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November 26

Chopin's modular forms

Chapter 11 of this. I can't help but wonder what they are. 67.117.145.149 (talk) 02:13, 26 November 2009 (UTC)[reply]

Do you have familiarity with modular forms? --PS T 02:53, 26 November 2009 (UTC)[reply]

I don't think that kind of modular form has anything to do with Frédéric Chopin. 67.117.145.149 (talk) 04:12, 26 November 2009 (UTC)[reply]

Perhaps this will help explain - apparently, the melodies consist of "tiny rhythmic and melodic units" (i.e. "modules"). Warofdreams talk 10:22, 27 November 2009 (UTC)[reply]

Thanks, that explains. I guess for that matter, a Puccini vocal solo performed in the Sydney Opera House

could be described as an aria under the curve. 67.117.145.149 (talk) 04:20, 28 November 2009 (UTC)[reply]

Conic sections

I've searched everywhere, but I can't find a proof showing that either an ellipse, parabola, or hyperbola can arise by cutting a cone. Can someone show me such a proof?

And on a related topic, if someone were to look at a circle at an angle, would it appear as an ellipse? I would think so, but I don't know how to show why... —Preceding unsigned comment added by 24.200.1.37 (talk) 02:22, 26 November 2009 (UTC)[reply]

What does your intuition suggest (with these sorts of questions, it is always best to attain an intuitive feel for the geometry; the proof merely encapsulates the intuitive feel in an algebraic language)? --PS T 02:28, 26 November 2009 (UTC)[reply]

With regards to the second question, a related question to think about is: what shape would one view, if he/she looked at the circle along the plane (that is, put his/her eyeline on the plane at a point outside the circle)? --PS T 02:33, 26 November 2009 (UTC)[reply]

I managed to answer my second question: apparently a plane intersecting a cylinder produces an ellipse as well (and I guess the proof for this will be closely linked to my first question).

As for my 'inuitive feel', I really don't have one: the definitions for these conic sections (locus of points equidistant from a line and a point, etc.) seem too far removed from cones. My only thought would be to show that a cone can be constructed by an infinite array of ellipses, parabolas, etc., but I don't know how to show this. —Preceding unsigned comment added by 24.200.1.37 (talk) 04:37, 26 November 2009 (UTC)[reply]

A cylinder can be thought of as a segment of cone that has been stretched out to infinity in both directions - the difference in angle between opposite sides of it tends to zero as it stretches to infinity. So cylinders behave very similarly to cones. --Tango (talk) 08:44, 26 November 2009 (UTC)[reply]

A proof is found in the article titled Dandelin spheres. Michael Hardy (talk) 04:55, 26 November 2009 (UTC)[reply]

The proof is really quite simple. Consider the cone in xyz-space given by the equation x² + y² − z² = 0. This cone has the z-axis as it's axis of symmetry (the line that passes straight down the middle and passes through the cone point). Consider the planes z = k for some k ≠ 0. The intersection of the plane z = k with the cone x² + y² − z² = 0 has the equation x² + y² = k², i.e. they are circles. Try the same for the planes y = k. The intersection of the plane y = k with the cone x² + y² − z² = 0 has the equation x² − z² = k², i.e. they are hyperbolae. Try the same for the planes z = x + k. The intersection of the plane z = x + k with the cone x² + y² − z² = 0 has the equation 2kx = y² − k², i.e. they are parabolae. Tilting these above planes will give ellipses in place of circles, and non-standard hyperbolae and parabolae. You just need to do a bit more work recognising the equations are equations os ellipses, hyperbolae and parabolae. It was important that k ≠ 0 so that we didn't get degenerate conics, e.g. a single point x² + y² = 0 or a pair or lines x² − z² = 0, or a single repeated line y² = 0. ~~ Dr Dec (Talk) ~~ 12:49, 26 November 2009 (UTC)[reply]

Top-in shuffle and deck ordering

Suppose you have a standard pack of 52 cards, and you shuffle them by picking the top card off the pack and putting it anywhere in the pack, including at the bottom and back on the top: "top-in shuffling" (so theoretically you might never actually change the order).

