Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.11 Application of Taylor Polynomials - 11.11 Exercises - Page 822: 39


$|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$

Work Step by Step

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: $|x_{n+1}-r|\leq \dfrac{M}{2K}|x_n-r|^2$
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