[math-fun] a strange class of algebraic numbers
Hello, I have been working on a simple idea that lead to something interesting. Some of you may know that the iteration Z(n+1)= Z(n)^2 + c for some simple c will lead to an algebraic number, in this case, many times to a 4'th degree algebraic number. This is quite puzzling for many reasons, first the convergence to a constant when it happens is geometric, but ??, since we know that the Mandelbrot set is self-similar , what if we take a series of very close rationals (the c's in the equation) , can it be that these numbers would also show some patterns in their binary ? or decimal expansion ? We can construct a series of rational values and from that get their algebraic equivalent when the formula Z(n+1) = Z(n)^2+c , does converge and then what..., The question is naive, there is no connection between these two things, but still I was very puzzled by the idea. So , as usual, since I can't completely understand this phenomena, at the least I made some tables of it, a lot of them, and then, I looked at these numbers, in binary in search of a pattern. By looking at those algebraic numbers I saw, or had the intuition that something is going on with some valuesand found something. Some algebraic numbers of degree 4 have a very definite and persistent pattern in their binary expansion, the pattern does not go to infinity BUT for some values, it can extend to the first 1000 billion binary digits. , much more than the ordinary approximation. These crazy numbers are chaotic in their continued fractiontoo, not onlythat, the log and the exp is also crazy. see for yourself : http://www.plouffe.fr/simon/On%20a%20strange%20class%20of%20algebraic%20numb... This pattern is astonishing. The one presented on page 1 of the article goes to at least the first 270 million bits, the numbers are f(n) = 1+1/4*(2*4^n+2*(16^n+1)^(1/2))^(1/2)/(2^n)^2, if n=8192, it gives a pattern that goes for 270 million bits, and for n=1048576, to 1000 billion digits. a very good question now : can this be generated with a cellular automata ? I submitted the article to the arxiv site. As far as I know, this is not known, I searched for some insights with the work of Douady and Hubbard, nothing found, interesting but not definite, maybe it needs a closer look, Best regards, Simon Plouffe
Very cool pix, but I don't understand French very well. At 06:42 PM 3/17/2014, Simon Plouffe wrote:
see for yourself :
http://www.plouffe.fr/simon/On%20a%20strange%20class%20of%20algebraic%20numb...
This pattern is astonishing.
Hello, The formula I have found is not a discovery, what is somewhat surprising is the order of approximation. Consider only the formula sqrt(sqrt(2^(2^n)+1)) is simple enough and not related to Mandelbrot, the surprising fact is that for small values of n, the pattern is extending for billions of binary digits from which a good half of them are 100 % predictable. Best regards, Simon Plouffe
Yes, I forgot to add : those numbers are easily approximable but : consider that there is a grey area where a number is quite recognisable and cannot be completely be expressible by rationals or irrationals, which is the case here. The approximation is quite good considering continued fractions but that does not explain the hundred of millions of bits that we can see instantly. respectfuly, Simon Plouffe 2014-03-18 15:58 GMT+01:00 Simon Plouffe <simon.plouffe@gmail.com>:
Hello,
The formula I have found is not a discovery, what is somewhat surprising is the order of approximation. Consider only the formula sqrt(sqrt(2^(2^n)+1)) is simple enough and not related to Mandelbrot, the surprising fact is that for small values of n, the pattern is extending for billions of binary digits from which a good half of them are 100 % predictable.
Best regards, Simon Plouffe
In the formulas for a strange algebraic number in the linked paper, such as X := 1 + sqrt(2*4^(2^13) + 2*sqrt((4^2)^2^13) + 1)) / 2^(14) , if we instead omit the “+1” in the inner square-root sign we would get: 1 + sqrt(2*2*4^(2^13)) / 2^(2^14) = 1 + 2 * 2^(2^13) / 2^(2^14) = 1 + 2^(1 - 2^13) which is very very very close to X. I would think this should go a long way towards explaining the base-2 bit pattern —Dan On Mar 17, 2014, at 6:42 PM, Simon Plouffe <simon.plouffe@gmail.com> wrote:
Hello,
I have been working on a simple idea that lead to something interesting. Some of you may know that the iteration Z(n+1)= Z(n)^2 + c for some simple c will lead to an algebraic number, in this case, many times to a 4'th degree algebraic number.
