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Albert Chan · (This post was last modified: 11-07-2021 11:09 AM by Albert Chan.)

This pi formula is very similar to zeta(2) alternating sum formula

(11-05-2021 03:55 PM)Albert Chan Wrote:
(10-31-2021 03:40 PM)Albert Chan Wrote: Let s = (-1)^n, x=n*n+n+1, T's = triangular number

ζ(2) ≈ (2 - 2/2^2 + 2/3^2 - ... + s*2/n^2)

- s/(x - 1/(x+4*T1 - 2^4/(x+4*T2 - 3^4/(x+4*T3 - 4^4/(x+4*T4 - ...
...
We can spend more time summing the alternating series, less time doing CF corrections.
In the end, we may come out ahead.

As an experiment, we cut down CF terms in half. (p pairs alternating series, do p CF terms)
Bonus: summing alternating series pairwise also made final sum more accurate.

Apply the same idea here, we have this for estimating PI

Code:

OPTION ARITHMETIC DECIMAL_HIGH

10 INPUT  PROMPT "p (pairs) = ":p

   LET x = (8*p+4)*p + 1.5

   LET a = 0

   LET b = 0

   FOR i = p TO 2 STEP -1

      LET t = i*i

      LET a = a + 4/((i-0.75)*(16*t-1))

      LET b = t*(4*t-1) / (x+4*(t+i)-b)

   NEXT i

   LET r = 1/(x-3/(x+8-b)) + a + 46/15   

   PRINT "Accurate digits ="; 1-LOG10(MAX(ABS(PI-r),1E-1000))  

   GOTO 10

END

p (pairs) = 100
Accurate digits = 308.79106444764398
p (pairs) = 101
Accurate digits = 311.85336518362854
p (pairs) = 300
Accurate digits = 921.2386113713281
p (pairs) = 301
Accurate digits = 924.30082763127136

For p=326, it reached 1000 digits full precision.