Graphics by P. Bourke, relating to the work of Bagula and Adamson
http://ozviz.wasp.uwa.edu.au/~pbourke/fractals/pascaltriangle/
Monday, August 31, 2009
Sunday, August 30, 2009
A164909 generated from the Dragon Curve
The Dragon Curve is A014577: 1, 1, 0, 1, 1,....
Sequence A164909 begins 1, 2, 3, 2, 3, 4, 3, 2, where the dragon curve terms are considered
codes: A164909 begins with "1", then using (1, 1, 0, 1, 1,...) as coding: 1 = add 1, 0 = subtract 1;
we obtain:
1...2...3...2...3...4...
.....1...1...0...1....1...
where rows of A088696 rows tend to A164909.
Sequence A164909 begins 1, 2, 3, 2, 3, 4, 3, 2, where the dragon curve terms are considered
codes: A164909 begins with "1", then using (1, 1, 0, 1, 1,...) as coding: 1 = add 1, 0 = subtract 1;
we obtain:
1...2...3...2...3...4...
.....1...1...0...1....1...
where rows of A088696 rows tend to A164909.
Binomial transform, an introduction
By operations, we use Pascal's triangle as an infinite lower triangular matrix: A007318 at OEIS,
http://www.tinyurl.com/4zq4q:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1;
...
= P, then multiply P * [any sequence as a vector....]. Say the sequence = (1, 2, 4, 8,...)
Then A007318 * [1,2,4,8,....] = [1, 3, 9, 27,....]; i.e. powers of 3.
The "reason" behind this is is that taking 1,3,3,...items with distinct labels, we create a binomial frequency of items such that given 2 items, (say, 1 and 2); we have one of the 1 and one of the 2 (as in row 1 of Pascal's triangle = (1, 1). Next row = (1, 2, 1), so with 3 items: (1, 2, 4), we have one 1, two 2's, and one 4, = (1, 2, 1) dot (1, 2, 4) = (1 + 4 + 4) = 9, a power of 3.
To create strings with these binomial frequencies of terms, we access the "Lengths of Infinite Farey Tree Continued fractions", in A088696, as follows:
1
1, 2,
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...; where these terms are based on a Gray code reflection principle: To get the next row, we bring down current row as the left half. ; say (1, 2, 3, 2), Now the "reflected" version of this =
(2, 3, 2, 1); but we add "1" to each term since each successive row introduces a new term. Then we append to (1, 2, 3, 2). That is, append (3, 4, 3, 2) to the right of (1, 2, 3, 2); getting:
(1, 2, 3, 2, 3, 4, 3, 2); representing a binomial frequency of terms labeled (1, 2, 3,...) since referring to Pascal's triangle, row 3 applies to 2^3 = 8 terms with a frequency of 4 distinct terms (1, 2, 3, 4); thus one 1, three 2's, three 3's, and one 4.
That's the "meaning" of the Binomial transform since the sums of these terms (1, 2, 3, 2, 3, 4, 3, 2) = 20, where the Binomial transform of (1, 3, 3...) = (1, 3, 8, 20,...).
However, our terms are (1, 2, 4, 8,...), not (1, 2, 3,...) so we index the (1, 2, 4, 8,..) terms with (1, 2, 3, 4,....), replacing the latter with the corresponding terms in (1, 2, 4, 8,...) getting:
1
1, 2
1, 2, 4, 2
1, 2, 4, 2, 4, 8, 4, 2
...where in the latter row we have one 1, three 2's, three 4's, and one 8, a binomial frequency., with sums of the rows = (1, 3, 9, 27,...); in other words, the binomial transform of
(1, 2, 4, 8,...) = (1, 3, 9, 27,...).
http://www.tinyurl.com/4zq4q:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1;
...
= P, then multiply P * [any sequence as a vector....]. Say the sequence = (1, 2, 4, 8,...)
Then A007318 * [1,2,4,8,....] = [1, 3, 9, 27,....]; i.e. powers of 3.
The "reason" behind this is is that taking 1,3,3,...items with distinct labels, we create a binomial frequency of items such that given 2 items, (say, 1 and 2); we have one of the 1 and one of the 2 (as in row 1 of Pascal's triangle = (1, 1). Next row = (1, 2, 1), so with 3 items: (1, 2, 4), we have one 1, two 2's, and one 4, = (1, 2, 1) dot (1, 2, 4) = (1 + 4 + 4) = 9, a power of 3.
