MY COMMENTS ON THIS ARTICLE
by LUIGI RIVARA (posted 10/17/10)
I read with big interest this article of Dr. Robert J. Wisner (also the article on the same subject
̢̢̮ââ¬Å¡Ã¬Ãâ¦Ã¢â¬ÅA disquisition of the square root of three̢̢̮ââ¬Å¡Ã¬Ã¯Ã¿Ã½).
It is very rewarding for a lover of mathematics as I am to find in Convergence article always very clear and very appealing.
The only point that was not clear to me in this article was relevant to the ̢̢̮ââ¬Å¡Ã¬Ãâ¦Ã¢â¬Årating system̢̢̮ââ¬Å¡Ã¬Ã¯Ã¿Ã½.
I understand from a personal E-mail interchanged with Prof. Wisner that in Diophantine approximation the goal is to get a good approximation with a minimal denominator, but following this line for sqrt(3) we can have
these approximations:
5/3 with a rating of (-2,1) and a value 5/3=1,66666666666667
Or 19/11 with a rating of (-2,2) and a value 19/11 = 1,72727272727273
But the real value of sqrt(3) is 1,732051 therefore 5/3 is far from the value
Following the explanation
̢̢̮ââ¬Å¡Ã¬Ãâ¦Ã¢â¬Åin Diophantine approximation the goal is to get a good approximation with a minimal denominator̢̢̮ââ¬Å¡Ã¬Ã¯Ã¿Ã½
there is no need for proceed along the ̢̢̮ââ¬Å¡Ã¬Ãâ¦Ã¢â¬Ågreek ladder̢̢̮ââ¬Å¡Ã¬Ã¯Ã¿Ã½ because the approximation with a minimal denominator is always the first rung
I like to share also a my explanation of the formula reported for the ̢̢̮ââ¬Å¡Ã¬Ãâ¦Ã¢â¬Ågreek ladder̢̢̮ââ¬Å¡Ã¬Ã¯Ã¿Ã½, always said that is not known how was found
For me this was the way
If we say that b/a = sqrt(N) we have
b^2/a^2 = N if we add to both said b/a we have
b/a + b^2/a^2 = N + b/a
b/a ( 1+b/a) = N + b/a
b/a = [N+ b/a] / (1+b/a) and with some manipulation
b/a = (a *N +b) /(a+b)
from which bn = an-1 * N + bn-1
an = an-1 + bn-1
Best reagards
Luigi Rivara