This is quite puzzling for many reasons, first the convergence to a constant when it happens is geometric, but ??, since we know that the Mandelbrot set is self-similar , what if we take a series of very close rationals (the c's in the equation) , can it be that these numbers would also show some patterns in their binary ? or decimal expansion ?
We can construct a series of rational values and from that get their algebraic equivalent when the formula Z(n+1) = Z(n)^2+c , does converge and then what...,
The question is naive, there is no connection between these two things, but still I was very puzzled by the idea.
So , as usual, since I can't completely understand this phenomena, at the least I made some tables of it, a lot of them, and then, I looked at these numbers, in binary in search of a pattern. By looking at those algebraic numbers I saw, or had the intuition that something is going on with some valuesand found something.
Some algebraic numbers of degree 4 have a very definite and persistent pattern in their binary expansion, the pattern does not go to infinity BUT for some values, it can extend to the first 1000 billion binary digits. , much more than the ordinary approximation. These crazy numbers are chaotic in their continued fractiontoo, not onlythat, the log and the exp is also crazy.
see for yourself :
http://www.plouffe.fr/simon/On%20a%20strange%20class%20of%20algebraic%20numb...
This pattern is astonishing. The one presented on page 1 of the article goes to at least the first 270 million bits,
the numbers are f(n) = 1+1/4*(2*4^n+2*(16^n+1)^(1/2))^(1/2)/(2^n)^2, if n=8192, it gives a pattern that goes for 270 million bits, and for n=1048576, to 1000 billion digits.
a very good question now : can this be generated with a cellular automata ?
I submitted the article to the arxiv site.
As far as I know, this is not known, I searched for some insights with the work of Douady and Hubbard, nothing found, interesting but not definite, maybe it needs a closer look,
Best regards,
Simon Plouffe
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... but that number 1 + 2^(1 - 2^13), in base 2 is just 1 position, no ? the simplest case like sqrt(sqrt(2^n+1)), has a pattern that I can't explain, well, so far, what rule is it ??? (in cellular automatas). here is a capture of the middle part 00000000000000000000011111111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000000000001101111111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000000001110011011111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000010010110001101111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000011011010010010010101111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000101010010100101111010000111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 01000100100010100000011100000101111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 10010011011111001100101110000110111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 11011110100010100101000111111111001011111111 not simple but : always the same. there should be a formula for this ! Best regards, Simon Plouffe
Simon: Try series( ( (1/x)^128+1)^(1/4),x=0,500); on Maple. Your binary expansion comes, for example, from the case x= 1/2. I think this goes a long way to explaining the pattern you are seeing. On 3/18/14, 11:34 AM, Simon Plouffe wrote:
... but that number 1 + 2^(1 - 2^13), in base 2 is just 1 position, no ?
the simplest case like sqrt(sqrt(2^n+1)), has a pattern that I can't explain, well, so far, what rule is it ??? (in cellular automatas).
here is a capture of the middle part
00000000000000000000011111111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000000000001101111111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000000001110011011111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000000010010110001101111111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000000011011010010010010101111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 00000101010010100101111010000111111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 01000100100010100000011100000101111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 10010011011111001100101110000110111111111111 11111111111111111111111111111111111111111111 00000000000000000000000000000000000000000000 11011110100010100101000111111111001011111111
not simple but : always the same. there should be a formula for this !
Best regards, Simon Plouffe
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participants (4)
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Dan Asimov -
Henry Baker -
Jeffrey Shallit -
Simon Plouffe