How to show that regula falsi has linear rate of convergence?

Question

How can we prove that regula falsi method has linear rate of convergence?

I know how to do so for the secant method but I am unable to derive it for regula falsi.

Any help is much appreciated. Thank you.

Answer 1

Assume first that the function has a root in $x=0$ and looks locally around $x=0$ like $f(x)=cx(x+d)$ with $c,d>0$ and the bracketing interval $[a,b]$ satisfying $-d<a<0<b$.

Now show that the function value at the mid-point $$ m=\frac{af(b)-bf(a)}{f(b)-f(a)}=\frac{ab·c(b-a)}{c(b-a)(b+a+d)}=\frac{ab}{a+b+d}<0 $$ is always negative (or show $m=a\frac{b}{b+(a+d)}>a>-d$), $$ f(m)=c·\frac{ab}{a+b+d}·\frac{ab+da+db+d^2}{a+b+d}=c·\frac{ab(a+d)(b+d)}{((a+d)+b)^2}<0, $$ as $a+d>0$. This means that the midpoint $m$ is always used to replace the left point $a$. For sufficiently small $a$, $$a_+=m\approx\beta a$$ with $\beta=\frac{b}{b+d}$ which establishes the linear convergence.

As $d$ is determined by the curvature of the function $f$, the convergence speed depends only on $b$, the farther $b$ is from $0$, the closer $β$ is to $1$, thus the slower the convergence. This also tells that any measure, no matter how crude, that decreases $b$ will substantially increase the speed of convergence.

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To translate this to the more general case, compare the quadratic approximation $$cx(x+d)=cdx+cx^2$$ with the quadratic Taylor polynomial $$f(x^*+x)=f(x^*)+f'(x^*)x+\frac12f''(x^*)x^2+o(x^2)$$ in the root $f(x^*)=0$.

We find $c=2f''(x^*)$ and $cd=f'(x^*)$, which of course only makes sense if both quantities are different from zero. Apply reflections on the $x$ and $y$ axes to get both derivatives and thus both of $c$ and $d$ to be positive.

Answer 2

In Comparative Analysis of Convergence of Various Numerical Methods by Robin Kumar and Vipan, the authors derive the convergence rate of some common numerical methods for solving equations.

In the specific case of the Regula-Falsi method, the authors assume the function is convex on an interval $]x_0,x_1[$ that contains the root, which implies one of the extreme points remains fixed. Then the linear convergence is derived in a similar way as it was in the secant method.

There are other references to derive the rate of convergence. You may try this question or this question and their associated references if you want other approaches.

Answer 3

A rather simple derivation may be done by releasing regula falsi degenerates into fixed-point iteration. Assume $f$ is convex in a neighborhood of the root. Let $b$ be the point which never changes in regula falsi. Let $x_n$ be the points that are updated each iteration. We then have a fixed-point iteration given by:

$$x_{n+1}=\frac{bf(x_n)-x_nf(b)}{f(x_n)-f(b)}=\Phi(x_n)$$

We then have the error given roughly by

$$x-x_{n+1}\simeq\Phi'(x)(x-x_n)$$

which, provided that $\Phi'(x)\in(0,1)$, means linear convergence. With the convexity assumption, one can show that $\Phi'(x)\in(0,1]$ is true, and assuming the root is simple one may further show that $\Phi'(x)\in(0,1)$.