To create strings with these binomial frequencies of terms, we access the "Lengths of Infinite Farey Tree Continued fractions", in A088696, as follows:
1
1, 2,
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...; where these terms are based on a Gray code reflection principle: To get the next row, we bring down current row as the left half. ; say (1, 2, 3, 2), Now the "reflected" version of this =
(2, 3, 2, 1); but we add "1" to each term since each successive row introduces a new term. Then we append to (1, 2, 3, 2). That is, append (3, 4, 3, 2) to the right of (1, 2, 3, 2); getting:
(1, 2, 3, 2, 3, 4, 3, 2); representing a binomial frequency of terms labeled (1, 2, 3,...) since referring to Pascal's triangle, row 3 applies to 2^3 = 8 terms with a frequency of 4 distinct terms (1, 2, 3, 4); thus one 1, three 2's, three 3's, and one 4.
That's the "meaning" of the Binomial transform since the sums of these terms (1, 2, 3, 2, 3, 4, 3, 2) = 20, where the Binomial transform of (1, 3, 3...) = (1, 3, 8, 20,...).
However, our terms are (1, 2, 4, 8,...), not (1, 2, 3,...) so we index the (1, 2, 4, 8,..) terms with (1, 2, 3, 4,....), replacing the latter with the corresponding terms in (1, 2, 4, 8,...) getting:
1
1, 2
1, 2, 4, 2
1, 2, 4, 2, 4, 8, 4, 2
...where in the latter row we have one 1, three 2's, three 4's, and one 8, a binomial frequency., with sums of the rows = (1, 3, 9, 27,...); in other words, the binomial transform of
(1, 2, 4, 8,...) = (1, 3, 9, 27,...).
Natural numbers, Gray code string
Refer to "Lengths of Infinite Farey tree continued fractions" where we have
1;
1, 2;
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2;
...Then consider as an infinite string: (1, 1, 2, 1, 2, 3, 2,....) as shown in sequence A088696, at http://www.tinyurl.com/4zq4q
These numbers may be considered codes for generating the binomial transform of sequences.
Refer to Binomial transform, an introduction
1;
1, 2;
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2;
...Then consider as an infinite string: (1, 1, 2, 1, 2, 3, 2,....) as shown in sequence A088696, at http://www.tinyurl.com/4zq4q
These numbers may be considered codes for generating the binomial transform of sequences.
Refer to Binomial transform, an introduction
Lengths of Infinite Farey Tree continued fractions
One half of the Infinite Farey tree, refer to http://www.tinyurl.com/4zq4q, then enter sequence
A088696:
A088696:
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
with the number of terms in each continued fraction representation generating A088696:
1
1 2
1 2 3 2
1 2 3 2 3 4 3 2
where the latter as a sequence may be considered the infinite linear string of the natural numbers based on the Gray code reflection principle. Refer to the entry, "Natural numbers, Gray code string"
Reflected Gray Codes
Refer to the post, "Gray code conversion rules" The following illustrates the reflection principle.
Take base 4, Gray code, using bits 0, 1, 2, 3:
0...000
1...011
2...002
3...003
...
4...013
5...010
6...011
7...012
.
8...022
9...023
10..020
11..021
. ...
Where the reflection rules for Gray code base N uses N terms at a time, refer to rightmost column. Bits are 0,1,2,3. Then the next string of 4 terms repeats last term of previous string (a 3), and continues with the same cycle: 3 ->0 -> 1 -> 2.
However, the next column uses 4 terms with 0, the next 4 with 1, the next 4 with 2's and the next with 3's. Next column going to the left uses 16 0's, 16 1's, etc.
Now refer to the blog, "Lengths of Infinite Farey tree continued fractions".
Take base 4, Gray code, using bits 0, 1, 2, 3:
0...000
1...011
2...002
3...003
...
4...013
5...010
6...011
7...012
.
8...022
9...023
10..020
11..021
. ...
Where the reflection rules for Gray code base N uses N terms at a time, refer to rightmost column. Bits are 0,1,2,3. Then the next string of 4 terms repeats last term of previous string (a 3), and continues with the same cycle: 3 ->0 -> 1 -> 2.
However, the next column uses 4 terms with 0, the next 4 with 1, the next 4 with 2's and the next with 3's. Next column going to the left uses 16 0's, 16 1's, etc.
Now refer to the blog, "Lengths of Infinite Farey tree continued fractions".
Cyclotomic third root of Unity, Gray code map, A164516.
http://www.research.att.com/~njas/sequences/index.html?q=a164516&language=english&go=Search
at http://www.tinyurl.com/4zq4q
then enter sequence A164516
at http://www.tinyurl.com/4zq4q
then enter sequence A164516
Origins of Chaos, Part II
Researchers:
In Part I we traced the origins of chaos to a Cartan-like set of matrices with (1,1,1,....) in the super and sub diagonals
(-1,0,0,0,...) in the main diagonal and stated that the cycle lengths for N using x^2 - 2 matched Euler's results for base 2 notation, 1/N.
Example: N = 7, cycle length = 3 since 3 is the "least exponent" a such that 2^a == 1 mod 7.; so a = 3 since 8 == 1 mod 7.
