   Chapter 4.8, Problem 21E ### 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.x3 = tan−1x

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=1, and the function x3=tan1(x).

Calculation:

Rewrite the given equation as f(x) = x3tan1(x).

Calculate the derivative of f(x).

f(x)=ddx(x3tan1(x))=ddx(x3)ddx(tan1(x))=3x211+x2

The value of f(x) at x1=1 is,

f(1)=3(1)211+(1)2=312=52=2.5

Calculate the value of f(x) at x1=1,

f(1)=(1)3tan1(1)=1(0.785398)=1+0.785398=0.214601

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=1,f(x1)=0.214601 and f(x1)=2.5,

x2=1(0.214601)(2.5)=1+(0.214601)(2.5)=1+0.0858400.914159

Calculate the value of f(x) at x2=0.914159,

f(0.914159)=(0.914159)3tan1(0.914159)=0.763952(0.740582)=0.763952+0.740582=0.023369

Calculate the value of f(x) at x2=0.914159,

f(0.914159)=3(0.914159)211+(0.914159)2=3×0.83566811.835686=2.5070600.544755=1.962304

Use Newton’s method to calculate x3.

Set n=2 and obtain the value of x3.

x3=x2f(x2)f(x2).

Substitute x2=0.914159, f(x)=0.023369 and f(x2)= 1.962304,

x3=(0.914159)(0.023369)(1.962304)=(0.914159)+(0.023369)(1.962304)=0.914159+0.0119080.902250

Calculate the value of f(x) at x3=0.902250,

f(0.902250)=(0.902250)3tan1(0.902250)=0.734481(0.734056)=0.734481+0.734056=0.000424

Calculate the value of f(x) at x3=0.902250,

f(0.902250)=3(0.902250)211+(0.902250)2=3×0.81405511.814055=2.4421650.544755=1.897410

Use Newton’s method to calculate x4.

Set n=3 and obtain the value of x3.

x4=x3f(x3)f(x3).

Substitute x3=0.902250, f(x3)=0.000424 and f(x3)=1.897410,

x4=(0.902250)(0.000424)(1.897410)=(0.902250)+(0.000424)(1.897410)=0.902250+0.0002230.902026

Calculate the value of f(x) at x4=0.902026,

f(0.902026)=(0.902026)3tan1(0.902026)=0.733934(0.733933)=0.733934+0.733933=0.0000006

Calculate thevalue of f(x) at x4=0.902026,

f(0.902026)=3(0.902026)211+(0.902026)2=3×0.81365011.813650=2.4409520.551374=1.889577

Use Newton’s method to calculate x5.

Set n=4 and obtain the value of x5.

x5=x4f(x4)f(x4).

Substitute x4=0.902026, f(x4)=0.0000006 and f(x4)=1.889577,

x5=(0.902026)(0.0000006)(1.889577)=(0.902026)+(0.0000006)(1.889577)=0.902026+0.00000030.902025

Substitute the value of x5 in the given function will give 0.

Therefore, the solution of the equation correct to six decimal places is x50.902025.

The value of f(x) at x1=0 is,

f(0)=3(0)211+(0)2=011=01=1

Calculate the value of f(x) at x1=0,

f(0)=(0)3tan1(0)=0+0=0

Use Newton’s method to calculate x2

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