Explanation Let f ( x ) = 1 x − c .then x n + 1 = x n − f ( x n ) f ' ( x n ) </ To determine b) To find for c = 10.3 , we have f ( x ) = 1 x − 10.3 and thus x n + 1 = 2 x n − 10.3 x n 2 .

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.7, Problem 31E

To determine

a)

To find out the Newton’s method for x−(f(x)f'(x))=2x−cx2.

To determine

b)

To find for c=10.3, we have f(x)=1x−10.3 and thus xn+1=2xn−10.3xn2.

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Chapter 4.7, Problem 30E

Chapter 4.7, Problem 32E

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Chapter 4 Solutions

Calculus - Standalone book

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

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a) To find out the Newton’s method for x − ( f ( x ) f ' ( x ) ) = 2 x − c x 2 .

Solution Summary: The author explains the Newton's method for 2x-cx2.

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a)

To find out the Newton’s method for x−(f(x)f'(x))=2x−cx2.

b)

To find for c=10.3, we have f(x)=1x−10.3 and thus xn+1=2xn−10.3xn2.

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Chapter 4 Solutions

a) To find out the Newton’s method for x − ( f ( x ) f ' ( x ) ) = 2 x − c x 2 .

Solution Summary: The author explains the Newton's method for 2x-cx2.

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Videos

a) To find out the Newton’s method for x−(f(x)f'(x))=2x−cx2.

b) To find for c=10.3, we have f(x)=1x−10.3 and thus xn+1=2xn−10.3xn2.

Want to see the full answer?

Chapter 4 Solutions

a)

To find out the Newton’s method for x−(f(x)f'(x))=2x−cx2.

b)

To find for c=10.3, we have f(x)=1x−10.3 and thus xn+1=2xn−10.3xn2.