September 2015 - mathfun@mailman.xmission.com

Re: [math-fun] Add 1 and reverse
by Warren D Smith 01 Oct '15

01 Oct '15

Actually, I'd suggest tabulating C so that the "Add C and reverse" operation yields large period mod 2^m (C likely will depend upon m) That way you'd get useful random number generator components, at least on future machines featuring a fast reverse operation.

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Re: [math-fun] Add 1 and reverse
by Warren D Smith 29 Sep '15

29 Sep '15

Kerry Mitchell: My initial calculations show that the orbit of 0 under "add 1 and reverse bits" is periodic: - if the number of bits (m) is odd, then the period is 2^m, - if the number of bits is even, then the period is 2^(m/2+1)-1. Verified from m=1 to 30 bits. --WDS: the m=even case seems provable: X is always a palindrome after an even #steps starting from X=0. Until finally we reach 1111..111 which takes (2^(m/2)-1)*2 steps, then next step yields 0. QED. the m=odd case is trickier.

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[math-fun] can vortices collide?
by Warren D Smith 29 Sep '15

29 Sep '15

A "vortex" is a point in the XY plane. It generates a vector field around it, which is (for a vortex located at the origin; for any other location please translate this field bodily) field = (-Y, X) * C / (X^2+Y^2) where the signed nonzero real number C is the "strength" of the vortex. The "laws of motion" of vortices are: Each vortex moves at a velocity which equals the sum of the fields generated by all the other vortices. Example: if there are exactly two vortices, they will move in two circular orbits about a common centerpoint located on the line segment between them, in such a way as to maintain fixed separation and fixed speeds. Why do we care? Because vortices that move according to the laws I just stated, generate a summed-field, which happens to be an exact solution of the "Euler equations" of inviscid, incompressible 2D fluid (except at the vortices themselves, where there are point singularities). QUESTION: Can vortices collide? I believe I can prove these theorems: 1. A 2-vortex collision is impossible; i.e. any collision must involve at least 3. 2. The collision must involve an infinite number of "circlings" as prologue. HOW TO SOLVE THE PROBLEM? You might be able to write down an exact solution. Or, you might be able to prove initial conditions yielding a collision, must exist, by using some topological argument.

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[math-fun] power question
by David Wilson 29 Sep '15

29 Sep '15

For any prime p and integer r, does there exist n such that n^n == r (mod p)?

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Re: [math-fun] "stealth" dark matter
by Warren D Smith 29 Sep '15

29 Sep '15

To E. Salamin: No, the "stealth dark matter" hypotheses need not imply the existence of "dark atoms." Let "macroscopic" mean "with length scale greater than the length scale of the dark-stealth-analogue of the strong force." There are by assumption no macroscopic charged particles made of dark stealrth matter. Hence, no dark electrons. Hence no dark atoms. There are "dark nuclei" which would be detectable with photons since thery contain charged microsopic components, but only very shortwave photons could detect them, i.e. very high energy gamma rays.

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[math-fun] The triviality of English (summary)
by Warren D Smith 29 Sep '15

