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A2x,=x,+z - 2x,+1 + x , = dn+2 - 2dn+1+ d, = [(A 1)' eh] d,, - + e, + eh = A(e, e , + ~ ) 2e, + 0 it follows that also (4-14) e:+O, + enen+,. w. n+ + We conclude that (A ek # 0 for all sufficiently large n, n > no,say. It follows that Aax, # 0 for n > no;hence the sequence {xk}is defined for n > no. We have AX, = Adn = (A + 8, - 1) d, and hence, subtracting s from (4-1I), Problem 12. " Give a quantitative interpretation of this statement, allowing you to answer the following question: As n m, how many steps are necessary (on the average) to reduce the error by a factor fs?

We are thus led to the algorithm considered in example 4. 8. In any 82 elements of numerical analysis interval [a, b] with 0 < a < d; ib the smallest value of the slope occurs at x = a, and it is easily seen that condition (iv) is satisfied for every b 2 Ha + c/a). Thus it follows that the sequence defined by (4-20) converges to d\/cfor every choice of x, > 0. Fomula (4-20) states that the new approximation is always the arithmetic mean of the old approximation and of the result of dividing c by the old approximation.

Since the construction ofp depends on the theory of the interpolating polynomial, we postpone the derivation of Muller's algorithm to chapter 10. (iv) Newton's method in rhe case F'(s) = 0. 7 under the assumption that F1(x) # 0, implying in particular that F1(s) # 0. Let us now consider the general situation where iteration where m 2 1. ]-lF(m)(~)hm-l+ O(hm) From this we find Thus, if m # 1, f'(s) # 0, and the convergence fails to be quadratic. However, the above analysis shows how to modify the iteration function in order to achieve quadratic convergence.

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