# To Derive: The algorithm for the square root of the given equation by Newton’s method.

### 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 23E

(a)

To determine

## To Derive: The algorithm for the square root of the given equation by Newton’s method.

Expert Solution

The derivation is xn+1=12(xn+axn)

### Explanation of Solution

Given:

The equation is x2a=0.

Formula used:

The Newton’s method is xn+1=xnf(xn)f(xn).

Calculation:

Let f(x)=x2a and obtain the derivative of f(x).

Let the derivative of f(x)=df(x)dx.

f(x)=d(x2a)dx=d(x2)dxd(a)dx=2x0=2x

To derive the xn+1 use the above mentioned formula.

xn+1=xnx2na2xn=2x2nx2n+a2xn=x2n+a2xn=12(xn+axn)

Hence, the derivation.

(b)

To determine

### To Compute: The root of 1000 correct to six decimal places by using part (a).

Expert Solution

The root correct to six decimal places is 31.6227762.

### Explanation of Solution

Given:

The equation is x2a=0.

Choose the initial point x1=30.

Calculation:

Calculate f(x1) at the point x1=30 and a=1000.

f(30)=(30)21000=9001000=100

Calculate f(x1) at the point x1=30.

f(30)=2×30=60

Substitute n=1 in the above mentioned result in part (a) and obtain x2

x2=12(x1+ax1)

Substitute x1=30,f(x1)=100 and f(x1)=60.

x2=30(100)60=30+10060=30+1.66666631.666666

Calculate f(x2) at the point x2=31.666666 and a=1000.

f(31.666666)=(31.666666)21000=1002.7777351000=2.777735

Calculate f(x2) at the point x2=31.666666.

f(31.666666)=2×31.666666=63.333332

Substitute n=2 in the above mentioned result in part (a) and obtain x3.

x3=12(x2+ax2)

Substitute x2=31.666666,f(x2)=2.777735 and f(x2)=63.333332.

x3=31.6666662.77773563.333332=31.6666660.04385931.622807

Calculate f(x3) at the point x3=31.622807 and a=1000.

f(31.622807)=(31.666666)21000=1000.0019241000=0.001924

Calculate f(x3) at the point x3=31.622807.

f(31.622807)=2×31.622807=62.666664

Substitute n=3 in the above mentioned result in part (a) and obtain x4.

x4=12(x3+ax3)

Substitute x3=31.622807,f(x3)=0.001924 and f(x3)=62.666664.

x4=31.6228070.00192462.666664=31.6228070.00003031.6227762

Thus, the root corrected to six decimal places is 31.6227762_.

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