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


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



