April 2005 - mathfun@mailman.xmission.com

[math-fun] (Fwd) [NOBNET 4625] Mathematical Games
by Bernie Cosell 09 May '05

09 May '05

I know that in addition to your likely being interested in this, I know that several of you have contributed to Mathematical Games over the years and so might want to get in touch with Ed... /Bernie\ ------- Forwarded message follows ------- Date: Mon, 8 Nov 2004 06:30:37 -0800 (PST) From: ed pegg <ed(a)mathpuzzle.com> Subject: [NOBNET 4625] Mathematical Games To: nobnet(a)iijnet.or.jp Martin Gardner just recently celebrated his 90th birthday. Also, maa.org is *finally* releasing the entire collection of Mathematical Games on CD rom next month. There were frightening amounts of copyright issues to work out, but they have been resolved. Catalog Code: TDG/FW04 ISBN 0-88385-545-3 List Price: $54.95 This is all 15 Mathematical Games books on a CD-Rom. I'd like to do a "behind the scenes" look at Mathematical Games for an article. Martin did lots, but he avidly used a lot of help from hundreds of people. I'd like to write an article about the background help. If you ever assisted in one of Martin's columns, I'd like to hear from you -- a note about what you did. I'll use these notes in my article. Ed Pegg Jr http://www.maa.org columnist http://www.mathpuzzle.com ------- End of forwarded message ------- -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] better posed problem? (small surprise)
by Phil Carmody 07 May '05

07 May '05

From: "Meeussen Wouter (bkarnd)" <wouter.meeussen(a)vandemoortele.com> > > it is a small surprise that, for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) , > w+w^7+w^8+w^18+w^19+w^25 = 0 > no cancellation of symmetric pairs, triples or quintuples in sight, > but w+w^7+w^19+w^25 = -w^13 (* four out of five *) > and w^8+w^18 = -w^28 = +w^13 (* two out of three *) > > could be called "cancellation by missing parts" ;-)) As (a) {0,6,12,18,24} cancels, (b) {0,10,20} cancels, and (c) {0,15} cancels. {5,15,25} cancels, so {5,6,12,18,24,25} cancels. The largest gap (mod 30) between terms is that between 25 and 5, so {0,1,7,13,19,20} is the symmetry-less set with minimal largest member that cancels. You've exploited the gap of 6 between 18 and 24 instead. 2*p*q is easily so minimised - any takers for zeta_210? 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!? Yahoo! Small Business - Try our new resources site! http://smallbusiness.yahoo.com/resources/

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RE: [math-fun] elliptic mean
by Daniel Asimov 02 May '05

02 May '05

<< Must a Mean associate? -- Rich Well, from ((x+y)/2 + z)/2 = x/4 + y/4/ + z/2 we see that even the arithmetic mean fails to associate. On the other hand, I have in the past worked with mean associates (though that was not obligatory). --Dan

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[math-fun] Laws of Average
by Schroeppel, Richard 30 Apr '05

30 Apr '05

I call the rule AB.CD = AC.BD Mid-Quarter-Swap. The axioms AA = A, AB = BA, and AB.CD = AC.BD are complete for identities involving the ordinary two-input Average. Distributivity, AB.C = AC.BC, follows easily. Dropping commutativity covers weighted averages with an unspecified weight W. Specific W will satisfy an additional axiom or two. For W = 2/3, Avg(A,B) = (A+2B)/3. additional axiom: AB.BC = AC.CB For W = 1/phi = .618..., the additional axiom is A.BC = C.BA, one of the Dozen Cousins of Associativity. If we introduce Cancellation, [ AX = AY -> X = Y ], into the regular Average axioms, then Inverse, [ AX=B has a solution X ] follows for finite systems, and we can proceed to define + . This leads to odd-order abelian groups. Three-input average axioms are AAA=A, ABC=BAC=BCA, ABC.DEF.G = ABD.CEF.G Contrast with Median (aka Majority): AAB=A, ABC=BAC=BCA, AB.CDE = ABC.ABD.E Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com on behalf of R. William Gosper Sent: Sat 4/30/2005 1:35 AM To: math-fun(a)mailman.xmission.com Subject: [math-fun] elliptic mean >Wrt Means, Gosper's objection was a bit more subtle: should we require that avg( avg(a,b), avg(c,d) ) = same thing with b and c swapped? Bingo. "Association" was a poor choice of words. Dyadic symmetry? >This *is* true for ordinary averages, and geometric, harmonic, etc. >means. And the resulting symmetric expressions make it obvious how to define avg(a,b,c). A nonobvious def in terms of dyadic avg is the root of avg(avg(a,b),avg(c,x)) = x . This does *not* work with elliptic mean. One plausible def is to iterate [a,b,c] <- [em(b,c),em(c,a),em(a,b)]. But I think em should be classified among the unscrupulous means. --rwg _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Triangles question
by lkmitch＠att.net 30 Apr '05