Similarly, 1/7 in base 2 = .001001001,... 3 cycle.
Similarly, we take X^3 - 2x^2 - x + 1 roots and using f(x), we find the 3 cycle; where for n-th degree equation, this corresponds to (2n+1) = N, so N = 7. Or, we can say given 2*Cos 2 Pi/7 = 1.24,....this has the 3 cycle using x^2 - 2. Thus
1. x^2 - 2 has a seed of 2*Cos 2Pi/N. and is cyclically chaotic. Using trig rules, we can manipulate x^2 - 2 and derive the other formulas for various trig values.
2. 2x^2 - 1 using seed Cos 2Pi/N.
3. 4x*(1-x); logistic map; using seed Sin^2 2Pi/N.
4. 2x / (1 - x^2) using Tan 2 Pi/N.
5. x^2 / (2 - x^2) using seed Sec 2 Pi/ N.
6. X^2 (x^2 - 2) using Csc. 2 Pi/N. and
7. (x^2 - 1 )/ 2x using seed Cot 2 Pi/N.
All 7 formulas are chaotic with (for example, N=7, the 3 cycle).
We can alternatively use Newton's method x1 = xo - f(xo)/f'x(xo) on x^2 = -1; resulting in #7, Cot.
PART III: Derivation of the M-set.
Bring up B. Mandelbrot's website, subsection: http://www.ncane.com/jm8d "Combinatorics in the M-Set - Lavaurs Algorithm" and notice the chart, where the 7-ths and 15-th are connected by "3" and ("4") respectively. The "3" is precisely the "3" cyclic period mentioned before for 7. Similarly, there's a "4" for the 15-th's since 2^4 == 16 == 1 mod 15.
For 31 we would have a "5" since referring to Table 48 in Beiler's book, for base 2, "5" is the "least exponent" mod 31. So Mandelbrot states "Similarly, the period 5 arcs connecting 5/31 to 6/31 and 7/31 to 8/31 collapse to the roots of 5-cycle midgets in the same limb".
Last, we note that according to Lavaur's algorithm, the M-set is precisely isomorphic with the logistic map; not suprising since we have the logistic map as #3 above.
We have thus traced the origins of Chaos to a particular variant of Pascal's triangle with repeated columns. Or, we can begin with the Cartan-like set of matrices as mentioned before : (1,1,1,...) in the super and subdiagonals and (-1,0,0,0,...) in the main diagonal. Alternatively, we can begin with using Newton's method on x^2 = (-1), then using trig, come up with the other 6 formulas. Chaos, at it's most fundamental level, originates with 2nd-degree equations and the results of our work bring us back to Euler with his observations regarding the cycle lengths of 1/N in base 2 notation. (generally, any base).; but the M-set and logistic maps incorporate the base 2.
..
Windows Live™: Keep your life in sync. Check it out.
In Part I we traced the origins of chaos to a Cartan-like set of matrices with (1,1,1,....) in the super and sub diagonals
(-1,0,0,0,...) in the main diagonal and stated that the cycle lengths for N using x^2 - 2 matched Euler's results for base 2 notation, 1/N.
Example: N = 7, cycle length = 3 since 3 is the "least exponent" a such that 2^a == 1 mod 7.; so a = 3 since 8 == 1 mod 7.
Similarly, 1/7 in base 2 = .001001001,... 3 cycle.
Similarly, we take X^3 - 2x^2 - x + 1 roots and using f(x), we find the 3 cycle; where for n-th degree equation, this corresponds to (2n+1) = N, so N = 7. Or, we can say given 2*Cos 2 Pi/7 = 1.24,....this has the 3 cycle using x^2 - 2. Thus
1. x^2 - 2 has a seed of 2*Cos 2Pi/N. and is cyclically chaotic. Using trig rules, we can manipulate x^2 - 2 and derive the other formulas for various trig values.
2. 2x^2 - 1 using seed Cos 2Pi/N.
3. 4x*(1-x); logistic map; using seed Sin^2 2Pi/N.
4. 2x / (1 - x^2) using Tan 2 Pi/N.
5. x^2 / (2 - x^2) using seed Sec 2 Pi/ N.
6. X^2 (x^2 - 2) using Csc. 2 Pi/N. and
7. (x^2 - 1 )/ 2x using seed Cot 2 Pi/N.
All 7 formulas are chaotic with (for example, N=7, the 3 cycle).
We can alternatively use Newton's method x1 = xo - f(xo)/f'x(xo) on x^2 = -1; resulting in #7, Cot.
PART III: Derivation of the M-set.
