   Chapter 3.8, Problem 30E

Chapter
Section
Textbook Problem

# (a) Use Newton’s method with x 1 = 1 to find the root ofthe equation x 3 − x = 1 correct to six decimal places.(b) Solve the equation in part (a) using x 1 = 0.6 as the initial approximation.(c) Solve the equation in part (a) using x 1 = 0.57 . (You definitely need a programmable calculator for this part.)(d) Graph f ( x ) = x 3 − x − 1 and its tangent lines at x 1 = 1 , 0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation.

To determine

a)

Find:

The root of the equation correct to 6 decimal places using Newton’s method.

Explanation

1) Concept:

Start with given initial approximation x1 . Evaluate fx1and f '(x1) and apply the Newton’s formula. After the first iteration we get x2. Repeat the process with x1  replaced by x2. Calculate the approximations x1, x2, x3,  until the successive approximations xn, xn+1  agree to the six decimal places.

2) Formula:

i) Newton’s formula for nth approximation

xn+1=xn-fxnf'xn for n=1,2,3,

ii) Power rule of differentiation:

ddxxn=nxn-1

3) Given

x1=1

4) Calculation:

x3-x-1=0

f'(x)=3x2-1

xn+1=xn-fxnf'xn

xn+1=xn-xn2-xn-13xn2-1

x2=x1-x13-x1-13x12-1

Substitute x1=1

x2=1-(1)3-1-13(1)2-1

x2= 1.5

For n=2

x3=x2-x23-x2-13x22-1

Substitute x2= 1

To determine

b)

Solve:

The equation using x1=0.6

To determine

c)

To Solve:

The equation using  x1=0.57

To determine

d)

Explain:

How newton’s method is sensitive to the value of initial approximation by graphing the function

fx=x3-x-1 and its tangent lines at x1=1, x1=0.6,  and x1=0.57

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