Analysis of the convergence order of the Illinois method

Question

While studying the order of convergence of the Illinois method, reading the original article by Dowell and Jarrat there's a passage that I don't understand.

Let $f:\left[a,b\right]\subset\mathbb{R}\to\mathbb{R}$ be a continuous functions with the property that $f\left(a\right)f\left(b\right)<0$. Dowell and Jarrat define the standard iterative scheme of the regula falsi method to solve $f\left(x\right)=0$, where $x\subset\left[a,b\right]$ as

$$x_{0} =a\,,\quad x_{1} =b\,,\quad x_{n+1} =\frac{x_{n}f\left(x_{n-1}\right)-x_{n-1}f\left(x_{n}\right)}{f\left(x_{n-1}\right)-f\left(x_{n}\right)}\,, $$ and replacing at each iteration $x_{n}$ by $x_{n+1}$ , if $f\left(x_{n}\right)f\left(x_{n+1}\right)>0$, otherwise replacing $x_{n-1}$ by $x_{n+1}$. Already here there's an issue: for successive iterations, one of the endpoints might be fixed, that is only one of the endpoints of a sub-interval of $\left[a,b\right]$ might change. In fact it is known, that the regula falsi method is prone to be “stuck” on one side of the root and one of the endpoints is fixed throughout or for various iterations. So the notation in the iteration scheme is at the very least misleading.

This is puzzling me because the authors have analyzed the asymptotic convergence rate by deriving the error evolution equation for a simple root $$e_{n+1}\propto e_{n}e_{n-1}\,.$$ This equation is found by considering that for two consecutive iterations each endpoint will be changed once.

I though that maybe the authors only considered such case because they were interested in studying the convergence order of the Illinois method. Contrary to the original regula falsi, the Illinois method wont get “stuck” on one side of the root, nonetheless it is not uncommon to have problems where only one endpoint is changed for various consecutive iterations.

My question is then, shouldn't a proper analysis of the asymptotic convergence order also consider the case when only one endpoint is changed for consecutive iterations, or is there a reason why those cases are not important for the final conclusion?

Answer 1

The following will repeat the analysis of the cited article under some simplifications. I'll leave it standing even if your question itself is not properly answered. For that I can only say that at some point the iteration will be close enough to the root so that a linear fraction approximation, the [1,1] Padé approximation, is close enough so that the 3-periodic UUM cycle for the linear fraction is also followed by the original function. This means that the suspected behavior can only occur in the initial steps of the iteration for a limited amount of iteration steps.

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Indeed, the convergence rate is not the golden ratio $\frac{1+\sqrt5}2$, as in the secant method and as implied by the error formula for the inverse linear interpolation. In fact, one can experimentally and on some model functions read off that the convergence rate is $\sqrt[3]3$, a cubic error reduction over 3 steps or function evaluations. This is comparable to the Halley method, or the Newton method enhanced by one simplified Newton step (using the previous derivative), as used in the Jenkins-Traub method.

See Regula falsi method, Bracketing root-finding methods: my modified Illinois method, Accuracy of approximation using linear interpolation for some numerical experiments.

A sketch of a proof can be found in https://www.overleaf.com/project/5f5c38eff940990001fd258d. As a simplified example of the situation, take a function that is nearly linear, $f(x)\approx x$, with just some curvature $f''(0)=-2ϵ$, for instance $$ f(x)=\frac{x}{1+\epsilon x} \implies c=\frac{af(b)-bf(a)}{f(b)-f(a)}=-\epsilon ab. $$ The last formula for the secant root demonstrates why the example function makes the further computations simpler.

Start the method with a bracketing interval $x_0<0<x_1$ so that the first secant root $x_2=-\epsilon x_0x_1>0$ replaces the right bracketing point. This triggers the Illinois correction formula for the next point computed for the bracketing interval $[x_0,x_2]$, \begin{align} x_3&=\frac{x_2f(x_0)/2-x_0f(x_2)}{f(x_0)/2-f(x_2)}\\ &= \frac{x_2x_0(1+ϵx_2)-2x_0x_2(1+ϵx_0)}{x_0(1+ϵx_2)-2x_2(1+ϵx_0)}\\ &= -x_0x_2\frac{1+ϵ(x_2-2x_0)}{x_0-2x_2-ϵx_0x_2}\\ &=-x_2\frac{1-ϵx_0(2x_0+ϵx_1)}{1+ϵ2x_1(1+ϵx_0)} \end{align} so in first approximation, $x_3=-x_2<0$ replaces the left boundary, the new bracketing interval $[x_3,x_2]$ is nearly symmetric to the root. Next $x_4=-ϵx_2x_3\approx ϵx_2^2>0$, so the replacement to $[x_3,x_4]$ is non-stalling. Thereafter $x_5=-ϵx_3x_4\approx ϵ^2x_2^3>0$ gives stalling behavior with the new interval $[x_3,x_5]$, establishing the 3-cycle and the order 3 distance reduction over the cycle, $x_{2+3(k+1)}=ϵx_{2+3k}^3$.