Math Forum Discussions

Domingo Gomez Morin

Posts: 215
Registered: 12/3/04

Duplication of the Cube.
Posted: Oct 7, 1994 10:27 PM

Plain Text

Continued Fraction of the Third Degree
Cube Root of 2

___
3 / 1
\/ 2 = 1 + --------------------------------------------------

1
3 + ------------------------
1
3 + ---
3
3 + -----------
3 + 1
3 + -----------------------------------
1
3 + -----------
3 + 1
3 + -------------------------
1
3 + ---
3
3 + ------------
3 + 1

###########################################################
Has anyone ever seen any reference about Continued Fractions of the
Third Degree?.
###########################################################

This kind of 'FRACTAL' or 'CONTINUED FRACTAL' comes from an extension
of the so simple 'Rational Process' I showed in my previous messages for
solving the duplication of the cube (Journal of Transf. Math., 1996).
-----------------------------------------------------------
Notice the relation between the fractions (see my previous messages on
'the duplication of the cube '):
4 5 6
- - -
3 4 5

9 11 14 (The armonies of Number!!!)
- -- --
7 9 11

20 25 32
-- -- --
16 20 25
. . .
. . .
. . .

and the two mean proportionals (a / x , x / y , y/2a) between two
given lines, showed by Hippocrates of Chios as solution for the
duplication of the cube (David Eugene Smith, History of Mathematics,
Vol. II, page 313).
------------------------------------------------------------

############################################################
Yes!, this 'CONTINUED FRACTAL' comes from an extremely simple
algorithm which could have been easily implemented since ancient times.
Believe it or Not!!!
oops...
############################################################

I hope someone could take the time to continue with the above fraction.
In fact, you can try to generate a similar continued fraction of the
fourth, fifth degree, etc... And... why not? Have some fun!, try to
compute the convergents of such 'CONTINUED FRACTALS'!!!.

Best regards.
Domingo