Iteration & Fixed Point

Transcription

1 Iteration & Fied Point As a method for finding the root of f this method is difficult, but it illustrates some important features of iterstion. We could write f as f g and solve g. Definition.1 (Fied Point) A fied point of a function g is a real number P such that P g P. Graphically we have The main problem with this method is that of finding an appropriate g that corresponds to our f. Definition. (Fied-point Iteration). The iteration p n 1 g p n for n, 1,... is called fied-point iteration. Theorem.1 Assume that g is a continuous function and that p n n is a sequence generated by fied-point iteration. If lim n p n P, then P is a fied point of g. Proof If lim n p n P then so does lim n p n 1 P lim n g p n (above). But g is continuous so P lim n p n 1 lim n g p n g lim n p n g P. In order for this to be of use we need to know when a function has a fied point. Theorem. Assume that g C a, b. Then 1. If the range of the mapping y g satisfies y a, b for all a, b, then g has a fied point in a, b.. Furthermore, suppose that g is defined over a, b and that a positive constant K 1 eists with g K 1 for all a, b, then g has a unique fied point P in a, b. Before we look at the proof we will see what the theorem means graphically. Roots 17

2 b b a a b a a b This does not satisfy the theorem because g is not in a, b for all a, b. (In some cases, we might fi this kind of problem with a change in interval). The function has a slope 1 so this does not satisfy the condition (there is a fied point - not unique) b a a b This is OK. We might note that even if the slope 1 if the function satisfies the first part of the theorem and is monotonic there will be a unique fied point. In the following we need the Mean Value Theorem which says if f C a, b and if f eists for all a, b then there is a unique number c a, b such that f f b f b c. b a Roots 18

3 Proof (of.) 1. If either a g a or b g b we have a fied point. Suppose neither of these are true, then, because of our assumption that g is in a, b, we must have both a g a b and a g b b. Now consider the function f g. We can see that f a a g a and f b b g b. As we saw in the bisection, the IVT says that there is a point P such that f P P g P or P g P, as required.. Suppose there are fied points P 1 and P in the interval. The conditions we have are consistent with the MVT so there is a point d such that g d g P g P 1 P P 1 1. P P 1 P P 1 i.e. if two fied points eist, g must be 1 at some point in a, b. We imposed a condition that g K 1 in a, b so we conclude that we can have only one fied point. As mentioned, the process used in the fied point method is iteration. i.e. to solve g we make an initial guess (say p ) and evaluate g p g(p ) g(p ) p 1 We find p 1 g p if p p 1 we quit, otherwise we repeat and evaluate p g p 1 or, in general p n 1 g p n. If we have no problems we will produce a sequence p, p 1, p,..., p n 1. Note that in the figure the slope is 1 for p 1. What would be the position of p? If it is less than p we will converge (slowly for a while) - if it is greater than p we will diverge. p Roots 19

4 4 p - pi If there is a unique fied point, this will converge to p. (In the graph the y-ais is distance from p - the graph shows the asymptotic approach to p.) Unlike bisection, which must converge once we have an interval with f a f b, the fied point method may diverge. Fied Point Algorithm The following is a version of the Matlab code that has been adapted to Scilab (along with some statements to output intermediate results): function [k, p, err, P] fipt(g, p, tol, ma1, form) // Input - g is the iteration function // - p is the initial guess for the fied-point // - tol is the tolerance // - ma1 is the maimum number of iterations // Output - k is the number of iterations that were carried out // - p is the approimation to the fied-point // - err is the error in the approimation // - P contains the sequence (pn).entries // NUMERICAL METHODS: Matlab Programs // (c) 4 by John H. Mathews and Kurtis D. Fink // Complementary Software to accompany the tetbook: // NUMERICAL METHODS: Using Matlab, Fourth Edition // ISBN: // Prentice-Hall Pub. Inc. // One Lake Street // Upper Saddle River, NJ 7458 P(1) p; printf( \n ) for k :ma1 P(k) evstr(g (P(k-1)) ); err abs(p(k)-p(k-1)); relerr err/(abs(p(k)) %eps); Roots

