2.3 The Fixed-Point Algorithm

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1 .3 The Fied-Point Algorithm 1. Mean Value Theorem: Theorem Rolle stheorem: Suppose that f is continuous on a, b and is differentiable on a, b. If f a f b, then there eists a number c in a, b such that f c 0. Theorem Mean Value Theorem: Suppose that f is continuous in a, b and is differentiable on a, b. Then there eists a number c in a, b such that f f b f a c. b a Note that: a. Rolle s Theorem is a special case of the Mean Value Theorem. b. B Rolle s Theorem, we know if f 0forall in a, b, then f a f b. On the other hand, the condition f a f b alone is not enough for us to determine if there eists a number c in a, b such that f c 0. Graphicall, Rolle s Theorem and the Mean Value Theorem can be described as follows f 0 f, f 1 0 c. When b is close to a ( b a is small), f c f b f a b a c 1.4, c. and f f b f a b a for a,b. Eample First show that the equation e 0 has a solution in 0,1. Then Determine if the solution is unique. Let f e.sincef 0 f e , b the Intermediate Value Theorem there eists a number c in 0, 1 such that f c 0. Therefore, the equation e 0hasa solution in 0, 1. Now let us check to see if f 0for in 0, 1, f 1 e 0 for all. So, f 0 for all in 0, 1 and therefore, f 0 onl once and the solution is unique. 1

2 Eample The graph of f cos 0. for in, is given below. Find graphicall all possible c in, satisfing the conclusion given in the Mean Value Theorem. Approimate f c for each c f Approimatel, c and c f f 1 f 1.5 c f f 1 f 0.5 c Comparison: f c , f c Fied-PointofaFunction: Definition A number p is said to be a fied point of a function g if g p p. Graphicall, a function has a fied point at p if its graph g intersects with the line intersect at p. For eample, the following graphs show that e and intersect at a power where is near e p p, when p 0.58 Hence, f e has a fied point near Some functions ma have more than one fied points and some functions ma not have a fied point. For eample, the function in (i) has no fied point because 1and do not intersect; and the function in (ii) has infinitel man fied points.

3 (i) 1 (ii) cos Algebraicall, we solve the equation g (or g 0 ) to determine if a function has an fied point over a given interval. Eample Find all fied points of g 1 1 and g cos if the eist. a. Set g 1 : 1, 1 0. Using the quadratic formula: no real solution. So, g 1 has no fied point for. b. Set g : cos, cos 0, n 1, n 1,,3,... So, g has infinitel man fied points for. 3. Eistence and Uniqueness of a Fied Point: Theorem Eistence and Uniqueness: Let g be continuous on a, b. i. If a g b for all in a, b,theng has a fied point p in a, b. ii. If, in addition, g eists on a, b and there eists a constant 0 K 1 such that g K for all in a, b, then p is unique. Note that: a. Both conditions: a g b for all in a, b and g K for all in a, b are sufficient conditions. So, in the case where the condition in i. does not hold, it is possible that g has a fied point; and in the case where the condition in ii. is not satisfied, it is also possible the fied point of g is unique. b. Because g is the slope of the tangent line to the curve g at, g K 1 means that the graph of g does not grow as faster than and not slower than. Proof of the eistence and uniqueness: i. If g a a or g b b, then p a or p b and g has a fied point. Now let g a a and g b b, and let h g. Since g is continuous on a, b, h is continuous on a, b. Observe that h a g a a 0andh b g b b 0. So, b the Intermediate Value Theorem, we know there eists a number c in a, b such that h c 0, that is 3

4 g c c 0org c c. So, c is a fied point of g in a, b. ii. Now let also g K for all in a, b where 0 K 1. Suppose that g has two fied points, sa p q in a, b. Then b the Mean Value Theorem, we know these eists a point c in q, p such that g g p g q c p q. g p g q Since g p p and g q q, p q p p q q 1. So, g c 1, this contradicts the given condition g 1 for all in a, b.so,gcannot have two fied points in a,b. Eample Let g for in 1,1. Determine if g has a fied point in 1, 1.Ifso, determine if the fied point is unique. Check conditions given in the theorem for the eistence and uniqueness: i. Observe that g min g andg ma g 1 g Since 1 g 1for all in 1, 1, g has a fied point in 1, 1. ii. Compute g 3.Since g 3 1, g has a unique fied point in 1, 1. 3 For g, we can solve its fied point p algebraicall. g Since , p is a unique fied point in 1, Check the graph of g : , -.-., Eample Let g 3 for in 0,1. Determine if g has a fied point in 0, 1. If so, determine if the fied point is unique. Check conditions given in the theorem for the eistence and uniqueness: i. Observe that g min g , and g ma g 0 1. Since 0 g 1, g has a fied point in 0, 1. ii. Compute g 3 ln3. g 3 ln3. Since g 0 ln3 1, there is no conclusion 4

