August 2006 - mathfun@mailman.xmission.com

[math-fun] Unabomber's Belongings To Be Auctioned
by Thane Plambeck 18 Aug '06

18 Aug '06

The book list starts here: http://www.thesmokinggun.com/archive/0811062tedk9.html -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] Poincare Conjecture & Perelman
by Schroeppel, Richard 17 Aug '06

17 Aug '06

[noted by Bill Cordwell] http://www.newscientist.com/article/dn9775-russian-mathematics-genius-shuns… Gregori Perelman, a colossus of mathematical theory, appears to have gone to ground. The world of mathematics is in uproar over rumours that its most prestigious prize will be turned down next week by one of its brightest stars. The Fields Medal, the equivalent of a Nobel Prize in mathematics, is awarded every four years to young mathematicians who have made the biggest impact in their fields. It is due to be presented by the King of Spain in a ceremony in Madrid on Tuesday 22 August. But Gregori Perelman, who has been widely tipped to receive it, has resigned his post at the Steklov Institute of Mathematics in St Petersburg, Russia, and gone to ground. "Nobody knows where he is," says Marcus du Sautoy, a mathematician at Oxford University in the UK. Perelman is thought to have become disillusioned with mathematics and disassociated himself from the field. Perelman achieved fame in the mathematics world for his work on the Poincaré Conjecture, one of topology's most celebrated problems. The conjecture, conceived by the French mathematician Henri Poincare in 1904, relates to the question of whether it is possible to deform a holed doughnut into a sphere by bending and stretching it - without cutting or tearing it. It turns out that there is no way to remove the "hole" in the doughnut and so it cannot be turned into a sphere. However, any shape that has no holes can always be deformed into a sphere. The Poincare Conjecture and a more general problem, called the Thurston Geometrization Conjecture, assert that the same is true for shapes in higher dimensions. Unprecedented refusal Perelman's proof of both problems, published in 2002, received widespread admiration for its inventiveness, even though mathematicians have yet to officially pronounce on its validity. "The consensus is that it is probably correct," says du Saute. The Poincare Conjecture is also famous as one of the Millennium Prize problems established by the Clay Mathematics Institute in Boston in 2000. The Institute is offering a prize of $1 million to the first correct proof. "Perelman doesn't seem to be interested in medals or money," du Sautoy notes. A refusal of a Fields Medal would be unprecedented. In 1966, the German mathematician Alexander Grothendieck refused to pick up his award in Moscow in protest against the Soviet Union's military intervention in Eastern Europe, although he did later accept it. But Grothendieck also became disillusioned with mathematics and left the field. He is now believed to be living as a hermit in Andorra.

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[math-fun] video of the AIM
by Thane Plambeck 16 Aug '06

16 Aug '06

http://video.google.com/videoplay?docid=-6956424081422740268 The architect's design for the American Institute of Mathematics, to be built in Morgan Hill, California. The facility is planned to have 167,000 square feet plus an underground parking garage. * * * * looks like there will be plenty of good corner offices. that center plaza thing looks *exactly* like a place in the alhambra, right? -- Thane Plambeck http://www.plambeck.org/ehome.htm

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[math-fun] more asinh representation
by Henry Baker 12 Aug '06

12 Aug '06

After thinking about it some more, the "binary" version of asinh doesn't seem to help very much in calculating/manipulating the representation. So it seems easier to stick with the traditional definition, which still has the advantage that for small x, asinh(x)~x. To extract the binary exponent from asinh(|x|), multiply by 1/ln(2)~1.442695041, convert to an integer, and subtract 1. So to compete with IEEE single precision floating point, I suggest the representation: asinhIEEE(x)=round(asinh(x*2^126)*2^23) as a 32-bit fixed point number. The "sign bit" is the high-order bit, the "exponent" is the next 8 bits, and the "mantissa" is 23 bits. The largest IEEE single precision float is ~2^128 ~ 3.4x10^38, so asinhIEEE(3.4*10^38) ~ 1,482,700,732 (easily representable in 31 bits) The smallest IEEE single precision normalized float is ~2^-126 ~ 1.18*10^-38, so asinhIEEE(1.18*10^-38) ~ 7,416,212 (a 23-bit number) The smallest IEEE single precision denormalized float is ~2^-149 ~ 1.4*10^-45, so asinhIEEE(1.4*10^-45) ~ 1 The "binary exponent" of |x| can be extracted as asinhIEEE(|x|)/2^23/ln(2)-127, i.e., shift to the right 23 bits, multiply by 1/ln(2) and subtract 127. So it appears that asinh can smoothly represent the IEEE "gradual underflow"/denorms.

