Another Ramanujan Trick

[Gerson W. Barbosa]

(09-30-2019 10:24 PM)Valentin Albillo Wrote:  .
Hi again, Gerson:

(09-30-2019 03:10 AM)Gerson W. Barbosa Wrote:  My second approximation uses 10 digits and produces 17 significant digits, but it requires four functions, twice as much compared to the first approximation you mention.

Funny, it seems that my counting methods and yours do differ. In your second approximation I count 13 digits (4,2,1,4,3,6,6,1,6,-6,-6,2,2) and 9 operations (3 additions, 2 divisions, 2 roots and 2 powers).

I would consider one 4 and the two instances of -6 as functions (fourth root and the reciprocals of sixth roots). But I have yet another trick up my sleeve:

10 DESTROY ALL
20 OPTION BASE 1 @ DIM A(35) @ COMPLEX B(34)
30 A(1)=3 @ A(34)=-1.E+12 @ A(35)=-2.4E+17
40 MAT B=PROOT(A)
50 DISP REPT(B(2))

The four significant digits in line 30 of the HP-71B program above yield 10 correct digits of pi.

Best regards,

Gerson.