How many iterations are required to get the cubic root of 10 accurate to 10 decimal places using the following formula:
a(n+1) ={A/a^2(n) + 2a(n)} / 3, where a =first guess, A=10 in this case, n =number of iteration from: 0, 1, 2, 3.......etc. This formula is for manually calculating (x)^1/3, where x is any number.
Thanks for help.
That is a very interesting formula!!.
Guess #1 =2.1
[10/2.1^2 + 2*2.1] / 3 =2.1558578987150415721844293272865........1st. iteration.
[10/2.155...^2 + 2*2.155...] / 3 =2.1544356293689498231197972784037......2nd iteration.
[10/2.154...^2 + 2*1.154.....] / 3 =2.1544346900322932740260161820501.....3rd. iteration.
So, let us see: 2.1544346900322932740260161820501^3 =10.000000000005702919684002568331,
accurate to 11 decimal places in 3 iterations!!!!!!. Pretty good!. It seems to converge quadratically.
