Example Graphs of f x are given below. (c) no zero in 2,2

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1 . - Bisection Method The idea of the Bisection Method is based on the Intermediate Value Theorem that ou studied in Calculus I. Recall that: Intermediate Value Theorem: If f is continuous on a,b and K is a number between f a and f b, then there eists a number c in a,b for which f c K. Note that: (i) If f a f b 0, then either f a or f b is less than 0 (K 0), and there eists a number c in a,b for which f c 0. The number c is a solution of the equation f 0. (ii) The condition in (i) is a sufficient condition, that is, if f a f b 0, it is still possible to have a zero in a,b. Eample Graphs of f are given below (a) one zero in, (b) two zeros in, (a) f f 0 and there is one zero in,. (b) f f but there are two zeros in,. (c) f f and there is no zero in,. (c) no zero in, Eample Let f 5. (a) Show that the equation f 0 has a solution in 0,. (b)findaninterval a, b that is different from 0, and f 0 has a solution in a,b. (a) Since f 0, f andf 0 f 0, b the Intermediate Value Theorem there eists a number c in 0, such that f c 0. (b) Since f 5 0andf f 0 0, the equation has a solution in,0. The Bisection Method: Let f be continuous on a, b,andf a f b 0. B the Intermediate Value Theorem, we know there eists a number c in a, b such that f c 0. Let be the solution of the equation f 0. The Bisection Method generates a sequence n as follows. Let a a and b b, compute a b (the average of a and b )and

2 a a a if f a f 0or if f f b 0. b b b Algorithm: For k,,..., having f a k f b k 0, compute k a k b k and a k a k a k k if f a k f k 0or if f k f b k 0. b k k b k b k The algorithm stops if f k 0or f k and k. Eample Solve the equation 0 for 0 b the Bisection Method. Compute and, and determine a,b and a,b. Let f and a,b 0,.Sincef 0, f and f 0 f 0, there is a solution in 0,.Leta 0andb. Compute 0, and f. Since f f 0, a andb. Compute and f.sincef f 0, 4 4 a andb. Convergence: Does n alwas converge to? Observe that Hence, b a b a b a n n b a. lim n n lim n n b a 0 lim n n. Rate of Convergence: How fast n converges to? Since n n b a n n b a K, n converges to as fast as n converges to 0, i.e., n O n. For a given 0, we can find N such that N as follows:

3 b a N N b a N ln ln b a ln ln b a ln N. ln Note that : i. Since N is derived from an upper bound of N, it is possible to have k N such that k. ii. N is independent of f and depends onl on a,b and given. Eample Find the number of iterations necessar to approimate within 0 4 b solving 0 using the Bisection Method. Let a,b,. Compute ln b a ln ln ln ln ln Let N 4. Check: using the Bisection Algorithm, we compute and However, from the following list, we see n n n How are N computed numericall b MatLab? MatLab Program: bisect.m For a given, an interval a, b, and a function f, the program does the following. () Check if f a f b 0, print out a message and stop the program if the condition fails. ln b a ln () Compute N (the smallest integer that is greater than or equal to ln ln b a ln ). ln () For k, compute k a k b k and a k a k a k k if f a k f k 0or if f k f b k 0. b k k b k b k The algorithm stops if f k 0ork N, and k. A hard cop of the program is attached at the end of the section. The function f can be either included in the program or as an input. It is called fun in the

4 program. Here are several eamples. Eample Approimate within 0 8 b solving the equation 0. Input in MatLab: bisect the leftend point a 0 the rightend point b epsilon 0^(-8) Here are the results obtained b bisect.m: N f N N 8 Real time 0 4

5 Eample Approimate "Fibonacci s forgotten number" within 0 8 b solving the equation 0 0. Input in MatLab: *.^ 0*-0; bisect the leftend point a 0 the rightend point b epsilon 0^(-8) Here are the results obtained b bisect.m: N f N N 8 Real time 0 Eample Solve Dr. Groetsch s equation ln Input in MatLab: sqrt(./(-))); bisect the leftend point a 0. the rightend point b epsilon 0^(-8) Here are the results obtained b bisect.m: N f N N 9 Real time for within Eercises: The graph of f for in 0, is given at the right. Let be the solution of the equation f 0for in 0,. (a) Compute graphicall and b the Bisection Method. 5 4 (b) What are a, b and a, b? (c) Find the smallest positive integer N such that N f

6 . (a) Show that the equation cos 0hasasolutionin 0,. (b) Compute, and numericall using the Bisection Method. State also a,b and a,b. (c) Find the smallest positive integer N such that N 0 8. (d) Use the MatLab program bisect.m to solve the equation cos 0in 0, within (a) Find an interval a,b in which the equation e 0 has a solution. Provide a reason to support our choice of a,b. (b) Using a,b obtained in (a), compute,and numericall using the Bisection Method. State also a,b and a,b. (c) Use the MatLab program bisect.m to solve the equation e 0in a,b within Approimate within 0 5 b solving 0 using the Bisection Method. 5. Using the Bisection Method, find all real solutions of the equation e 4 0 within 0 6. For each real solution, give first the interval a,b that ou are using. 6