Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Textbook Question

Chapter 4.7, Problem 23E

(a) Apply Newton’s method to the equation x² − a = 0 to derive the following square-root algorithm (used by the ancient Babylonians to compute $a$ ):

$x n + 1 = 1 2 ( x n + a x n )$

(b) Use part (a) to compute $1000$ correct to six decimal places.

(a)

Expert Solution

To determine

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

Answer to Problem 23E

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

Explanation of Solution

Given:

The equation is $x2−a=0$ .

Formula used:

The Newton’s method is $xn+1=xn−f(xn)f′(xn)$ .

Calculation:

Let $f(x)=x2−a$ and obtain the derivative of $f(x)$ .

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

$f′(x)=d(x2−a)dx=d(x2)dx−d(a)dx=2x−0=2x$

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

$xn+1=xn−x2n−a2xn=2x2n−x2n+a2xn=x2n+a2xn=12(xn+axn)$

Hence, the derivation.

(b)

Expert Solution

To determine

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

Answer to Problem 23E

The root correct to six decimal places is $31.6227762$ .

Explanation of Solution

Given:

The equation is $x2−a=0$ .

Choose the initial point $x1=30$ .

Calculation:

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

$f(30)=(30)2−1000=900−1000=−100$

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 $f′(x1)=60$ .

$x2=30−(−100)60=30+10060=30+1.666666≈31.666666$

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

$f(31.666666)=(31.666666)2−1000=1002.777735−1000=2.777735$

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 $f′(x2)=63.333332$ .

$x3=31.666666−2.77773563.333332=31.666666−0.043859≈31.622807$

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

$f(31.622807)=(31.666666)2−1000=1000.001924−1000=0.001924$

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 $f′(x3)=62.666664$ .

$x4=31.622807−0.00192462.666664=31.622807−0.000030≈31.6227762$

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

Chapter 4.7, Problem 22E

Chapter 4.7, Problem 24E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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(a) Apply Newton’s method to the equation x 2 − a = 0 to derive the following square-root algorithm (used by the ancient Babylonians to compute a ): x n + 1 = 1 2 ( x n + a x n ) (b) Use part (a) to compute 1000 correct to six decimal places.

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Answer to Problem 23E

Explanation of Solution

Answer to Problem 23E

Explanation of Solution

Chapter 4 Solutions