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[math-fun] How to Compute This Probability?
by Robert Baillie 20 Jul '06

20 Jul '06

Hypothetical situation: On July 7, a reporter (let's call her Eileen Byrne) asks Donald Rumsfeld questions on 9 topics, in this order: 1. North Korea 2. William Perry's op-ed on striking NK 3. diplomatic efforts on NK 4. stability of Iraq 5. political pressure in U.S. for withdrawal 6. U.S. troops' morale 7. American journalists giving away state secrets 8. Iraq War is being fought at home, in the media 9. Hamdan decision On July 8, a different reporter (let's call her Monica Crowley) asks Donald Rumsfeld questions on the same 9 topics, in the same order, except with items 6 and 9 reversed. Let's assume that there are only these 9 topics to choose from. What is the probability that these "reporters" chose the order of their questions independently? (That is, either the same order, or at most one switch of two questions?) Bob Baillie

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[math-fun] GF2 polynomial multiplication
by Schroeppel, Richard 19 Jul '06

19 Jul '06

(continuing a thread from June 29) I mentioned a half dozen ways of doing GF2 polynomial multiplication. One of them is to represent the polynomial by its value at several places. [Do the multiplication of the values at each place, then recover the polynomial product (or its values) by interpolation.] Gene points out that GF2 has a limited repertoire of places. FTAW offers the fix of removing the problem to GF[2^k]. I'm mentioning here that you can actually remove the problem to Z, and just use ordinary integers to do the problem, then clip the coefficients back down to 0-1 values. Another option is to do the problem with coefficients moved into Mod P or Mod N, or even a collection of Mods. You can even use mod 2+i, or phi, or 1+w (w=cbrt(1)); the last two shade over into GF[2^k]. Rich ---------- To make this work for GF(2), you have to reinterpret them as polynomials over GF(2^k) for some sufficiently large k. Franklin T. Adams-Watters -----Original Message----- From: Eugene Salamin <gene_salamin(a)yahoo.com> --- "Schroeppel, Richard" <rschroe(a)sandia.gov> wrote: ... > Another way to represent polynomials is by giving the values at enough > points. ... But for polynomials over GF2, the only possible values are 0 and 1. Gene

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[math-fun] Lagrangian for electrodynamics
by Eugene Salamin 19 Jul '06

19 Jul '06

Can one of the advanced theoretical physics people here answer this question. Is there a Lagrangian, either classical or quantum, that describes the dynamics of the electromagnetic field together with charged particles, electrons if we need to be specific? From what I see, there is a Lagrangian in which the field variables are varied, with fixed charges and currents, to give Maxwell's equations. A second Lagrangian with particle coordinates varied, for fixed fields, gives the Lorentz force law. And finally a third Lagrangian (see Landau & Lifschitz, "Classical Theory of Fields") tries to meld the two together, but is only valid to order (v/c)^2. I am looking for One Lagrangian that rules them all, one that determines the exact theory of electrodynamics. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] Philately
by Richard Guy 19 Jul '06

19 Jul '06

Apologies if this is too far off topic, but I'm going this afternoon to the launching of a commemorative Canadian postage stamp celebrating the 100th anniversary of the founding of the Alpine Club of Canada. If the response is not too overwhelming, and can be made before 15:00 Mountain Time today, Wednesday, 2006-07-19, I will endeavor to send first day cards or envelopes, or whatever is available, to any philatelically inclined munster or seqfan who sends a snailmail address. R.

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Re: [math-fun] Another physics question
by dasimov＠earthlink.net 18 Jul '06

18 Jul '06

<< The idea was to give a time-independent example of energy, but explaining the difference between mass and weight may not be clear to beginners, especially why acceleration has to be involved. >> One pedagogic loophole the keeps acceleration only implicit, would be to define work (energy E) as determined by the time t it takes to bring a (frictionless) body of mass m up to speed s (along a straight line): E = ms/t. This has a visceral aspect to it, always a good thing in teaching. --dan

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RE: Re: [math-fun] Buried Cable problem
by Torgerson, Mark D 18 Jul '06

