For a given function is there a general procedure to find an initial value for x_1 such that Newton's method bounces back and forth between two values forever?

Cyrus Munoz

Cyrus Munoz

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2022-08-30

For a given function is there a general procedure to find an initial value for x 1 such that Newton's method bounces back and forth between two values forever?

Answer & Explanation

Jonathan Bailey

Jonathan Bailey

Beginner2022-08-31Added 10 answers

For function f ( x ) and x n + 1 = x n 1 f ( x n ) , simply set x n = x n + 2 , which yields x = ( x 1 f ( x ) ) 1 f ( x 1 f ( x ) ) .. Solve for x and you have all points that either oscillate between two values or remain stationary. Such points are not guaranteed to exist for all f.
In fact, if you set x n = x n + k with any k and solve for x, the value will pass through at most k points before arriving back at the original x.

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