   Chapter 4.8, Problem 26E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# 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.cos(x2 − x) = x4

To determine

To find: The solutions of the equation correct to eight decimal places.

Explanation

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Given:

f(x)=cos(x2x)x4

Calculation:

f(x)=cos(x2x)x4

Differentiate with respect to x,

f(x)=(2x1)sin(x2x)4x3

Then,

By Newton’s method,

xn+1=xncos(xn2xn)xn4(2xn1)sin(xn2xn)4xn3=xn+cos(xn2xn)xn4(2xn1)sin(xn2xn)+4xn3

In figure 1,

It is clear that 1st zero is

x1=0.75

Substitute n=1 in xn+1=xn+cos(xn2xn)xn4(2xn1)sin(xn2xn)+4xn3,

x2=x1+cos(x12x1)x14(2x11)sin(x12x1)+4x13

Substitute x1=0.75 in x2=x1+cos(x12x1)x14(2x11)sin(x12x1)+4x13,

x2=0.75+cos((0.75)2(0.75))(0.75)4(2(0.75)1)sin((0.75)2(0.75))+4(0.75)3=0.73514521

Substitute n=2 in xn+1=xn+cos(xn2xn)xn4(2xn1)sin(xn2xn)+4xn3,

x3=x2+cos(x22x2)x24(2x21)sin(x22x2)+4x23

Substitute x2=0.73514521 in x3=x2+cos(x22x2)x24(2x21)sin(x22x2)+4x23,

x3=0.73514521+(cos((0.73514521)2(0.73514521))(0.73514521)4)((2(0.73514521)1)sin((0.73514521)2(0.73514521))+4(0.73514521)3)=0.73485921

Substitute n=3 in xn+1=xn+cos(xn2xn)xn4(2xn1)sin(xn2xn)+4xn3,

x4=x3+cos(x32x3)x34(2x31)sin(x32x3)+4x33

Substitute x3=0

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