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The question of which method is more efficient, Newton's method or the secant method, was answered by Jeeves. He showed that if the effort required to evaluate f(x)' is less than 43 percent of the effort required to evaluate f(x), then Newton's method is more efficient. Otherwise, the secant method is more efficient.

The question of which method is more efficient, Newton's method or the secant method, was answered by Jeeves. He showed that if the effort required to evaluate f(x)' is less than 43 percent of the effort required to evaluate f(x), then Newton's method is more efficient. Otherwise, the secant method is more efficient.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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Example for y =x - cos(x) on [0.0,4.0] for eps = 1.0e-6 initial point is 1.0 iterations root 0.73909 5 for bisectional method it takes 22 iterations, but for Newton only 4 iterations. The question of which method is more efficient, Newton's method or the secant method, was answered by Jeeves. He showed that if the effort required to evaluate f(x)' is less than 43 percent of the effort required to evaluate f(x), then Newton's method is more efficient. Otherwise, the secant method is more efficient.
Transcribed Image Text:Example for y =x - cos(x) on [0.0,4.0] for eps = 1.0e-6 initial point is 1.0 iterations root 0.73909 5 for bisectional method it takes 22 iterations, but for Newton only 4 iterations. The question of which method is more efficient, Newton's method or the secant method, was answered by Jeeves. He showed that if the effort required to evaluate f(x)' is less than 43 percent of the effort required to evaluate f(x), then Newton's method is more efficient. Otherwise, the secant method is more efficient.
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