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RE : [math-fun] Re: Prime's position in S
by Eric Angelini 12 Aug '06

12 Aug '06

Hello Dean, a warm thanks for the computations and the Mathematica code. I will complete OEIS's ecord A121053 at the end of the month. Best and thank you again ! É. ________________________________ De: math-fun-bounces+eric.angelini=kntv.be(a)mailman.xmission.com de la part de Dean Hickerson Date: sam. 12/08/2006 0:01 À: seqfan(a)ext.jussieu.fr; math-fun(a)mailman.xmission.com Objet : [math-fun] Re: Prime's position in S Mostly to Eric Angelini: > Hello SeqFan and math-fun, > a somehow self-describing sequence here, with a question: > http://www.cetteadressecomportecinquantesignes.com/PrimePos.htm > Best, > Éric A. In the limit, exactly half of the terms are prime. Here's a formula, found empirically, for a(n) for n>=5: Let pi(n) be the number of primes <= n and p(n) be the n'th prime. Then: if n is prime or (n is composite and n+pi(n) is even) then a(n) = p(floor((n+pi(n))/2)); if n is composite and n+pi(n) is odd and n+1 is composite then a(n) = n+1; if n is composite and n+pi(n) is odd and n+1 is prime then a(n) = n+2. Also, for n>=5, n is in the sequence iff either n is prime or n+pi(n) is even. I'm sure this can all be proved by induction on n; I'll leave it to you to check all of the cases if you want to. It follows from this that, for n>=4, the number of primes among a(1), ..., a(n) is exactly floor((n+pi(n))/2. Since pi(n)/n -> 0 as n -> infinity, this is asymptotic to n/2. Here's some Mathematica code that can be used to check this. a[n] is computed from your definition; b[n] is computed from the formula above: $RecursionLimit=Infinity; a[1]=2; inseq[n_]:=False; (* inseq[n] will be True if n is in the sequence *) inseq[2]=True; a[n_]:=a[n]= Module[{m}, a[n-1]; For[m=1, inseq[m] || ((inseq[n] || m==n) && !PrimeQ[m]) || (m<n && !PrimeQ[a[m]]), m++, Null]; inseq[m] = True; Return[m] ]; b[n_]:= If[n<=4, {2,3,5,1}[[n]], If[PrimeQ[n] || EvenQ[n+PrimePi[n]], Prime[Floor[(n+PrimePi[n])/2]], If[PrimeQ[n+1], n+2, n+1] ] ]; a/@Range[1000] == b/@Range[1000] (* Should return True *) countprimes[n_]:=Length[Select[a/@Range[n],PrimeQ[#]&]]; cp[n_]:=Floor[(n+PrimePi[n])/2]; countprimes/@Range[4,1000] == cp/@Range[4,1000] (* Should return True *) Eric, feel free to use any or all of this in your sequence entry. Dean Hickerson dean(a)math.ucdavis.edu _______________________________________________ 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] Unabomber's Belongings To Be Auctioned
by Daniel Asimov 11 Aug '06

11 Aug '06

That book list is odd for its absence of math books -- he had resigned an asst. prof. position at Berkeley (apparently without explanation), which is probably correlated with this -- and the absence of any books on the chemistry of explosives (my guess is that some authority decided it would be unwise to spread those book around. Another possibility is that he already knew all he needed to know . . . he was known to enjoy setting off bombs as a high school student). * [begin: more than you wanted to know alert] * I had one real encounter, and one second-order one, with Kaczynski. As a grad student in Berkeley in '70, my then girlfriend and I hired him to take a series of photos of us on the Berkeley campus (as I later learned, this was not long after he resigned). Then about 15 years later, when I was teaching computer graphics in the Berkeley (EE)CS dept. at Berkeley, one of my very best students picked up a package he saw in a computer room and it exploded, severely injuring one of his hands and one of his eyes. The student was already an Air Force captain, and a few days later received an acceptance to enter the astronauts program -- both of which career paths were closed to him after the bomb. (But he's now a successful professor of electrical and computer engineering at U. Colorado, Boulder.) * [end: more than you wanted to know alert] * --Dan

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[math-fun] Re: Prime's position in S
by Dean Hickerson 11 Aug '06

