Evaluation of ζ(2) by the definition (sort of) [HP42S & HP71B]

10292021, 02:16 PM
Post: #12




RE: Evaluation of ζ(2) by the definition (sort of) [HP42S & HP71B]
(10262021 09:47 AM)Albert Chan Wrote: It would be better if we had more coefficients to confirm the pattern. We can also sum odd terms: sum(1/(2k1)^2, k=1..inf) = pi^2/8 = 3/4*ζ(2) ζ(2) = 4/3 * sum(1/(2k1)^2, k=1..inf) If we sum only n terms, correction term ≈ ∫(1/(2x1)^2, x=n+1/2 .. inf) = 1/(4n) Doing numerically for the continued fraction form, we get a very similar form as above. 1/(4*n + 1/(2*1.5*n + 16/(8*2.5*n + 81/(2*3.5*n + 256/(8*4.5*n + ... ))))) Example, estimate ζ(2) with n=3 lua> n = 3 lua> 1 + 1/3^2 + 1/5^2 1.1511111111111112 lua> 1/(4*n + 1/(2*1.5*n + 16/(8*2.5*n + 81/(2*3.5*n)))) 0.08258933029740148 lua> 4/3 * (1.1511111111111112 + 0.08258933029740148) 1.6449339218780168 lua> pi*pi/6 1.6449340668482264 

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