On Lagrange algorithm for reduced algebraic irrationalities

In 2015, we proved that the calculation of another successive coefficient of the continued fraction expansion of a given algebraic number $\alpha$ requires the calculation of two values of the minimal polynomial corresponding to the residual fraction. In the talk, we prove the similar assertion where we replace the minimal plynomial of the residual fraction to the minimal polynomial of the initial number $\alpha$.
Since all the roots are different, we introduce the following notation
$$ \delta(\alpha)=\min_{2\le j\le n}\left|\alpha^{(1)}-\alpha^{(j)}\right|>0, $$
Theorem. Let $\alpha=\alpha_{0}$ be the real root of an irreducible integer polynomial and let $\alpha = \alpha^{(1)}$, $\alpha^{(2)}$,..., $\alpha^{(n)}$ be its rest roots. Next, suppose that $\alpha$ has the following expansion into continued fraction:
$$ \alpha\,=\,\alpha_{0}\,=\,q_{0}\,+\,\cfrac1{q_{1}\,+\,\cfrac{1}{\ddots+\cfrac{1}{q_{k}+\cfrac{1}{\ddots}}}}\,. $$
Define the index $m_{0}=m_{0}(\alpha,\varepsilon)$ by the inequalities
\begin{equation*} \frac{2(n-1)}{Q_{m_0-1}\delta(\alpha) }\,<\,\varepsilon. \end{equation*}
Then the relations: $q_{m}\,=\,q_{m}^{*}$, if
$$ (-1)^{m}f_{0}\left(\frac{q_{m}^{*}P_{m-1}+P_{m-2}}{q_{m}^{*}Q_{m-1}+Q_{m-2}}\right)\!>\!0\quad \text{and}\quad (-1)^{m}f_{0}\left(\frac{(q_{m}^{*}+1)P_{m-1}+P_{m-2}}{(q_{m}^{*}+1)Q_{m-1}+Q_{m-2}}\right)\!<\!0, $$
$q_{m}\,=\,q_{m}^{*}+1$, if
$$ (-1)^mf_{0}\left(\frac{(q_{m}^{*}+1)P_{m-1}+P_{m-2}}{(q_{m}^{*}+1)Q_{m-1}+Q_{m-2}}\right)>0, $$
$q_{m}\,=\,q_{m}^{*}-1$, if
$$ (-1)^mf_{0}\left(\frac{q_{m}^{*}P_{m-1}+P_{m-2}}{q_m^*Q_{m-1}+Q_{m-2}}\right)<0, $$
hold for any $m>m_{0}$; here
\begin{equation*} q_{m}^{*}\,=\,\left[\frac{f'_{0}\left(\frac{P_{m-1}}{Q_{m-1}}\right)}{Q_{m-1}^2 \left|f_{0}\left(\frac{P_{m-1}}{Q_{m-1}}\right)\right|}-\frac{Q_{m-2}}{Q_{m-1}}\right]. \end{equation*}