January 2003 - mathfun@mailman.xmission.com

Re: [math-fun] Copyright extension
by Henry Baker 15 Jan '03

15 Jan '03

I think that these works now all belong to Walt Disney. ;-) At 04:55 PM 1/15/03 EST, asimovd(a)aol.com wrote: >The Supreme Court just blessed the law of 1998 extending the 75 years of >copyright protection to 95 years. > >(See <http://www.nytimes.com/2003/01/15/business/15CND-COPY.html>.) > >This isn't great news for The People, who benefit by having works fall into >the public domain. > >Question: What happens to works that received their copyright between 1909 >and 1922 ? These fell into the public domain before the 1998 law, but were >copyrighted less than 95 years ago. Do they stay in the public domain or >revert to the copyright holders? > >--Dan

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[math-fun] OEIS: base 2 rev rev lex order fixed points
by Joerg Arndt 14 Jan '03

14 Jan '03

Didn't find them in the OEIS. Btw. where does the EOIS live by now ? (I recall it was asked not to submit new sequences at the old site). Reversed binary words of the words in reversed lex order ('.' for zero, asterisk at fixed points): 0: .... = 0 * 1: ...1 = 1 * 2: ..11 = 3 3: ..1. = 2 4: .1.1 = 5 5: .111 = 7 6: .11. = 6 * 7: .1.. = 4 8: 1..1 = 9 9: 1.11 = 11 10: 1.1. = 10 * 11: 11.1 = 13 12: 1111 = 15 13: 111. = 14 14: 11.. = 12 15: 1... = 8 16: 1...1 = 17 17: 1..11 = 19 18: 1..1. = 18 * 19: 1.1.1 = 21 Step through the words via: static inline ulong next_lexrev(ulong x) // Return next word in (reversed) lex order. { ulong x0 = x & -x; if ( 1==x0 ) { x ^= x0; x0 = x & -x; x0 ^= (x0>>1); } else x0 >>= 1; x ^= x0; return x; } static inline ulong prev_lexrev(ulong x) // Return previous word in (reversed) lex order. { ulong x0 = x & -x; if ( x & (x0<<1) ) x ^= x0; else { x0 ^= (x0<<1); x ^= x0; x |= 1; } return x; } fixed points are the seq: 0, 1, 6, 10, 18, 34, 60, 66, 92, 108, 116, 130, 156, 172, 180, 204, 212, 228, 258, 284, 300, 308, 332, 340, 356, 396, 404, 420, 452, 514, 540, 556, 564, 588, 596, 612, 652, 660, 676, 708, 780, 788, 804, 836, 900, 1026, 1052, 1068, 1076, 1100, 1108, 1124, 1164, 1172, 1188, 1220, 1292, 1300, 1316, 1348, 1412, 1548, 1556, 1572, 1604, 1668, 1796, 2040, 2050, 2076, 2092, 2100, 2124, 2132, 2148, 2188, 2196, 2212, 2244, 2316, 2324, 2340, 2372, 2436, 2572, 2580, 2596, 2628, 2692, 2820, 3064, 3084, 3092, 3108, 3140, 3204, 3332, 3576, 3588, 3832, 3960, 4024, 4056, 4072, 4098, 4124, 4140, 4148, 4172, 4180, 4196, 4236, 4244, 4260, 4292, 4364, 4372, 4388, 4420, 4484, 4620, 4628, 4644, 4676, 4740, 4868, 5112, 5132, 5140, 5156, 5188, 5252, 5380, 5624, 5636, 5880, 6008, 6072, 6104, 6120, 6156, 6164, 6180 ... These are the words where the bit count is equal zero and to 2**(i0) where i0 is the index of the lowest set bit: 0: ........... 1: ..........1 6: ........11. 10: .......1.1. 18: ......1..1. 34: .....1...1. 60: .....1111.. 66: ....1....1. 92: ....1.111.. 108: ....11.11.. 116: ....111.1.. 130: ...1.....1. 156: ...1..111.. 172: ...1.1.11.. 180: ...1.11.1.. 204: ...11..11.. 212: ...11.1.1.. 228: ...111..1.. 258: ..1......1. 284: ..1...111.. 300: ..1..1.11.. 308: ..1..11.1.. 332: ..1.1..11.. 340: ..1.1.1.1.. 356: ..1.11..1.. 396: ..11...11.. 404: ..11..1.1.. 420: ..11.1..1.. 452: ..111...1.. 514: .1.......1. 540: .1....111.. 556: .1...1.11.. [-snip-] 1556: ..11....1.1.. 1572: ..11...1..1.. 1604: ..11..1...1.. 1668: ..11.1....1.. 1796: ..111.....1.. 2040: ..11111111... 2050: .1.........1. 2076: .1......111.. 2092: .1.....1.11.. 2100: .1.....11.1.. 2124: .1....1..11.. 2132: .1....1.1.1.. 2148: .1....11..1.. [-snip-] 4644: .1..1...1..1.. 4676: .1..1..1...1.. 4740: .1..1.1....1.. 4868: .1..11.....1.. 5112: .1..1111111... 5132: .1.1......11.. 5140: .1.1.....1.1.. 5156: .1.1....1..1.. 5188: .1.1...1...1.. 5252: .1.1..1....1.. 5380: .1.1.1.....1.. 5624: .1.1.111111... 5636: .1.11......1.. 5880: .1.11.11111... 6008: .1.111.1111... 6072: .1.1111.111... 6104: .1.11111.11... 6120: .1.111111.1... 6156: .11.......11.. 6164: .11......1.1.. 6180: .11.....1..1.. ulong is_lexrev_fixed_point(ulong x) // Return whether x is a fixed point in the prev_lexrev() - sequence { if ( x & 1 ) { if ( 1==x ) return 1; else return 0; } else { ulong w = bit_count(x); if ( w != (w & -w) ) return 0; if ( 0==x ) return 1; return ( (x & -x) & w ); } } Hope C isn't that much hated here I am not allowed to post snippets like this. I wrote the equivalent routines for mixed radix lex order words, there seem to be no nontrivial fixed points for base =!=2. (Still have to check the code and this assertion). all the best, jj -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.

