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 5E
To determine

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

x3+2x4=0 where x1=1 .

Expert Solution

Answer to Problem 5E

The third approximation is x3=1.1797 .

Explanation of Solution

Given information:

The equation is given as:

  x3+2x4=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,

x3+2x4=0 and x1=1

This implies −

  f(x)=x3+2x4f'(x)=3x2+2

Now, let n=1 ,

  x2=x1f(x1)f'(x1)

  x2=x1x13+2x143x12+2

  x2=(1)(1)3+2(1)43(1)2+2x2=11+243+2x2=1345

  x2=1(15)x2=1+15x2=65x2=1.2

The second approximation is x2=1.2 .

Let n=2 ,

  x3=x2f(x2)f'(x2)

  x3=x2x23+2x243x22+2

  x3=(1.2)(1.2)3+2(1.2)43(1.2)2+2x3=1.21.728+2.444.32+2x3=1.24.12846.32

  x3=1.2(0.1286.32)x3=1.2(0.020253164)x3=1.179746836

The third approximation with four decimal places is x3=1.1797 .

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