approximating irrational roots of algebraic equations with the Pierce expansion

Question

Let $ p (x) = 1 - x \lfloor \frac{1}{x} \rfloor $ then the Pierce expansion of a real number $x \in {R}$ is expressed by \begin{equation} x_1 = \sum_{n = 1}^{\infty} (- 1)^{n + 1} \prod_{m = 1}^n q_m (x) = q_1 (x) (1 - q_2 (x) (1 - q_3 (x) (1 - q_4 (x) (1 - \ldots)))) \end{equation} where \begin{equation} q_n (x) = \frac{1}{p_n (x)} = \lfloor \frac{1}{p^{(n)} (x)} \rfloor \end{equation} and \begin{equation} p_n (x) = \frac{1}{\lfloor \frac{1}{p^{(n)} (x)} \rfloor} \end{equation} is the greatest positive integer such that \begin{equation} x_{n + 1} = 1 - x_n p_n > 0 \end{equation} For example, the iterates of $p (x)$ on $x = \gamma$ generates the sequence \begin{equation} \ \end{equation} so that \begin{equation} [q_0 (\gamma), \ldots, q_9 (\gamma)] = [1, 2, 6, 13, 21, 24, 225, 615, 17450, 23228] \end{equation} therefore \begin{equation} \begin{array}{ll} \gamma & = 1 - \frac{1}{2} + \frac{1}{2 \cdot 6} - \frac{1}{2 \cdot 6 \cdot 13} + \frac{1}{2 \cdot 6 \cdot 13 \cdot 21} + \ldots .\\ & = 1 - \frac{1}{2} + \frac{1}{12} - \frac{1}{156} + \frac{1}{3276} \ldots . \end{array} \end{equation} \begin{equation} \begin{array}{llllll} n & {Equation} p_{n, 3} x_n^3 + p_{n, 2} x_n^2 + p_{n, 1} x_n - p_{n, 0} = 0 & {Solution} & p^{(n - 1)} (x_1) & {Recurrence} & q_n\\ 1 & x_1^3 - 5 x_1 + 2 = 0 & x_1 = \sqrt{2} - 1 & p^{(0)} (x_1) & = \frac{1}{2} (1 - x_2) & 2 = \lfloor \frac{5}{2} \rfloor\\ 2 & - x_2^3 + 3 x_2^2 + 17 x_2 - 3 = 0 & x_2 = 3 - 2 \sqrt{2} & p^{(1)} (x_1) & = \frac{1}{5} (1 - x_3) & 5 = \lfloor \frac{17}{3} \rfloor\\ 3 & x_3^3 + 12 x_3^2 - 452 x_3 + 64 = 0 & x_3 = 10 \sqrt{2} - 14 & p^{(2)} (x_1) & _{} = \frac{1}{7} (1 - x_4) & 7 = \lfloor \frac{452}{64} \rfloor\\ 4 & - x_4^3 + 87 x_4^2 + 21977 x_4 - 111 = 0 & x_4 = 99 - 70 \sqrt{2} & p^{(3)} (x_1) & = \frac{1}{197} (1 - x_5) & 197 = \lfloor \frac{21977}{111} \rfloor \end{array} \end{equation} therefore the solution is approximated by \begin{equation} \begin{array}{ll} \tilde{x}_1 & = \frac{1}{2} - \frac{1}{2 \cdot 5} + \frac{1}{2 \cdot 5 \cdot 7} - \frac{1}{2 \cdot 5 \cdot 7 \cdot 197} + O \left( \frac{1}{2 \cdot 5 \cdot 7 \cdot 197} \right)\\ & = \frac{1}{2} - \frac{1}{10} + \frac{1}{70} - \frac{1}{13790} + O \left( \frac{1}{13790} \right)\\ & = \frac{408}{985} + O \left( \frac{1}{13790} \right)\\ & = 0.41421319796954314 \ldots + O (0.0000725 \ldots)\\ x_1 & = 0.41421356237309504 \ldots\\ \tilde{x}_1 - x_1 & = 0.0000000364403551 \ldots\\ \frac{1}{13790} & = 0.00007251631617 \ldots \end{array} \end{equation}

Answer 1

The sequence $q_n$ is equal to \begin{equation} q_n = \lfloor \frac{p_{n, 1}}{p_{n, 0}} \rfloor \end{equation} where $p_{n, m}$ is the coefficient of the $m$-th power of $x$ in the $n$-th transformed equation.