 \rightarrow X \sim \mbox{Bin}(n,0.278635)$$
 * {| class="wikitable" border="1"


 * + E
 * 18.66852 || 1083.331
 * 48.33148 || 2804.669
 * }
 * $$\Chi^2=2.427138$$
 * Is that correct? How do I get degrees of freedom and test whether the value is significant?  It Is Me Here   t / c 12:00, 9 February 2012 (UTC)
 * That looks about right. The degrees of freedom is the number of cells (4) minus the number of parameters in your probability distribution that you got from the data (in this case, 3: the probability of someone getting in, the number of white applicants and the number of black applicants), so that's 4-3=1 (there's always 1 degree of freedom for this kind of 2-by-2 contingency table). You test for significance in the same way as for any hypothesis test - look up the critical value for the confidence level you are interested in (it's always a one-tailed test for chi-squared) and compare it to the test statistic you have calculated. If your value is greater than the critical value, you reject the null hypothesis. --Tango (talk) 22:19, 10 February 2012 (UTC)
 * Is that correct? How do I get degrees of freedom and test whether the value is significant?  It Is Me Here   t / c 12:00, 9 February 2012 (UTC)
 * That looks about right. The degrees of freedom is the number of cells (4) minus the number of parameters in your probability distribution that you got from the data (in this case, 3: the probability of someone getting in, the number of white applicants and the number of black applicants), so that's 4-3=1 (there's always 1 degree of freedom for this kind of 2-by-2 contingency table). You test for significance in the same way as for any hypothesis test - look up the critical value for the confidence level you are interested in (it's always a one-tailed test for chi-squared) and compare it to the test statistic you have calculated. If your value is greater than the critical value, you reject the null hypothesis. --Tango (talk) 22:19, 10 February 2012 (UTC)
 * That looks about right. The degrees of freedom is the number of cells (4) minus the number of parameters in your probability distribution that you got from the data (in this case, 3: the probability of someone getting in, the number of white applicants and the number of black applicants), so that's 4-3=1 (there's always 1 degree of freedom for this kind of 2-by-2 contingency table). You test for significance in the same way as for any hypothesis test - look up the critical value for the confidence level you are interested in (it's always a one-tailed test for chi-squared) and compare it to the test statistic you have calculated. If your value is greater than the critical value, you reject the null hypothesis. --Tango (talk) 22:19, 10 February 2012 (UTC)

Consider the Bayesian approach. See Bayesian inference.

The probability for obtaining 1089 out of 1089+2799 is
 * $$p_1\approx\frac{1089+1}{1089+1+2799+1}\pm\sqrt{\frac{\frac{1089+1}{1089+1+2799+1}\frac{2799+1}{1089+1+2799+1}}{1089+1+2799+1+1}}$$

or
 * $$p_1\approx 0.280206\pm 0.007200$$

The probability for obtaining 13 out of 13+54 is
 * $$p_2\approx\frac{13+1}{13+1+54+1}\pm\sqrt{\frac{\frac{13+1}{13+1+54+1}\frac{54+1}{13+1+54+1}}{13+1+54+1+1}}$$

or
 * $$p_2\approx 0.202899\pm 0.048067$$

So
 * $$p_1-p_2\approx 0.280206-0.202899\pm\sqrt{ 0.007200^2 + 0.048067^2}$$

or
 * $$p_1-p_2\approx 0.077307\pm 0.048603$$

Zero is only 0.077307/0.048603=1.59058 standard deviations away from $$p_1-p_2$$. That is not significant. Bo Jacoby (talk) 00:43, 11 February 2012 (UTC).

Froda's Theorem Flawed ?
The proof assumes that all points of discontinuity in the interval can be enumerated: Let x1,x2,...xn be the points of discontinuities of f. This assumes that the set of points of discontinuities is countable (since if they can be enumerated with the natural numbers) yet this is exactly what needs to be proven. One cannot assume the theorem is true and claim it shows the theorem is true. A correct proof would be to assume the set of points of discontinuity is not countable and show that this would contradict that f is continuous. If I can either develop or find a proper proof of the theorem I will submit it for review. Feedback would be appreciated. - Noah Vieira — Preceding unsigned comment added by Vieiranoa (talk • contribs) 16:47, 6 February 2012 (UTC)


