Chapter 8.8, Problem 33E

To determine

To Prove: The value of $f^{″}\left(x\right)$ exists in the interval of $I$ that contains $r,x_n,x_{n+1},$ and $f^{″}\left(x\right)\le M$ , also $f^{\prime }\left(x\right)\ge K$ for all the values of $x\in I$ then $\left|x_{n+1}-r\right|\le \frac{M}{2K}{\left|x^n-r\right|}^2$ .

Explanation of Solution

Given:

The equation is $f\left(x\right)=0$ and form the initial approximation for the $x_i$ such that the obtained successive approximation is,

  $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$

Calculation:

Consider the Taylor’s inequality is,

  $\left|R_n\right|\le \frac{M}{\left(n+1\right)!}\left|x-a\right|$

Then,

  $\begin{array}{l}\left|R_n\right|\le \frac{M}{\left(1+1\right)!}\left|r-x_n\right|\\ \left|R_n\right|\le \frac{M}{2}\left|r-x_n\right|\\ \left|R_1\right|\le \frac{M}{2}\left|r-x_n\right|\\ R_1=f\left(x\right)-T_1\end{array}$

Consider the first order polynomial is,

  $T_1=f\left(a\right)+f^{\prime }\left(a\right)\left(x-4\right)$

Then,

  $\begin{array}{l}{R}_1=f\left(x\right)-f\left(a\right)-f^{\prime }\left(a\right)\left(x-a\right)\\ R_1=f\left(r\right)-f\left(x_n\right)-f^{\prime }\left(a\right)\left(r-x_n\right)\end{array}$

By the transformation of the newton’s equation,

  $f\left(x_n\right)=f^{\prime }\left(x_n\right)\left(x_n-x_n+1\right)$

Then,

  $f\left(x_n\right)=f^{\prime }\left(x_n\right)\left(x_n-x_n+1\right)$

Then,

  $\begin{array}{l}{R}_1=-f^{\prime }\left(x_n\right)\left(x_n-x_{n+1}\right)-f^{\prime }\left(x_n\right)\left(r-x_n\right)\\ R_1=-f^{\prime }\left(x_n\right)\left(x_n-x_{n+1}+r-x_n\right)\\ R_{{}_1}=-f^{\prime }\left(x_n\right)\left(r-x_{n+1}\right)\end{array}$

Consider the absolute value is,

  $\begin{array}{l}\left|R_1\right|=\left|-f^{\prime }\left(x_n\right)\left(r-x_{n+1}\right)\right|\\ \left|R_1\right|=\left|f^{\prime }\left(x_n\right)\left(r-x_{n+1}\right)\right|\end{array}$

Since, the value of $f^{\prime }\left(x\right)\ge K$ . Then,

  $\left|R_1\right|=K\left|r-x_{n+1}\right|$

Then, plug in the Taylor’s inequality is,

  $\left|x_{n+1}-r\right|\le \frac{M}{2K}\left|x_n-r\right|$

Hence, proved.