11 Aug '06

Mostly to Eric Angelini: > Hello SeqFan and math-fun, > a somehow self-describing sequence here, with a question: > http://www.cetteadressecomportecinquantesignes.com/PrimePos.htm > Best, > �ric A. In the limit, exactly half of the terms are prime. Here's a formula, found empirically, for a(n) for n>=5: Let pi(n) be the number of primes <= n and p(n) be the n'th prime. Then: if n is prime or (n is composite and n+pi(n) is even) then a(n) = p(floor((n+pi(n))/2)); if n is composite and n+pi(n) is odd and n+1 is composite then a(n) = n+1; if n is composite and n+pi(n) is odd and n+1 is prime then a(n) = n+2. Also, for n>=5, n is in the sequence iff either n is prime or n+pi(n) is even. I'm sure this can all be proved by induction on n; I'll leave it to you to check all of the cases if you want to. It follows from this that, for n>=4, the number of primes among a(1), ..., a(n) is exactly floor((n+pi(n))/2. Since pi(n)/n -> 0 as n -> infinity, this is asymptotic to n/2. Here's some Mathematica code that can be used to check this. a[n] is computed from your definition; b[n] is computed from the formula above: $RecursionLimit=Infinity; a[1]=2; inseq[n_]:=False; (* inseq[n] will be True if n is in the sequence *) inseq[2]=True; a[n_]:=a[n]= Module[{m}, a[n-1]; For[m=1, inseq[m] || ((inseq[n] || m==n) && !PrimeQ[m]) || (m<n && !PrimeQ[a[m]]), m++, Null]; inseq[m] = True; Return[m] ]; b[n_]:= If[n<=4, {2,3,5,1}[[n]], If[PrimeQ[n] || EvenQ[n+PrimePi[n]], Prime[Floor[(n+PrimePi[n])/2]], If[PrimeQ[n+1], n+2, n+1] ] ]; a/@Range[1000] == b/@Range[1000] (* Should return True *) countprimes[n_]:=Length[Select[a/@Range[n],PrimeQ[#]&]]; cp[n_]:=Floor[(n+PrimePi[n])/2]; countprimes/@Range[4,1000] == cp/@Range[4,1000] (* Should return True *) Eric, feel free to use any or all of this in your sequence entry. Dean Hickerson dean(a)math.ucdavis.edu

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Re: [math-fun] Prime's position in S
by N. J. A. Sloane 10 Aug '06

10 Aug '06

Eric - nice sequence! I will add it to the OEIS right now - it will be A121053. Of course further terms, comments, etc., will be welcomed. Neil

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[math-fun] Prime's position in S
by Eric Angelini 10 Aug '06

10 Aug '06

Hello SeqFan and math-fun, a somehow self-describing sequence here, with a question: http://www.cetteadressecomportecinquantesignes.com/PrimePos.htm Best, Éric A.

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[math-fun] more binary asinh
by Henry Baker 09 Aug '06

09 Aug '06

I've done more analysis on "binary" asinh and its analogy with floating point -- especially the analogy with "gradual underflow"/"denormalized" arithmetic. as2(x):=asinh(x)/log(2) which is the inverse of s2(x):=(2^x-2^-x)/2. I looked at the 16 bits on either side of the binary point. For x>256, this function is log2(2x) for the 16 bits below the binary point -- i.e., as2(2^8) ~ 9, to 16 bits of precision. For x<2^-5, this function is x/log(2) ~ 1.442695x, to 16 bits of precision. So, asinh(x) is like Kahan's "gradual underflow" floating point numbers, wherein the exponent doesn't go below 0, and the mantissa starts shifting out to the right. Unfortunately, with as2(x) for small x, we don't get x itself, but 1.442695x, but we do get the correct binary exponent (plus 1). We can get x itself with the following function: asl2(x):=asinh(x*log(2))/log(2) so that for x<2^-5, we get asl2(x) ~ x. Unfortunately, with asl2(x), the exponent is now screwed up so that it is no longer an integer for x=2^k, k large. For x>2^9, we can recover the correct exponent by subtracting log2(2ln2)=log2(ln4)~0.4712336270551, so that asl2(2^k)-0.4712336270551 ~ k, for k>9, to at least 16 bits of precision. So, with these asinh functions, we can either get the exponent right, or the mantissa right, but not both at the same time. --- So, asinh(x) seems to combine both the exponent and the mantissa into the same number, but requires that both the exponent and mantissa be brought into "common terms" before being combined. Unfortunately, the "common term" requires either adding or multiplying by a transcendental number, which is a bit inconvenient. To compete with IEEE single precision, we would encode x as as2ieee(x):=asinh(x*2^127*log(2))/log(2) (the exponent of 2 might be off by +-1, but you get the idea.) Thus, as2ieee(x) wouldn't start "denormalizing" until approx the same precision that IEEE single float would. The major difference between asinh encoding and IEEE single float is that the majority of precision (for most numbers) is in the "exponent", not the "mantissa". In this way, asinh encoding is more like Wang's "LOCI" logarithmic encoding for his 1960's-era calculators.

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