Explanation Given information: f ( x ) = x ⋅ e 2 x Formula used: Newton’s method, x n + 1 = x n − f ( x n ) f ' ( x n ) Proof: Differentiating f ( x ) with respect to x, it gives f ' ( x ) = x ⋅ ( 2 e 2 x ) + 1 ⋅ e 2 x = ( 1 + 2 x ) ⋅ e 2 x Using formula of x n + 1 , we get x n + 1 To determine (ii) To compute: The x 1 , x 2 , ... , x 10 using x 0 = -3 and x 0 = 7 and to analyze both the results.

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.8, Problem 32E

To determine

(i)

To show:

The sequence obtained by applying Newton’s method to f(x) satisfies xn+1=2xn22xn+1.

To determine

(ii)

To compute:

The x1,x2,...,x10 using x0 = -3 and x0 = 7 and to analyze both the results.

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Chapter 4.8, Problem 31E

Chapter 4.8, Problem 33E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ 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 Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.1 - 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(i) To show: The sequence obtained by applying Newton’s method to f ( x ) satisfies x n + 1 = 2 x n 2 2 x n + 1 .

Solution Summary: The author analyzes the sequence obtained by applying Newton's method to f(x).

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(i)

To show:

The sequence obtained by applying Newton’s method to f(x) satisfies xn+1=2xn22xn+1.

(ii)

To compute:

The x1,x2,...,x10 using x0 = -3 and x0 = 7 and to analyze both the results.

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Chapter 4 Solutions

(i) To show: The sequence obtained by applying Newton’s method to f ( x ) satisfies x n + 1 = 2 x n 2 2 x n + 1 .

Solution Summary: The author analyzes the sequence obtained by applying Newton's method to f(x).

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(i) To show: The sequence obtained by applying Newton’s method to f(x) satisfies xn+1=2xn22xn+1.

(ii) To compute: The x1,x2,...,x10 using x0 = -3 and x0 = 7 and to analyze both the results.

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Chapter 4 Solutions

(i)

To show:

The sequence obtained by applying Newton’s method to f(x) satisfies xn+1=2xn22xn+1.

(ii)

To compute:

The x1,x2,...,x10 using x0 = -3 and x0 = 7 and to analyze both the results.