Convergence of Regula Falsi (false position) method

Question

Let $f$ be a continuous function such that $f(a)f(b)<0$.Let $a_0=a$ and $b_0=b$ $$x_1:=a_0-\frac{b_0-a_0}{f(b_0)-f(a_0)}f(a_0)$$ If $f(a_0)f(x_1)<0$, set $a_1=a_0$ and $b_1=x_1$

If $f(x_1)f(b_0)<0$, set $a_1=x_1$ and $b_1=b_0$ $$x_2:=a_1-\frac{b_1-a_1}{f(b_1)-f(a_1)}f(a_1)$$ If $f(a_1)f(x_2)<0$, set $a_2=a_1$ and $b_2=x_2$

If $f(x_2)f(b_1)<0$, set $a_2=x_2$ and $b_2=b_1$

having defined $a_1,a_2,...,a_{n-1},b_1,b_2,...,b_{n-1}$ like above $$x_n=a_{n-1}-\frac{b_{n-1}-a_{n-1}}{f(b_{n-1})-f(a_{n-1})}f(a_{n-1})$$

Now $\{x_n \}$ is a sequence $[a,b]$, By Bolzano Weierstrass theorem, there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ which converges to a point $p \in [a,b]$, I want to prove that $f(p)=0$,

If $f(p)<0$, then there exists a $\delta>0$ such that $f(x)<0$ $\forall \in [p-\delta,p+\delta]$

Since $\{x_{n_k} \}$ converges to $p$, there exists $k \in \mathbb{N}$ such that $|x_{n_k}-p|<\delta$ $\forall k \geq K$

$$S_a:=\{k \geq K : x_{n_k}=a_{n_k}\}$$

$$S_b:=\{ k \geq K : x_{b_k}=b_{n_k} \}$$

here, I am stuck, I know that one of the sets defined above is empty but not able to prove it,

Answer 1

First of all, the method has similarities to the bisection method, only that the midpoint is chosen differently and thus there is no guarantee that the interval length converges to zero. But it remains that the sequence $(a_n)$ is monotonically increasing, while $(b_n)$ is monotonically decreasing, and both sequences bound each other.

Thus both sequences have limits $a_*$ and $b_*$. If $f(a_*)f(b_*)<0$, then the secant root $s$ between them is a proper midpoint, at a positive distance to both end points. This is impossible, as the same would be true for some interval $[a_n,b_n]$ with $n$ so large that the distances $a_n$ to $a_*$ and $b_n$ to $b_*$ are very much smaller than the distance to $s$.

One would also have to make sure that the difference in the function values is small enough so that the secant roots are close enough together.