## Calculus 8th Edition

$|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
Consider $f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r)$ Let us suppose that $f(r)=0$ $f(r)=f(x_n) +f'(x_n)(r-x_n)+R_1(r) \implies f(x_n) +f'(x_n)(r-x_n)+R_1(r)=0$ This gives: $x_n-r-\dfrac{f(x_n)}{f'(x_n)}(r-x_n)=\dfrac{R_1(r)}{f'(x_n)}$ $\implies |x_{n+1}-r|=|\dfrac{R_1(r)}{f'(x_n)}| ...(a)$ Need to use Newton's formula. $|R_1{r}|\leq |\dfrac{f''(r)}{2!}||r-x_n|^2$ Thus, the equation (a) can be arranged as: $|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$