5 p P(k); printf(form, real(p), imag(p), err, relerr) if (err tol) (relerr tol) then break; end; end; if k ma1 then disp( maimum number of iterations eceeded ) end; P P ; endfunction Note that the stopping condition is a combination of absolute and modified relative error. We will look at how good this is with a function that has a fied point at a small number. If we use function y fismall() y -.5* 1^(-); endfunction g.5 1, then the solution of P g P.5P 1 is P If we use the sci-file from the book, we find getf(file_loc fipto.sci ); [k, p, err, P] fipto( fismall, 1^(-),.5, 1,... %1.6e %1.6e %1.6e %1.6e\n ) 5.e-1.e 5.e e-5 P 1.D- * 1..5 err 5.D-1 p 5.D-1 k Not very accurate!!! What happened? The answer is, of course that the desired tol is.5 but our p is 1 1 leading to p This gives p 1 p so the answer is within our acceptable error but is inaccurate with respect to the correct solution. The problem is the use of err tol in the termination. This is always satisfied. Roots 1

6 To check this, we modify the fipt file to remove that condition and use if (relerr tol), break; end Then getf(file_loc fiptom.sci ); [k, p, err, P] fiptom( fismall, 1^(-),.5, 1,... %1.6e %1.6e %1.6e %1.6e\n ) 5.e-1.e 5.e e-5 7.5e-1.e.5e e-5 6.5e-1.e 1.5e e e-1.e 6.5e e-6 P 1.D- * err 6.5D- p 6.875D-1 k 5. This is better, but it gives the answer as 6.875D-1 but we know the answer is so we have just 1 digit correct. Why? The answer lies in the definition of relerr. Mathews has defined it to be relerr err/(abs(p(k)) %eps) with %eps.e-16 We find that the eps (which is used to guard against division by zero) swamps the approimation. The values are err relerr If we revise the definition of relerr to eclude eps, and write the stopping condition as if err tol*abs(p) then break; end; we find [k, p, err, P] fipt( fismall, 1^(-),.5, 1,... %1.6e %1.6e %1.6e %1.6e\n ) 5.e-1.e 5.e-1 1.e 7.5e-1.e.5e e-1 6.5e-1.e 1.5e-1.e e-1.e 6.5e e e-1.e 3.15e e e-1.e 1.565e e e-1.e 7.815e e e-1.e 3.965e e e-1.e e e e-1.e e e e-1.e e e-4 Roots

7 e-1.e.44146e e e-1.e 1.73e e e-1.e e e e-1.e e e e-1.e e e e-1.e e e e-1.e e e e-1.e e e-6 P 1.D- * err 1.97D-6 p 6.667D-1 k. which is the epected result. Note that the outout has been reformated to save space. This will often be done Recall that in bisection we had the problem of finding the point(s) at which the curves y 3 y 1 4 intersect (it took iterations). Consider using the fied point to find the root of f One of the difficulties that makes the fied point method difficult (impossible?) to use is finding the correct choice of function g. Roots 3

8 We try various possiblities - Attempt 1: First, the obvious: g f If we iterate on it starting at 1.5 which looks to be close to one root - [k, p, err, P] fipt( g1, 1.5,.5, 1,... % 1.6e %1.6e %1.6e %1.6e\n ); -8.75e-1.e.375e.71486e 6.734e.e 7.674e e e.e e e e 8.e 1.755e 8 1.5e e 4.e e 4 1.e e 7.e e 7 1.e e 16.e.8713e 16 1.e -1.#INDe.e 1.#QNANe 1.#QNANe 1.#QNANe.e 1.#QNANe 1.#QNANe maimum number of iterations eceeded If you look at the values for real(p), you see that the value is increasing quite rapidly. To understand why, look at the graph (with equal scaling on the aes) of d d g g 1 - equal scaling g 1 - -ais stretched Diverges. Slope too large. Roots 4

9 Attempt : Net we could rewrite as 4 1 from which we have 4 1 or g 1 4 1/ [k, p, err, P] fipt( g, 1.5,.5, 1,... %1.6e % 1.6e %1.6e %1.6e\n ); e-1.e e e e.e.1841e e e e e e.7536e e e e e e e e.38466e e e e.1877e e e e.96998e e e e.5651e e e e maimum number of iterations eceeded Note that, for the first time, the imaginary part of p is non-zero. This occurs because 1 4 if A look at the iteration process shows that the large slope causes the process to go well past 1.5. The line from the.997,.997 point never intersects g resulting in a comple point of intersection. Diverges: Slope too large. Roots 5

