Creating digits of pi

[Gerson W. Barbosa]

(05-13-2018 05:09 PM)SlideRule Wrote:  I find … … a useful publication.

BEST!
SlideRule

I don't know it, but I am sure it is. Thanks for the book suggestion!

Regarding the latest kind of series I've been experimenting with, I've found out that Wolfram Alpha is a good verifying tool. Just submit them to W|A and it will tell what it evaluates to. For instance,

sum(k=1, ∞, (-1)^(k + 1)/((6k - 5)(6k - 3)(6k - 1))) = π/48

and

sum(k=1, ∞, 1/((8k - 7)(8k - 5)(8k - 3)(8k - 1))) = π(2 - √2)/192

I'll stick to the former because I intend to use it in the hp 50g with no external libraries. Also, it can be implemented with only one division and one multiplication per term.

Best regards,

Gerson.

--------------------------

>LIST
10 DESTROY ALL
12 INPUT K @ IF K=0 THEN END
15 S=0 @ Z=2*MOD(K,2)-1 @ W=Z
20 A=K*(K*(216*K-324)+138)-15
25 N=K*K @ M=2*K-1
30 FOR I=1 TO K
35 S=S+Z/A @ Z=-Z
40 N=N-M @ M=M-2
45 E=648*N+30 @ A=A-E
50 NEXT I
55 DISP 48*S;48*(S-W/(K*K*(432*K+600)))
60 GOTO 12
>RUN
? 1
 3.2  3.15348837209 
? 2
 3.13073593074  3.13893265205
? 3
 3.14521556875  3.14240262923
? 4
 3.13998510127  3.14127376106
? 5
 3.14243720856  3.14174155638
? 10
 3.14148304058  3.14158060156
? 100
 3.14159254249  3.14159265208
? 372
 3.14159265144  3.14159265359
? 1500
 3.14159265356  3.14159265359

I have yet to find the correcting continued fraction. The second value uses only the first correction term, which is not good enough for my purpose.