Bring up B. Mandelbrot's website, subsection: http://www.ncane.com/jm8d "Combinatorics in the M-Set - Lavaurs Algorithm" and notice the chart, where the 7-ths and 15-th are connected by "3" and ("4") respectively. The "3" is precisely the "3" cyclic period mentioned before for 7. Similarly, there's a "4" for the 15-th's since 2^4 == 16 == 1 mod 15.
For 31 we would have a "5" since referring to Table 48 in Beiler's book, for base 2, "5" is the "least exponent" mod 31. So Mandelbrot states "Similarly, the period 5 arcs connecting 5/31 to 6/31 and 7/31 to 8/31 collapse to the roots of 5-cycle midgets in the same limb".
Last, we note that according to Lavaur's algorithm, the M-set is precisely isomorphic with the logistic map; not suprising since we have the logistic map as #3 above.
We have thus traced the origins of Chaos to a particular variant of Pascal's triangle with repeated columns. Or, we can begin with the Cartan-like set of matrices as mentioned before : (1,1,1,...) in the super and subdiagonals and (-1,0,0,0,...) in the main diagonal. Alternatively, we can begin with using Newton's method on x^2 = (-1), then using trig, come up with the other 6 formulas. Chaos, at it's most fundamental level, originates with 2nd-degree equations and the results of our work bring us back to Euler with his observations regarding the cycle lengths of 1/N in base 2 notation. (generally, any base).; but the M-set and logistic maps incorporate the base 2.
..
Windows Live™: Keep your life in sync. Check it out.
Friday, August 28, 2009
Origins of Chaos, part 1.
The most fundamental level of chaos may be found in the infinite set of Cartan-like matrices
with (1,1,1,...) in the super and subdiagonals and (-1,0,0,0,...) in the main diagonal. For example,
[-1, 1, 0;
1, 0, 1;
0, 1, 0]
.
with charpoly x^3 + x^2 - 2x - 1 = 0, has e-vals/roots relating to the Heptagon,
2 * 2Cos (k)*2Pi/7; k = 1, 2, 3.; with constants as f(x), x^2 - 2 cyclic with a 3 period..
The charpolys corresponding to the foregoing matrices = A065941, http://tinyurl.com/4zq4q;
as follows, with signs: (++--++...):
.
1;
1, 1;
1, 1, 1;
1, 1, 2, 1;
1, 1, 3, 2, 1;
1, 1, 4, 3, 3, 1;
1, 1, 5, 4, 6, 3, 1;
...
getting:
1;
x + 1;
x^2 + x - 1;
x^3 + x^2 - 2x - 1;
x^4 + x^3 - 3x^2 - 2x + 1;
x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1;
...with the following trigonometric identities:
(given x = 2 Cost 2A), then
.
sin A / sin A = 1;
sin 3A / sin A = (x + 1);
sin 5A / sin A = (x^2 + x - 1)
sin 7A / sin A = (x^3 +x^2 - 2x - 1);
...etc. above; continued.
with (1,1,1,...) in the super and subdiagonals and (-1,0,0,0,...) in the main diagonal. For example,
[-1, 1, 0;
1, 0, 1;
0, 1, 0]
.
with charpoly x^3 + x^2 - 2x - 1 = 0, has e-vals/roots relating to the Heptagon,
2 * 2Cos (k)*2Pi/7; k = 1, 2, 3.; with constants as f(x), x^2 - 2 cyclic with a 3 period..
The charpolys corresponding to the foregoing matrices = A065941, http://tinyurl.com/4zq4q;
as follows, with signs: (++--++...):
.
1;
1, 1;
1, 1, 1;
1, 1, 2, 1;
1, 1, 3, 2, 1;
1, 1, 4, 3, 3, 1;
1, 1, 5, 4, 6, 3, 1;
...
getting:
1;
x + 1;
x^2 + x - 1;
x^3 + x^2 - 2x - 1;
x^4 + x^3 - 3x^2 - 2x + 1;
x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1;
...with the following trigonometric identities:
(given x = 2 Cost 2A), then
.
sin A / sin A = 1;
sin 3A / sin A = (x + 1);
sin 5A / sin A = (x^2 + x - 1)
sin 7A / sin A = (x^3 +x^2 - 2x - 1);
...etc. above; continued.
Online calculators
1. WolframAlpha
2. http://wims.unice.fr/wims/en_tool~linear~matrix.html
or by subject, enter "matrix calculator" into the field
3. Richard Mathar's online calculator, at
http://www.ncane.com/7mh
2. http://wims.unice.fr/wims/en_tool~linear~matrix.html
or by subject, enter "matrix calculator" into the field
3. Richard Mathar's online calculator, at
http://www.ncane.com/7mh
OEIS, Online Encyclopedia of Integer Sequences
OEIS, Online Encyclopedia of Integer Sequences
at http://www.tinyurl.com/4zq4q
at http://www.tinyurl.com/4zq4q
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