- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] constructing closures of groups
by James Propp 06 Oct '04

06 Oct '04

Here's a funny way to construct the reals (or rather a redundant version of the reals, together with an extra element called plus-or-minus-infinity): First, given a sequence of subsets A_1,A_2,... of some set X, say that the sequence "converges" if, for all x in X, *either* x belongs to A_n for all n sufficiently large *or* x belongs to X\A_n for all n sufficiently large. We define the limit of such a sequence of sets to be the set of all x in X for which x belongs to A_n for all n sufficiently large. That is, we are using the discrete topology on the power set 2^X. Now let S be the set of rationals strictly between -1 and 1. Then, for any sequence of rational numbers r_n, we can look at the translates S+r_n = {s+r_n: s in S}; if this sequence converges (in the sense defined above), then it converges to some set T. The sets T constitute our initial model of the extended real numbers, which we can then fix by throwing out a countable set of redundancies (much as we fix the decimal system by throwing out .999..., or by throwing out 1.000..., or by decreeing them to be the same). Putting it differently: the extended real line (with some redundancy) can be identified with the orbit-closure of S under the additive action of Q on 2^Q, using the discrete topology on 2^Q. (Plus-or-minus-infinity is the empty set.) If we put in a kludge by modifying the definition of convergence so as to require that at least one element of X belongs to A_n for all n sufficiently large, then we effectively prevent the sets A_n from wandering off to plus-or-minus infinity, so we end up constructing a redundant version of the real numbers, without an extra element. Alternatively, we can let X = Q/Z instead of Q, with S equal to the set of elements of Q/Z with fractional part strictly between 0 and 1/2 (say). Then we don't have to worry about the translates of S wandering away, so the orbit-closure construction constructs R/Z from Q/Z (with some redundancy but with no extra points "at infinity"). Then we can construct R from R/Z (as a wreath-product). (1) Is this orbit-closure point of view useful for anything? (And is it new?) You can use it to define the p-adic completion of Q if you change the setup slightly (replacing the set S by a function, namely the p-adic valuation function, and by letting Q act on this function in the obvious way). You can also try this approach when Q is replaced by any group. Can other interesting completions of groups be gotten this way? Also: (2) Has anyone ever needed to use a number system that contains two or three copies of each rational number and one copy of each irrational number? (You can get such a system from Dedekind cuts in a fairly obvious way, but if you want your cuts to be closed under algebraic operations, you need to enlarge your notion of a cut to include pairs in which the left-set includes everything less than some rational number q, and the right-set includes everything greater than q, but q itself is in neither half.) I can't think what this number system is good for, but it must be good for describing something. (3) Order ideals (sets that are closed under "under") in Q that aren't the empty set, or all of Q, come in three flavors: ideals of the form {r: r < q} for some rational number q; ideals of the form {r: r \leq q} for some rational q; and ideals of the form {r: r < x} for some irrational number x. Non-trivial order ideals in R come in two just flavors: those of the form {x: x < y} for some y, and those of the form {x: x \leq y} for some y. Are there any ordered sets (or "generalized ordered sets" of some kind --- I choose to be purposefully vague here) that look something like the continuum, but have the property that non-trivial order ideals come in only one flavor? If there were such a continuum, it might be weird in some respects, but it might in other ways match the naive properties of the physical world better than the real numbers do. After all, when we cut a piece of paper into two pieces, it makes no sense to ask whether the dividing line got included in the left piece or the right piece! (4) How does one's picture of the real line change if one adopts some version of constructivism? (Does the constructivists' real line have the sort of properties I am looking for in (3)?) Jim Propp

1 0

Re: [math-fun] moore-penrose inverse
by Mike Stay 28 Sep '04

28 Sep '04

I wrote: > I'd like to have an algebraic rep for S like the > formula for the q-derivative, but I have no idea how to go > about it. Sorry--this question has an obvious answer if q is a constant (S x^n = x^(n+1)/[n]), but in this case q is a function. Better stated, the operator I'm trying to find a one-sided inverse for is f(x^(p^j)) - f(x) Df = -----------------, x^(p^j) - x i.e., the deformation is a Frobenius map. Then x^(np^j-1) - 1 Dx^n = -------------- x^(n-1) = [n] x^(n-1). x^(p^j-1) - 1 Thanks, Mike ----- Original Message Follows ----- > I've been playing with the q-derivative operator > Df = (f(qx)-f(x))/(qx-x) > over a finite field. It's obviously singular, since > adding a constant to f makes no difference to the > derivative. > > Given a matrix rep for D, I can use Gauss-Jordan > elimination to find a matrix rep for an integral operator > S such that DSf = f iff there exists some F such that DF = > f. I'd like to have an algebraic rep for S like the > formula for the q-derivative, but I have no idea how to go > about it. > > Can anyone help? Thanks, > > -- > Mike Stay > staym(a)clear.net.nz > http://www.cs.auckland.ac.nz/~msta039 -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

