June 2010 - mathfun@mailman.xmission.com

[math-fun] a problem from martin gardner
by Thane Plambeck 05 Jun '10

05 Jun '10

I found a short note from Martin Gardner to David A Klarner in my garage (long story; anyway...) The letter is dated 4 Oct 1983. This is what it says (in part) * * * *begin quotation * * * Do you know whether an n-rowed triangle-shaped array of dots like this * * * * * * * * * * ..... can be tiled by single triangles shaped like this * * * or by 1x3 straight lines like this * * * It's been driving me nuts for two weeks! Is there a coloring argument? Martin ******* end quotation ***** -- Thane Plambeck tplambeck(a)gmail.com http://thaneplambeck.typepad.com/

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[math-fun] Is this possible?
by Hans Havermann 05 Jun '10

05 Jun '10

Quadruples {a,b,c,d}, a>b>c>d>0, where a+b, a+c, a+d, b+c, b+d, c+d are all squares, a+b+c, a+b+d, a+c+d, b+c+d are all cubes, and a+b+c+d is a fourth power.

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Re: [math-fun] iter_exp()
by Steve Witham 05 Jun '10

05 Jun '10

>From: "Christian G. Bower" <bowerc(a)usa.net> > > > > iter_exp( z ) = lim > > n -> oo exp^n( z0 + z0^(z-n) ) > > > > iter_exp( z + 1 ) = exp( iter_exp( z ) ) > > > > Which is nice, but there are no z's where iter_exp(z) = 0 or 1 or e > > or e^e... >I don't think there are z's that map to 0, but remember exp is not 1-1 >in the complex plane. Find a place where iter_exp(z) = 2 pi i and that >yields a place where iter_exp(z) = 1 > >I find one around 1.511 - .217i > >Giving us 1 at around 2.511 - .217i Yes, I was premature...1 and so e, e^e, etc., all in a row. I haven't plotted the curve that connects those, nor the curves where z is closer to the real line, past about e^e. The wobbles in the imaginary part there exceed +/- pi, and soon after that the paths must be really nuts, and then hopelessly nuts, etc. Also the fixed points can be reached in similar ways. I've avoided branches and cuts my whole life until now. I think I can map the inverse function which is 1-M, so it won't cause overlaps, and maybe get an interesting fractal, but I still need to wrap my head around the branches (or vice-versa) first. --Steve

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Re: [math-fun] Probability puzzle
by Dan Asimov 03 Jun '10

03 Jun '10

Andy wrote: << So returning to the original question, the chance that the length of the left-hand piece is <= x is the chance that both points are >= x, which is (1-x)^2. So the answer to Dan's question is the value where (1-x)^2 = 1/2. So the best guess is 1 - (sqrt(2) / 2). >> Which is the answer I had in mind. That didn't slow you down at all, Andy. But if I ever knew that the median minimizes expected absolute error, I had entirely forgotten, so also thanks to Andy for a nice explanation of why that's true -- and solves this problem a good deal faster than the way I used. Here's the discrete version of the puzzle, where I don't know if the median trick is helpful: ---------- Suppose two numbers are chosen from {1,2,3,...,52} without replacement, independently at random. What is the best real number guess G (in closed form) for the value of the smaller number, if once again best means having least expected absolute error in the long run? ---------- --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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[math-fun] Probability puzzle
by Dan Asimov 03 Jun '10

03 Jun '10

Suppose two points are chosen independently at random from [0,1], cutting it into three pieces. What's the best guess G, in closed form, for the size of the left-hand piece, if best is defined as minimizing the average distance between G and the actual size of the left-hand piece? --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] Probability puzzle
by Dan Asimov 03 Jun '10

03 Jun '10

If I told you that, you'd know the answer, wouldn't you? --Dan Fred wrote: << What's the difference between that and the mean left-hand piece size? WFL << On 6/3/10, Dan Asimov <dasimov(a)earthlink.net> wrote: Suppose two points are chosen independently at random from [0,1], cutting it into three pieces. What's the best guess G, in closed form, for the size of the left-hand piece, if best is defined as minimizing the average distance between G and the actual size of the left-hand piece? >> >> ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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[math-fun] Cartesian product isn't associative
by Bill Gosper 02 Jun '10

02 Jun '10

My tutor Julian points out an inconvenient fact: R x (R x R) = {(x,(y,z))} and (R x R) x R = {((x,y),z)} and neither = Euclidean 3Space {(x,y,z)} = R x R x R . What is the approved way to sweep this under the rug? --rwg Earlier quasiperiodic Somos results: http://arxiv.org/pdf/math/0703470

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[math-fun] Wouter has uncovered a startling Somos paradox
by Bill Gosper 01 Jun '10

01 Jun '10

Re my first one--"period 8"/3^(n/4)^2. I assume he won't mind being quoted to the List, since he simply took me at my word, if a bit heavy-handedly: WM>----------------------------------------------------- I don't get it: you say it's 3^(n^2/16) time a (fixed) period 8 but I find 3^n {1,1,-1,0,1,-1,-1,0} to be the denominator of: {s[1],s[2],s[3],s[4]}={1,1,-1,-2/3}; s[n_]:=s[n]=(s[n-1]*s[n-3]+s[n-2]^2)/s[n-4] /. Indeterminate->0 s/@ Range[48] FactorInteger[%]//ColumnForm or am I reading you all haywireishly? Wouter. ------------------------------------------------------<MW I.e., he got 1, 1, -1, -2/3, 1/3, 1/9, -1/27, 0, 1/243, -1/729, -1/2187, 0, 1/19683, 1/59049, -1/177147, 0, 1/1594323, -1/4782969, -1/14348907, 0,... whereas I dodged 0/0 by plugging x=-2/3 into 1, 1, -1, x, 1 + x, -1 - x + x^2, -1 - x - x^3,... getting 1, 1, -1, -2/3, 1/3, 1/9, -1/27, 0, 1/243, -1/729, -1/2187, 2/19683, 1/59049, -1/531441, -1/4782969, 0, I guess Wouter should have been a little more cautious about 0/0. --rwg

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[math-fun] Martin Gardner handshake puzzle
by Dan Asimov 01 Jun '10

01 Jun '10

Yesterday's NY Times printed three Martin Gardner puzzles from his Mathematical Games column, and a very pleasant one of them is this handshake question: 1. My wife and I recently attended a party at which there were four other married couples. Various handshakes took place. No one shook hands with himself or herself or with his or her spouse, and no one shook hands with the same person more than once. After all the handshakes were over, I asked each person, including my wife, how many hands he or she had shaken. To my surprise each gave a different answer. How many hands did my wife shake? (May 1973) --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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