30 Apr '05

Is the "distance from the corners" question equivalent to this? Take two pairs of Pythagorean triples, (a1, b1, c1) and (a2, b2, c2). Let the point in question be at the origin, with square corners at (a1, b1), (-a2, b1), (-a2, -b2), and (a1, -b2). Then the problem is to find triples such that (a1, b2, c3) and (a2, b1, c4) are Pythagorean triples too. I looked at that problem briefly for triples up to 20,000 on a side. The closest I came was a square of side 4063 from triples (3, 4, 5) and (4060, 4059, 5741). In that case, the distance to the quadrant 2 corner was 4060.002 and the distance to the quadrant 4 corner was 4059.001. This is a neat problem visually: draw all the arcs of integer radius from each of the four corners and look for points where four arcs intersect. After minutes of laborious effort, I have decided that the problem has no solution. :-) Kerry -------------- Original message from "Daniel Asimov" <dasimov(a)earthlink.net>: -------------- > I heard this question in the immediately > equivalent version: > > < > Is there a point in the plane at rational > distances from each corner of a unit square?

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RE: [math-fun] Triangles question; elliptic mean
by Schroeppel, Richard 30 Apr '05

30 Apr '05

This one is easy: start with the unit square in R2 and a point in the center. Move the center point up in the Z direction until it reaches any convenient rational or integer value. --- Wrt Means, Gosper's objection was a bit more subtle: should we require that avg( avg(a,b), avg(c,d) ) = same thing with b and c swapped? This *is* true for ordinary averages, and geometric, harmonic, etc. means. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Daniel Asimov Sent: Friday, April 29, 2005 3:59 PM To: math-fun Subject: Re: [math-fun] Triangles question I heard this question in the immediately equivalent version: < Is there a point in the plane at rational distances from each corner of a unit square? > My former thesis advisor, Moe Hirsch, has remarked, "If we don't know that, we don't know *anything*!" Here's a wider class of questions: Suppose we have a planar unit square in R^n for any n >= 2. Then is there a point of R^n at rational distance from each corner of the square? And if that fails: Suppose we have a planar unit square in Hilbert space H = {f: Z+ -> R : sum of all f(k)^2 < oo }. Then is there a point of H at rational distance from each corner of the square? (Here d(f,g) = sum of all (f(k)-g(k))^2.) --Dan ----------------------------------------------------------------- Richard Guy writes: << What we'd REALLY like to know is: Is there a point at integer distances from each corner of a square with integer edge? >> _______________________________________________ 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] Axioms for mean
by Eugene Salamin 30 Apr '05

30 Apr '05

--- "Schroeppel, Richard" <rschroe(a)sandia.gov> wrote: > > Wrt Means, Gosper's objection was a bit more subtle: > should we require that > > avg( avg(a,b), avg(c,d) ) = same thing with b and c swapped? > > This *is* true for ordinary averages, and geometric, harmonic, etc. > means. > > Rich This fails for the AGM. So if we accept the AGM as legitimate, then we must reject this as an axiom. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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Re: [math-fun] Triangles question
by Daniel Asimov 29 Apr '05