We let p_n be the probability that, after n iterations, the cards are found in increasing order: first, I need to show that p_n converges to some p as n -> infinity, regardless of the initial ordering, and find this p, apparently using the property of aperiodicity. I know of a theorem that if we have an irreducible and aperiodic markov chain, with an invariant distribution 'a' (aP=a), then for all initial distributions, P(x_n=j)->a_j, and for transition matrix P, Pⁿ_{i j}->a_j as n->infinity, for all i. However, I'm not sure how to show (or find) there exists an invariant distribution for the ordering of the cards, since surely there are 52! possible orderings, which would lead to a very big P!

Next, i've shown that, at least until the bottom card reaches the top, the ordering of cards inserted beneath it is uniformly random, and I now want to show that for all n, $|p_{n}-p|\leq {\frac {52(1+\log(52))}{n}}$ - I'm told "the Chebyshev inequality $P(Y>n)\leq {\frac {\mathbb {E} (Y)}{n}}$ may be useful" - though I'm not sure that actually is the standard Chebyshev inequality!

Now I can get a 52(1+log(52))/n, by noting that when the original bottom card is at position k from the bottom, the waiting time for a new card to be inserted below it is about 52/k-1 (is this actually true, rigorously?), and so summing these for k=2,3,...52, and adding '1' for the additional waiting time to be reinserted once reaching the top, we get 52(1/1+1/2+1/3+1/4+....1/51+1/52), which (by looking at the integral of 1/x from 2 to 52) is <= 52(1+log(52)), I believe - and this completely reshuffles the pack. So, presumably that's where the E(Y) on the RHS of the inequality comes from, but what about the left? I can't see how to get $|p_{n}-p|$ out of the left - if we let Y be the number of iterations it takes to reshuffle the deck completely, then that would give us the corresponding E(Y), but how is the probability Y>n linked to $|p_{n}-p|$ ? To be honest, I can't even really think of how to interpret $|p_{n}-p|$ on its own, let alone how to involve it in the rest of the problem!

Any help would be greatly appreciated, and many thanks - 131.111.8.97 (talk) 04:09, 26 November 2009 (UTC)[reply]

Isn't it just obviously 1/52! (and how else could it be invariant of the initial ordering)? —Preceding unsigned comment added by 67.117.145.149 (talk) 04:18, 26 November 2009 (UTC)[reply]

I agree. If you start off with an unbiased deck and you do things to do that don't involve looking at any of the cards (which I assume is the case), then there is no way you can end up with anything but an unbiased deck. --Tango (talk) 04:23, 26 November 2009 (UTC)[reply]

I assume the initial ordering is known. Otherwise the problem obviously doesn't make sense. We know intuitively that for any starting order the probability in the limit is 1/52!, but the OP wants to prove it. Rckrone (talk) 06:12, 26 November 2009 (UTC)[reply]

In the case that Y ≤ n, the probability of the deck being ordered is p since we know that it's actually shuffled. Otherwise the deck has some probability of being ordered that you don't know, call it q. So p_n = (1 - Pr(Y>n))p + Pr(Y>n)q = p + Pr(Y>n)(q - p). Since |q - p| < 1, we get that |p_n - p| < Pr(Y>n). Hopefully that makes some intuitive sense too. As the chance of the deck not having been totally shuffled decreases, the closer the behavior of the deck will be to a truly shuffled deck. Rckrone (talk) 06:51, 26 November 2009 (UTC)[reply]

A clarification (maybe this was causing the confusion?): p isn't defined here as the limit of p_n, rather it's taken as the probability of a shuffled deck being in order (which happens to be 1/52!), and then it's proved that p is the limit of p_n. Rckrone (talk) 07:06, 26 November 2009 (UTC)[reply]

That makes much more sense, thanks so much! :) 82.6.96.22 (talk) 11:43, 26 November 2009 (UTC)[reply]