Re:Going beyond the Golden Ratio
by extremelearning on Tuesday March 12, 2019 @10:48AM
(#19368647)
Yes.
Consider π. For any S>0, you can construct an infinite number of rational approximations that have a score of less than S.
But for any quadratic irrational (surd), as the depth of the corresponding continued fraction increases, the score will converge (in an alternating manner) to a critical score, S.
This means that for any score S < S, there is only a finite set of rational approximations that have a score of less than S.
For example, in figure 3, for S=0.4 < 1/√5 ≃ 0.447, there is only one fraction that gives a score of less than S=0.4.
Consider π. For any S>0, you can construct an infinite number of rational approximations that have a score of less than S.
But for any quadratic irrational (surd), as the depth of the corresponding continued fraction increases, the score will converge (in an alternating manner) to a critical score, S.
This means that for any score S < S, there is only a finite set of rational approximations that have a score of less than S.
For example, in figure 3, for S=0.4 < 1/√5 ≃ 0.447, there is only one fraction that gives a score of less than S=0.4.
Hope that helps!