Fixed-Point Iterations (2.2)

Transcription

1 Fied-Point Iterations (.). Mean Value Theorem: a. Rolle s Theorem : Suppose that f is continuous on a, b and is differentiable on a, b.iff a f b, then there eists a number c in a, b such that f c. b. 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 forall 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. Graphicall, Rolle s Theorem and the Mean Value Theorem can be described as follows f f, f c. When b is close to a ( b a is small), f c f b f a b a c.4, c. and f f b f a b a for a,b. Eample First show that the equation e has a solution in,. Then Determine if the solution is unique. Let f e.sincef f e.63, b the Intermediate Value Theorem there eists a number c in, such that f c. Therefore, the equation e hasa solution in,. Now let us check to see if f for in,, f e for all. So, f for all in, and therefore, f onl once and the solution is unique.

2 Eample The graph of f cos. 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.. Approimatel, c.9 and c f f f. c..9 f f f. c f Comparison: f c.446, f c.37. Fied-Point of a Function: 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 and the line intersect at p. e p p, when p.8 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 and the function in (ii) has infinitel man fied points.

3 (i) (ii) cos Algebraicall, we solve the equation g (or g ) to determine if a function has an fied point over a given interval. Eample Find all fied points of g and g cos if the eist. a. Set g :,. Using the quadratic formula: 4 3 no real solution. So, g has no fied point for. b. Set g : cos, cos, n, n,,3,... So, g has infinitel man fied points for. 3. Eistence and Uniqueness of a Fied Point: Let g be continuous on a, b. i. If a g b for all in a, b, then g has a fied point p in a, b. ii. If, in addition, g eists on a, b and there eists a constant K 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 means that the graph of g does not grow as faster than and not slower than. Proof: 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 andh b g b b. So, b the Intermediate Value Theorem, we know there eists a number c in a, b such that h c, that is g c c org c c. 3

4 So, c is a fied point of g in a, b. ii. Now let also g K for all in a, b where K. 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. So, g c, this contradicts the given condition g for all in a, b.so,gcannot have two fied points in a,b. Eample Let g 3 for in,. Determine if g has a fied point in,.ifso, determine if the fied point is unique. Check: i. Observe that g min g 3 andg ma g g. Since g for all in,, g has a fied point in,. ii. Compute g 3.Since g 3, g has a unique fied point in,. 3 For g, we can solve its fied point p algebraicall. g Since 3 3, p is a unique fied point in,.. Check the graph of g : , -.-., Eample Let g 3 for in,. Determine if g has a fied point in,. If so, determine if the fied point is unique. Check: i. Observe that g min g 3, and g ma g. Since g, g has a fied point in,. ii. Compute g 3 ln3. g 3 ln3. Since g ln3, there is no conclusion about the uniqueness. From the graph of g below, we can see that g has a unique fied point p. in,. But we 4

5 cannot solve p algebraicall. How can solve a fied point numericall? , in, 4. The Fied-Point Iteration: It is an algorithm to find a fied-point of a function over an interval assuming the fied point is unique. Algorithm: Given g, and a, b, choose p in a, b and compute p,p,..., as follows: p n g p n for n,,... Implement the algorithm in a programming language which does the following: Input g, interval a, b, p in a, b, and K ma, and compute p n g p n for n,,... The program terminates if a. p n p n and then p p n ;or b. p n b or p n a, and the program fails; or c. n K ma. The following two eamples show graphicall how the Fied-Point Iteration works.. Fied Point Iterations: {p n }, g()=(-e + -)/3, p = 6 Fied Point Iterations: {p n }, g()=(-.), p =.4.4 (p,g(p )) (p,p )=(g(p ),g(p )).3.3 (p,g(p )) 4 (p,g(p )) =. (p,p ) 3 (p,g(p )).. (p,p ).. p p p p p (i) Clearl, the Fied-Point iteration finds the fied point p in (i) and diverges in (ii). (ii). Convergence and the Rate of Convergence:

6 Questions: Assume that g has a unique fied point p in a, b and p is in a, b.let p n g p n, n,,... i. Under what condition(s), does p n converge to p? ii. If lim n p n p, what is the rate of converge? Fied-Point Theorem: Let g be continuous on a, b and a g b. Suppose that g eists for all in a, b,and Then lim n p n p for an p in g K for all in a, b where K. a, b,and p n p K n ma p a, b p and p n p K n K p p, for all n,,... Proof: Let p be in a, b and p n be generated b the Fied-Point Iteration. Observe that p n p g p n g p. B the Mean Value Theorem, we know there eists a number c in a, b such that g p n g p p n p g c or g p n g p g c p n p. Since g for all in a, b, p n p g p n g p g c p n p g c p n p K p n p K K p n p K p n p... K n p p lim n p n p lim n K n p p. Therefore, lim n p n p. Since p p p a or p p b p, p n p K n p p K n ma p a, b p. Observe that p n p n g p n g p n K p n p n K g p n g p n K g p n g p n 3... K n p p So for an m n, p m p n p m p m p m p m... p n p n p m p m p m p m... p n p n K m p p K m p p... k n p p K n K m n K m n... K p p Since lim m p m p, 6 K n Km n K p p