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[math-fun] Re: q-differential equations
by Mike Stay 13 Jan '03

13 Jan '03

Ah ha! The values of q that satisfy the definition of a q-derivative over GF(p^n) are q=x^(p^k - 1) unless q isn't a generator. -- Mike Stay staym(a)clear.net.nz

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[math-fun] q-differential equations over a finite field
by Mike Stay 13 Jan '03

13 Jan '03

I've been playing with q-derivatives over a finite field. It turns out that not all values of q give a derivative that's linear. Over GF(p^n), setting q=x^(p-1) works: Let x^a + x^b = x^c. Then we want D(x^a)+D(x^b) = D(x^a+x^b) = D(x^c). (qx)^a - x^a (qx)^b - x^b (qx)^a + (qx)^b - (x^a + x^b) ------------ + ------------ = ----------------------------- qx - x qx - x qx - x If q = x^(p-1) then we have (x^pa + x^pb) - (x^a + x^b) (x^a + x^b)^p - (x^a + x^b) = --------------------------- = --------------------------- (*) qx - x qx - x x^pc - x^c (qx)^c - x^c = ---------- = ------------ qx - x qx - x At (*) we can only do that because all the inner coefficients on the pth line of Pascal's triangle are divisible by p and therefore congruent to zero. Other values of q work, too, but I haven't looked at them enough to see if there's an intuitive reason for them yet. Once we have a working derivative, we can start doing differential equations. For example, what's the solution to f'(x) = f(x)? Normally it's f(x) = exp(x), but over GF(2^3) mod x^3-x-1 it's f(x) = x^5: Setting q=x, the q-derivative is (xx)^5 - x^5 x^10 - x^5 x^3 - x^5 011 - 111 ------------ = ---------- = --------- = --------- (xx) - x x^2 - x x^2 - x 110 100 x^2 = --- = --- = x^(-2) = x^5 110 x^4 With more variables in our polynomials we can start looking at physics over a finite field: we can solve the wave equation or Schroedinger's equation (pick units so hbar = c = m = 1): instead of complex phases, we get phases mod p. Do the squares still sum to 1? Can we do quantum computation this way? -- Mike Stay staym(a)clear.net.nz

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[math-fun] Re: galois fields, 2-var polynomials
by Mike Stay 12 Jan '03

