Cube root [HP-35]

[Albert Chan]

We can use Pade approximation to speedup convergence.
Example, simple Pade[1,1] gives cubic convergence.

Pade[(1+z)^(1/n), {z,0,1,1}] = \(\large{2n + (n+1)z \over 2n + (n-1)z}\)

\(\large\sqrt[n]k = x \left(\sqrt[n] {1 + ({k \over x^n} - 1)}\right)
≈ x \left({k(n+1)\;+\; (n-1)x^n \over k(n-1)\;+\;(n+1)x^n } \right) \)

For cube roots, we have: \(\large\sqrt[3]k ≈ x \left({2k\;+\; x^3 \over k\;+\;2x^3 } \right) \)

Example, cube root of 0.1 (guess x=0.5):

0.5
0.46428 57142 85714 28571
0.46415 88833 67588 52147
0.46415 88833 61277 88924 10076 35091 94543
0.46415 88833 61277 88924 10076 35091 94465 76551 34912 50112 ... (98 correct digits)

Above formula is *exactly* the same as Halley's method, H_f(x), with f(x) = x^n - k
see Method of obtaining Third Order Iterative Formulas