March 2014 - mathfun@mailman.xmission.com

[math-fun] Mathematica 9 can (rather impressively) do that Stirling error infinite product
by Bill Gosper 30 Mar '14

30 Mar '14

in http://www.tweedledum.com/rwg/idents.htm that Dick Askey once deemed "wild". Unfortunately, using that bleeping useless Glaisher symbol instead of Zeta'[-1], which it bashes to 1/12 - Log[Glaisher]. This is like bashing Zeta[3] to Apéry. Zeta'[-1] is also simply related to Porter's constant, so why doesn't Mma call it (some function of) Porter? Since it comes up in Hyperfactorial[1/2] and BarnesG[1/2], I used to advocate the symbol π_1, where π_0 := π, but, so far at least, Mma doesn't support Hyperfactorial[n,k] := 1^1^k 2^2^k ... n^n^k. In the past, and on that page, I used Zeta'[2], but that is almost always clumsier. Best for that identity is probably Product[n!*(n/E)^-n/Sqrt[2*π*(n + 1/6)], {n, ∞}] == E^(-(1/12)+2 (Zeta'[-1]) (2 π)^(1/4) Sqrt[(1/6)!]. Can someone emit a rule to melt Glaishers? --rwg

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[math-fun] Continuous argument
by Fred Lunnon 30 Mar '14

30 Mar '14

Preparing a seminar on binomial coefficient extension --- avoiding the nuclear option? --- for a computer science audience, I prepared plots of re, im, abs, arg parts of the gamma function qua function of a complex variable. It struck me that the arg plot would be much improved if one could avoid it snapping back to the principal value when passing thru' angles (+/1)pi --- proceeding instead continuously in whichever direction happened to be current at that point. How can arg() be hacked to achieve such an effect? [ Note that arg(gamma(z)) really is discontinuous around the poles at non-positive integer z --- but that is not an issue here, since special measures must be taken in that neighbourhood anyway. Rather appropriate subject line for the list, what? ] Fred Lunnon

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[math-fun] reversing the order of bits in a word
by Warren D Smith 29 Mar '14

29 Mar '14

Let w be a 64-bit word ("unsigned long long," in C language). The following C instruction sequence w ^= w>>32; w ^= w<<32; w ^= w>>32; w ^= w>>16; w ^= w<<16; w ^= w>>16; w ^= w>>8; w ^= w<<8; w ^= w>>8; w ^= w>>4; w ^= w<<4; w ^= w>>4; w ^= w>>2; w ^= w<<2; w ^= w>>2; w ^= w>>1; w ^= w<<1; w ^= w>>1; will overwrite w with its bit-order-reversal. In general for a (2^k)-bit long word this will use 3*k shifts and 3*k XOR operations and with zero(?) or one(?) extra registers consumed for temp storage and no funny masking constants needed. -- Warren D. Smith http://RangeVoting.org

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[math-fun] errors by WDS corrected
by Warren D Smith 29 Mar '14

29 Mar '14

arggh, sorry for my incorrect word bit-reversal hack. It was "so obviously correct, I did not need to test it." Famous last words... ---------- Also, in my "Knight distance" post, I made this minor error: (7,7) was actually correctly handled by my formula, and did not need to be an exception. Apparently the only exceptions are (0,1) and (2,2). Here is C-language code -- which I now actually did test -- for Knight distance where dx = abs(x2-x1), dy=abs(y2-y1). It is conjecturally correct for all integers, but it is proven correct only for the finite set tested (i.e. an 8x8 chessboard): Ki = max(dx,dy); L1 = dx+dy; if(Ki<3){ if(L1==1){ return(3); } if(L1==4){ return(4); } //omit next line if bi-infinite chessboard: if( (InCorner(a) || InCorner(b)) && dx==1 && dy==1 ){ return(4); } } if(L1 & 1){ //different square colors, so KnightDist=odd F = ( 4+max(2*L1, 3*Ki) ) / 12; F *= 2; F++; //divide rounds downward }else{ //KnightDist=even F = ( 10+max(2*L1, 3*Ki) ) / 12; F *= 2; //divide rounds downward } return(F); //knight distance It is rather annoying that I do not have a proof of correctness. One would think there would be trivial inductive demonstration, where the base case is the fact it works when L1<50, or whatever. -- Warren D. Smith http://RangeVoting.org

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Re: [math-fun] errors by WDS corrected
by Henry Baker 29 Mar '14

29 Mar '14

"Beware of bugs in the above code; I have only proved it correct, not tried it. --Knuth" http://www-cs-faculty.stanford.edu/~uno/faq.html At 11:52 AM 3/28/2014, Warren D Smith wrote: >arggh, sorry for my incorrect word bit-reversal hack. >It was "so obviously correct, I did not need to test it." >Famous last words...