12 Jan '03

I think what I described is a commutative ring. Are there irreducible polynomials in two variables that would give a representation of GF(p^nm)? Or are rings the only thing you can get out of a construct like that? -- Mike Stay staym(a)clear.net.nz

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[math-fun] galois fields, 2-var polynomials?
by Mike Stay 12 Jan '03

12 Jan '03

Galois fields talk about polynomials in one variable modulo some irreducible polynomial in that variable. When we go up to two variables, what kinds of structures emerge? Do they map onto larger fields somehow, or rings, or what? I'm looking at something like a polynomial that's nth order in x, mth order in y, with coefficients modulo a prime. If you hold one variable or the other constant, you can treat it like a galois field and reduce it modulo some irreducible polynomial in that variable. -- Mike Stay staym(a)clear.net.nz

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[math-fun] "On Quaterions and Octonions" finally available.
by asimovd＠aol.com 10 Jan '03

10 Jan '03

At long last, AK Peters has informed me that "On Quaternions and Octonions", by our own John Conway, and Derek Smith, has been published and will be available at Amazon next week. The blurb: << On Quaternions and Octonions John Conway/Derek Smith This book investigates the geometry of the quaternion and octonion algebras. Following a comprehensive historical introduction, the special properties of 3- and 4-dimensional Euclidean spaces are illuminated using quaternions, leading to enumerations of the corresponding finite groups of symmetries. The second half of the book discusses the less familiar octonion algebra, concentrating on its remarkable "triality symmetry" after an appropriate study of Moufang loops. The arithmetics of the quaternions and octonions are also described, and the book concludes with a new theory of octonion factorization. Topics covered include: history; the geometry of complex numbers; quaternions and 3-dimensional groups; quaternions and 4-dimensional groups; the Hurwitz integral quaternions; the composition algebras, Moufang loops; octonions and 8-dimensional geometry; integral octonions; and the octonion projective plane. Year: 2003 ISBN: 1-56881-134-9 Price: $29 160 pages. Hardcover. >> --Dan

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[math-fun] galois field in exponent
by Mike Stay 09 Jan '03

09 Jan '03

I thought this up today. Has anyone ever seen anything like this? It's reminiscent of eliptic curves, but it's not the same as far as I can tell. Take a_n, b_n as elements of a prime-order multiplicative subgroup of a field. Let g be the generator of the subgroup, and lg be discrete log base g. For example, nonzero elements of GF(2^3), g=x. Then we can form a kind of "super-polynomial" A(y) = (a_n)^(y^n) (a_(n-1))^(y^(n-1)) ... (a_0) which we can rewrite as A(y) = g^( (y^n) lg a_n + (y^(n-1)) lg a_(n-1) + ... + lg a_0 ) Since the logs are all mod 7, the exponent is a polynomial in GF(7) and can be given an irreducible modulus with degree d yeilding a representation of GF(7^d). Multiplication of elements in the field is done by exponentiation: A(y) (X) B(y) = A(y)^(lg B(y)) = B(y)^(lg A(y)) Since each of the a_n are polynomials in x over GF(2^3), A(y) is really A(x,y), so it's more fun than just GF(7^d). Since we can stack two of these, we can stack N of them; pick some prime-order subgroup of 7^d and let the log of those elements be the coefficients of a polynomial in z. What happens when you add these things? -- Mike Stay staym(a)datawest.net staym(a)clear.net.nz

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[math-fun] Analytic function with Fibonacci coefficients
by asimovd＠aol.com 09 Jan '03

09 Jan '03

Just for fun, find f(z) = sum from k = 0 to oo of (F_n z^n), and its radius of convergence where F_n is the nth Fibonacci number (F_0 = 0, F_1 = 1,...). --Dan

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[math-fun] 1.24 trillion digits of Pi
by Simon Plouffe 09 Jan '03

09 Jan '03

hello, there is more, Kanada used Machin formula and another one(?) to verify the computation apparently and made basic statistics about the distribution of digits (0 to 9), that's about it. see the article of Peterson in : http://www.sciencenews.org/20021214/mathtrek.asp see also : http://www.super-computing.org/ I find it surprising that he took that formula which is by far not the best. Simon Plouffe

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