August 2015 - mathfun@mailman.xmission.com

Re: [math-fun] A transcendental puzzle
by Keith F. Lynch 14 Aug '15

14 Aug '15

Dan Asimov <dasimov(a)earthlink.net> wrote: > Do you mean normal to a specific base? No. To all integer bases B > 1. > I'm also interested in normal to factorial base, which seems more > natural than normal to base b or even to all bases. Scott Huddleston <c.scott.huddleston(a)gmail.com> wrote: > How do you define normal to factorial base? I'm assuming that by "factorial base" Dan means this: http://en.wikipedia.org/wiki/Factorial_number_system#Fractional_values Since the Nth place has N allowed values, defining normality is tricky. I'd say if you divide those N values into B equal bins in each case, to create a base-B number, if that base-B number is normal for all integers B > 1, I'd say that counts as normal. For instance suppose the Nth digit of your factorial base number is the Nth prime mod N (Sloane's A004648): 0, 1, 2, 3, 1, 1, 3, 3, 5, 9, 9, 1, 2, 1, 2, ... Divide each of them by N: 0/1, 1/2, 2/3, 3/4, 1/5, 1/6, 3/7, 3/8, 5/9, 9/10, 9/11, 1/12, 2/13, 1/15, 2/15, ... Bin these fractions into, say, deciles (B=10), i.e. the first decimal digit of the fraction: 0, 5, 6, 7, 2, 1, 4, 3, 5, 9, 8, 0, 1, 0, 1, ... If 0.056721435980101... is normal in base 10, and similarly with all other values of B, I'd say the original number is normal to the factorial base. This also works for defining what's meant by the normality of an infinite list of real numbers that are between 0 and 1. However, I think a more natural form of normality than base factorial is the continued fraction form of a number. It's normal if the coefficients have the expected distribution, in which each coefficient appears less often than the next smallest coefficient, and in which the geometric mean of the coefficients approaches Khinchin's constant. Granted, this isn't quite the same thing. For example quadratic irrationals are all probably normal in every integer base, but *none* of them are normal in the continued fraction sense.

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[math-fun] The quest to be normal
by Warren D Smith 14 Aug '15

14 Aug '15

The quest to be normal =======Warren D Smith====Aug 2015======== http://mathworld.wolfram.com/ChampernowneConstant.html constructs a stupid but normal (in radix 10) real number: concatenate all the positive integers in order to get its decimal expansion. I say that construction is "stupid" because it occurs to me that there is a much simpler-to-prove-normal construction: 1. Let K=1. 2. Concatenate all the K-digit numbers in order. (Including numbers starting with 0, e.g. "02" is a "2-digit number.") 3. Double K. 4. Go back to step 2. The radix-10 normality of my number is trivial to see. For Champernowne's number it is much less trivial. Incidentally, sum(n=0..B-1)of n*B^(-n-1) = B^(-B) * (B^B-B^2+B-1) / (B-1)^2 so that if we write L_N = sum(n=0..N-1)of 2^n * B^(2^n) with L_0 = 0, then my number which is normal to radix B is given by MyNumber = sum(k=0,1,2,...)of B^(-B-L_k) * (B^B-B^2+B-1) / (B-1)^2. NEW(?) PROBLEM: Construct a number which is normal to every radix B>=2 simultaneously. (It is easy to see an uncountable infinity of such numbers exist, but I'm unaware of anybody actually constructing any example.) Here is a candidate, constructed as a mixed-radix number to radix B where B increases. 1. Let K=1 and B=2 and N=2. 2. Concatenate all the K-digit radix-B numbers in order. (Including numbers starting with 0, e.g. "01" is a "2-digit number.") 3. Multiply K by N. 4. Increment N. 5. Replace B by B^(N^2). 6. Multiply B by the Nth prime. 7. Go back to step 2. I think this number is normal simultaneously in every radix of the form (2*3*5*7*11*13*17*...*P)^k with P prime and k>0. Is it, and if so is that enough to prove normality to all radices B>=2? end.

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[math-fun] Game question
by David Wilson 13 Aug '15

13 Aug '15

Consider a game in which two players, A and B, each choose distinct integers by turn. A's object is to maximize the length of the longest A.P. among his selected integers. B's object is to limit the length of A's longest A.P. Show that B cannot prevent A from obtaining an A.P. of length 3. Can B prevent A from obtaining an A.P. of some length N? What is the smallest such N?

