(09-13-2014 11:11 AM)Don Shepherd Wrote: Yes, Thomas, that yields 3.141592654 after 33 iterations.
Don, thanks for taking the time to confirm this. What bothers me a little is that it needs 33 iterations since that's what I get when the result is rounded to 10 significant digits:
Code:
PreAns Ans
0: 0.000000000 2.000000000
1: 1.414213562 2.828427125
2: 1.847759065 3.061467459
3: 1.961570561 3.121445152
4: 1.990369453 3.136548491
5: 1.997590912 3.140331157
6: 1.999397637 3.141277251
7: 1.999849404 3.141513801
8: 1.999962351 3.141572940
9: 1.999990588 3.141587725
10: 1.999997647 3.141591422
11: 1.999999412 3.141592346
12: 1.999999853 3.141592577
13: 1.999999963 3.141592634
14: 1.999999991 3.141592649
15: 1.999999998 3.141592652
16: 1.999999999 3.141592653
17: 2.000000000 3.141592654
18: 2.000000000 3.141592654
19: 2.000000000 3.141592654
20: 2.000000000 3.141592654
21: 2.000000000 3.141592654
22: 2.000000000 3.141592654
23: 2.000000000 3.141592654
24: 2.000000000 3.141592654
25: 2.000000000 3.141592654
26: 2.000000000 3.141592654
27: 2.000000000 3.141592654
28: 2.000000000 3.141592654
29: 2.000000000 3.141592654
30: 2.000000000 3.141592654
31: 2.000000000 3.141592654
32: 2.000000000 3.141592654
33: 2.000000000 3.141592654
Where does it start to deviate?
Cheers
Thomas