Wikipedia:Reference desk/Archives/Mathematics/2009 March 17

= March 17 =

Elliptic and Parabolic equations
I am still confused what exactly are elliptic and parabolic differential equations, and why exactly are they called so. Could someone explain? deeptrivia (talk) 00:32, 17 March 2009 (UTC)


 * The operator of an elliptic pde, like the laplacian, has Fourier transform x^2 + ... + y^2 like a circle or ellipse. The operator of a parabolic pde, like the heat operator, has Fourier transform  x^2 - y, like a parabola.  The operator of a hyperbolic pde, like the wave operator, has Fourier transform x^2 - y^2 like a hyperbola. JackSchmidt (talk) 04:20, 17 March 2009 (UTC)

Two point boundary value problems
I want to get as much information as I can about nonlinear two-point third order boundary value problems (e.g., y ' ' ' + a*y' + b*cos(y) = 0, y(0) =y0, y(1) = y1, y'(1) =yp1). Some things I would like to know are: existence of solutions, uniqueness of solutions, analytical solutions (if possible), approximations, numerical solutions, range of the solution space, etc. Could anyone point me to some references on this? Thanks! deeptrivia (talk) 01:22, 17 March 2009 (UTC)

Surface name?
Does the undulating sheet like curve/surface formed by the equations $$ \sin(x)cos(y)\ $$ or $$ \sin(y)cos(x)\ $$, etc. have a paricular name? --Leif edling (talk) 07:00, 17 March 2009 (UTC)
 * You may talk about the equations $$\scriptstyle z=\sin(x)\cos(y)\ $$ or $$\scriptstyle z=\sin(y)\cos(x)\ $$. No, I do not know of particular names. Bo Jacoby (talk) 10:40, 17 March 2009 (UTC).
 * Oops! Well $$\scriptstyle z=\sin(x)\cos(y)\ $$ was what I meant. --Leif edling (talk) 16:58, 17 March 2009 (UTC)

Chess problem
Consider a game of chess in which Black gives White queen odds: that is to say, the game is played exactly like a standard game of chess, except that Black starts without his queen. Given skillful play by both sides, White will win, of course, but for how many moves can Black delay checkmate? —Preceding unsigned comment added by 75.24.76.213 (talk) 08:39, 17 March 2009 (UTC)
 * I'd be amazed if this is known. In fact, I don't think anyone has even proved that white can force a win in this situation. Algebraist 13:08, 17 March 2009 (UTC)
 * On this page (warning: old and hard to load) dedicated to "difficult" computer problems, there was a note on the position where Black has only a Queen. While the page was about 12 years old, it pointed out that while computers could find mates in 12, it was suspected that there might be a mate in 11, but they couldn't prove or disprove it.  While I suspect that today's computers could answer this question one way or the other, it shows how hard it is to "prove" a win.


 * I don't doubt that Queen odds is lost with best play, but it will be some time before it's proven, let alone "Mate in ?" determined. (BTW, the main site of that Queen problem has quite a few interesting positions and commentary.)  Baccyak4H (Yak!) 15:38, 17 March 2009 (UTC)


 * So if it's so hard to solve, it would perhaps make an interesting game. I have seen board games with unequal armies, but never one in which the win condition for the weaker side was simply to survive for a given number of moves. —Preceding unsigned comment added by 75.37.236.230 (talk) 20:04, 17 March 2009 (UTC)


 * White has a winning position... But finding a winning line is almost as hard as solving chess... For the record, after a minute or two my engine can find nothing but a very comfortable position for white (+11.28, depth of 20 half-moves). Pallida  Mors  15:35, 17 March 2009 (UTC)
 * It depends on how well each side plays. I don't know whether you can define "equal ability". -- PS T  23:29, 17 March 2009 (UTC)
 * We aren't talking about 'equal ability', we're talking about perfect play, which is easy to define. Algebraist 23:38, 17 March 2009 (UTC)