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Re: [math-fun] A transcendental puzzle
by Keith F. Lynch 13 Aug '15

13 Aug '15

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > I had a simpler example in mind, namely one that follows from *only* > the following facts about algebraic numbers: > -- The algebraic numbers form a field of characteristic zero; > -- x and exp(x) cannot both be algebraic. As did I. The solution I was thinking of is the unique real solution to x = exp(-x). Approximately 0.5671432. Decades ago, by the same reasoning, I realized that the unique real solution to x = cos(x) must be transcendental. Of course that means cos(cos(x)), cos(cos(cos(x))), cos(cos(cos(cos(x)))), etc., are all also transcendental, since they're all the same number. Approximately 0.7390851. (Puzzle: Is the solution to x = cos(x) still transcendental if you do it in degrees instead of radians?) Dan Asimov <asimov(a)msri.org> wrote: > Why did God make it so hard to prove transcendence? That's nothing compared to normality. Again "nearly all" real numbers are normal, but it's not easy to prove that any specific number is normal. Pi, for instance, was proven irrational in the 18th century and transcendental in the 19th, but the 20th didn't accomplish anything with it except increase the number of known digits from a few hundred to a few hundred billion. And the 21st has done nothing except increase the number of known digits by another factor of 100 or so. Maybe the 22nd century will prove (or disprove) that it's normal. Or maybe pi's normality is literally unknowable. Are any specific numbers known to be normal? That's not clear to me. Wikipedia says, without references: .... This theorem established the existence of normal numbers. In 1917, Wacl/aw Sierpinski showed that it is possible to specify a particular such number. Becher and Figueira proved in 2002 that there is a computable absolutely normal number, however no digits of their number are known. What is the specification of Sierpinski's number? And what is the algorithm for computing Becher/Figueira's number, and why have no digits been computed? Thanks.

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[math-fun] Transcendent boringness
by Warren D Smith 12 Aug '15

12 Aug '15

x * exp(A*x) = B if A,B nonzero algebraic, has only transcendental roots x. Proof (which must have been known to RC Schroeppel): if x algebraic, then A*x is, hence exp(A*x) isn't, hence B isn't, except B is, hence x transcendental. Extension: And now if we take the ln() of both sides ln(x) = ln(B) - A*x is transcendental if B=1. And... exp(A*x) = B*x^P similarly has transcendental roots x, and ln(x) is transcendental, for any nonzero rational power P. In particular exp(x) = -x has a transcendent real root x = -0.567143 and then x, exp(x)=-x, exp(exp(x))=1/x, all transcendent.

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[math-fun] A transcendental puzzle
by Keith F. Lynch 12 Aug '15

12 Aug '15

Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > However, I did slip up again: gamma has not actually been proven > transcendental! I should have employed exp(1), ... Indeed, exp(x) is transcendental for all non-zero algebraic x. It follows from that that if exp(x) is algebraic, x must be transcendental. Hence, for instance, the natural log of 2 must be transcendental. But since the transcendental numbers are of a higher cardinality than the algebraic numbers, it must be that in "most" cases x and exp(x) are both transcendental. Puzzle: Give an example where x and exp(x) are both known to be transcendental. I'll give a solution in a week if nobody posts one sooner.

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[math-fun] Integer-sided 4-scalenohedra with rational volume
by Ed Pegg Jr 12 Aug '15

