Golden Number Series: Fibonacci, etc.

The Fibonacci series can be generated using a self-looping formula:

An+2 = An+1 + An

Beginning with the initial values for An+1 and An, of:
A0 = 0 and A1 = 1

So,
00 + 11 = 12 ® 00, 11, 12
11 + 12 = 23 ® 00, 11, 12, 23
12 + 23 = 34 ® 00, 11, 12, 23, 34
23 + 34 = 55 ® 00, 11, 12, 23, 34, 55
34 + 55 = 86 ® 00, 11, 12, 23, 34, 55, 86
55 + 86 = 137 ® 00, 11, 12, 23, 34, 55, 86, 137

....................

Quadratic f » An+1 ÷ An

So,
Quadratic f1 = 0.61803398874989.... » 0.6153846.... = 86 ÷ 137

and its multiplicative reciprocal (which is also f2),
Quadratic f-1 » An ÷ An-1

is,
Quadratic f2 = 1.61803398874989.... » 1.625 = 137 ÷ 86

f1 and f2 satisfy the quadratic equation:
0 = x2 ± x – 1, if one of the f's are negated.


 
Let's do things a little differently.
Let's make this self-looping routine two dimensional:

An+1 = An + Bn
Bn+1 = An

Beginning with the initial values for An and Bn, of:
A0 = 1, B0 = 0

So,
A1 = 1 = 10 + 00
B1 = 1 = 10

n ® 0, 1
A ® 1, 1
B ® 0, 1


A2 = 2 = 11 + 11
B2 = 1 = 11

n ® 0, 1, 2
A ® 1, 1, 2
B ® 0, 1, 1


A3 = 3 = 22 + 12
B3 = 2 = 22

n ® 0, 1, 2, 3
A ® 1, 1, 2, 3
B ® 0, 1, 1, 2


A4 = 5 = 33 + 23
B4 = 3 = 33

n ® 0, 1, 2, 3, 4
A ® 1, 1, 2, 3, 5
B ® 0, 1, 1, 2, 3


A5 = 8 = 54 + 34
B5 = 5 = 54

n ® 0, 1, 2, 3, 4, 5
A ® 1, 1, 2, 3, 5, 8
B ® 0, 1, 1, 2, 3, 5


A6 = 13 = 85 + 55
B6 =  8 = 85

n ® 0, 1, 2, 3, 4, 5,  6
A ® 1, 1, 2, 3, 5, 8, 13
B ® 0, 1, 1, 2, 3, 5,  8


A7 = 21 = 136 + 86
B7 = 13 = 136

n ® 0, 1, 2, 3, 4, 5,  6,  7
A ® 1, 1, 2, 3, 5, 8, 13, 21
B ® 0, 1, 1, 2, 3, 5,  8, 13

....................



The Regular, Five-Sided Pentagon Contains
The Roots of the Golden, Quadratic Polynomial
Within the Ratios Between its Diagonals and Side

Quadratic f1 » Bn ÷ An
Quadratic f2 » An ÷ Bn

So,
Quadratic f1 = 0.61803398874989.... » 0.6190476... = 137 ÷ 217
Quadratic f2 = 1.61803398874989.... » 1.6153846... = 217 ÷ 137



So far, we've been dealing with Quadratic f.
Now let's deal with the Cubic. We are leaving the domain
of traditional mathematics at this point. There is no
known f satisfying any Cubic polynomial (but of course,
I live to refute that!):

A self-looping routine that is two dimensional:

An+1 = An + Bn + Cn
Bn+1 = An + Bn
Cn+1 = An

Beginning with the initial values for An, Bn, and Cn, of:
A0 = 1, B0 = 0, C0 = 0

So,
A1 = 1 = 10 + 00 + 00
B1 = 1 = 10 + 00
C1 = 1 = 10

n ® 0, 1
A ® 1, 1
B ® 0, 1
C ® 0, 1


A2 = 3 = 11 + 11 + 11
B2 = 2 = 11 + 11
C2 = 1 = 11

n ® 0, 1, 2
A ® 1, 1, 3
B ® 0, 1, 2
C ® 0, 1, 1


A3 = 6 = 32 + 22 + 12
B3 = 5 = 32 + 22
C3 = 3 = 32

n ® 0, 1, 2, 3
A ® 1, 1, 3, 6
B ® 0, 1, 2, 5
C ® 0, 1, 1, 3


A4 = 14 = 63 + 53 + 33
B4 = 11 = 63 + 53
C4 =  6 = 63

n ® 0, 1, 2, 3,  4
A ® 1, 1, 3, 6, 14
B ® 0, 1, 2, 5, 11
C ® 0, 1, 1, 3,  6


A5 = 31 = 144 + 114 + 64
B5 = 25 = 144 + 114
C5 = 14 = 144

n ® 0, 1, 2, 3,  4,  5
A ® 1, 1, 3, 6, 14, 31
B ® 0, 1, 2, 5, 11, 25
C ® 0, 1, 1, 3,  6, 14


A6 = 70 = 315 + 255 + 145
B6 = 56 = 315 + 255
C6 = 31 = 315

n ® 0, 1, 2, 3,  4,  5,  6
A ® 1, 1, 3, 6, 14, 31, 70
B ® 0, 1, 2, 5, 11, 25, 56
C ® 0, 1, 1, 3,  6, 14, 31

 
A7 = 157 = 706 + 566 + 316
B7 = 126 = 706 + 566
C7 =  70 = 706

n ® 0, 1, 2, 3,  4,  5,  6,   7
A ® 1, 1, 3, 6, 14, 31, 70, 157
B ® 0, 1, 2, 5, 11, 25, 56, 126
C ® 0, 1, 1, 3,  6, 14, 31,  70

....................




