C.3 The 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

\[ y = F(x) = f(a_n) + \frac{f(b_n)-f(a_n)}{b_n-a_n}(x-a_n) \nonumber \]

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

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

Equation C.3.1 fale position method.

Set \(c_n=\frac{a_n f(b_n) - b_n f(a_n) }{f(b_n)-f(a_n)}\text{.}\) If \(f(c_n)\) has the same sign as \(f(a_n)\text{,}\) then

\[ I_{n+1}=[a_{n+1},b_{n+1}]\quad\text{with}\quad a_{n+1}=c_n,\ b_{n+1}=b_n \nonumber \]

and if \(f(c_n)\) and \(f(a_n)\) have opposite signs, then

\[ I_{n+1}=[a_{n+1},b_{n+1}]\quad\text{with}\quad a_{n+1}=a_n,\ b_{n+1}=c_n \nonumber \]

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