Question

In Section 4.6 we considered Newton's method for approximating a root \({\rm{r}}\)of the equation\({\rm{f(x) = 0}}\), and from an initial approximation \({{\rm{x}}_{\rm{1}}}\) we obtained successive approximations\({{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{, \ldots }}\), where

\({{\rm{x}}_{{\rm{n + 1}}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{ - }}\frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{{{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}\)

Use Taylor's Formula with \({\rm{n = 1,a = }}{{\rm{x}}_{\rm{n}}}\)and\({\rm{x = r}}\)to show that if \({{\rm{f}}^{''}}{\rm{(x)}}\) exists on an interval \({\rm{I}}\)containing \({\rm{r,}}{{\rm{x}}_{\rm{n}}}\), and\({{\rm{x}}_{{\rm{n + 1}}}}\), and , then

(This means that if \({{\rm{x}}_{\rm{n}}}\)is accurate to d decimal places, then\({{\rm{x}}_{{\rm{n + 1}}}}\)is accurate to \({{\rm{x}}_{{\rm{n + 1}}}}\)about \({\rm{2d}}\)decimal places. More precisely, if the error at stage n is at most\({\rm{1}}{{\rm{0}}^{{\rm{ - m}}}}\), then the error at stage \({\rm{n + 1}}\)is at most\({\rm{(M/2K)1}}{{\rm{0}}^{{\rm{ - 2m}}}}\).)

Short answer

Hence, the given statement is verified.

Step-by-step solution

Concept Introduction

In any term finite Taylor series approximation, Taylor's inequality is an estimate result for the value of the remainder term. Indeed, if there exists a real number satisfying on some interval for any function that meets the hypotheses of Taylor's theorem, the remainder satisfies.

Step 2:Calculation

Consider the given values and simplify,

To begin, we have:

\(\begin{aligned}{}{\rm{f(x) = }}{{\rm{T}}_{\rm{1}}}{\rm{(x) + }}{{\rm{R}}_{\rm{1}}}{\rm{(x)}}\\ \Rightarrow {\rm{f(r) = }}{{\rm{T}}_{\rm{1}}}{\rm{(r) + }}{{\rm{R}}_{\rm{1}}}{\rm{(r)}}\end{aligned}\)

We also have the following:

\(\begin{aligned}{}{{\rm{T}}_{\rm{1}}}{\rm{(x) = f(a) + }}{{\rm{f}}'}{\rm{(a)(x - a)}}\\ \Rightarrow {{\rm{R}}_{\rm{1}}}{\rm{(x) = f(x) - f(a) - }}{{\rm{f}}'}{\rm{(a)(x - a)}}\end{aligned}\)

Now, let's use the substitutions \({\rm{a = }}{{\rm{x}}_{\rm{n}}}\)and\({\rm{x = r}}\), as indicated. Furthermore, for any \({{\rm{f}}^{''}}{\rm{(x)}}\) exist on an interval I containing \({\rm{r,}}{{\rm{x}}_{\rm{n}}}\), and\({{\rm{x}}_{{\rm{n + 1}}}}\), and\(\left| {{{\rm{f}}^{''}}{\rm{(x)}}} \right|{\rm{£ M,}}\left| {{{\rm{f}}'}{\rm{(x)}}} \right| \ge {\rm{K}}\) for all \({\rm{x}} \in {\rm{I}}\).

Using the fact that r is a root of the equation f(x)=0, we may derive the following:

\(\begin{aligned}{}{{\rm{R}}_{\rm{1}}}{\rm{(r) = f(r) - f}}\left( {{{\rm{x}}_{\rm{n}}}} \right){\rm{ - }}{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)\left( {{\rm{r - }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ = - f}}\left( {{{\rm{x}}_{\rm{n}}}} \right){\rm{ - }}{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)\left( {{\rm{r - }}{{\rm{x}}_{\rm{n}}}} \right){\rm{ = }}{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)\left( {{{\rm{x}}_{\rm{n}}}{\rm{ - r}}} \right){\rm{ - f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)\\\frac{{{{\rm{R}}_{\rm{1}}}{\rm{(r)}}}}{{{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{\rm{ = }}\left( {{{\rm{x}}_{\rm{n}}}{\rm{ - r}}} \right){\rm{ - }}\frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{{{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{ - }}\frac{{{\rm{f}}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{{{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}{\rm{ - r = }}{{\rm{x}}_{{\rm{n + 1}}}}{\rm{ - r}}\;\;\;\left( {{\rm{ Since }}\left| {{{\rm{f}}'}{\rm{(x)}}} \right|{\rm{ > K > 0}}} \right)\\\left| {\frac{{{{\rm{R}}_{\rm{1}}}{\rm{(r)}}}}{{{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)}}} \right|{\rm{ = }}\left| {{{\rm{x}}_{{\rm{n + 1}}}}{\rm{ - r}}} \right| \Rightarrow \left| {{{\rm{R}}_{\rm{1}}}{\rm{(r)}}} \right|{\rm{ = }}\left| {{{\rm{x}}_{{\rm{n + 1}}}}{\rm{ - r}}} \right|{\rm{ \times }}\left| {{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)} \right|\end{aligned}\)

Using Taylor's inequality, we now obtain the following:

\(\begin{aligned}{}{\rm{ |}}{{\rm{R}}_{\rm{1}}}{\rm{(r)|}} \le \frac{{\rm{M}}}{{{\rm{2!}}}}{\left| {{\rm{r - }}{{\rm{x}}_{\rm{n}}}} \right|^{\rm{2}}}\\\left| {{{\rm{x}}_{{\rm{n + 1}}}}{\rm{ - r}}} \right|{\rm{ \times }}\left| {{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)} \right|{\rm{ \& £ }}\frac{{\rm{M}}}{{\rm{2}}}{\left| {{{\rm{x}}_{\rm{n}}}{\rm{ - r}}} \right|^{\rm{2}}}\\\left| {{{\rm{x}}_{{\rm{n + 1}}}}{\rm{ - r}}} \right| \le \frac{{\rm{M}}}{{{\rm{2}}\left| {{{\rm{f}}'}\left( {{{\rm{x}}_{\rm{n}}}} \right)} \right|}}{\left| {{{\rm{x}}_{\rm{n}}}{\rm{ - r}}} \right|^{\rm{2}}} \le \frac{{\rm{M}}}{{{\rm{2K}}}}{\left| {{{\rm{x}}_{\rm{n}}}{\rm{ - r}}} \right|^{\rm{2}}}\end{aligned}\)

Therefore, we use the definition of \({{\rm{R}}_{\rm{1}}}{\rm{(x)}}\) and Taylor's inequality, we can prove the inequality with some mathematical manipulation.