CR - 5D

The 5-D Case

We obtain the 5-D convergent ratio vector by successive transformations of a base vector by the 5-D triangular matrix:

000011
000112
00111×3
011114
111115

This leads to the sequence

5-D Sequence
1 5 15 55 190 671 2353 8272 29056 s1
2 9 29 105 365 1287 4516 15873 55759 s2
3 12 41 146 511 1798 6314 22187 77946 … s3
4 14 50 175 616 2163 7601 26703 93819 s4
5 15 55 190 671 2353 8272 29056 102091 s5


The Matrix Aspect

While in the 3-D case the convergent ratio led to s/m, s/l, m/l and inverse relations, six core matrices and four equations, the 5-D case creates many more.  A core group of 20 5×5 matrices together with the 5×5 zero and identity matrices is required to build the matrix field for this convergent ratio.  They comprise the solutions of the 5-D equations.  Since negative solutions are included, there are 8 quintic equations to consider.

Here is a table of the core 5-D matrices:

Table of 5-D Core Matrices
s1→s5 (s5/s1)
00001
00011
00111
01111
11111
s1→s4 (s4/s1)
00010
00101
01011
10111
01111
s1→s3 (s3/s1)
00100
01010
10101
01011
00111
s1→s2 (s2/s1)
01000
10100
01010
00101
00011
s5→s1 (s1/s5)
000-11
00-110
0-1100
-11000
10000
s5→s2 (s2/s5)
00-110
0-11-11
-11-110
1-1100
01000
s5→s3 (s3/s5)
0-1100
-11-110
1-11-11
01-110
00100
s5→s4 (s4/s5)
-11000
1-1100
01-110
001-11
00010
s4→s3 (s3/s4)
00-101
0-1001
-10010
00100
11000
s3→s5 (s5/s3)
0-1010
-10001
00001
10010
01100
s2→s4 (s4/s2)
-10100
00010
10001
01001
00110
s3→s2 (s2/s3)
1001-1
011-10
01000
1-1010
-10001
s2→s5 (s5/s2)
10-101
00001
-10110
00110
11001
s4→s1 (s1/s4)
-1-1010
-1-1001
00-101
10000
0110-1
s3→s4 (s4/s3)
-110-11
1-1010
00010
-111-11
10010
s2→s3 (s3/s2)
-1011-1
00100
11-101
10001
-10110
s4→s5 (s5/s4)
0110-1
1110-1
11000
00001
-1-1011
s4→s2 (s2/s4)
-1-1001
-1-1-111
0-1001
01000
1110-1
s3→s1 (s1/s3)
-111-10
1001-1
10000
-110-11
0-1010
s2→s1 (s1/sub>/s2)
11-1-11
10000
-1011-1
-10100
10-101


Matrices by Size

Component Relations

There are many interesting patterns in these matrices, especially when compared to the 3-D and other matrix fields.  The triangular matrix transforms s1 to s5 and this being the greatest transformation, we could call it the biggest matrix.  The matrices in the first row can be consecutively subtracted from each other to produce the second row.  Then those in the second row can be consecutively added to produce the third row (except the last).  





 

The 5-D Equations

It is simple enough to arrange the solutions and negatives in size order but then selecting 8 groups of 5 is not as easy as for the 3-D case.

Plan B: To obtain the 5-D equations we make use of the relations above to express all terms as functions of a single one.  Here we will use s1/s5 because this method worked in the 3-D case.

(s1 + s2 + s3 + s4 + s5) / (s1 + s2 + s3 + s4 + s5) = 1

= s1 / (s1 + s2 + s3 + s4 + s5)

+ s2 / (s1 + s2 + s3 + s4 + s5)

+ s3 / (s1 + s2 + s3 + s4 + s5)

+ s4 / (s1 + s2 + s3 + s4 + s5)

+ s5 / (s1 + s2 + s3 + s4 + s5)

= s5 / (s1 + s2 + s3 + s4 + s5) × (s1/s5 + s2/s5 + s3/s5 + s4/s5 + s5/s5)

Now s5 / (s1 + s2 + s3 + s4 + s5) = s1 / s5

And s1 / (s1 + s2 + s3 + s4 + s5) = s1 / s5 × s5 / (s1 + s2 + s3 + s4 + s5) = (s1 / s5)2

Moreover s4 / s5 = (s2 + s3 + s4 + s5) / (s1 + s2 + s3 + s4 + s5)

= (s1 + s2 + s3 + s4 + s5) / (s1 + s2 + s3 + s4 + s5) - s1 / (s1 + s2 + s3 + s4 + s5)

= 1 - (s1 / s5)2

And s2 / s5 = s2 / s1 × s1 / s5

= (1 - (s1 / s5)2) + 1) × s1 / s5  (because s2/s1 = s4/s5 + 1)

= (2 - (s1 / s5)2) × s1 / s5

Finally s3 / s5 = s3 / s2 × s2 / s5

= 1 / (1 - (s1 / s5)) × s1 / s5 × (2 - (s1 / s5)2)

= (s1 / s5 / (1 - (s1 / s5)) × (2 - (s1 / s5)2)>

At this stage we substitute r for s1 / s5 to obtain:

r(r + r(2 - r2) + r(2-r2)/(1 - r) + (1 - r2) + 1) = 1

or r5 - r4 - 4 r3 + 3 r2 + 3r - 1 = 0



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