1 0

[math-fun] moore-penrose inverse
by Mike Stay 28 Sep '04

28 Sep '04

I've been playing with the q-derivative operator Df = (f(qx)-f(x))/(qx-x) over a finite field. It's obviously singular, since adding a constant to f makes no difference to the derivative. Given a matrix rep for D, I can use Gauss-Jordan elimination to find a matrix rep for an integral operator S such that DSf = f iff there exists some F such that DF = f. I'd like to have an algebraic rep for S like the formula for the q-derivative, but I have no idea how to go about it. Can anyone help? Thanks, -- Mike Stay staym(a)clear.net.nz http://www.cs.auckland.ac.nz/~msta039

1 0

[math-fun] minimum area convex lattice polygons (yes, again 8-))
by Chris Landauer 27 Sep '04

27 Sep '04

hihi, all - well, barring a few power failures and such, i've been running these programs on one or more machines (up to five here at home)continuously since last december, and i have some preliminary results to report (more complete values will be ready in a few more weeks) minimum area convex lattice polygons with n sides i separated them into odd and even n, since the even case is much simpler (for even n, minimum area polygons are centrally symmetric, so the search range is much smaller) i ran systematic searches and random searches, all with a strict a priori bound on the maximum edge length (to make the search range finite) because the systematic searches were run with an a priori bound on edge length, they do not necessarily find the actual minimum area for a given n (a larger bound could find a smaller area polygon, which happens occasionally for lower values of n, and also for n=27 and n=29 in going from bound r=7 to bound r=8) the random search programs found all best known values for even n<=50, but there are low maximum edge bounds for n=48 and n=50 (marked with '??') - the random searches were run for a fixed amount of time (24hrs for larger n) the random search programs found most best known values for odd n<=51, but there are several low maximum edge bounds (marked with '??'), and there was no systematic search for odd n>=47 (there aren't enough edges possible for edge bound r=7 for any n>=48), so the random search program cg produced the best known values because the edge length bound (columns marked 'max edge bound') for smaller n is much larger than the smallest maximum edge length found for the minimum area polygons found, i am pretty confident that these best known values are the best values (except for the ones marked with '??') more later, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org even n systematic max random (if different) min min longest edge min min longest n area perim edge bound area perim edge 04 1 4 1 50 06 3 8 2 50 08 7 12 2 50 10 14 18 3 50 12 24 24 3 50 14 40 30 3 50 16 59 36 3 50 18 87 44 4 50 20 121 52 4 50 22 164 62 5 50 24 210 72 5 40 26 274 80 5 30 28 345 90 5 20 30 430 98 5 14 32 523 108 5 14 34 632 118 5 14 36 749 128 5 12 38 890 138 5 12 40 1039 148 5 10 42 1222 162 7 10 44 1412 176 7 10 46 1620 192 7 10 48 1838 208 8 09 ?? 50 2088 218 8 09 ?? odd n systematic max random (if different) min min longest edge min min longest n area perim edge bound area perim edge 03 0.5 4 2 50 05 2.5 8 2 50 07 6.5 12 3 50 09 10.5 16 3 50 11 21.5 22 3 50 13 32.5 28 4 50 15 51.5 34 4 40 17 75.5 42 5 30 19 106.5 50 5 20 21 144.5 60 5 10 23 193.5 66 5 10 25 248.5 76 5 09 27 312.5 88 7 08 ?? 29 391.5 96 7 08 ?? 31 485.5 104 5 07 33 589.5 114 5 07 35 704.5 124 6 07 ?? 37 843.5 136 6 07 ?? 39 994.5 144 6 07 ?? 41 1170.5 156 6 07 ?? 43 1369.5 168 6 07 ?? 45 1605.5 180 6 07 1596.5 192 7 (cg, r10) 47 -- 07 1833.5 204 7 (cg, r08) 49 -- 07 2062.5 214 7 (cg, r08) 51 2315.5 224 7 (cg, r08) (cg is one of the random search programs, and r is the edge length bound)

2 1

[math-fun] Re: Partitioning 3-space into straight lines, exactly one in each direction
by Dean Hickerson 19 Sep '04

