The Cosmic Game ( PDFDrive.com )

Cosmic Game © Douglass A. White, 2012 v151207 206
n: φ n ± φ -n Sum = Lucas Integer
0 1 + 1 002
1 1.618033989 - 0.618033989 001
2 2.618033989 + 0.381966011 003
3 4.2360679944 - 0.236067977 004
4 6.854101966 + 0.145898034 007
5 11.090169944 - 0.090169944 011
6 17.94427191 + 0.05572809 018
7 29.034441854 - 0.034441854 029
8 46.978713764 + 0.021286236 047
9 76.013155617 - 0.013155617 076
10 122.991869381 + 0.008130619 123
This is another way of generating the Lucas series that shows how it relates to the powers
of φ using good old Egyptian reciprocal math. If you run the ratios on the reciprocals
backwards, you also get φ. We already know how to calculate φ on a regular Senet
Board of integer unit squares. If we have another Senet Board laid out in φ units relative
to your integer unit board, we can work them together. For example, if we want to find
φ 4 , we know that φ 2 is φ + 1, and we can calculate φ on our unit Senet Board and add 1
unit to it to get φ 2 . Then we multiply this by 3 by simply marking off 3 string intervals
of φ 2 . Next we subtract a 1-unit interval, and lo and behold, we have an interval that is
φ 4 in length. An Egyptian could do this on his Senet Board with a piece of string as fast
as you can by pecking on your electronic calculator.
Now we will use φ to calculate an Egyptian approximation to 4 √10. The value of φ is
about 1.618. Recall that the altitude of the Benben pyramidion on our φ pyramid is √φ,
which is about 1.272. We need to make a line segment that is √30 units long. We can
do this, because we know how to get √10 = 3.162 and √3 = 1.732 on our Senet Board.
The product of these will be √30 = 5.477. The Egyptian scribe can find this several
ways. He can measure his two lengths and then calculate the product arithmetically, or
he can mechanically use % as his standard unit and measure off 1.732 units with his Oper
unit standard string or measuring rod that he has calibrated like a meter (cubit) stick.
Once he has √φ and √30, he draws a right triangle with these dimensions for its two legs.
The hypotenuse is √(30 + φ) = √31.618, which is very close to the square root of
10×3.162. He measures that to be about 5.623 and then divides that by % = 3.162. He
gets 5.623/3.162 = 1.7783, which is a pretty accurate approximation of 4 √10, as you will
see if you enter 10 on your calculator and punch the √ key twice. Our ancient scribe has
found the answer using nothing more than the Benben stone on the Great Pyramid, his
Senet Board, a piece of string, and some pen scratch calculations. Later we will show
you yet another very simple averaging method that an Egyptian scribe could use for
quickly finding square roots of any number, whole number or otherwise. For an ancient
scribe, using geometry might have been faster than calculating numerically on papyrus.