7 p p n m lim p m p n m lim K n Km n K p p K n K p p Note that : a. n p n p p n p K K p p, for all n,,...implies that K n K p p.sop n p n p O K n. converges to pwith the rate of convergence of O K n, i.e., b. For a given, we can estimate the number N of iterations needed to approimate p b p N.Thatis, find N such that K N K p p K N K b a KN K b a Since K, ln K. So, N ln ln K b a ln K Nln K ln. K b a Eample Determine whether or not the function has a fied point in the given interval. If so, determine if the Fied-point Iteration 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. (i) g 3 e,, (ii) g 3 /,, (i) Check the range of g : From the graph of g on,, wehave g g 3 e g 3 e So, g 3 e has a fied point in,. Check the maimum value of g : g 3 e. From the graph of g, g g 3 K for all in,. 3 So, g has a unique fied-point in, and the sequence p n generated b the Fied-Point Iteration converges to p. Estimate the number N of iterations: N ln ln , N. ln 3 Use the Fied-Point Iteration to solve the fied point in, and p p n where 7

8 p n 9. p n p n. b. Check the range of g : Observe that g min g 8, g ma g So, g for all in,. Hence, g has a fied point in,. Check the maimum value of g : g 3. 3 g 3 3 From the graph of g, g 3 g forsome in,. So we cannot conclude the sequence p n generated b the Fied-Point Iteration converges to p. Use the Fied-Point Iteration to solve the fied point in, and p p n where p n p n. p, p , p ,...,p , p Fied-Point Iteration for Solving The Equation: f Let be a solution of the equation f. To solve using the Fied-Point Iteration, a function g needs to be defined first such that is a fied point of g, thatis, g. 8

9 Eample Consider solving 3. 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. Determine if the sequence p n generated b the Fied-Point Iteration with the function g. Consider a, b,.sincef f, the equation has a solution in,. A naive choice of g : since 3, we can let g 3. Check the range of g : g min g andg ma g, so, g andghas a fied point in,. Check the maimum value of g : g 3, g 3 forsome in,. So, it is not certain b the Fied-Point Theorem if p n converges to p. Observe that p n n,,,,.... Rewrite the equation 3 as 3, 3.Let g 3. Check the range of g : g min g, g ma g, so, g andg has a fied point in,. Check the maimum value of g : g 3. Since 3 g is not defined at, g is unbounded. So, it is not certain if p n converges to p. Observe that p n n,,,, Rewrite the equation 3 as 3 andthen or g.. Let Check the range of g : g min g, and g ma g,so, g andghas a fied point in,..6 Check the maimum value of g : g, g g From the graph of g above, we see g.7 K. So, p is unique in, and p n converges to p. Estimate the number N of iterations needed:

10 N ln ln , N 36. ln.7 Use the Fied-Point Iteration to solve the fied point in, 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, where f 4 3. a. g 3 /4 b. g a. Set 4 3. Then /4. b. Check if Then So, Eample The following four methods are proposed to compute 7 /. Rank them in order, based on their apparent speed of convergence, assuming p. (i) p n 7 p 3 / n (ii) p n p n p n 7 (iii) The function g for each iteration is: p n p n p n p n p n 7 (iv) p 4 n p n p n 7 p n (i) g 7 3 / 7 / (iii) g (iv) g (ii) g

11 g() 8 6 graphs of g, g, g3 and g4 =g () 4 =g () = =g 3 () =g 4 () All four functions have a fied point on a, b. Compute g : (i) g 7 / 4 3 (ii) g (iii) g (iv) g 4 4 Check the value of g at p 7 / : g p 7 / p p g p 3p 4 p g 3 p 4 8 p g 4 p p p So, the sequences p n generated b the Fied-Point Iteration using g and g do not converge. The sequences p n generated b the Fied-Point Iteration 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 / p using g 4, 7 / p Eample Show that if A is an positive number, then the sequence defined b n n A, for n, n

12 converges to A whenever. Let lim n n. Then lim n n lim n n A n A A A A.