Doubt in derivation of Regular falsi method

Question

I have doubt in derivation. Pls clarify

Let points be $(a, f(a)), (c,0), (b,f(b))$

$y-f(a) = m (x-a) = \frac{f(b)-f(a)}{b-a}(x-a)$

Substituting y=0, x=c,

$0-f(a) = \frac{f(b)-f(a)}{b-a}(c-a) \implies \fbox {c=a-f(a) ( $\frac{b-a}{f(b)-f(a)}$)}$ ---->(1)

But if i use geometry in uploaded figure

$c = a+ f(a) cot \theta$

where $\cot \theta = (1/slope ) = (1/tan \theta) = \frac{(f(b)-f(a))}{(b-a)}$

here $\theta = \angle ACB =$ slope of line AC = $ \tan^{-1}\frac{(f(b)-f(a))}{(b-a)}$

this means $c=a+f(a)/\tan \theta = \fbox {$a+f(a) \frac{b-a}{f(b)-f(a)}$}$ ---> (2)

If we compare (1) and (2) equations, i am getting sign change i..e, a+() and a-()

But in textbook regular falsi method final equation is given by (1) and not (2).

Pls point me to where i went wrong.

Answer 1

You did not consider that $f(a)$ is negative. Thus your formula should be $$ c=a+(-f(a))\cot θ $$ which gives back the first variant of the midpoint formula.

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This is not only the regula falsi formula, this secant root formula is also used in the secant method, Dekker's, Brent's method etc.