CODE IN MATLAB:

(a)

(b)

(simple iteration) Find the smallest positive root of the equation 1 cos(x) + 1 +e-2x7 Ite-2x = 0 by using the simple iteration with Xo = 3 and g(x) = arccos(-1/(1+e-2x)). What is the approximated value of the root? - (bisection method) Consider finding the root of f(x) = x6 X 1 on [1,2]. (a) Show that the function has a unique root $ on [1, 2]. Implement the bisection method to get the approximate value of the root. (b) Find the approximation of with high precision. This can be done by either Newton's method or bisection method with lots of iterations. Then compute _In(en+1) In(en) " In(en) In(en-1)' where en = \Xn (to compute en in the code, use the accurate approximation of & instead of the real & in the expression). Prove that if that for some q> 1 (recall definition 1.7), lim en = 0, lim not exists and equal to , n >00 nto en then we mush have limnyoorn = q, provided that we further assume u + 0 for q> 1 and u (0,1) for q = 1. (The hint of proof is to consider -_ [In(en+1) q ln(en)] [In(en) q ln(en-1)] In - 9 In(en) In(en-1) and show the limit is 0 for the case q > 1.) Due to this reason, people usually use rn as an indicator for the order of convergence in the experiments. Now plot rn vs. n and uses this to show liment i = > 0 no e cannot be true for any q > 1. This roughly says that the bisection method is not converging with an order q > 1. (c) Plot In(en) vs. n. (Although it is not a straight line, the plot is still close to a line, which implies that we still approximately get en ~ Cun for some u (0,1).)