12 Aug '15

8 copies of any triangle can make a convex 4-scalenohedron. Some triangles can make 3 distinct convex 4-scalenohedra. If integer triangles are used, some of the convex 4-scalenohedra have rational volume. I haven't yet found a triangle that admits two integer volumes. So far, all non-integral volumes have a denominator of 3. Here are some solutions I've found so far, with the triangle sides followed by the volume. 05,07,08 -- 96 07,13,14 -- 384 09,21,22 -- 3200/3 15,18,23 -- 4928/3 14,19,25 -- 2112 16,19,23 -- 2400 11,31,32 -- 2400 19,21,28 -- 2400 17,20,23 -- 3360 13,43,44 -- 4704 22,25,27 -- 6720 15,57,58 -- 25088/3 25,27,32 -- 8736 21,37,42 -- 31360/3 24,39,43 -- 39200/3 17,73,74 -- 13824 28,35,37 -- 15840 27,38,39 -- 49280/3 31,34,37 -- 17472 37,40,53 -- 18816 33,37,46 -- 18816 42,43,59 -- 19584 28,41,47 -- 20064 19,91,92 -- 21600 35,39,56 -- 24480 41,47,62 -- 26880 35,42,43 -- 28224 38,49,61 -- 31680 37,43,46 -- 32256 38,43,55 -- 34944 39,49,58 -- 38400 44,49,63 -- 38688 41,46,63 -- 42240 32,67,71 -- 42336 37,58,61 -- 47424 43,52,61 -- 50400 45,53,58 -- 59136 43,54,67 -- 59136 41,59,70 -- 61248 49,53,54 -- 65280 49,63,92 -- 69600 49,59,68 -- 76032 55,68,87 -- 82656 61,65,86 -- 86400 56,61,77 -- 86400 51,62,63 -- 267520/3 50,67,87 -- 94080 47,71,76 -- 97440 47,80,83 -- 110880 51,79,88 -- 117600 61,63,76 -- 119040 59,67,74 -- 126720 59,67,76 -- 131040 64,65,71 -- 137088 51,86,87 -- 412160/3 59,77,86 -- 149760 65,66,79 -- 150528 71,73,76 -- 184800 69,79,92 -- 198912 67,80,87 -- 205920 I've also found various bipyramids and other 8-triangle shapes with rational volumes. 03,03,04 -- 32/3 08,09,09 -- 896/3 17,17,24 -- 384 11,11,12 -- 672 12,19,19 -- 1632 16,33,33 -- 15872/3 20,27,27 -- 18400/3 20,31,31 -- 7392 20,51,51 -- 39200/3 43,43,60 -- 16800 33,33,40 -- 54400/3 28,51,51 -- 73696/3 24,73,73 -- 27264 57,57,80 -- 89600/3 41,41,48 -- 35328 36,83,83 -- 68256 56,57,57 -- 257152/3 59,59,60 -- 98400 67,67,84 -- 145824 72,89,89 -- 252288 There's probably something of interest in these somewhere. --Ed Pegg Jr

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[math-fun] A cleverness for FullSimplify
by Bill Gosper 12 Aug '15

12 Aug '15

FullSimplify[EllipticK[1/2 - 1/2 Sqrt[(-1 + 2 z)^2]] + EllipticK[1/2 (1 + Sqrt[(-1 + 2 z)^2])]] should give EllipticK[1 - z] + EllipticK[z] because it's an even function of sqrt(2z-1), so the branching cancels out. Generalizing slightly, a strange variant of Legendre's relation: EllipticE[1/2 (1 + Sqrt[(-1 + 2 z)^2])] EllipticK[1/2 - 1/2 Sqrt[(-1 + 2 z)^2]] + EllipticE[1/2 - 1/2 Sqrt[(-1 + 2 z)^2]] EllipticK[1/2 (1 + Sqrt[(-1 + 2 z)^2])] - EllipticK[1/2 - 1/2 Sqrt[(-1 + 2 z)^2]] EllipticK[1/2 (1 + Sqrt[(-1 + 2 z)^2])] == Pi/2 Less obviously, (1/2 + 1/(-1 + E^Sqrt[z^2])) Sqrt[z^2] == (1/2 + 1/(-1 + E^z)) z --rwg (MathIsFun)

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[math-fun] transcendental^3
by Warren D Smith 12 Aug '15

12 Aug '15

>RC Schroeppel: >Take x as the root of x e^x = 1 near .567143. >Then log(x), x, and e^x are all transcendental. --WDS: I agree! Nice find.

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[math-fun] Combinatorial game theory and checkers
by James Propp 12 Aug '15

12 Aug '15

Combinatorial game theory has been applied to Go (see Berlekamp and Wolfe's book "Mathematical Go: Chilling Gets the Last Point") and chess (see Elkies' article "On numbers and endgames: Combinatorial game theory in chess endgames", arXiv:math/9905198). Has it ever been applied to checkers? Jim

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