R x n. 2 R We simplify this algebraically, obtaining 2x n x n 1 x n x n

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1 Math 42 Homework 4. page 3, #9 This is a modification of the bisection method. Write a MATLAB function similar to bisect.m. Here, given the points P a a,f a and P b b,f b with f a f b,we compute the point c, that lies on the line connecting P a and P b. Then we determine within which interval a,c or c,b a root must lie by checking the signs of f a, f c, f b. You will need to develop the formula for c and insert this formula in place of c (a b)/2 in the function bisect.m. The value of c as computed must (why?) serve as your next approximation x n of the actual root x (as opposed to x n a n b n 2 as we used in bisection). Apply your function, as well as bisection, to the problem of determing the root x of f x x 2 with starting values a and b 4. Does this method improve on bisection? Does b n a n as is true in bisection? Explain why with a graphical argument. The discussion of the secant method gives the appropriate formula for the value of x c that lies on the line passing through a, f a and b, f b : c b m f b f a af b bf a f b where m or c b a f b f a As opposed to the secant method, however, here we discard either a or b so that the root still lies between c and whichever previous point we keep. We need to make a few changes in the bisection program. First the value of c needs to be the "next guess" and not (a b)/2 because a and b do not approach each other in this method. The values of c are stored in a column array P. The stopping criterion had to be changed so as to compare successive values of c and not b-a, for the same reason. Next we need to change the values of u or v, depending on whether c replaces a or b, respectively. That (luckily for me) was not an issue in the bisection method because the sign of w is always the same sign as that of the function value (f(a) or f(b)) that it replaces. Finally, it is helpful if we provide the array P as one of the outputs.

2 function [x,p] bisect2(f,a,b,tol) %function x bisect2(f,a,b,tol) %f is either a variable name assigned with an inline function %of the name of a function M-file, in single quotes u feval(f,a);v feval(f,b); %evaluates the function f(a) and f(b) if u, x a;return;end %checking if we got lucky if v, x b;return;end if u*v %checking if we made a mistake with our a,b disp( function values are not opposite sign ); return; end P [a;b]; %keep a record of root approximations while abs(p(end)-p(end-)) tol %adjustment from bisection c (a*v-b*u)/(v-u);w feval(f,c); if u*w ; b c;v w; %need to replace function values as well as points elseif v*w a c;u w; %need to replace function values as well as points else x c; %w must be opposite sign of u or v unless w return end P [P;c]; %tack on next approximation end P x c; When applied to the problem of f x x 2, a, b 4 we obtain (some values are left off): [x,p] bisect2(f,,4,.e-8); P

3 What happens is that the value of b never changes - we always have b 4. The "generic" picture in this problem is shown below. The value of c is always to the left of the root at. b-a does not approach because b never changes, but c does approach the root x. y x Because of this behavior, the line that connect a and b is a poor approximation of what is happening near the root. While the values of c converge to the root, the convergence is only linear, and in this case is worse than the bisection method. We can see this numerically by calculating in P the ratio of each error to the error of its predecessor: E -P; E(2:end)./E(:end-) ans The error thus satisfies e n. 6e n whereas, "on average", e n. 5e n in the bisection method. 2. page 7, # The equation to consider here is x 2 R or x 2 R, so that we are to find the positive root of f x x 2 R. We have f x 2x and Newton s method becomes x n x n f x n f x n x n x n 2 R We simplify this algebraically, obtaining 2x n x n x n x n R x n 2 2x R n 2 2x n 2 x n x R n. 3

4 3. page 8, #7, do by hand calculation We have f x x 5 x 3 3, f x 5x 4 3x 2 so if x n we have f x n 3, f x n 2, x n x n f x n f x n page 8 #23 We need to find an f x such that Newton s method for f x reduces to x n x n 2 x n R. A moment s thought brings you to f x x R so that f x x 2 and Newton s method is x n x n x R x n x n Rx 2 n x n 2 x n R as required. x2 n We run the experiment: x.2;x [x];for n :,x x*(2-4*x);x [X;x];end E.25-X %the array of errors E We see that the number of significant digits roughly doubles each time. For a more accurate illustration of quadratic convergence, we divide each error (of those errors that are not zero to MATLAB s precision) by the square of its predecessor. The result should approximately be constant, in fact approximately equal to f x 2 f x. We obtain: E(2:5)./E(:4).^2 ans These are clearly close to 4. Now recall f x x 2 and so f x 2 x 3 and at x /4 we have 2 f x f x

5 5. What happens when Newton s method is applied to determining the root x of the function f x tan x (in MATLAB atan(x) ) if we begin at x? What about x 2? Explain what happened graphically and calculate the exact value r with the property that Newton s method will converge in this problem when x r and diverge when x r. (Hint: Find an equation for r and then calculate its value numerically ) We just run Newton s method as requested: f inline( atan(x) ); fp inline(./( x.^2) ); x ;X [x];for n :5,x x-f(x)/fp(x);x [X;x];end x 2;X [x];for n :5,x x-f(x)/fp(x);x [X;x];end.e * Newton s method converges for x and diverges for x 2. You can see this graphically by following the tangent lines that give x n from x n. Starting at x :.5.5 x 2 x Starting at x 2: 5

6 x 2 x Where is the transition point from convergence to divergence? This occurs when you get a cycle. If x x, you will find because tan x is an odd function, that x 2 x x and we have a cycle. The equation x x results in x x tan x and replacing x by x, we obtain x x x 2 tan x or 2 x 2x x 2 tan x. We can solve this using bisection (a and b 2 should work because the first choice converges and the second diverges, but you can certainly check it) or by Newton s method. However Newton s method requires that we take a derivative (ugh) so we ll use bisection. f inline( 2*x-( x.^2)*atan(x) ); x bisect(f,,2,.e-); x x Finally, let s check by running Newton s method starting at this x. Remember to redefine f and fp back to the original problem. f inline( atan(x) );fp inline(./( x.^2) ); [x];for n :,x x-f(x)/fp(x);x [X;x];end

7 What you see towards the end is instability caused by the fact that our initial starting point and numerical computations are only approximate. Eventually x n will fall slightly to one side or other of the transition point. In this case it appears that the iterations are starting to diverge. By n 25, the divergence is apparent. 6. page 22, #8 Besides doing the problem, find a "reasonable looking" starting value for x which actually causes Newton s method to diverge - check the graph of f x Easy enough: f inline( x.^3-x.^2-x- );fp inline( 3*x.^2-2*x- ); x 2;X [x]; for n :5, x x-f(x)/fp(x);x [X;x];end But watch out: if we choose x we obtain x ;X [x]; for n :5, x x-f(x)/fp(x);x [X;x];end Warning: Divide by zero. Inf NaN NaN NaN NaN What happened? We blundered into a point where f x, sending Newton s method off to infinity. You can see this on the graph of f x. Now it is probably true that as long as we can avoid the two local extrema during the iterations, and for almost any starting value, although Newton s method may wander around quite a bit it will eventually come back and "find" the root. But the reasonably looking x starting value is just exactly the wrong choice. 7

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