10 Attempt 3: Net we could rewrite as from which we have g / [k, p, err, P] fipt( g3, 1.5,.5, 1,... %1.6e %1.6e %1.6e %1.6e\n ); e.e.1346e e e.e e e e.e e e e.e e e e.e e e e.e e e e.e e e e.e.973e e e.e e e e.e e e e.e.745e e e.e e e e.e e e e.e e e e.e e e e.e e e e.e 4.98e e-6 Converged a bit faster than bisection with 17 iterations d d g We were lucky. slope 1 only for Converges: Cycles in g 3 iterations Roots 6

11 Attempt 4: Net we could rewrite as from which we have g [k, p, err, P] fipt( g4, 1.5,.5, 1,... %1.6e % 1.6e %1.6e %1.6e\n ); e.e e e e.e e e e.e e e e.e e e e.e e e e.e e e-6 1/ d d g Converges: Cycles in taking 6 iterations g 4 Roots 7

12 Attempt 5: The net function choice is not obvious - we will see where it comes from later on. g [k, p, err, P] fipt( g5, 1.5,.5, 1,... %1.6e % 1.6e %1.6e %1.6e\n ); e.e e e e.e e e e.e 3.96e e e.e 5.497e e d d g Converges:Steps in.taking 4 iterations g 5 Roots 8

13 We can see how these work by looking at straight lines (g in red) If g 1 we have a cycle. If g 1 diverges g 1 with g g 1 with g Theorem.3 (Fied-point Theorem). Assume that 1. a. i. g, g C a, b, ii. K is a positive constant, iii. p a, b, and iv. g a, b for all a,b.. If g K 1 for all a,b, then p n g p n 1 will converge to the unique fiedpoint P a, b. In this case P is said to be an attractive fied point. 3. If g 1 for all a, b, then the iteration p n g p n 1 will not converge to P. In this case P is said to be an repelling fied point and the iteration ehibits local divergence. Corollary.1 If g satisfies the hypotheses of Theorem.3 then bounds for the error involved in using p n to approimate P are given by and For eample, consider P p n K n P p for all n 1 P p n K n 1 K p 1 p for all n 1 Roots 9

14 g / g / 3 On 1, this is negative. g 3 1/.113 g The Corollary does not seen to apply, but (as we noted earlier) if we consider the following we can restrict ourselves to 1, 1.5 with slope.5, Then we find that the maimum slope is k From the Corollary the error would be p n p k n 1 k p p n n.6186 Recall from the output from Attempt 3 p and p 1.5, p , p , p , etc. so by the above Theoretical Actual p 1 p p p p 3 p etc. Roots 3

15 For any f the difficult part is determining a g for g, such that the iteration will converge. This requires that we find a g that has a small slope in a, b - the smaller the slope, the faster the convergence. This also involves determining the interval a,b and a good initial guess. e.g. Consider a function such as f e 1 Because f 1 e 1 and f e we know that there is a root in 1,. We might try the obvious so in 1, g e, e e 1 g e 1.35, g This does not satisfy the conditions. A less obvious choice would be In 1, g 5, 3. e 1 ln 1 g g 1 The slope appears to be OK but we need to check that g 1, Because g is increasing we have ln 3 g ln g This would appear to satisfy the requirements. [k, p, err, P] fipt( fi_ep, 1.3,.5, 1,... %1.6e %1.6e %1.6e %1.6e\n ); e.e e e e.e e e e.e e e e.e e e e.e e e e.e e e e.e 6.155e e e.e e-4.816e e.e.1454e e e.e e e e.e e e e.e e e-5 or Roots 31

16 e.e e e e.e e e e.e e e e.e e e-6 Note that if g is not in a, b we might still be able to adjust the interval to a, b for which g a, b. Advantages of fied point- It might be very fast (could also be slow). Disadvantages- Hard to find the appropriate g and a,b combination. May diverge or cycle. Roots 3