Equations (3(a)-(b)) and (4(a)-(b)) from "Numerical Experiments on Application of Richardson Extrapolation With Nonuniform Grids" (DOI) provide the following solution to a nonlinear equation:
My question: How do they get from (2) to (3)?
To provide a little more detail, for various reasons (that aren't really relevant to the question), I am using the following expressions for $a_{1}$, $a_{2}$, and $a_{3}$:
$$ \begin{matrix} a_{1}=1 \\ a_{2}=r_{2}^p \\ a_{3}=r_{3}^{2p} \end{matrix} $$
Additionally, instead of the dependent variable being $F$, I'm using $w$. When the dependent variable and the constant ($C$) are eliminated from equations (2(a)-(c)) the following nonlinear equation is derived that needs to be solved for $p$:
$$ \frac{w_{3}-w_{2}}{w_{2}-w_{1}}=r_{3}^p \frac{\left (\frac{r_{2}}{r_{3}} \right )^p-r_{3}^p}{1-r_{2}^p} \tag{2.1}\label{eq21} $$
Unless I've made a mistake somewhere, using the notation I have above, I believe equations (3(a)-(b)) and (4(a)-(b)) become the following:
$$ p=\left | \frac{\ln \left |\frac{w_{2}-w_{3}}{w_{1}-w_{2}} \right |}{\ln \left (r_{2} \right )}+q\left ( p \right ) \right | \tag{3.1a}\label{eq31a} $$ $$ q(p)=\frac{\ln\left (\frac{\left (\frac{r_{3}^2}{r_{2}} \right )^p-s}{r_{2}^p-s} \right )}{\ln \left ( r_{2} \right )} \tag{3.1b}\label{eq31b} $$ $$ s=1 \cdot \text{sgn} \left ( \frac{w_{3}-w_{2}}{w_{2}-w_{1}} \right ) \tag{3.1c}\label{eq31c} $$
In short, How do I get from \eqref{eq21} to \eqref{eq31a}, \eqref{eq31b}, and \eqref{eq31c}?
