January 2007 - mathfun@mailman.xmission.com

[math-fun] late follow-up
by Michael Reid 28 Feb '07

28 Feb '07

some time ago, i wrote: > > Before people go too far looking for such charts, I have a question: is > > it possible to have 4 points in a plane, no 3 colinear, so that all 6 > > distances are integers? And with the additional constraint that no 2 > > distances are equal? > > it is possible to have arbitrarily many points in the plane, > no three collinear, such that all distances between them are > integers, and no two such distances are equal. there hasn't been any response, except for dan's query "why is this true?". since no one has answered, here is my solution. rational points on the unit circle x^2 + y^2 = 1 are (rationally) parametrized by (x, y) = P(t) = ((1 - t^2) / (1 + t^2), 2t / (1 + t^2)) . let A = P(0) = (1, 0), B = P(oo) = (-1, 0) and O = (0, 0) ; then the parameter t is tan(<PBA) = tan(<POA / 2) . it is easy to show that the distances PA and PB are rational if and only if t^2 + 1 is the square of a rational number, which is the case if and only if s = tan(<PBA / 2) is rational. we get a new parametrization of the unit circle by this parameter s : P(s) = ((1 - 6s^2 + s^4) / (1 + s^2)^2, 4s(1 - s^2) / (1 + s^2)^2) for rational s , such points have rational distance to both A and B , and the points obtained are dense in the unit circle. moreover, by ptolemy's theorem, any two such have rational distance to each other. this gives an infinite subset of the plane such that all the pairwise distances are rational, and one can also include the center of the circle. i do not know if this is a maximal subset of the plane with this property, although i think that it is likely. it remains to get a set with all distances different. this can be done in several ways. one way is to choose an infinite sequence {P_j} of points in the above set such that each point is on the top half of the circle, and such that <P_(j+1)OA < (<P_jOA) / 2 . for an explicit example, choose s = 2^(-j) , j = 0, 1, 2, ... to get the points P_j = ((2^(4j) + 6(2^(2j)) + 1) / (2^(2j) + 1)^2 , 4(2^j)(2^(2j) - 1) / (2^(2j) + 1)^2) . another way is to choose z to be the complex number corresponding to P(s) , and then to take the complex numbers z^(2^k) , k = 1, 2, 3, ... . (this will work as long as s is not 0, 1, -1 or oo , so that z is not a root of unity.) for a *finite* set, we can get all distances to be integers simply by scaling. i do not know if it's possible to have an infinite set, not all on a line, such that all pairwise distances are integers. i also do not know about dan's follow-up question: | And what about the higher-dimensional analogues: | | Q3: What is the maximum number of points in R^n, no hyperplane | (i.e., n-1 dimensional plane) containing n+1 of them, such that all | interpoint distances are integers? but i note that even with the weaker condition that no hyperplane contains all of the points, the question is still non-trivial! mike

7 26

[math-fun] New proof of Euclid's Theorem
by Paul R. Pudaite 09 Feb '07

09 Feb '07

In the December 2006 issue of Am. Math. Monthly , Filip Saidak presents "A New Proof of Euclid's Theorem." Here's a concise rendition: Let f(x) = x^2 + x, let g[0] be a positive integer, and for n = 1, 2, 3, ..., let g[n] = f(g[n-1]). Then g[n] has at least n different prime factors. I omit the proof for those who might enjoy working it out for themselves. The case of g[0] = 1 seems somewhat canonical. I'm curious about which primes appear as factors in the sequence of g[n] (2, 6, 42, 1807, ...). This sequence begins: 2, 3, 7, 13, 43, 73, 139, 181, 547, 607, 1033, 1171, ... Paul

4 3

[math-fun] Polynomial names
by Kerry Mitchell 31 Jan '07

31 Jan '07

Hi all, In terms of polynomial type names, what comes after cubic, quartic, quintic, ?, ?, ? Specifically, I'd like terms up through 10th order, if they exist. Thanks, Kerry -- lkmitch(a)gmail.com www.fractalus.com/kerry

2 2

[math-fun] modeling trochea and lunks
by N. J. A. Sloane 29 Jan '07

29 Jan '07

Dear Funsters and SeqFans, This is neither fun nor to do with sequences, I'm afraid. But a friend of a friend has been talking to me about making a 3-D model of the trochea and lungs, modeling the flow of air. This sounds like classical numerical analysis, and I know there are experts on this list. Does anyone know of previous work on this problem, or on similar problems? If so please write to me off-line. Thanks! Neil (njas(a)research.att.com)

1 0

[math-fun] photonic crystal & 2nd law
by Schroeppel, Richard 26 Jan '07

26 Jan '07

This note from Tom Knight got deflected by the list software. Sorry TK. -- Rich ------------------ Subject: Re: [math-fun] Ultraviolet catastrophe & Maxwell's demon This can't work, as you point out. As with all complex situations, this one will be difficult to fully analyze, but the problem almost certainly lies in the finite temperature of the photonic crystal. The crystal will come into equilibrium with the black body temperature, and then will emit isotropically a photon spectrum identical to the original black body. The reason we have the second law of thermodynamics is that you don't have to fully analyze such a situation to realize it can't work. How embarrassing.

