   Chapter 4.8, Problem 20E ### 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 solutions of the equation correct to six decimal places. ln x = 1 x − 3

To determine

To find: All the solutions of the equation correct to six decimal places.

Explanation

Formula used:

The newton’s formula is, xn+1=xnf(xn)f(xn).

Given:

The initial approximation x1=0.6, and the function is ln(x)=1x3.

Calculation:

The given equation is ln(x)=1x3.

Rewrite the given equation and equate to zero ln(x)1x3.

Thus, f(x) = ln(x)1x3.

Calculate the derivative of f(x).

f(x)=ddx(ln(x)1x3)=ddx(ln(x))ddx(1x3)=1x+1(x3)2

The value of f(x) at x1=0.6 is,

f(x)=10.6+1(0.63)2=10.6+15.76=1.666666+0.173611=1.840277

Calculate the value of f(x) at x1=0.6,

f(0.6)=ln(0.6)1(0.63)=0.5108251(2.4)=0.510825+1(2.4)=0.510825+0.416666

On further simplification of the above equation is,

f(0.6)=0.510825+0.416666=0.094158

Use Newton’s method to calculate x2.

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

Set n=1 and obtain the value of x2.

x2=x1f(x1)f(x1)

Substitute x1=0.6,f(x1)=0.094158 and f(x1)=1.840277,

x2=0.6(0.094158)(1.840277)=0.6+(0.094158)(1.840277)=0.6+0.0511650.651165

Calculate the value of f(x) at x2=0.651165,

f(0.651165)=ln(0.651165)1(0.6511653)=0.428992+12.348834=0.428992+0.425743=0.003249

Calculate the value of f(x) at x2=0.651165,

f(x)=10.651165+1(0.6511653)2=10.651165+1(2.348835)2=10.651165+15.517025=1.535709+0.181257

On further simplification of the above equation is,

f(x)=1.535709+0.181257=1.716966

Use Newton’s method to calculate x3.

Set n=2 and obtain the value of x3.

x3=x2f(x2)f(x2).

Substitute x2=0.651165 , f(x)=0.003249 and f(x2)= 1.716966,

x3=(0.651165)(0.003249)(1.716966)=(0.651165)+(0.003249)(1.716966)=0.651165+0.0018920.653057

Calculate the value of f(x) at x3=0.653057,

f(0.653057)=ln(0.653057)1(0.6530573)=0.4260901(2.346943)=0.426090+1(2.346943)=0.426090+0.426086

On further simplification of the above equation is,

f(0.653057)=0.426090+0.426086=0.000003

Calculate the value of f(x) at x3=0.653057,

f(x)=10.653057+1(0.6530573)2=1.531259+1(2.346943)2=1.531259+0.181549=1.712808

Use Newton’s method to calculate x4

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