18 Jul '06

For a more practical solution On one end link all the wires in pairs labeling them P1, P2, etc. (If there are an odd number of wires label the leftover uniquely) On the other end identify the pairs, them label the wires in the pairs P1a P1b, P2a P2b, etc. (and identify the odd guy if it exists.) Tie all the 'a' wires together, keeping the 'b' wires separate. March back over to the other side and separate and label a's and b's -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Michael Kleber Sent: Monday, July 17, 2006 8:14 PM To: math-fun Subject: Re: Re: [math-fun] Buried Cable problem Much thanks to Franklin, who has been good enough to come up with *both* of the solutions to the cable problem. Now I can get on with my extended discussion. The executive summary of this longwinded message: The triangle number solution is really cool, but sadly, the much simpler daisy-chaining solution is better. I would dearly love to see a modification of the puzzle, for which the triangle number solution is the only one. Please please please. Okay, now for the details. When we first heard the problem, both (funster-lurker) Thomas Colthurst and then I independently arrived at the triangle-number solution, as did some colleagues at the Broad when I started to spread the problem around. The quick summary of that solution: say the number of wires is triangular, eg n=10: draw the triangle * * * * * * * * * * First patch together all wires in the same row; from the other end of the cable, you can tell the size of the equivalence class each wire is in. Now patch together all wires in the same column, and trek back to the beginning. The sizes of a wire's two equivalence classes act as unique coordinates. This can be generalized beyond triangle numbers, too. In the 10-wire version above, the coordinates given to the wires end up being 44; 43,34; 42,33,24; 41,32,23,14. All of these pairs sum to one of 8,7,6,5. So if we had, say, three more wires: * * * then we could simultaneously apply this technique to them, getting coordinates 22;12,21; there's no problem confusing the two triangles with each other. In general, then, we can use this method for n wires any time n can be written as a sum of triangular numbers such that if we use T(r) in the sum, we can't use any T(s) for r<s<=2r. Chewing on this last week, I was delighted to prove that this condition holds for all but finitely many n: there are only 39 numbers which cannot be so written; the largest is 720. (Yes, NJAS, I'll get the list to you eventually, but let this discussion play out first please!) So I was feeling pleased with my clever technique and my proof, and all was well until I shared the problem with Broad colleague Jade Vinson (another mathematician, of holyhedron fame). Jade promptly came up with a daisy-chaining method, as Franklin has already described; one construction works for all odd n, and a tiny variation works for all even n>2. This leaves the beautiful triangle number solution looking too clever by half: it's both harder and less universal. Thus my appeal to math-fun: please find a puzzle, of similarly elegant statement, to which the triangle solution is the only one. There certainly should be hope, because the daisy-chaining method queries the state of the system in a fundamentally different way from the triangle technique: for triangles, you need only figure out *how many* other wires each one is associated with, while you can only daisy-chain if you can learn the *identity* of an associated wire. Can anyone rescue "All but 39 values of n" from the dustbin of too-clever puzzle solution history? --Michael Kleber On 7/17/06, franktaw(a)netscape.net <franktaw(a)netscape.net> wrote: > Here's an entirely different approach, which works for all n > 2. > > The case where n is odd is simpler. Link all the wires in pairs, with > 1 left over. At the other end, identify all the pairs. Now link the > singleton to one of pair, link the other member of that pair to one of > a different pair, etc., until the last pair will have one wire > unlinked. Now identify these pairs at the other end. The originally > isolated wire is already identified; the wire it is now linked to can > be identified, which identifies the other member of that pair, etc. > > For n even, link all but 2 wires in pairs. At the other end, link one > of the previously unconnected wires to one of a pair, and chain > through the remaining pairs as before. The other previously > unconnected wire remains unconnected. First identify the wire which > was never connected; the remaining wires then reduce to the previous case. > > Note that this solves the problem never connecting more than 2 wires > together, so we could solve the problem with switchboards instead of > the patch cord. > > Franklin T. Adams-Watters > -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Another physics question
by Jason Holt 18 Jul '06

