# The approximation of the given number, 100 100 . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 4.7, Problem 12E
To determine

## To calculate: The approximation of the given number,  100100 .

Expert Solution

The approximation is x3=2.01346125 .

### Explanation of Solution

Given information:

The equation is given as:

100100 .

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 number,

100100

The number can be expressed as : −

x100100=0f(x)=x100100f'(x)=100x99

Let the initial approximation x1=1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1x1100100100x199

x2=(1)((1)100100100(1)99)x2=(1)(1100100)x2=(1)(99100)x2=1+99100x2=199100

x2=1.99

The second approximation is x2=1.99 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2x2100100100x299

x3=(1.99)((1.99)100100100(1.99)99)

Now,

(1.99)100=(199100)100(199100)100=log(199100)100=100log(199100)=100[log(199)log(100)]=100[2.2988530762]=100(0.298853076)=29.88530764

x3=(1.99)(29.88530764100100(29.88530764))x3=(1.99)(70.114692362988.530764)x3=(1.99)(0.023461258)

x3=1.99+0.023461258x3=2.013461258

The third approximation with eight decimal places is x3=2.01346125 .

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