I posted this question over on Math Stack Exchange (link), but have not received a response. I'm wondering if it's too complicated for that audience, so I'm posting it here in the hopes that someone here may be able to help me more.

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:

Relevant section of the paper

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}?


I just discovered this (link) paper which identifies the type of equation I'm looking at here as a "transcendental equation" (see equation 10).

  • $\begingroup$ In Equation 3b) there is one more ")" than "("... $\endgroup$
    – user35593
    Nov 12, 2020 at 18:26
  • $\begingroup$ Yes, I noticed that as well. I think it is a typo, but the error is repeated in Equation (4b). I believe the version reproduced in my question (i.e. equation (3.1b)) accurately represents what was intended. $\endgroup$
    – tlewis3348
    Nov 12, 2020 at 18:35
  • $\begingroup$ Furthermore equation 3a) is an equation for $n$ but it contains $n$ itself as an argument so the equations are not really solved $\endgroup$
    – user35593
    Nov 12, 2020 at 18:35
  • $\begingroup$ This is a solution to a nonlinear equation. An initial value of $f(n)=q(p)=0$ is used, the value of $n$ or $p$ is calculated, that value is used to update the value for $f(n)$ or $q(p)$, and the process is continued until the value of $n$ or $p$ remains relatively constant. My problem is I want to get a similar answer with a slightly different version of equation (2.1), and I don't know how to do that. $\endgroup$
    – tlewis3348
    Nov 12, 2020 at 18:43
  • $\begingroup$ In equation $(3.1a)$ what is $q(p)$ supposed to be? It appears out of nowhere. $\endgroup$
    – Somos
    Nov 13, 2020 at 1:16

1 Answer 1


Taking the ratio of 2b) and 2a) resp. 2c) and 2b) and taking the logarithm yields \begin{eqnarray} \ln\left(\frac{F-F_1}{F-F_2}\right)&=&n\cdot \ln\left(\frac{a_1}{a_2}\right)\\ \ln\left(\frac{F-F_2}{F-F_3}\right)&=&n\cdot \ln\left(\frac{a_2}{a_3}\right) \end{eqnarray} Taking again the ratio yields $$ \ln\left(\frac{F-F_1}{F-F_2}\right)ln\left(\frac{a_2}{a_3}\right)=\ln\left(\frac{F-F_2}{F-F_3}\right)\ln\left(\frac{a_1}{a_2}\right) $$ It follows that $$ (F-F_1)^{\ln(a_2)}(F-F_2)^{\ln(a_3)}(F-F_3)^{\ln(a_1)}=(F-F_1)^{\ln(a_3)}(F-F_2)^{\ln(a_1)}(F-F_3)^{\ln(a_2)}, $$ i.e. an equation with the only unknown $F$. If this is solved we can solve for $n$ using $$ n=\frac{\ln\left(\frac{F-F_1}{F-F_2}\right)}{\ln\left(\frac{a_1}{a_2}\right)}=\frac{\ln(F-F_1)-\ln(F-F_2)}{\ln(a_1)-\ln(a_2)}. $$ Then one can solve for $C$ using one of the equations 2a),2b) resp. 2c).

  • $\begingroup$ But the ultimate point of all this is to calculate what $F$ is knowing what $a_i$ and $F_i$ are. In other words, I can't calculate what $F$ is without first being able to calculate what $n$ is. Note that equation (3c) gives $F$ in terms of $n$. $\endgroup$
    – tlewis3348
    Nov 12, 2020 at 21:12
  • $\begingroup$ I got equation (2.1) by solving for $F$ in (2(a)-(c)), setting (2a) = (2b) and (2b) = (2c), solving for $C$ in both, and setting both expressions for $C$ equal to each other. This is fairly straightforward to me and there are several other places that give similar expressions (though they're somewhat different do to different definitions of $a_i$). The problem is, I can't get from equation (2.1) to equations (3.1). $\endgroup$
    – tlewis3348
    Nov 12, 2020 at 21:19

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