Fixed-Point Iteration To find a solution to p = g(p) given an initial approximation po: INPUT initial approximation po; tolerance TOL; maximum number of iterations No. OUTPUT approximate solution p or message of failure. Step 1 Seti 1. Step 2 While i No do Steps 3-6. Step 3 Set p = g(po). (Compute pi.) Step 4 If p-pol < TOL then OUTPUT (p); (The procedure was successful.) STOP. Step 5 Set i=i+1. Step 6 Set pop. (Update po.) Step 7 OUTPUT ("The method failed after No iterations, No =', No); (The procedure was unsuccessful.) STOP. Example Solution Show that Theorem 2.3 does not ensure a unique fixed point of g(x) = 3* on the interval [0, 1], even though a unique fixed point on this interval does exist. Solution g'(x)=-3-* In 3 <0 on [0, 1], the function g is strictly decreasing 1 g(1) == g(x) ≤1 = g(0), for 0≤x≤1. ≤ on [0, 1]. So Thus, for x = [0, 1], we have g(x) = [0, 1]. The first part of Theorem 2.3 ensures that there is at least one fixed point in [0, 1]. However, g'(0) In 3 -1.098612289, = so g'(x) 1 on (0, 1), and Theorem 2.3 cannot be used to determine uniqueness.