How do you derive the secant method formula from the equation below?

Question

The Secant Method forumula is;

$$ x_{i+1}=x_i - \frac{f(x_i)(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})}.$$

Derive the formula from the equation below;

$$\frac{f(x_i)}{f(x_{i-1})}=\frac{x_i-x_{i+1}}{x_{i-1}-x_{i+1}}.$$

Answer 1

$$ x_{i+1}=x_i - \frac{f(x_i)(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})} \\ \overset{(1)}\iff \frac{f(x_i)(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})}=x_i -x_{i+1} \\ \overset{(2)}\iff f(x_i)(x_i-x_{i-1})=(x_i -x_{i+1})(f(x_i)-f(x_{i-1})) = (x_i -x_{i+1})f(x_i)- (x_i -x_{i+1})f(x_{i-1}) \\ \overset{(3)}\iff f(x_i)(x_{i+1}-x_{i-1})=f(x_i)(x_i-x_{i-1}-x_i+x_{i+1})=-(x_i -x_{i+1})f(x_{i-1}) \\ \overset{(4)}\iff \frac{f(x_i)}{f(x_{i-1})}=\frac{x_{i}-x_{i+1}}{x_{i-1}-x_{i+1}} $$

(1) subtract $x_i$ on both sides

(2) multiply both sides by $(f(x_i)-f(x_{i-1}))$

(3) subtract $f(x_i)(x_i-x_{i+1})$ on both sides

(4) divide both sides by $-(x_{i+1}-x_{i-1})f(x_{i-1})$, note that if $(x_{i+1}-x_{i-1}) = 0$ or $f(x_{i-1})=0$ then the original problem makes no sense so we may suppose that they are different from $0$.