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 7E
To determine

To calculate: The third approximation to the root of equation,

x5x1=0 where x1=1 .

Expert Solution

Answer to Problem 7E

The third approximation is x3=3.3489 .

Explanation of Solution

Given information:

The equation is given as:

  x5x1=0

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,

x5x1=0 and x1=1

This implies −

  f(x)=x5x1f'(x)=5x41

Now, let n=1 ,

  x2=x1f(x1)f'(x1)

  x2=x1x15x115x141

  x2=(1)((1)5(1)15(1)41)x2=(1)(11151)x2=1(14)x2=414x2=34

  x2=0.75

The second approximation is x2=0.75 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2x25x215x241

  x3=(0.75)((0.75)5(0.75)15(0.75)41)x3=(0.75)(0.2373040.7511.582031251)x3=0.75(1.5126960.58203125)

  x3=0.75(2.59899447)x3=0.75+2.59899447x3=3.34899447

The third approximation with four decimal places is x3=3.3489 .

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