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[math-fun] reversing the order of bits in a word
by Warren D Smith 29 Mar '14

29 Mar '14

Knuth gave the following absolutely incredible algorithm, here as C which I tested: //Don Knuth incredible algorithm for reversing a 64-bit word x: uint64 y; x = ((x>>1) & U64(0x5555555555555555)) | ((x & U64(0x5555555555555555)) << 1); y = (x ^ (x>>4)) & U64(0x0300C0303030C303); x ^= y ^ (y<<4); y = (x ^ (x>>8)) & U64(0x00C0300C03F0003F); x ^= y ^ (y<<8); y = (x ^ (x>>20)) & U64(0x00000FFC00003FFF); x ^= y ^ (y<<20); x = (x>>34) | (x<<30); return(x); Good luck figuring it out. And what are the 8, 16, and 32-bit versions of it (if any)?

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Re: [math-fun] reversing the order of bits in a word
by Bill Gosper 28 Mar '14

28 Mar '14

TK> I installed that instruction. It was a slight variant of the PDP-6/10 rotate combined instruction, which rotated two adjacent accumulators as a single 72 bit word. The bit reverse version shifted one accumulator right and the other left, while passing the bit across the boundary. <TK And it was called CIRCulate! Damn, I'd completely forgotten! I think Rich used it while counting the 5x5 magic squares. --rwg

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[math-fun] Stop signs
by James Propp 28 Mar '14

28 Mar '14

What is the officially approved way to manufacture a stop sign? Is a perfectly made stop sign really a regular octagon, or is it something slightly different? (My hope is that the answer varies from state to state.) Jim Propp

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Re: [math-fun] Stop signs
by Bill Gosper 28 Mar '14

28 Mar '14

On Mar 28, 2014, at 2:58 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: I also seem to recall that there were some adjustments to the octagon that give a convex shape with an even lower packing density than the octagon. I don't know if the inf over all convex shapes S of the sup over all density-defined packings' densities is realizable by a specific shape (though I would expect so) or by a specific packing of that shape (no idea). Does anyone out there know more about this? Reinhardt conjecture: http://arxiv.org/abs/1103.4518 -Veit "In 1934, Reinhardt asked for the centrally symmetric convex domain in the plane whose best lattice packing has the lowest density." Doesn't this disregard, e.g., oddgons and Reuleaux triangles? --rwg

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[math-fun] nonrandom numbers...
by Warren D Smith 28 Mar '14

28 Mar '14

For some reason Plouffe seems to be ignoring me. Anyhow, here are some specific examples in various radices. (Incidentally, radices other than 2 such as 3, might make nicer image files.) in MAPLE9: f := (x) -> (4/Pi)*EllipticE(2*sqrt(x)) + ((8*x-2)/Pi)*EllipticK(2*sqrt(x)); which is a transcendental function. Why be algebraic when you can be transcendent? [I think it is known that f(rational) is going to yield transcendental numbers if not rational. I have not carefully verified that... theorems on this topic reviewed in this Waldschmidt paper: http://www.math.jussieu.fr/~miw/articles/pdf/SurveyTrdceEllipt2006.pdf ] evalf( f(10^(-10)), 60 ); 1.00000000010000000000250000000025000000003906250000765625000 evalf( f(10^(-100)), 6000 ); #first 6000 digits in radix 10 nonrandom evalf( f(10^(-1000)), 600000 ); #first 600000 digits in radix 10 nonrandom evalf( f(10^(-10000)), 60000000 ); #first 60000000 digits in radix 10 nonrandom etc. Note this verifies my earlier-posted claim the length of the nonrandomness goes as the SQUARE of N for f(10^(-N)). Of course if you prefer binary to base 10 then use f(2^(-100)) and the like. Recently on mathfun somebody mentioned the fact that the 5th-degree algebraic function y=f(x) defined as the solution y of the Bring-Jerrard irreducible quintic y^5 + y + x = 0 has a nice known Maclaurin series with integer coefficients which grow roughly simply-exponentially. Also that's mentioned here: http://en.wikipedia.org/wiki/Bring_radical#Series_representation So anyhow, why be a wimp and use 4th-degree or 2nd-degree algebraic numbers, when you can be cooler and use 5th-degree? So, MAPLE9 again... f := (x) -> solve( y^5 + y + x, y ); #warning, finds 5 roots, we only want the real root evalf( f(10^(-10)), 1400 ); #verifies the first 1400 digits are nonrandom evalf( f(10^(-100)), 140000 ); #longer nonrandom numbers evalf( f(10^(-1000)), 14000000 ); And basically the same thing works for ANY-degree algebraic numbers. For example, say you like Plouffe want degree=4? Try f := (x) -> solve( y^4 - y + x, y ); #warning, finds 4 roots, we only want the least real root evalf( f(10^(-10)), 1000 ); #verifies the first 1000 digits are nonrandom evalf( f(10^(-100)), 100000 ); #longer nonrandom numbers evalf( f(10^(-1000)), 10000000 ); Want degree=3? Try f := (x) -> solve( y^3 + y + x, y ); #warning, finds 3 roots, we only want the real root evalf( f(10^(-10)), 500 ); #verifies the first 500 digits are nonrandom Want quadratic? f := (x) -> sqrt(1-4*x); evalf( f(10^(-100)), 16000 ); evalf( f(10^(-1000)), 1600000 ); etc.

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