29 Apr '05

I heard this question in the immediately equivalent version: < Is there a point in the plane at rational distances from each corner of a unit square? > My former thesis advisor, Moe Hirsch, has remarked, "If we don't know that, we don't know *anything*!" Here's a wider class of questions: Suppose we have a planar unit square in R^n for any n >= 2. Then is there a point of R^n at rational distance from each corner of the square? And if that fails: Suppose we have a planar unit square in Hilbert space H = {f: Z+ -> R : sum of all f(k)^2 < oo }. Then is there a point of H at rational distance from each corner of the square? (Here d(f,g) = sum of all (f(k)-g(k))^2.) --Dan ----------------------------------------------------------------- Richard Guy writes: << What we'd REALLY like to know is: Is there a point at integer distances from each corner of a square with integer edge? >>

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RE: [math-fun] elliptic mean
by Daniel Asimov 29 Apr '05

29 Apr '05

This is an excellent question that I wish I knew the answer to. Perhaps the axioms should also cover the case of the mean of n points as well as two: Then the first axiom might be revised to say the domain of a mean is any finite subset of points of most generally a metric space. One other axiom should be that the output is a continuous function of the input. In recent unpublished research, I describe how to take a kind of mean of finite -- and many other -- subsets of any compact (Riemannian) manifold X. For example if X is the usual spherical surface S^2, the mean of any two non-antipodal points is the unique midpoint of the shortest arc between them, and the mean of two antipodal points is the equator with the uniform measure on it. (And the mean of the equator is its pair of poles.) For x,y in R+, there are of course the p-means m_p for -oo <= p <= oo: m_p(x,y) = lim as q -> p of ((x^p + y^p)/2)^(1/q) for any {x,y} \subset R+. I wonder if there is a nice set of axioms for a (symmetric) mean on R+ such that the means that satisfy these axioms are exactly the means of the form m(x,y) = g^(-1)((g(x) + g(y))/2) with g required to be continuous and increasing on R+ (and throw in the means that are limits of these means). --Dan -------------------------------------- Marc writes: << But what is the Official Definition of "mean" anyway? Looking at mathworld, it would seem that we must have x M y = y M x x min y <= x M y <= x max y ax M ay = a(x M y) >>

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[math-fun] RE: Unsolved? problem
by msubbara 29 Apr '05

29 Apr '05

Dear Richard: It is indeed a good idea to collect Murray Klamkin problems. I am particularly fond of one his problems-but it did not appear in Math journals, but in the book by L.A.Graham : Ingenious Mathematical problems and methods, Dover. Murray's problem,which appears as #72 in the book,reads: find the smallest number such that if the last digit is removed and placed at the beginning to become the first digit, this new number is nine times the original one. The solution given in the book itself is the astronomical fijure 10,112,359,550,561,797,752,808,988,764,044,943,820,224,719. What would be the solution if we do similar operation with the last two digits and ask '99 times the original number", or replace the last three digits similarly to get 999 times the original number etc. Subbarao >===== Original Message From Richard Guy <rkg(a)cpsc.ucalgary.ca> ===== >I'm collecting Murray Klamkin >problems and solutions and am >currently going thru Math Mag. > >I came across Problem 886, >Math Mag 48(1975) 57--58 >[nothing to do with Murray] >which isn't properly stated >but should read as in OEIS >A003508 : > >a(n) = a(n-1) + 1 + sum of >distinct prime factors of >a(n-1) that are < a(n-1). > >This leads to >1,2,3,4,7,8,11,12,18,24,30,41,42,55,... > >The original problem asked that >if you start elsewhere, e.g., > >5,6,12, ... or >9,13,14,24, ... or >10,18, ... or >15,24, ... > >do you always merge with the >original sequence? Evidently > >91,112,122,186,... takes a little >while. > >Has anyone ... Can anyone prove >Charles Trigg's guess ? R.

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