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Re: [math-fun] Copyright extension
by Henry Baker 16 Jan '03

16 Jan '03

There appear to be at least two issues: * Whether the extension actually induces more creativity * Whether the extension exceeds "limited" It is clear that no amount of extending can coerce Walt Disney to create more cartoons -- he's dead, and so are all the other authors and artists whose work is at issue. So this doesn't pass the "smell test". It is also clear that _retroactively_ extending can't coerce people to create more, since they couldn't possibly have known about it at the time. So this doesn't pass the "smell test", either. There used to be a legal notion of "no perpetuities", which led to limitations such as 100 year bonds, etc. In England, I believe that leases run 99 years for the same reason. At 95 years, we're pretty close to bumping up against that limit, so unless people's average lifespans quickly get extended by modern medicine, it may be difficult for Congress to extend this again. Re people's lifetimes--we may already be at the point where an author may never "die". What if we found some living DNA for Walt Disney and cloned him? Perhaps we could coerce a bit more creativity out of him... It's already a cinch that we'll see Ted Williams out on the baseball fields again. At 10:06 AM 1/16/03 -0500, Bernie Cosell wrote: >On 16 Jan 2003 at 9:44, Michael Kleber wrote: > >> Bernie Cosell wrote: >> >> > Second, as a general rule, the Supreme Court very rarely intrudes on >> > Congress's >> > discretion to "interpret" things, and so it was always unlikely that >> > SCOTUS >> > would get involved in the hairsplitting of "100 yrs is OK, but 110 is >> > not, and >> > life +70 is too long, but maybe life +50 isn't" -- they would view >> > that, quite >> > properly, IMO, as a political question best resolved by Congress rather >> > than a >> > judicial matter cast in concrete by a court decision. >> >> But let's be mathematicians for a moment. Eventually something must >> deal with the fact that the constitutionsays copyright terms cannot be >> infinite but can, de facto, be unbounded. So while each extension of >> the copyright term is itself constitutional, congress could >> hypothetically >> extend it for 20 years every 20 years, and that has the same effect as >> a single clearly unconstitutional action. > >Why is it clearly unconstitutional? Being mathematicians again, since there is >a difference between "finite but unbounded" and "not finite" why couldn't the >former be constitutional even if the latter is not? > > /Bernie\ > >-- >Bernie Cosell Fantasy Farm Fibers >mailto:bernie@fantasyfarm.com Pearisburg, VA > --> Too many people, too few sheep <--

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Re: [math-fun] Copyright extension
by Henry Baker 16 Jan '03

16 Jan '03

It's very likely that everyone on the Court and everyone in the Senate who voted for this thing will not survive until 2020. It's called "passing the buck to the next generation". It seems to be done a lot -- e.g., the Social Security problem. At 09:44 AM 1/16/03 -0500, Michael Kleber wrote: >But let's be mathematicians for a moment. Eventually something must >deal with the fact that the constitutionsays copyright terms cannot be >infinite but can, de facto, be unbounded. So while each extension of >the copyright term is itself constitutional, congress could >hypothetically >extend it for 20 years every 20 years, and that has the same effect as >a single clearly unconstitutional action. > >How can this get resolved? Surely it is the domain of the supreme >court -- >or maybe of the direct limit of all supreme courts :-)...

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[math-fun] Copyright extension
by asimovd＠aol.com 16 Jan '03

16 Jan '03

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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Re: [math-fun] Copyright extension
by Henry Baker 16 Jan '03

16 Jan '03

It's not right-wing v. left-wing -- Ruth Bader Ginsburg wrote the majority decision, and she's not normally considered right-wing. I think that the Supremes were much more worried about things like the balance of trade -- Hollywood is one of the few US industries that has a positive trade balance with the rest of the world. As usual, the prospect of very real and imminent damage to a particular group v. very dispersed and not-well-understood gains for society at large, meant that they would try for damage control. Also, the Supremes have grandchildren & great grandchildren who want to opportunity to go to Disneyland -- where would Disneyland be without Mickey? ;-) At 03:21 PM 1/15/03 -0800, Steve Gray wrote: >----- Original Message ----- >From: <asimovd(a)aol.com> >To: <math-fun(a)mailman.xmission.com> >Sent: Wednesday, January 15, 2003 1:55 PM >Subject: [math-fun] Copyright extension > >> 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. > >GRAY: > And in about 2020, if the congress then is as disgustingly corrupt as >the present one, they will extend it to maybe 200 years. (Why fool around >with small increases?) And if the Supremes are as dominated by right-wing >ideologues as they are now, the extension will be blessed again.

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[math-fun] judit polgar
by Thane Plambeck 16 Jan '03

16 Jan '03

She took the lead today in the prestigious Corus Chess tournament at Wijk Aan Zee. She's by far the strongest woman chess player ever. This tournament has the current world champion(s) Kramnik and Ponomariov in it. She beat ex world Champion Karpov today, finding a novelty "over the board" as we chess junkies say http://www.corusgroup.com/coruschess/ Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/home.htm

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[math-fun] rectangle folding puzzle
by Scott Huddleston 15 Jan '03

15 Jan '03

Let ABCD be the vertices of a paper rectangle, in cyclic order, with |BC| >= |AB|. Fold the paper so vertex A lies on side BC. Let X and Y be the points where the fold intersects the boundary of ABCD. What is the minimum value of |XY|, and what fold(s) achieve it? Enjoy, - Scott

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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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