1 0

[math-fun] Topswops
by Richard Guy 24 Jan '07

24 Jan '07

Neil & others, There's something wrong with A000375 & A000376 (or with me). The first says A000375 Topswops (1): shuffle n cards labeled 1..n. If top card is m, reverse order of top m cards. a(n) is max number of steps before top card is 1. 0, 1, 2, 4, 7, 10, 16, 22, 30, 38, 51, 65, 80, 101, 113, 139 The second says A000376 Topswops (2): shuffle n cards labeled 1..n. If top card is m, reverse order of top m cards. a(n) is max number of steps before deck is sorted. 1, 2, 4, 7, 10, 16, 22, 30, 38, 51, 63, 80, 101, 112, 130, 159 In the ordinary way, the deck will not get sorted, since once 1 has come to the top, no more sorting will take place. Somewhat artificially, when 1 comes to the top, you could put it aside and then reduce all the numbers on the other cards by 1 and repeat. It would be good to have some examples of what is going on. With 4 cards there are just two perms which require 4 flips: 3142 --> 4132 --> 2314 --> 3214 --> 1234 2413 --> 4213 --> 3124 --> 2134 --> 1234 In these cases the deck finishes up sorted. Is this always the case if you start with a perm which requires the max number of flips? If so, then the two sequences should be the same. R.

3 2

Re: [math-fun] Ultraviolet catastrophe & Maxwell's demon
by Eugene Salamin 24 Jan '07

24 Jan '07

The photonic crystal lowers the emissivity of the black body at those wavelengths that are blocked. Incandescant lamps emit mostly in the infrared. If the emissivity is lowered in the infrared, while keeping it high in the visible, then the lamp will more efficiently emit useful light. There are no paradoxes here. Gene ----- Original Message ---- From: Henry Baker <hbaker1(a)pipeline.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Tuesday, January 23, 2007 9:41:01 AM Subject: [math-fun] Ultraviolet catastrophe & Maxwell's demon (I apologize if this is more physics than math.) I attended a talk last night by an MIT professor who is working with "photonic crystals", which have the property that they can selectively allow photons of certain wavelength/energy to propagate, while disallowing other photons. These devices are non-dissipative, so that the non-selected photons simply bounce back. One type of crystal is made of a conductor with small holes, and the size of the holes determines which wavelengths can get through. (Think Faraday cage.) This professor suggests that one can put such a photonic crystal "filter" in front of a "blackbody", and thereby allow only the most energetic photons to leave. _Presumably_, the photons that bounce back into the blackbody exchange energy with the blackbody to reconstitute the classic blackbody radiation spectrum, so that the only photons leaving the blackbody are those with high energies. This is not a perpetual motion machine of the first type, because energy is conserved, but isn't it a perpetual motion machine of the second type? Why doesn't this idea violate the non-existence of Maxwell's demon? ____________________________________________________________________________________ Do you Yahoo!? Everyone is raving about the all-new Yahoo! Mail beta. http://new.mail.yahoo.com

1 0

[math-fun] Combing the Primes...
by Eric Angelini 24 Jan '07

24 Jan '07

... doesn't produce anything worth clicking here, I guess: http://www.cetteadressecomportecinquantesignes.com/CombZePrimz.htm Best, É.

1 0

[math-fun] Continued logarithm expansion
by Daniel Asimov 23 Jan '07

23 Jan '07

With everyone clamoring for an explanation of the sequence I mentioned in a recent post, I can't continue to ignore all the minds thirsting for knowledge. Starting from x in (0,1), set N(1) = floor(-ln(x)) and x(1) = frac(-ln(x)) (where frac(c) := c - floor(c)). Continue in the same way: N(k+1) = floor(-ln(x(k)) and x(k+1) = frac(-ln(x(k)). Then the continued logarithm expansion of x is by definition N(1), N(2), N(3), . . . . According to Mathematica, the continued log expansion of pi/4 is this: 0,1,0,0,1,0,2,0,1,1,2,2,1,0,0,1,0,1,0,2,0,0,1,0,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,2,1,0,1,1,0,0,0,0,0,3,1,0,1,1,4,4,1,0,1,1,0,1,1,0,0,2,1,0,1,0,1,0,4,4,0,0,0,1,0,1,0,3,1,0,0,1,0,1,2,2,0,1,3 (first 100 terms, anyhow). I am not totally sure of their accuracy; I got Mathematica to create the exact expression for N(k), 1 <= k <= 100, and then asked Mathematica to evaluate it: If the quoted sequence is accurate, this idea does not appear in OEIS. (Otherwise, it may well.) ---------------------------------------------------------------------------------------------------------- In[1]:= x[0] = Pi/4 In[2]:= For[k=1,k<=100,k=k+1,{tmp=-Log[x[k-1]],n[k]=Floor[tmp],x[k]=tmp-n[k]}] In[3]:= For[k=1,k<=100,k=k+1,Print[k,n[k] ]] ---------------------------------------------------------------------------------------------------------- Note: This would be slightly more analogous to a continued fraction expansion if the function replacing 1/x -- that is, -ln(x) -- instead of mapping (0,1) to (0,oo), mapped (0,1) to (1,oo). This is easily achieved by using 1 - ln(x) in lieu of -ln(x). The only difference would be that all the terms of the sequence of integers would be greater than they are by 1: 1,2,1,1,2,1,3,1,2,2,3,3,2,1,1,2,1,2,1,3,1,1,2,1,3,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,3,2,1,2,2,1,1,1,1,1,4,2,1,2,2,5,5,2,1,2,2,1,2,2,1,1,3,2,1,2,1,2,1,5,5,1,1,1,2,1,2,1,4,2,1,1,2,1,2,3,3,1,2,4 (This is also not in OEIS.) --Dan

1 0

[math-fun] topswops
by Joshua Zucker 23 Jan '07

23 Jan '07

Hi folks, I'm amateurishly working on the topswops problem (as in A000375 and A000376 of OEIS). Does anyone know anything about the state of the art in lower and upper bounds for terms beyond those given in the OEIS? I'd like to poke around and improve some of the lower bounds if I can, by actually constructing permutations. Thanks, --Joshua Zucker

1 0