Expected value of length of a chord
If I were to randomly construct a chord in the unit circle, what is the probability that the length of the chord is greater than $$ \sqrt{3} $$? What is the expected value of the length of the chord? AMorris (talk)  &#x25CF;  (contribs)  09:23, 17 March 2009 (UTC)
 * The answer depends on the random method of construction of the chord. You may for example (A) pick a random point inside the unit circle as the center of the chord, or you may (B) pick a random point on the circumference of the circle as one end point of the chord and pick a random arc between 0 and 360 degrees for defining the other. In case A the chord is greater than $$\scriptstyle \sqrt{3}$$ when the midpoint is inside a circle of radius 1/2, and the probability is 1/4. In case B the the chord is greater than $$\scriptstyle \sqrt{3}$$  when the arc is between 120 and 240 degrees, and the probability is 1/3. So your question is not well posed and the answer is not well defined.  Bo Jacoby (talk) 10:28, 17 March 2009 (UTC).
 * (ec) It depends what you mean by "randomly" - see Bertrand's_paradox. AndrewWTaylor (talk) 10:30, 17 March 2009 (UTC)

Or in other words, it depends on the probability distribution of the random chord. Michael Hardy (talk) 20:22, 18 March 2009 (UTC)

Trapezoid solving an angle
I don't know how to solve for angle MNC. I know that it is an isosceles trapezoid and I know that angle B is 88 degrees. That means that angle A is 88 degrees, and angles C and D are 92 degrees. But I don't know how to solve for angle MNC, could you guys help me out? BTW- M is the midpoint of AD, and N is the midpoint of BC. I just drew the horizontal line in there. Thanks for the help! --71.98.25.121 (talk) 23:31, 17 March 2009 (UTC)

By what you are saying MNCD is also an isosceles trapezoid and the MNC=ABC=88 degrees. Dauto (talk) 00:28, 18 March 2009 (UTC)


 * Please remember that we do not do homework here, although it is customary to provide guidance or a hint if the OP has shown an effort to solve the problem themselves (as this poster has done). A few helpful comments or directed questions would have been better.  ("Give a man to fish...")  -- Tcncv (talk) 01:09, 18 March 2009 (UTC)


 * Yes, thank you for your post. I am aware that you do not do homework here. My assignment included over 40 problems; I only asked about one part of a problem I had already tried but didn't understand. Thank you for your explanation Dauto. --71.98.25.121 (talk) 01:37, 18 March 2009 (UTC)


 * My apologies, I did not mean to be condescending, and I think you misunderstand me – my note was not directed at your question, but at the way it was answered. Questions such as yours are welcome, especially since you had shown that you had already made a good faith effort at working it yourself (which I acknowledged above).  My point was that a helpful hint that allows you to find the solution is preferable to giving the answer outright.  At least that is my understanding of our policy here.  And yes, Dauto is a regular contributor of much useful information here, such as the detailed analysis in the Probability in Rummikub topic above.  -- Tcncv (talk) 02:39, 18 March 2009 (UTC)

Since AM and BN have equal lengths and MAB and NBA have equal angles, the distances of the two points M and N below the line AB must be equal; hence the line MN is parallel to AB. Two lines parallel to each other must both meet a line crossing both of them at the same angle. That should answer your question. Michael Hardy (talk) 20:20, 18 March 2009 (UTC)


 * Thank you for your responses. I'm guessing Michael Hardy's response is the "correct" one. And I'm sorry, Tcncv, for being so rude. I was having a bad day. Not much of an excuse, but I shouldn't have taken it out on you. --71.117.36.70 (talk) 21:02, 18 March 2009 (UTC)


 * No Problem. I suspect I am not alone in that I enjoy seeing questions such as yours come along.  They add variety and give us amateurs a chance to exercise our minds while making ourselves useful.  Thank you.  -- Tcncv (talk) 00:23, 19 March 2009 (UTC)