This recursive sequence appears to converge to 3 both when a1 = 1 or when a1 = 15. When a1 = 15, the sequence is bounded above by 7.8 and bounded below by 3. The series converges quite quickly as the sequence only needs 7 iterations of an for python’s round off error to round an to 3.0 when a1 = 15.When a1 = 1, the sequence is bounded above by 5 and below by 3. The sequence is decreasing infinitely, but is never less than 3. When a1 = 1, the series converges quickly as well, but the round off error occurs on the 6th iteration rather than the 7th, meaning that the series converges quicker than when a1 = 15. It seems that the fact that a1 = 1 rather than a1 = 115 has very little significance to the series because they both converge to 3.
Below is an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =15.
Below is an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =1.
Below is a program that performs 16 iterations of an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =15. To perform iterations of an array of the iterations of a(n+1) =(a(n) +9/a(n))/2, a(1) =1, I simply changed an1 = 15 to an1 = 1 in line 14.
Below is a graph of the iterations of an when a1 = 1. The series converges to 3.
Below is a graph of the iterations of an when a1 = 15.
