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 24E

(a)

To determine

To Derive: The square root of the given reciprocal equation by Newton’s method

Expert Solution

Explanation of Solution

Given:

The equation is 1xa=0.

Formula used:

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

Calculation:

Let f(x)=1xa and obtain the derivative of f(x).

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

f(x)=d(1xa)dx=d(1x)dxd(a)dx=1x20=1x2

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

xn+1=xn1xna1xn2=xn+xnaxn2=2xnaxn2

Hence, the derivation.

(b)

To determine

To Compute: The root of 11.6984 correct to six decimal places by using part (a).

Expert Solution

Answer to Problem 24E

The root correct to six decimal places is 0.588788_.

Explanation of Solution

Given:

The equation is 1xa=0.

Choose the initial point x1=0.5.

Calculation:

Calculate f(x1) at the point x1=0.5 and a=1.6984.

f(0.5754)=10.51.6984=21.6984=0.301600

Calculate f(x1) at the point x1=0.5.

f(0.5)=1(0.5)2=4

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

Substitute n=1 to obtain x2=2x1ax12.

Substitute x1=0.5,f(x1)=0.301600 and f(x1)=4.

x2=2×0.51.6984×(0.5)2=11.6984×0.25=10.4246000.5754

Calculate f(x2) at the point x2=0.5754 and a=1.6984.

f(0.5754)=10.57541.6984=1.7379211.6984=0.039521

Calculate f(x2) at the point x2=0.5754.

f(0.5754)=1(0.5754)2=3.020370

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

x3=2x2ax22

Substitute x2=0.5754,f(x2)=0.039521 and f(x2)=3.020370.

x3=0.57540.039521(3.020370)=31.666666+0.039521(3.020370)=0.5754+0.0130840.588484

Calculate f(x3) at the point x3=0.588484 and a=1.6984.

f(0.588484)=10.5884841.6984=1.6992791.6984=0.000879

Calculate f(x3) at the point x3=0.588484.

f(0.588484)=1(0.588484)2=2.887549

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

x4=2x3ax32

Substitute x3=0.588484,f(x3)=0.000879 and f(x3)=2.887549.

x4=0.5884840.000879(2.887549)=0.588484+0.000879(2.887549)=0.588484+0.0003040.588788

Thus, the root corrected to six decimal places is 0.588788_.

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