19 Sep '04

> Recently someone asked on sci.math.research > whether R^3 could be foliated by straight lines, > with exactly one line in each direction. That was me. Unfortunately, I borrowed the word "foliate" from an earlier posting, unaware that it implied some sort of continuity. > If we drop the continuity requirement for a > foliation, the problem becomes this one: > > QUESTION: Is R^3 the disjoint union of > [bi-infinite] straight lines L_d such that each > direction d occurs exactly once? That's the question I was trying to ask. Thanks to William Thurston for the proof that this follows from the axiom of choice. I wonder if there's a constructive proof. Dean Hickerson dean(a)math.ucdavis.edu

2 1

[math-fun] Partitioning 3-space into straight lines, exactly one in each direction
by Daniel Asimov 18 Sep '04

18 Sep '04

Recently someone asked on sci.math.research whether R^3 could be foliated by straight lines, with exactly one line in each direction. Someone else soon provided a proof that this is impossible (using homotopy theory). If we drop the continuity requirement for a foliation, the problem becomes this one: QUESTION: Is R^3 the disjoint union of [bi-infinite] straight lines L_d such that each direction d occurs exactly once? (I.e., each line through the origin of R^3 is parallel or equal to exactly one of the L_d's.) My guess is yes, but it's just a guess. --Dan

2 1

[math-fun] Inverter program for the Plouffe Inverter (new version of sept. 18 2004).
by Simon Plouffe 18 Sep '04

18 Sep '04

Hello, I made a new version of my inverter program. The file is at http://www.lacim.uqam.ca/%7Eplouffe/inverter.txt (563 kbytes). The text file is to be used with Maple, versions 9 and higher. There are 2 new funtions, pilll and pialg. They both use the new PSLQ routine, it needs a real number of at least 24 digits and up to 110 digits. The new function pilll uses 127 different tables of real numbers. The file can be downloaded and it is FREE, it can be copied at will as long as you mention the source. NOTE : the functions pismart() does not work this weekend because they are reconfiguring the machines at UQAM, please be patient. It should be back to normal in the next few days. Simon Plouffe Explanations on how to use it : ########################################################################################################## ########################################################################################################## # Plouffe's Inverter Maple interface, version of Sept. 18 2004. # Copyright (c) 1986-2004 by Simon Plouffe # # This program is to be used with Maple 7 or 8 on any platform. # A live connection is needed in order to work, preferably an ADSL, Cable modem, LAN or 56k modem. # This program makes a link to the website http://pi.lacim.uqam.ca/eng and can make request # to the website within a maple session. # The program is self-contained and DO not need anything else than maple and a connection to the internet. # There are 3 functions ; pilook, pismart and pidev (other functions are under development). # pilook makes a request to the Plouffe's Inverter and returns 1 answer (is there is any). # pismart is a personal souped-up version of the pilook function that makes 140 requests to the inverter. # the input needs to be a real floating point number or something that maple can evaluate to a real # number. The working precision is 24 digits, 16 are customary in order to return valid answers. # To make it to work : open a maple session, type read `inverter.txt`; # The name of the file can be anything. # Type any real number and watch it run. # pismart makes 200 variations on the input ans tries to see if there would not be any collision # against the tables. There are currently 201 million constants in the Inverter. # pidev expands the input x into 17 different development like the continued fractions, binary expansion, # Fibonacci representation and tries to find rational polynomial representation of x. # # Examples, pilook(X); minimal precision is 16 digits, this is a lookup to # Plouffe's inverter and returns the shortest answer (in characters). # # pilll(X); precision has to be from 24 to 110 digits, will try if the number # X is a linear combination of known constants like Pi, gamma, sqrt(2). # There are 127 tables of constants within the program. # # pismart(X); minimal precision is 16 digits, preferably 24. This portion # tries simple combination of the number X and simple functions. # There are 200 lookups made to the Inverter, uses the <socket> # function of Maple to make a request using <http://....>. # pidev(X); minimal precision is 16 digits. That portion expands the number # in various bases like base 2, Fibonacci base, Egyptian fraction etc # and tries if the result is a sequence which is a rational polynomial. # There are 17 different tests in this portion. # For example you may try pidev(tan(1.0)); to see the result. # # pialg(X); Minimal precision is preferably 24 digits. This routine will try # to find if X is the root of a polynomial equation of up to [sqrt(Digits)]. # For example with 64 digits will test if X is a root of a 8'th degree pol. # # piall(X); That one tries them all, pismart, pidev, pialg and pilll. The necessary # precision is adjusted within the routine. # # This program is FREE and can be distributed to anyone as long as the source is mentioned. ########################################################################################################### # # Plouffe's Inverter # # 1XS1?. .. # ?XXXXXXXSSXSSOIOXXXXXXXSSIIOSOC # ;;?1IC1?;;;: .:;;:. . # SXXXXXXXXXXXXXXXXXSXXXSOSOSOCIII1 # :?!!!!;!?!;::..::;;;;;;;!!!;!?. # ?IXXXXXXXXXXXSXSSSSSSSSSOSSSXXX=. # ?XXXXXXXXXXXXXXXXXXSSSSSCXX1IIICI? # =X :? S? : # 11 ;1 X: .:: # I= = X ;. :::. # ??;OI?=?;; ? C . # :;;;S=!!!?!: ?. ?? ;?? # I . : :S !! # ? ! O! # ! :? :C # .? !; !: # .! !. ?. # ;; =; : # !:; ; : # !;? !. # ?.I. !. # ?:XI :! # ;?XX. .???!;!!=I? ?! ;! # ;.XXC .!!?!;!!!!;!. ?O=?!! # ; OXX: :;; ;1XXX # ;C?. :::....... !SSO?!? # ?OII1CCOOO=. ?CXXX # !!!;!;;;;;: !? # # original drawing by Yves Chiricota (1995), Universite du Quebec a Chicoutimi ##############################################################################

