Cube root [HP-35]

[Albert Chan]

(03-16-2020 10:54 PM)Albert Chan Wrote:  One issue is optimal range of k, to maximize convergence rate.
In other words, with guess = k^y, y≠1/3, should we convert \(\sqrt[3]{27} = 10 \sqrt[3]{0.027}\) ?

At the break-even point, errors should match, but with opposite sign:

\(\sqrt[3]k - k^y = 10(k/1000)^y - \sqrt[3]k\)
\(2 \sqrt[3]k = (1 + 10/1000^y) k^y \)
At break-even, \(\large k = (1/2 + 5/1000^y)^{1 \over 1/3\;-\;y}\)

I made an error assuming better guess speedup Pade convergence.
Plotting the errors suggest my pade setup prefer over-estimated guess.

For guess x₀ = √√k = k^(1/4), optimal k is before break-even point, at k ≈ √1000
Thus, optimal k should be in range [0.031623, 31.623)

\(\large\sqrt[3]k ≈ x \left({2k\;+\; x^3 \over k\;+\;2x^3 } \right) \)

Example, try \(\quad\sqrt[3]{27} =\; 3\)
x₀ = √√(27)   → x₃ = 2.9999 99999 99999 99934     // error =  +66 ULP
x₀ = 10√√.027 → x₃ = 3.0000 00000 00000 00709     // error = -709 ULP

Example, try \(\quad\sqrt[3]{32}\) = 3.1748 02103 93639 89495 ...
x₀ = √√(32)   → x₃ = 3.1748 02103 93639 89237     // error = +258 ULP
x₀ = 10√√.032 → x₃ = 3.1748 02103 93639 89710     // error = -215 ULP

Edit: rough estimate, with k in range [0.031623, 31.623), x₀ = √√k :

\(\large\sqrt[3]k ≈ x_1 = {½\;+\; x_0 \over ½\;+\; 1/x_0}\)     // relative error < 1.5%