Evaluation of ζ(2) by the definition (sort of) [HP-42S & HP-71B]

[Albert Chan]

(11-03-2021 12:38 AM)Albert Chan Wrote:  Note: I did not add the b initial correction, for fair comparison.
...
For n=478, it almost reached 1000 digits full precision (1 ULP error, last digit = 8 instead of 9)

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 = F₂ / F₄

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 = F₁₈ / F₂₀
We might as well replace top coef with 1/ϕ^2 = 2-ϕ

The others are not as important. Patch with initial b:

LET b = ((n+2.0472)*n+1.0118) * (3-SQR(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.