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Math 128a, Homework 8 1. Let Φ(x) = will converge. 1 (2 3 due November 6. − e + x ). Determine an interval [a, b] on which the fixed-point iteration x 2 2. Let f (x) = x3 − x2 − x and x0 = 10. This polynomial has three real roots. So in addition to Newton’s method, in order to find the largest root we can also apply the “double-step” method described on page 293. Apply both methods, and write down the number of iterations each one takes before finding the largest zero with precision < 10−6 . Apply the methods only to this one function, i.e. you don’t have to write a procedure to implement them (although you can). 3. (a) If a function f has a double zero at ξ, i.e. f (x) = (x − ξ)2 g(x) with g(ξ) 6= 0, we showed that Newton’s method is locally convergent only linearly, with constant 0.5. Show that if we know that the root is a double root, a modified Newton’s method xn+1 = xn − 2 · f (xn )/f 0 (xn ) converges quadratically. (b) Let f (x) = x3 − x2 − x + 1. Then f has a double root at 1. Let x0 = 1.1. Using the facts about the order of convergence of the methods, how many iterations would you expect Newton’s method to take to converge to the root within 10−6 ? How many iterations would you expect the method of part (a) to take? (c) Now run these two methods, and see how many iterations are needed in reality to achieve this tolerance. Again, since you only need to run these methods for one function, you can do this “by hand” if you want to. 4. Problem 5.7. 5. Write Matlab functions to implement the bisection method, secant method, the method of false position, and if you have time, Newton’s method. It makes sense to implement these as a single function with the method fed in as a parameter. Your procedure should take as a parameter a function and an interval such that the function has different signs at the endpoints of the interval. It should stop when the zero is found to within a prescribed absolute tolerance. Test your procedure on the function f (x) = tan(πx) − 6, with the initial interval [0, 0.48]. Note that the function has in this interval a root ξ = (1/π) arctan 6 ≈ 0.447431543. Run the procedure for 10, 20, and 50 iterations, and compare the results of the different methods. Discuss. 1