The Regular, Seven-Sided Heptagon Contains
The Roots of the Golden, Cubic Polynomial
Within the Ratios Between its Diagonals and Side

Cubic f1 » Cn ÷ An
Cubic f2 » An ÷ Bn
Cubic f3 » Bn ÷ Cn

So,
Cubic f1 = 0.445041868.... » 0.4458598... =  707 ÷ 1577
Cubic f2 = 1.246979604.... » 1.2460317... = 1577 ÷ 1267
Cubic f3 = 1.801937736.... » 1.8          = 1267 ÷  707

f1, f2, and f3 satisfy the cubic equation:
0 = x3  x2 – 2x ± 1, if every other f is negated,
odd or even numbered, doesn't matter.

So,
f1, –f2, f3

or,
–f1, f2, –f3

will satisfy the golden cubic.



The pairings of A, B, C, D, etc.... that generate approximations
of f may not be so obvious at this point, so I will go over the
next one --- the quartic --- in the infinitely, golden series:

n ® 0, 1, 2,  3,  4,  5,   6,   7, ...
A ® 1, 1, 4, 10, 30, 85, 246, 707, ...
B ® 0, 1, 3,  9, 26, 75, 216, 622, ...
C ® 0, 1, 2,  7, 19, 56, 160, 462, ...
D ® 0, 1, 1,  4, 10, 30,  85, 246, ...




The Regular, Nine-Sided Nonagon (Enneagon) Contains
The Roots of the Golden, Quartic Polynomial
Within the Ratios Between its Diagonals and Side

The 'B' triangle in the diagram above is equilateral:
the lengths of its three sides are equal to one another
and all of its internal angles are equal to 60°. So,
the pairings of line segments must be able to pair two
'B' segments together.

Quartic f1 » Dn ÷ An
Quartic f2 » Bn ÷ Bn
Quartic f3 » An ÷ Cn
Quartic f4 » Cn ÷ Dn

So,
Quartic f1 = 0.347296355.... » 0.3479490... = 2467 ÷ 7077
Quartic f2 = 1               » 1            = 6227 ÷ 6227
Quartic f3 = 1.532088886.... » 1.5303030... = 7077 ÷ 4627
Quartic f4 = 1.879385242.... » 1.8780487... = 4627 ÷ 2467

You'll notice by the nature of the pairings and the
outcome of their ratios that f1 appears to be approaching
zero as its limit with each higher degree polynomial
depicted, while fn (the largest 'n' of each set of f's)
seems to approach the integer 2 as its limit. You may also
notice that only odd-sided polygons contain pairings of
diagonals with each other and with the outer edges that
satisfy these number series constructions which are the
same constructions depicted in HE Huntley's "Reflecting
Beams of Light". Only prime-sided polygons will be
completely unique. All others will contain some unique
roots of a polynomial (whose coefficients are all integers)
plus some roots of a former prime- or composite- sided
polygon.

±f1, f2, ±f3, and f4 will satisfy the quartic equation:
0 = x4 ± x3 – 3x2  2x + 1



The above diagram depicts a nine-sided nonagon (or,
enneagon) which contains three unique roots for the
quartically, golden polynomial listed above, plus one more
root (the integer: 1) carried over from the equilateral
triangle (since a nine-sided polygon is divisible by three,
it must contain a three-sided polygon). The equilateral
triangle, although not covered here (for its simplicity),
contains only one root and that root of unity (1) does not
emanate from any sequence of numbers. Unity is the only
number in its series. But it does result from dividing the
length of the side of an equilateral triangle with the
length of its base:




The Regular, Three-Sided Triangle Contains
The Root of the Golden, Linear Polynomial
0 = x  1 when (f,x) = ± 1
Within the Ratio Between its Base and Side


Lastly, and fifth, in this series of examples is the
pairings of A, B, C, D, and E that generate approximations
of f for the fifth-degree (quintic) polynomial of the
infinitely, golden series:

n ® 0, 1, 2,  3,  4,   5,   6,    7, ...
A ® 1, 1, 5, 15, 55, 190, 671, 2353, ...
B ® 0, 1, 4, 14, 50, 175, 616, 2163, ...
C ® 0, 1, 3, 12, 41, 146, 511, 1798, ...
D ® 0, 1, 2,  9, 29, 105, 365, 1287, ...
E ® 0, 1, 1,  5, 15,  55, 190,  671, ...




The Regular, Eleven-Sided Undecagon (Hendecagon) Contains
The Roots of the Golden, Quintic Polynomial
Within the Ratios Between its Diagonals and Side

Quintic f1 » En ÷ An
Quintic f2 » Cn ÷ Bn
Quintic f3 » An ÷ Cn
Quintic f4 » Bn ÷ Dn
Quintic f5 » Dn ÷ En

So,
Quintic f1 = 0.284629677.... » 0.2851678... =  6717 ÷ 23537
Quintic f2 = 0.830830026.... » 0.8312528... = 17987 ÷ 21637
Quintic f3 = 1.309721468.... » 1.3086763... = 23537 ÷ 17987
Quintic f4 = 1.682507066.... » 1.6924882... = 21637 ÷ 12787
Quintic f5 = 1.918985947.... » 1.9180327... = 12877 ÷  6717

±f1, f2, ±f3, f4, and ±f5 will satisfy the quintic equation:
0 = x5  x4 – 4x3 ± 3x2 + 3x  1

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