Page 210 - Applied Numerical Methods Using MATLAB

PROBLEMS 199
4.2 Bisection Method and Fixed-Point Iteration
Consider the nonlinear equation treated in Example 4.2.

f(x) = tan(π − x) − x = 0 (P4.2.1)

Two graphical solutions of this equation are depicted in Fig. P4.2, which
can be obtained by typing the following statements into the MATLAB
command window:

>>ezplot(’tan(pi-x)’,-pi/2,3*pi/2)
>>hold on, ezplot(’x+0’,-pi/2,3*pi/2)

(a) In order to use the bisection method for finding the solution between
1.5 and 3, Charley typed the statements shown below. Could he get the
right solution? If not, explain him why he failed and suggest him how
to make it.

>>fp42 = inline(’tan(pi-x)-x’,’x’);
>>TolX = 1e-4; MaxIter = 50;
>>x = bisct(fp42,1.5,3,TolX,MaxIter)

(b) In order to find some interval to which the bisection method is applica-
ble, Jessica used the MATLAB command “find()” as shown below.
>>x = [0: 0.5: pi]; y = tan(pi-x) - x;
>>k = find(y(1:end-1).*y(2:end) < 0);
>>[x(k) x(k + 1); y(k) y(k + 1)]
ans = 1.5000 2.0000 2.0000 2.5000
-15.6014 0.1850 0.1850 -1.7530

This shows that the sign of f(x) changes between x = 1.5 and 2.0
and also between x = 2.0 and 2.5. Noting this, Jessica thought that she
might use the bisection method to find a solution between 1.5 and 2.0
by typing the following command.

>>x=bisct(fp42,1.5,2,TolX,MaxIter)
Check the validity of the solution—that is, check if f(x) = 0 or not—by
typing

>>fp42(x)
If her solution is not good, explain the reason. If you are not sure about
it, you can try plotting the graph in Fig. P4.2 by typing the following
statements into the MATLAB command window.

>>x = [-pi/2+0.05:0.05:3*pi/2 - 0.05];
>>plot(x,tan(pi - x),x,x)