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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Section
BuyFindarrow_forward

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.

sin x = x2 − 2

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 is sin(x)=x22.

Calculation:

Rewrite the given equation as f(x)=sin(x)x2+2.

Calculate the derivative of f(x).

f(x)=ddx(sin(x)x2+2)=ddx(sin(x))ddx(x2)+ddx(2)=cos(x)2x0=cos(x)2x

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

f(1)=cos(1)2×(1)=cos(1)+2=0.540302+2=2.540302

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

f(1)=sin(1)(1)2+2=0.8414701+2=0.841470+1=0.158530

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)=0.158530 and f(x1)=2.540302,

x2=1(0.158530)(2.540302)=10.0624051.062405

Calculate the value of f(x) at x2=1.062405,

f(1.062405)=sin(1.062405)(1.062405)2+2=0.8735281.128704+2=2.002232+2=0.002232

Calculate the value of f(x) at x2=1.062405,

f(1.062405)=cos(1.062405)2×(1.062405)=0.486772+2.124810=2.611582

Use Newton’s method to calculate x3

Set n=2 , and obtain the value of x3.

x3=x2f(x2)f(x2).

Substitute x2=1.062405 , f(x)=0.002232 and f(x2)= 2.611582,

x3=(1.062405)(0.002232)(2.611582)=(1.062405)+(0.002232)(2.611582)=1.062405+0.0008541.061550

Calculate the value of f(x) at x3=1.061550,

f(1.061550)=sin(1.061550)(1.061550)2+2=0.8731121.126888+2=2.0000004+2=0.0000002

Calculate thevalue of f(x) at x3=1.061550,

f(1.061550)=cos(1.061550)2×(1.061550)=0.487519+2.123100=2.610619

Use Newton’s method to calculate x4

Set n=3 , and obtain the value of x4.

x4=x3f(x3)f(x3).

Substitute x3=1.061550, f(x3)=0.0000002 and f(x3)=2.610619,

x4=(1.061550)(0.0000002)(2.610619)=1.061550+0.000000071.061549

Calculate the value of f(x) at x4=1.061549,

f(1.061549)=sin(1.061549)(1.061549)2+2=0

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