   Chapter 3.8, Problem 24E

Chapter
Section
Textbook Problem

# 23-26 Use Newton’s method to find all the solutions of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. x 5 − 3 x 4 + x 3 − x 2 − x + 6 = 0

To determine

Find:

All the solutions of the equation correct to eight decimal places using Newton’s method.

Explanation

1) Concept:

Start with initial approximation x1  which is obtained from graph. 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 eight 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) Calculation:

The equation is x5-3x4+x3-x2-x+6=0

Consider the function fx=x5-3x4+x3-x2-x+6

We find the x1  value using the graph.

From the graph we get initial roots are -1, 1.3 and 2.7

fx=x5-3x4+x3-x2-x+6

f'(x)=5x4-12x3+3x2-2x-1

xn+1=xn-fxnf'xn

xn+1=xn-x5-3x4+x3-x2-x+65x4-12x3+3x2-2x-1

i) First root

x1= -1

x2=x1-x15-3x14+x13-x12-x1+65x14-12x13+3x12-2x1-1

x2=-1-(-1)5-3-14+-13--12-(-1)+65(-1)4-12-13+3-12-2(-1)-1

x2 -1.04761905

fx2= (-1.04761905)5-3-1.047619054+-1.047619053--1.047619052-(-1.04761905)+6=-0.07508733

Evaluate f'x2

5-1.047619054-12-1.047619053+3-1.047619052- 2(-1.04761905)-1 =24.20756286

x3=-1.04761905-(-0.07508733) 24.20756286

x3 -1.04451724

Evaluate f(x3)

fx3= -1.044517245-3-1.044517244+-1.044517243--1.044517242-(-1.04451724)+6=-0.00033989

Evaluate f'x3

5-1.044517244-12-1.044517243+3-1.044517242- 2(-1.04451724)-1 =23.98868929

x4=-1.04451724-(-0.00033989) 23.98868929

x4-1.04450307

Evaluate f(x4)

fx4= -1.044503075-3-1.044503074+-1.044503073--1.044503072-(-1.04450307)+6=0.00000003

Evaluate f'x4

5-1.044503074-12-1.044503073+3-1.044503072- 2(-1.04450307)-1 =23.987692655

x4=-1.04450307-(0.00000003 ) 23.987692655

x5-1.04450307

x4-1.04450307 x5

ii) Second root:

x1= 1.3

x2=x1-x15-3x14+x13-x12-x1+65x14-12x13+3x12-2x1-1

x2=1

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