18 Jul '06

I also had a physics question recently. I was trying to explain e=mc^2 to my family, and was going to try to give an intuitive example of just how much energy there is in 300,000,000^2 kg*m^2/s^2. I realized that impulse (or momentum), measured in kg*m/s, is much easier for me to conceptualize and explain: a baseball travelling at 30mph has a certain amount of kinetic energy that anyone can experience personally just by catching it. It was much more awkward to try to explain pushing a 1kg weight exactly hard enough to accelerate it 1m/s^2, but only until it had travelled 1m. So if both units can be used to express energy, why do we speak of impulse as so different from energy, and is there an intuitive way to explain the conversion from the 30mph baseball's energy into kg*m^2/s^2 without calculus? I still remember the test question in 10th grade physical science in which I was asked to calculate the work required to carry a 10kg weight across a 10m room. I argued that, at least in the ideal sense, it required no work (if you're infinitely patient), since there was no change in gravitaional potential energy, but the teacher wouldn't budge. I suppose I've resented work ever since. Oh yeah, and maybe Mike Stay could also briefly summarize quantum gravity for us in a way that makes it clear just how that principle works. I was wondering about that, too. -J

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[math-fun] primes in pi; buried cable
by Schroeppel, Richard 18 Jul '06

18 Jul '06

This is from Eric Weisstein ... Date: Mon, 17 Jul 2006 08:42:54 -0500 (CDT) From: "Eric W. Weisstein" <eric(a)weisstein.com> X-X-Sender: eww(a)wopr.wolfram.com To: Sequence Fans Mailing List <seqfan(a)ext.jussieu.fr> cc: Math-Fun mailing list <math-fun(a)CS.Arizona.EDU> Subject: primes in pi: 1, 2, 6, 38, 16208, 47577, 78073, ... Anyone curious about prime number searches may be (slightly) interested to learn that the concatenation of the first 78073 digits of pi (including the initial 3) forms a probable prime. For the complete list of known indices of primes in the digits of pi (as listed above), see http://www.research.att.com/~njas/sequences/A060421. I believe at 78073 digits, pi(78073) becomes the 65th largest known PRP (cf. http://www.primenumbers.net/prptop/prptop.php) Cheers, -Eric --------- I saw the buried cable problem in one of Gardner's puzzle columns, long long ago. I've always been curious if the solver is required to cleanup his patch cord. Rich

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[math-fun] Stupid physics question
by Henry Baker 17 Jul '06

17 Jul '06

According to Noether, energy conservation is equivalent to time-invariance & other conservation laws are equivalent to other symmetries. So here's the question: the laws of physics, as we currently understand them, imply that _information_ is conserved -- we can run physics movies backwards, so long as enough (i.e., _all_) detail is retained. This appears to be the case even in the face of black holes and quantum spookiness. So what is the corresponding symmetry?

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RE: [math-fun] Physics question
by Andy Latto 17 Jul '06

17 Jul '06

> -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of Henry Baker > Sent: Sunday, July 16, 2006 9:23 PM > To: David Wilson > Cc: math-fun > Subject: Re: [math-fun] Physics question > > > There seems to be another analog to this system: semiconductors. > > The electrons act a little like a gas & bounce around, while the > places they leave are called "holes". However, they aren't identical: > electrons have higher mobility than holes, which is why NPN > transistors > are faster than PNP transistors. When I was very young, one of my favorite books was _Many_Moons_, by James Thurber. One of the things the Royal Astronomer (or possibly the Royal Mathematician) brags to the king that he has done is calculating the Speed of Darkness. This was just meant to be silly, but I was intrigued by it. If a room is brightly lit by a lightbulb, and you turn the bulb off, it gets dark, but not instantaneously. It will be some amount of time before the far corner of the room is dark, so there is a meaningful speed of darkness. At the time, I was proud of myself for figuring out that this speed would be exactly the speed of light. While that's true if you're only considering direct illumination, if you take into account the fact that some of the reason that the far corner is light is not through direct illumination, but via reflections off of the walls, the matter becomes more complicated. Andy Latto andy.latto(a)pobox.com >

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