 * It works from that there can only be a finite number of discontinuities greater than a given amount in a monotone function between two points on a closed interval because the monotone function can only change by a finite amount. So for each size the number bigger is countable and you just count the lot in decreasing size order. Dmcq (talk) 19:51, 6 February 2012 (UTC)
 * This makes sense, but I was a little confused too. I think that the number of jumps larger than some alpha being finite must have to do with the fact that, for a monotone function, (finite) left and right limits must exist for every point in the domain. Reading over the definition again, I conclude that e.g. Tan(x) is not monotonic on [-pi/2, pi/2]. Is that right? SemanticMantis (talk) 01:11, 7 February 2012 (UTC)
 * It is monotonic, but it isn't defined at the ends of that closed interval. The interval being closed is important because it means the function is bounded by the values at the two ends. Dmcq (talk) 11:44, 7 February 2012 (UTC)
 * Anyway I thought Froda's theorem didn't depend on the function being monotonic but what's there talks about monotonicity and looks a bit silly. It should talk about essential discontinuities as well where either the left or right limits don't exist - there the ones that can be uncountable. That turns it into a decent sized theorem. So yeah I think something there is flawed. Dmcq (talk) 12:02, 7 February 2012 (UTC)
 * The proof that the discontinuities of a monotone function are countable is much easier. Just pick a rational number within each jump.   Sławomir Biały  (talk) 12:06, 7 February 2012 (UTC)
 * True - but that's not the theorem I believe. I just had a look at the talk page to say something like this and somebody else has said it and more nine months ago with no reply. Dmcq (talk) 12:10, 7 February 2012 (UTC)
 * I'm not disagreeing. It would be strange to refer to this almost trivial result by a moniker.   Sławomir Biały  (talk) 12:30, 7 February 2012 (UTC)

Maple advice
Hi. I'm working on a presentation, and I'd like to make animated graphs. I've only used Maple a little bit, and in looking at the documentation, I'm a bit overwhelmed by my options for making plots. I'd like to describe the graphs I'm trying to create, and see if anyone can suggest the easiest Maple packages to use to do it. If there's another program that would be even easier, I'd be open to that, but I only have until Friday to learn it and make it happen.

The idea is this: I've got a curve in space given by three parametric equations, and a point on the curve. I've got the parametric equations for my space curve already, and I know how I want the sets of 2, 3 and 4 points to approach the fixed point. I know how to write down the equations for all the lines, planes and spheres that I need.
 * 1) As two points independently approach my fixed point, the line joining them approaches the tangent line to the space curve at my fixed point.
 * 2) As three points independently approach my fixed point, the plane containing them approaches the osculating plane to the space curve at my fixed point.
 * 3) As four points independently approach my fixed point, the spherical surface containing them approaches the osculating sphere to the space curve at my fixed point.

What I don't know is how to make it all happen with Maple, or another user-friendly software package. Can anyone help me with this? Thanks very much in advance for any tips. -GTBacchus(talk) 20:59, 6 February 2012 (UTC)


 * Use the animate3d, for example:
 * with(plots):
 * animate3d([x,y,x^2+k*y^2],x=-1..1,y=-1..1,k=-1..1);
 * will give you an animate where k is the parameter. You can add commands like frames=100 to tell Maple how many frames to make, i.e. how slowly to animate:
 * animate3d([x,y,x^2+k*y^2],x=-1..1,y=-1..1,k=-1..1,frames=100,style=wireframe);

— Fly by Night  ( talk )  17:22, 7 February 2012 (UTC)

Sums of unit vectors of prime factions of pi?
Consider the set V = (Vp where p is prime and V is the unit vector from 0,0 to (cos(pi/p),(sin(pi/p)). Is the following statement true: The only sum of vectors in Vp (each vector occurs a non-negative number of times) (which go from 0,0 to xt, yt) where xt and yt are rational are the multiples of V2?Naraht (talk) 21:24, 6 February 2012 (UTC)
 * It is true. Instead of vectors in R2, consider these objects as complex numbers in the complex plane. Then each Vp represents a primitive 2pth root of unity ζ2p. A rational vector in R2 corresponds to an element of the field Q[i] = Q(ζ4). All the elements ζ2p are linearly independent from one another over Q (see cyclotomic field), so there's no non-trivial rational linear combination of them which is contained in Q[i], unless it's just a multiple of ζ4. Rckrone (talk) 03:10, 7 February 2012 (UTC)
 * Thanx!Naraht (talk) 21:06, 9 February 2012 (UTC)