29 Sep '15

The triviality of English Dan Asimov Adam P. Gaucher Michael Kleber (Posed problem 25 sept 2015) Andy Latto Wouter Meeussen Warren D Smith Allan Wechsler Regard English as a left-cancellative and right-cancellative multiplicative semigroup with identity, i.e. obeying the relations XY=ZY or YZ=YX implies X=Z, and having an element "1" such that 1X=X1=X. If any two different-meaning words which sound the same are "equivalent" we shall show each letter of the alphabet (as well as space and apostrophe) generates the trivial group, i.e. all alphabet letters (and space and apostrophe) equal the identity element. A=1: AISLE=ISLE B=E=1: PLUM=PLUMB, CLIME=CLIMB C=S=L=1: SCENT=CENT=SENT, CELL=SELL=CEL=SEL, MILL=MIL D=A=1: AD=ADD, REDD=READ=RED E=1: AID=AIDE, AR=ARE G=1: GNU=NU, GNOME=NOME, SIGN=SINE H=W=1: HOUR=OUR, O=OH=OWE I=U=1: BAIT=BATE, MAID=MADE, AID=ADE, FIR=FUR K=W=1: KNIGHT=NIGHT, SCULL=SKULL, KNOT=NOT, KNOW=NO M=1: MUMM=MUM, DAMNED=DAMMED N=1: DAMN=DAM, IN=INN, SUNN=SUN, CONN=CON O=1: TO=TOO P=1: PSI=SIGH, COUP=COO, WOO=WHEW R=1: TOR=TORR, CURR=CUR T=1: BUT=BUTT, BAT=BATT U=1: BY=BUY, GUILT=GILT W=1: WRAP=RAP, WRUNG=RUNG, WRITE=RITE=RIGHT, TWO=TO, SO=SOW, WHO=HOO X=1: LOX=LOCKS, TAX=TACKS, FAUX=FOE A=1: ROAD=ROWED E=L=U=1: WHOLLY=HOLEY=HOLY, SEW=SOW, NU=NEW, PER=PURR F=1: TOUGH=TUFF, STAPH=STAFF, PHI=FI, PHAT=FAT J=1: GIN=JINN, JEANS=GENES L=1: WOOD=WOULD O=1: SUE=SOU Q=1: QI=KI, CUE=QUEUE V=1: LUVVY=LOVEY=LUVVIE, BEVY=BEVVY Y=N=1: HYMN=HIM, BITE=BYTE, LIE=LYE Z=1: DAYS=DAZE, PHIZ=FIZZ, FAT=PHAT, FAZE=PHASE miscellaneous other non-obvious relations: REIGN=RAIN, HEIR=AIR, WHETHER=WEATHER, COLONEL=KERNEL space=1: CATWALK=CAT_WALK, BE_CALM=BECALM aprostrophe=1: CANT=CAN'T A weaker result had been shown by Jean-Francois Mestre, Rene Schoof, Lawrence Washington, Don Zagier: http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 Quotients Homophones des Groupes Libres, Experimental Mathematics 2,3 (1993) 153-155. They also claimed there was an unpublished proof by Herbert Gangl & Norbert Schappacher of the triviality of German; speculated Japanese katakana was relation-free; and gave as proof of the triviality of French SOIE=SOI, SERRE=SERT=CERF, CE=SE, BALLE=BAL, LAID=LAIT, HAUT=AU, PARLENT=PARLE, ALLEZ=ALLER, SEPT=CET, CHAMPS=CHANT, FARD=PHARE, AN=EN, MAIS=METS, BAYER=BAILLER, SANG=CENT, JET=GEAI, ABBESSE=ABAISSE, KHAN=QUAND, LACQUE=LAC, COQ=COKE, POT=PEAU, WATT=OUATE, VAGUONS=WAGON. (They also showed various French accent marks are trivial.) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Should dark matter falling into a black hole
by Bill Gosper 28 Sep '15

28 Sep '15

radiate anything but gravity? --rwg

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Re: [math-fun] Something linguistic
by Warren D Smith 27 Sep '15

27 Sep '15

Aha, BEVY=BEVVY and LOVEY=LUVVIE indeed seem to fill much-needed gaps in the triviality proofs, and also useful are AISLE=ISLE, SCENT=CENT, BITE=BYTE The Frenchies' paper also mention OF_OURS=AVOWERS, but sorry, that to my ear is distinguishable -- which if accepted would also prove the space character=1 (which anyhow would follow rom, ION=I_ON, CATWALK=CAT_WALK, SEEN_ISLE=SENILE, or some such).

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[math-fun] "stealth" dark matter?
by Warren D Smith 27 Sep '15

27 Sep '15

http://arxiv.org/abs/1503.04205 advances a new hypothesis. There are dark matter analogues of quarks and gluons allowing the dark-quarks to form dark-neutrons. The dark strong force is stronger than the usual strong force causing the dark neutrons to be more strongly bound. (Actually this paper's setup is not the exact analogue of quarks, mathematically, but close enough for my present purposes.) The interesting thing about this class of hypotheses is, the dark quarks actually would be charged particles and hence highly visible with ordinary photons, i.e. naively not "dark" at all. However, due to the unbreakable binding into dark-neutrons, which are uncharged, they effectively are invisible EXCEPT to very high energy photons, or in the early universe when it was very hot, there presumably were charged dark particles too. This allows new phenomena to happen that old-style dark matter would not have permitted, giving them more freedom to fit their baloney to observations. This kind of dark matter also ought to be observable even today in interactions with high energy particles. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Something linguistic
by Warren D Smith 27 Sep '15

27 Sep '15

The Mestre et al paper handles V using chivvy = chivy or leitmotiv = leitmotif but neither would be accepted under the "too much logic forbidden" rule that you are not allowed to use alternate spellings of the same word (which the rest of their proof obeys), and under that rule V still is unresolved. (Stephen=Steven? Who's he? I never knew anybody whose name was spelled in two different allowable ways, anyhow.) They then claim German also is trivial (without proof, attributed to Herbert Gangl) but that Japanese katakana seems to be completely relation-free. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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