Is there a simple fixed point iteration for the Delian constant (cube root of $2$) in base $10$ or $16$?

Question

Im looking for a simple iteration that computes the cube root of $2$ in base $10$(decimal) or base $16$(hexadecimal).

Probably the simplest way is iterating a small polynomial with simple rational coefficients of the form $\frac{n}{10}$ or $\frac{n}{16}$.

Newton and Householder methods do not provide that I think. And neither do continued fractions I assume. I considered median methods but without succes.

I know this case for the pythagoras constant $\sqrt 2$ in hexadecimal(base $16$):

$$\frac{1}{2}x(3 - \frac{1}{2} x^2)(1 + \frac{3}{32}(x^2 - 2)^2)(1 - \frac{1}{64}(x^2 - 2)^3)(1 + \frac{16+11}{2048}(x^2 - 2)^4)$$

is going to $\sqrt 2$ with speed $\frac{3 x^5}{4}$.

So Im motivated to find the apparantly harder cube root of $2$.

edit

I found a solution but it converges slowly.

Are there solutions that converge faster ??

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Answer 1

I found $2$ results.

$$z - \frac{(z^3 - 2)(z+1)}{10}$$

for decimal and

$$z - \frac{(z^3 - 2)(z+1)}{16}$$

for hexadecimal.

In fact they can be ajusted to the base you want to use.

For bases larger than $6$ :

$$z - \frac{(z^3 - 2)(z+1)}{a}$$

for base $a$.

They clearly fail for starting point at $-1$ but for real numbers between $\frac{-3}{4}$ and $2$ they usually work.

the julia set picture :