BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4.7, Problem 15E
To determine

To calculate: The roots of the equation,

  (x2)2=lnx .

Expert Solution

Answer to Problem 15E

The roots are

  x21.333333

  x30.816537

  x41.262929

Explanation of Solution

Given information:

The equation is given as:

  (x2)2=lnx .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

  xn+1=xnf(xn)f'(xn) and so on.

Calculation:

Consider the equation,

  (x2)2=lnx

Now,

  (x2)2lnx=0f(x)=(x2)2lnxf'(x)=2(x2)1x

Let the initial approximation x1=1

Now, at n=1

  x2=x1f(x1)f'(x1)

  x2=x1(x12)2lnx12(x12)1x1

  x2=(1)((12)2ln(1)2(12)11)x2=(1)((1)202(1)1)x2=(1)(121)

  x2=1(13)x2=1(0.333333333)x2=1.333333333

The second approximation is x2=1.333333 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2(x22)2lnx22(x22)1x2

  x3=(1.333333)((1.3333332)2ln(1.333333)2(1.3333332)11.333333)x3=(1.333333)(0.4444448880.2876818220.6666660.750000187)x3=1.333333(0.732126711.416666187)

  x3=1.3333330.5167955x3=0.816537499

The third approximation with six decimal places is x3=0.816537 .

Let n=3 ,

  x4=x3f(x3)f'(x3)

  x4=x3(x32)2ln(x3)2(x32)1x3

  x4=(0.816537)((0.8165372)2ln(0.816537)2(0.8165372)10.816537)x4=(0.816537)(1.400584672(0.202683052)2.3669261.224684246)x4=0.816537(1.6032677243.591610246)

  x4=0.816537+0.446392457x4=1.262929457

The fourth approximation with six decimal places is x4=1.262929 .

Therefore, the roots with six decimal places are :-

  x21.333333

  x30.816537

  x41.262929

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!