5 about the uniqueness. From the graph of g below, we can see that g has a unique fied point p 0.55 in 1,1. But we cannot solve p algebraicall. How can solve a fied point numericall? , in 0, 1 4. The Fied-Point Algorithm: The Fied-Point Algorithm is an algorithm that finds the fied-point of a function over an interval assuming the fied point is unique in this interval. Algorithm Fied-Point Algorithm: Given g, and a, b, choose p 0 in a, b and compute p 1,p,..., as follows: p n g p n 1 for n 1,,... Implement the algorithm in a programming language which does the following:input g, interval a, b, p 0 in a, b, and K ma, and compute p n g p n 1 for n 1,,... The program terminates if i. p n p n 1 and then p p n ;or ii. p n b or p n a, and the program fails; or iii. n K ma. The MatLab program fipt.m implements the Fied-Point Algorithm to find p n with input (1) the function gfun for g ; () an initial approimation p 0 to the fied-point; (3) an accurac requirement ; and (4) the maimum number for iterations K ma. The following two eamples show graphicall how the Fied-Point Algorithm works. 5

6 0.5 Fied Point Iterations: {p n }, g()=(-e + -)/3, p 0 =0 6 Fied Point Iterations: {p n }, g()=(-0.5), p 0 = (p 0,g(p 0 )) (p 1,p 1 )=(g(p 0 ),g(p 0 )) (p 1,g(p 1 )) 4 (p 1,g(p 1 )) = 0.5 (p,p ) 3 (p 0,g(p 0 )) (p 1,p 1 ) p 0 p p 1 1 p 0 p (1) () Clearl, the Fied-Point Algorithm finds the fied point p in (1) and diverges in (). 5. Convergence and the Rate of Convergence of the Fied-Point Algorithm: Questions: Assume that g has a unique fied point p in a, b and p 0 is in a, b.let p n g p n 1, n 1,,... a. Under what condition(s), does p n converge to p? b. If lim n p n p, what is the rate of converge? Fied-Point Theorem: Theorem Fied-Point Theorem: Let g be continuous on a, b and a g b. Suppose that g eists for all in a, b, and g K for all in a, b where 0 K 1. Then lim n p n p for an p 0 in a, b, and p n p K n ma p 0 a, b p 0 and p n p K n 1 K p 1 p 0, for all n 1,,... Proof: Letp 0 be in a, b and p n be generated b the Fied-Point Algorithm. Observe that p n p g p n 1 g p. B the Mean Value Theorem, we know there eists a number c in a, b such that g p n 1 g p p n 1 p g c or g p n 1 g p g c p n 1 p. Since g 1 for all in a, b, 6

7 p n p g p n 1 g p g c p n 1 p g c p n 1 p K p n 1 p K K p n p K p n p... K n p 0 p 0 lim n p n p lim n K n p 0 p 0. Therefore, lim n p n p. Since p 0 p p 0 a or p 0 p b p 0, Observe that So for an m n 1, p n p K n p 0 p K n ma p 0 a, b p 0. p n 1 p n g p n g p n 1 K p n p n 1 K g p n 1 g p n K g p n g p n 3... K n p 1 p 0 p m p n p m p m 1 p m 1 p m... p n 1 p n Since lim m p m p, p p n p m p m 1 p m 1 p m... p n 1 p n K m 1 p 1 p 0 K m p 1 p 0... K n p 1 p 0 K n K m n 1 K m n... K 1 p 1 p 0 K n 1 Km n 1 K m lim p m p n m lim K n p 1 p 0 1 Km n 1 K p 1 p 0 K n 1 K p 1 p 0 Note that : a. Rate of convergence: p n p K n 1 K p 1 p 0, for all n 1,,...implies that p n p K n 1 1 K p 1 p K g p 0 p 0. So p n converges to pwith the rate of convergence of O K n, i.e., p n p O K n. b. Order of convergence: From p n p g c n 1 p n 1 p K p n 1 p,wehave p n p p n 1 p K. Hence, p n converges to p linearl 1 with an asmptote error constant K. c. The smallest possible number of iterations: For a given, we can estimate the number N of iterations needed to approimate p b p N. That is, find N such that K N 1 K p 1 p 0 K N 1 K b a KN 1 b K Nln K ln a b K a ln ln 1 K Since 0 K 1, ln K 0. So, N b a ln K. 7

