Practical Numerical Analysis: Sheet 3 Solutions

Transcription

1 Practical Numerical Analysis: Sheet 3 Solutions 1. We need to compute the roots of the function defined by f(x) = sin(x) + sin(x 2 ) on the interval [0, 3] using different numerical methods. First we consider the bisection method. Recall in this method we choose values a and b such that f(a) and f(b) have opposite sign. We then set c = (a + b)/2 (i.e. the midpoint of the interval [a, b]), compute f(c) and then repeat on the half of the interval where the sign changes. Thus a bisection algorithm function can be written as follows: % B i s e c t i o n method % Here we assume t h a t f ( a ) and f ( b ) have o p p o s i t e s i g n % and are non zero. % I t would be b e t t e r to make a check of t h i s and return % an e r r o r i s t h i s i s not the case. function [ xnew, i t s ]= b i s e c t ( a, b, t o l ) f a=f ( a ) ; fb=f ( b ) ; i t s =0; c=(a+b ) / 2 ; while ( abs ( f c )> t o l && i t s <2000) i f f c fa >0 % f ( a ) and f ( c ) have the same s i g n so r e p l a c e a with c a=c ; f a=f c ; % r e p l a c e b with c b=c ; fb=f c ; c=(a+b ) / 2 ; i t s=i t s +1; xnew=c ; We also recall that we can predict the number of steps the bisection algorithm requires for the roots to converge to a tolerance tol. This is given by n = log(b a) log(tol) log(2) For the regula falsi algorithm we make a more intelligent prediction of c by defining c = af(b) bf(a) f(b) f(a) 1..

2 Thus the regula falsi code is an almost trivial change to the bisection code. For the Illinois algorithm we use the same definition of c as in regula falsi except that if the algorithm goes right twice we halve f(b) so that c = af(b)/2 bf(a) f(b)/2 f(a). There is an analogous expression for c which uses f(a)/2 is we go left twice. For this method we thus need to keep track of whether we stepped left or right last. We can do this with a flag as shown in the code below. % I l l i n o i s a l g o r i t h m % f l a g keeps a check on whether we stepped l e f t or r i g h t at the % p r e v i o u s i t e r a t i o n function [ xnew, i t s ]= i l l i n o i s ( a, b, t o l ) f a=f ( a ) ; fb=f ( b ) ; i t s =0; c=(a fb b f a ) / ( fb f a ) ; flag =0; while ( abs ( f c )> t o l && i t s <2000) i f f c fa >0 % r e p l a c e a with c a=c ; f a=f c ; i f ( flag == 1) c=(a fb/2 b f a ) / ( fb/2 f a ) ; c=(a fb b f a ) / ( fb f a ) ; flag= 1; % r e p l a c e b with c b=c ; fb=f c ; i f ( flag == 1) c=(a fb b f a /2)/( fb f a / 2 ) ; c=(a fb b f a ) / ( fb f a ) ; flag =1; 2

3 i t s=i t s +1; xnew=c ; Finally, we use Newton s method for root finding where the iterates are defined as x k+1 = x k f(x k) f (x k ). This can be evaluated very simply in Matlab as % Newton s method to s o l v e f ( x)=0 % The f u n c t i o n s f and fp return the v a l u e of the f u n c t i o n and i t s % d e r i v a t i v e function [ xnew, i t s ]=newt ( xold, t o l ) f x o l d=f ( xold ) ; i t s =0; while ( abs ( f x o l d )> t o l ) xnew=xold f x o l d / fp ( xold ) ; xold=xnew ; f x o l d=f ( xold ) ; i t s=i t s +1; xnew=xold ; First we plot the function f(x) on the interval [0, 3] to see how many roots there are and roughly where they lie in order to get initial guesses for the rootfinding methods. A plot of f(x) is shown in Figure f(x) x Figure 1: A plot of f(x) = sin(x) + sin(x 2 ). 3

4 We then run our rootfinding algorithms to get the results shown in Tables 1 and 2. The results show that, as expected, each method converges to the same set of roots. The regula falsi method improves on the bisection method (in terms of iteration counts) and the Illinois method improves on regula falsi. The predicted number of iterations for the bisection method is a slight overestimate of the actual number of iterations required but this is probably because the theoretical prediction is based on the error in the root whereas the algorithm terminates when the residual (namely f(c) ) is less than the tolerance. Table 2 also demonstrates fast convergence of Newton s method to the three roots. [-0.1, 1, 1e-10] [1, 2.2, 1e-10] [2.2, 3, 1e-10] Bisection root Bisection its Bisection predicted its Regula falsi root Regula falsi its Illinois root Illinois its Table 1: Roots found by the rootfinding algorithms based on the bisection method from starting guesses a, b and with tolerance tol given as a triple [a, b, tol]. x 0 = 0.1 x 0 = 2 x 0 = 2.5 Newton root Newton its Table 2: Roots found by Newton s method with a tolerance of 1e-10. 4

5 2. In this question we are required to compute a Newton fractal for the function f(z) = z 3 1. We can either use Newton s method with starting guesses in a region of the complex plane, or we can rewrite this as a system of the form x 3 3xy 2 1 = 0, 3x 2 y y 3 = 0, or g(x) = 0 with x = (x, y) T. In the former case the Newton iterates are given by whereas for the system we have z n+1 = z n f(z n) f (z n ), x n+1 = x n J(x n ) 1 g(x n ), where J(x n ) is the Jacobian of g evaluated at the point x n. We then colour the starting guess according to which root we converge to. The codes for this are on the course material webpage. The code newt frac.m computes the Newton fractal for f(z) = z 3 1 using the system approach while newt fracz.m uses complex numbers (the latter is much faster). The resulting fractal is shown in Figure 2. The code newt frac2.m computes the Newton fractal for f(z) = z 4 1 and the resulting fractal is shown in Figure 3. Figure 2: The Newton fractal for f(z) = z

6 Figure 3: The Newton fractal for f(z) = z