1 0

Re: [math-fun] An Amazing Anagram
by Phil Carmody 15 Sep '04

15 Sep '04

From: Thane Plambeck <thane(a)best.com> > Subject: Re: [math-fun] An Amazing Anagram I fail to see what examples of simple anagrams, and the Shakespeare one that came before it, have to do with maths. For those desperate for anagrams related to the 2004 election, I started a thread in a location where anagrams are not off-topic, in fact they _are_ the topic - the newsgroup alt.anagrams - findable at your local newsserver, with the subject line "RFG (request for grams) - 2004 election" or at: http://groups.google.com/groups?threadm=873c1jux5d.fsf%40nonospaz.fatphil.o… Please feel free to join in there. (Although the base phrase was changed slightly, do check out Rick Rothstein's reply.) However, anagrams and mathematics are not entirely unrelated. For example there are interesting combinatorial algorithms related to anagrams, as demonsrated in recent years by Noam Elkies and others (historically Mike Keith has had a mathematical bent and tackled similar ideas too). In particular, given sets of phrases S = { s0, s1, .... sm }, T = { t0, t1, ... tn }, find the substets S', T' of S and T respectively such that S' ~ T' (where '~' is the anagram equivalence operator). Or, given one set of phrases S = { s0, s1, ... sn }, find two disjoint subsets S', T', of S such that S' ~ T'. For small sets, in particular if S' and T' are deliberately to be kept small (thus reducing the chance of finding anagrams) the combinatorial expansion is quite manageable. For example Noam's ITALY + GABON = LIBYA + TONGA BELARUS + INDIA = LIBERIA + SUDAN would drop out quite quickly as there are <10^5 country name pairs. For larger sets, the combinatorial explosion guarantees a vast number of results, but the sparseness means that a naive brute-force search would be unlikely to find any. e.g. Noam's http://groups.google.com/groups?selm=abchdf%24up8%241%40news.fas.harvard.edu cannot be found by just chance. On another mathematical note - algorithms for finding anagrams can be applied to the problem of finding amicable pairs! (Just with a theoretically infinite-sized alphabet, but we're mathematicians, and that shouldn't scare us, surely?) Phil ===== When inserting a CD, hold down shift to stop the AutoRun feature In the Device Manager, disable the SbcpHid device. http://www.cs.princeton.edu/~jhalderm/cd3/ __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

1 0

[math-fun] the volume of an inflated square of rectangular bag.
by Simon Plouffe 15 Sep '04

15 Sep '04

Hello everybody, I always been puzzled by this fact : I just went to the grocery store to buy plastic garbage bags. They sell it in various sizes in liters but the format is in inches like 26 x 36 inches. That model if I recall has a volume of 121 liters. But actually if I have a square very thin plastic bag and I inflate the bag with air what is the shape of an inflated bag? (square of rectangular). Is this problem has been addresed before? What is the volume? of an inflated 1 meter x 1 meter bag. for sure the surface is 2 square meters. Any idea someone? Simon Plouffe

3 2

[math-fun] strong law of small numbers ? surely not!
by wouter meeussen 12 Sep '04

12 Sep '04

the count of plane partitions of n that can be extended in exactly 3 ways to a plane partition of n+1 by adding one element equals Anum=A007425 = 1,3,3,6,3,9,3,10,6,9,3,18,3,9,9,15 .. or Inverse Moebius transform applied twice to all 1's sequence. or tau_3(n) *** the count of solid partitions of n that can be extended in exactly 4 ways to a solid partition of n+1 by adding one element equals Anum=A007426 1,4,4,10,4,16,4,20,10,16 .. or Inverse Moebius transform applied thrice to all 1's sequence or tau_4(n) *** ?? duh? W.

1 1