## SectionC.3The false position (regula falsi) method

Let $f(x)$ be a continuous function and let $a_1\lt b_1$ with $f(a_1)$ and $f(b_1)$ being of opposite sign.

As we have seen, the bisection method generates a sequence of intervals $I_n=[a_n,b_n]\text{,}$ $n=1,2,3,\cdots$ with, for each $n\text{,}$ $f(a_n)$ and $f(b_n)$ having opposite sign (so that, by continuity, $f$ has a root in $I_n$). Once we have $I_n\text{,}$ we choose $I_{n+1}$ based on the sign of $f$ at the midpoint, $\frac{a_n+b_n}{2}\text{,}$ of $I_n\text{.}$ Since we always test the midpoint, the possible error decreases by a factor of 2 each step.

The false position method tries to make the whole procedure more efficient by testing the sign of $f$ at a point that is closer to the end of $I_n$ where the magnitude of $f$ is smaller. To be precise, we approximate $y=f(x)$ by the equation of the straight line through $\big(a_n,f(a_n)\big)$ and $\big(b_n,f(b_n)\big)\text{.}$

The equation of that straight line is

\begin{equation*} y = F(x) = f(a_n) + \frac{f(b_n)-f(a_n)}{b_n-a_n}(x-a_n) \end{equation*}

Then the false position method tests the sign of $f(x)$ at the value of $x$ where $F(x)=0\text{.}$

\begin{align*} \amp F(x) = f(a_n) + \frac{f(b_n)-f(a_n)}{b_n-a_n}(x-a_n) =0 \\ \amp \iff x= a_n - \frac{b_n-a_n}{f(b_n)-f(a_n)} f(a_n) = \frac{a_n f(b_n) - b_n f(a_n) }{f(b_n)-f(a_n)} \end{align*}

So once we have the interval $I_n\text{,}$ the false position method generates the interval $I_{n+1}$ by the following rule. 1 The convergence behaviour of the false position method is relatively complicated. So we do not discuss it here. As always, we invite the interested reader to visit their favourite search engine.