ROOTFINDING. We assume the interest rate r holds over all N in +N out periods. h P in (1 + r) N out. N in 1 i h i

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1 ROOTFINDING We want to find the numbers x for which f(x) = 0, with f a given function. Here, we denote such roots or zeroes by the Greek letter α. Rootfinding problems occur in many contexts. Sometimes they are a direct formulation of some physical situtation; but more often, they are an intermediate step in solving a much larger problem. An example with annuities Suppose you are paying into an account an amount of P in per period of time, for N in periods of time. The amount you are deposited is compounded at at interest rate of r per period of time. Then at the beginning of period N in +1, you will withdraw an amount of P out per time period, for N out periods. In order that the amount you withdraw balance that which has been deposited including interest, what is the needed interest rate? The equation is h P in (1 + r) N in 1 i h i = P out 1 (1 + r) N out We assume the interest rate r holds over all N in +N out periods.

2 As a particular case, suppose you are paying in P in = $1, 000eachmonthfor40years. Thenyouwishto withdraw P out = $5, 000 per month for 0 years. What interest rate do you need? If the interest rate is R per year, compounded monthly, then r = R/1. Also, N in =40 1 = 480 and N out =0 1 = 40. Thus we wish to solve " µ R " 480 µ 1# = R # What is the needed yearly interest rate R? Theanswer is.9%. How did I obtain this answer? This example also shows the power of compound interest.

3 THE BISECTION METHOD Most methods for solving f(x) =0areiterative methods. We begin with the simplest of such methods, one which most people use at some time. We assume we are given a function f(x); and in addition, we assume we have an interval [a, b] containing the root, on which the function is continuous. We also assume we are given an error tolerance ε>0, andwewantanapproximaterooteα in [a, b] forwhich α eα ε We further assume the function f(x) changes sign on [a, b], with f(a) f(b) < 0

4 Algorithm Bisect(f,a,b, ε). Step 1: Define c = 1 (a + b) Step : If b c ε, accept c as our root, and then stop. Step 3: Ifb c>ε, then check compare the sign of f(c) tothatoff(a) andf(b). If sign(f(b)) sign(f(c)) 0 then replace a with c; and otherwise, replace b with c. ReturntoStep1. Denote the initial interval by [a 1,b 1 ], and denote each successive interval by [a j,b j ]. Let c j denote the center of [a j,b j ]. Then α c j bj c j = c j a j = 1 ³ bj a j Since each interval decreases by half from the preceding one, we have by induction, µ 1 n α c n (b 1 a 1 )

5 EXAMPLE Find the largest root of f(x) x 6 x 1=0 accurate to within = With a graph, it is easy to check that 1 <α<. We choose a =1,b = ; then f(a) = 1,f(b) = 61, and the requirement f(a) f(b) < 0issatisfied. The results from Bisect are shown in the table. The entry n indicates the iteration number n. n a b c b c f(c)

6 Recall the original example with the function. h f(r) =P in (1 + r) N in 1 i h i P out 1 (1 + r) N out Checking, we see that f(0) = 0. Therefore, with a graph of y = f(r) on[0, 1], we see that f(x) < 0if we choose x very small, say x =.001. Also f(1) > 0. Thus we choose [a, b] =[.001, 1]. Using ε = yields the answer with an error bound of eα = α c n for n = 0 iterates. We could also have calculated this error bound from 1 (1.001) =

7 Suppose we are given the initial interval [a, b] =[1.6, 4.5] with ε = How large need n be in order to have α c n ε Recall that α c n µ 1 n (b a) Then ensure the error bound is true by requiring and solving µ 1 n (b a) ε µ 1 n ( ) Dividing and solving for n, wehave µ.9 n log = Therefore, we need to take n =16iterates.

8 ADVANTAGES AND DISADVANTAGES Advantages: 1.Italwaysconverges.. You have a guaranteed error bound, and it decreases with each successive iteration. 3. You have a guaranteed rate of convergence. The error bound decreases by 1 with each iteration. Disadvantages: 1. It is relatively slow when compared with other rootfinding methods we will study, especially when the function f(x) has several continuous derivatives about the root α.. The algorithm has no check to see whether the ε is too small for the computer arithmetic being used. [This is easily fixed by reference to the machine epsilon of the computer arithmetic.] We also assume the function f(x) iscontinuouson the given interval [a, b]; but there is no way for the computer to confirm this.