Another Ramanujan Trick

[Albert Chan]

(09-29-2019 01:28 PM)ttw Wrote:  Continued fractions (to anything) give a sequence of p(i))/q(i) of numerators and denominators.
The error in the ith term is less than 1 part in q(i)*(q(i)+q(i-1)). (Assuming I'm not off 1 in counting.)

The statement is true, but you can tighten the bound with error < \(\Large{1 \over q_{i}q_{i+1} } \)

(97; 2, 2, 3, 1) = 2143 / 22
(97; 2, 2, 3, 1, 16539) = 35444733 / 363875

Pi^4 must be between 2 convergents.

gap = 1 / (22 * 363875) = 1 / 8005250

To get first fraction for Pi^4 with estimate better than 2143/22
Convergent before 2143/22 is 1656/17, solve for semi-convergent with less |error|

XCas> fsolve(pi^4 - (1656+2143k)/(17+22k) = 2143/22 - pi^4, k) → k ≈ 8269.54

Semi-convergent error switched sign when k=8270. To confirm:

XCas> float(pi^4 .- [2143/22, 17724266/181957]) → (1.24912e-7, -1.24898e-7)