8 Eample Determine whether or not the function has a fied point in the given interval. If so, determine if the Fied-point Algorithm will converge to the fied point. In the case when it converges, estimate the number of iterations possibl needed to approimate the fied point within (1) g 1 3 e, 0,1 () g /, 0, (1) Check the range of g : From the graph of g on 0,1, wehave0 g (1) g 1 3 e () g 1 3 e So, g 1 3 e has a fied point in 0,1. Check the maimum value of g : g 1 3 e. From the graph of g, g g K 1 for all in 0,1. 3 So, g has a unique fied-point in 0,1 and the sequence p n generated b the Fied-Point Algorithm converges to p. Estimate the number N of iterations: 8 N ln10 5 ln , let N 11. ln 1 3 Use the Fied-Point Algorithm to solve the fied point in 0,1 and p p n where p n p n Using MatLab program fipt.m, we have the following. 1/3*(-ep().^); fipt input initial point p0 0 input the tolerance for stopping criterion, e:.0001,10^(-8) 10^(-5) input the maimum number of iterations 100 Algorithm converges with number of iterations ans 9 fied point p Now we check the values of g at p i : [p gfun(p)]

9 As ou see, g p 9 p 9. () Check the range of g : Observe that g min g , g ma g 0 1 So, 0 g for all in 0,. Hence, g has a fied point in 0,. Check the maimum value of g : g g From the graph of g, g g 1forsome in 0,. So we cannot conclude the sequence p n generated b the Fied-Point Algorithm converges to p. Using MatLab program fipt.m, we have the following. 1/*sqrt(10-.^3); fipt input initial point p0 0 input the tolerance for stopping criterion, e:.0001,10^(-8) 10^(-5) input the maimum number of iterations 100 Algorithm converges with number of iterations 18 fied point p Check some values of p i and g p i : [p(13:18,1) gfun(p(13:18,1))]

10 g p 18 p Fied-Point Algorithm for Solving The Equation: f 0 Let be a solution of the equation f 0. To solve using the Fied-Point Algorithm, a function g needs to be defined first such that is a fied point of g, thatis, g. Eample Consider solving Find an interval a, b on which the equation has a solution. Find a function g such that the fied point of g is the solution of the equation: f 0. Determine if the sequence p n generated b the Fied-Point Algorithm with the function g. Consider a, b 1, 0.Sincef 1 f , the equation has a solution in 1,0. (1) A naive choice of g : since 1 3, we can let g 1 3. Check the range of g : g min g 0 1 andg ma g 1 0, so, 1 g 0andghas a fied point in 1,0. Check the maimum value of g : g 3, g 3 1forsome in 1,0. So, it is not certain b the Fied-Point Theorem if p n converges to p. Observe that p n n 0 0, 1, 0, 1,.... () Rewrite the equation 3 1 0as 3 1, 3 1.Let g 3 1. Check the range of g : g min g 0 1, g ma g 1 0, so, 1 g 0andg has a fied point in 1,0. Check the maimum value of g : g Since 3 1 g is not defined at 1, g is unbounded. So, it is not certain if p n converges to p. Observe that p n n 0 0, 1, 0, 1,.... (3) Rewrite the equation 3 1 0as 3 1 andthen 1 1 or 1 1. Let g 1 1. Check the range of g : g min g 0 1, and g ma g 1 1,so, 1 g 0andghas a fied point in 1,0. 10

11 Check the maimum value of g : g 1, g g 1 From the graph of g above, we see g 0.7 K 1. So, p is unique in 1,0 and p n converges to p. Algebraicall, let h g. h 1 3 0, 1 for h reaches its maimum at and hence, K Estimate the number N of iterations needed: N ln10 5 ln , let N 36. ln 0.7 N ln10 5 ln , let N 30. ln 0.65 Use the Fied-Point Algorithm to solve the fied point in 0,1 and p p n where n 7 and p p n p n Eample Show that each of the following functions has a fied point at p precisel when f p 0, where f 4 3. (1) g 3 1/4 () g (1) Set Then /4. () Check if , then So, Eample The following four methods are proposed to compute 7 1/5. Rank them in order, based on their apparent speed of convergence, assuming p

