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 23E

(a)

To determine

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

Expert Solution

Answer to Problem 23E

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

Answer to Problem 23E

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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