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

11032021, 11:28 PM
Post: #20




RE: Evaluation of ζ(2) by the definition (sort of) [HP42S & HP71B]
(11032021 12:38 AM)Albert Chan Wrote: Note: I did not add the b initial correction, for fair comparison. Fibonacci numbers pops out again, when I tried to estimate initial b XCas> X := n*n+n+1 XCas> T(k) := k*(k+1)/2 XCas> b2 := (n+2)^4/(X+4*T(n+2)) :; XCas> b1 := (n+1)^4/(X+4*T(n+1)  b2) :; XCas> expand(partfrac(b1)[1]) 3/8*n^2 + 23/32*n + 29/128 Note that 3/8 = F_{2} / F_{4} Rinse and repeat, add all the way to b10, we get this initial b: 2584/6765*n^2 + 159050/203401*n + 7976465719/20640116475 Again, 2584/6765 = F_{18} / F_{20} We might as well replace top coef with 1/ϕ^2 = 2ϕ The others are not as important. Patch with initial b: LET b = ((n+2.0472)*n+1.0118) * (3SQR(5))/2 n = 100 Accurate digits = 216.76041181744226 n = 101 Accurate digits = 218.85708591701648 n = 102 Accurate digits = 220.95367741927092 n = 400 Accurate digits = 844.54363777628652 n = 401 Accurate digits = 846.63481219743616 n = 402 Accurate digits = 848.72598292186942 For n=475, it reached 1000 digits full precision. 

« Next Oldest  Next Newest »

User(s) browsing this thread: 1 Guest(s)