12 g() The function g for each iteration is: (1) p n 1 7 p 3 n 1 p n 1 1/ () p n p n 1 p 5 n 1 7 p n 1 (3) p n p n 1 p 5 n 1 7 (4) p 4 n p n 1 p 5 n p n 1 (1) g / 1 7 1/ (3) g (4) g () g graphs of g1, g, g3 and g4 =g () 4 =g 1 () = 0 =g 3 () =g 4 () All four functions have a fied point on 1,. Compute g : Check the value of g at p 7 1/5 : 1 (i) g / (ii) g (iii) g (iv) g g 1 p p p 1/ g p 1 3p 14 p g 3 p p 5 g 4 p p p

13 So, the sequences p n generated b the Fied-Point Algorithm using g 1 and g do not converge. The sequences p n generated b the Fied-Point Algorithm using g 3 and g 4 converge and the third sequence converges faster than the the 4th one. The testing results show that using g 3, 7 1/5 p using g 4, 7 1/5 p Eample Show that if A is an positive number, then the sequence defined b n 1 n 1 A, for n 1, n 1 converges to A whenever 0 0. Let lim n n. Then lim n n lim n 1 n 1 A n 1 1 A 1 A A A. Eample Let p n be generated b the Fied-point Algorithm with the function g and let p be the fied point of g such that lim n p n p. Determine the orders of convergence and the asmptotic error constants of the sequence p n in the cases where (1) g p 0; and () g p 0 but g p 0. (1) Assume that g p 0. B the Mean Value Theorem, we have p n 1 p g p n g p g c n p n p g c n p n p where c n is between p n and p. lim n p n 1 p p n p lim n g c n g p. So, the order of convergence is 1 and the asmptotic error constant g p. () Assume that g p 0 but g p 0. B the Talor Theorem, we have for p, Then g g p g p 1! g p n g p g p 1! p g! p n p g n! p n 1 p g p n g p g p g p n! p n 1 p p n p 1 g p n, p,where is between and p. p n p g p g p n! lim n p n p g p 1 p n 1 p p n p 1 So, the order of convergence is and the asmptotic error constant 1 g p. g p p n p g p n p n p 13

14 Eercises: 1. The graph of f for 3 5 is given at the left. Find graphicall all possible c in 3, 5 that satisfies the conclusion given b the Mean Value Theorem f. The graph of g for in 0, 1 is given at the left. Let p 0 0. Compute graphicall p 1, p, p 3 generated b the Fied-point Algorithm g, Use the Fied-Point Algorithm to approimate the solution of the equation 3 1 0on 1, within 10 5 with p 0 1. Give g first and eplain how ou choose g. 4. Consider solving the equation cos 0for in 0,. a. Show that the equation cos 0 has a unique solution in 0, b two steps: (i) show the equation has a solution in 0, b the Intermediate Value Theorem; (ii) show the solution is unique b Rolle s Theorem. b. Approimate the solution of the equation within 10 8 b the following methods: (i) the Bisection Method (bisect.m); (ii) the Newton Method (newton.m) with p 0 0; (iii) the Fied Point Method (fipt.m) with g cos and p 0 0. Report the number of iterations for each method. Rank the methods based on numbers of iterations. c. Estimate (the best ou can) the asmptotic error constant for the Newton Method. Does match with the performance of the Newton Method? 14

15 5. Consider the function g a. Show that g has a unique fied point on the real line. b. Can we show g has a unique fied point using the theorem for the eistence and uniqueness? Eplain. c. What is the order of convergence if we use the Fied-Point Algorithm to find this fied-point? Show our work in detail. 6. Consider the function g e. a. Show that g has a unique fied point on the interval 0,1. b. Approimate the fied point of g within 10 8 using the Fied-Point Algorithm. c. Estimate (algebraicall) the number of iterations to approimate the fied point within 10 8 (using the formula given in Notes c. after the Fied-Point Theorem. 7. The following four methods are proposed to compute 3 1. Rank them in order, based on their apparent speed of convergence, assuming p p n 1 1 p (1) p n n 1 () p 1 n p n 1 p 3 n 1 1 (3) p n p n 1 p 4 n 1 1p n 1 p n 1 1 (4) p n 1 p n 1 3p n 1 15