Weneed to show that x k I Assume x I x x then x 1 x x x 2 1 2 f 00 \u03be f x x x x

# Weneed to show that x k i assume x i x x then x 1 x x

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Weneed to show that x k I . Assume x 0 I , | x 0 - x * | ≤ then | x 1 - x * | = | x 0 - x | 2 1 2 f 00 ( ξ 0 ) f 0 ( x 0 ) ≤ | x 0 - x * || x 0 - x * | M ( ) ≤ | x 0 - x * | M ( ) < | x 0 - x * | ≤ Now assume that x k I then | x k +1 - x * | = | x k - x | 2 1 2 f 00 ( ξ k ) f 0 ( x k ) 2 M ( ) < x k +1 I So now | x k +1 - x * | | x k - x * | 2 M ( ) = ≤ | x k - x * | M ( ) . . . | x - 0 * | ( M ( ) k +1 0 , k → ∞ , x k x * as k → ∞ The need for a good initial guess x 0 for Newton’s method should be emphasized. In practice, this is obtained with another method, like bisection. 12.6 The Secant Method Sometimes it could be computationally expensive or not possible to evaluate the derivative of f . The following method, known as the secant method,
12.6. THE SECANT METHOD 199 replaces the derivative by the secant: x k +1 = x k - f ( x k ) f ( x k ) - f ( x k - 1 ) x k - x k - 1 , k = 1 , 2 , . . . (12.18) Note that since f ( x * ) = 0 x k +1 - x * = x k - x * - f ( x k ) - f ( x * ) f ( x k ) - f ( x k - 1 ) x k - x k - 1 , = x k - x * - f ( x k ) - f ( x * ) f [ x k , x k - 1 ] = ( x k - x * ) 1 - f ( x k ) - f ( x * ) x k - x * f [ x k , x k - 1 ] ! = ( x k - x * ) 1 - f [ x k , x * ] f [ x k , x k - 1 ] = ( x k - x * ) f [ x k , x k - 1 ] - f [ x k , x * ] f [ x k , x k - 1 ] = ( x k - x * )( x k - 1 - x * ) f [ x k ,x k - 1 ] - f [ x k ,x * ] x k - 1 - x * f [ x k , x k - 1 ] = ( x k - x * )( x k - 1 - x * ) f [ x k - 1 , x k , x * ] f [ x k , x k - 1 ] If x k x * , then f [ x k - 1 ,x k ,x * ] f [ x k ,x k - 1 ] 1 2 f 00 ( x * ) f 0 ( x * ) and lim k →∞ x k +1 - x * x k - x * = 0, i.e. the sequence generated by the secant method would converge faster than linear. Defining e k = | x k - x * | , the calculation above suggests e k +1 ce k e k - 1 (12.19) Let’s try to determine the rate of convergence of the secant method. Starting with the ansatz e k Ae p k - 1 or equivalently e k - 1 = ( 1 A e k ) 1 /p we have e k +1 ce k e k - 1 ce k 1 A e k 1 p ,
200 CHAPTER 12. NON-LINEAR EQUATIONS which implies A 1+ 1 p c e 1 - p + 1 p k . (12.20) Since the left hand side is a constant we must have 1 - p + 1 p = 0 which gives p = 1 ± 5 2 , thus p = 1 + 5 2 = 1 . 61803 (12.21) gives the rate of convergence of the secant method. It is better than linear, but worse than quadratic. Sufficient conditions for local convergence are as in Newton’s method. 12.7 Fixed Point Iteration Newton’s method is a particular example of a functional iteration of the form x k +1 = g ( x k ) , k = 0 , 1 , . . . with the particular choice of g ( x ) = x - f ( x ) f 0 ( x ) . Clearly, if x * is a zero of f then x * is a fixed point of g , i.e. g ( x * ) = x * . We will look at fixed point iterations as a tool for solving f ( x ) = 0. Example 33. Suppose we want to solve x - e - 1 = 0 in [0 , 1] . Then if we take g ( x ) = e - x , a fixed point of g corresponds to a zero of f . Definition 22. Let g is defined in an interval [ a, b ] . We say that g is a contraction or a contractive map if there is a constant L with 0 L < 1 such that | g ( x ) - g ( y ) | ≤ L | x - y | , for all x, y [ a, b ] . (12.22) If x * is a fixed point of g in [ a, b ] then | x k - x * | = | g ( x k - 1 ) - g ( x * ) | L | x k - 1 - x * | L 2 | x k - 2 - x * · · · L k | x 0 - x * | → 0 , as k → ∞
12.7. FIXED POINT ITERATION 201