?In Section 4.8 we considered Newton's method for approximating a solution \(r\) of the

Chapter 11, Problem 39

(choose chapter or problem)

In Section 4.8 we considered Newton's method for approximating a solution \(r\) of the equation \(f(x)=0\), and from an initial approximation \(x_{1}\) we obtained successive approximations \(x_{2}, x_{3}\),..., where

                                                       \(x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f\left(x_{n}\right)}\)

Use Taylor's Inequality with \(n=1, \alpha=x_{n}\), and \(x=r\) to show that if \(f^{\prime \prime}(x)\)

 exists on an interval \(I\) containing \(r, x_{n}r, x_{n}r, x_{n}\), and \(x_{n+1}\), and \(\left.\right|^{\prime \prime}(x)|\leqslant M,| f^{\prime}(x) \mid \geqslant K\)for all \(x \in I\), then

                                                         \(\left|x_{n+1}-r\right| \leqslant \frac{M}{2 X}\left|x_{n}-r\right|^{2}\)

[This means that if \(x_{n}\) is accurate to \(d\) decimal places, then \(x_{n+1}\) is accurate to about \(2 d\) decimal places. More precisely, if the error at stage \(n\) is at most \(10^{-m}\), then the error at stage \(n+1\) is at most \((M / 2 K) 10^{-2 m}\).]

Equation Transcription:

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Text Transcription:

r

f(x) = 0

x_1

x_2, x_3

x_{n+1} = x_{n} - frac{f(x_n)}{f(x_n}

n = 1, alpha = x_n

x = r

f’’ (x)

I

r, x_n

x_n+1

|f’’ (x)| leqslant M,|f^prime(x)| geqslant K

x \in I

|x_{n+1} - r| leqslant frac{M}{2K} |x_{n}-r|^2

x_n

d

x_n+1

2d

n

10^-m

n + 1

(M/2K) 10^-2m