Solving Nonlinear Equations

We have given the following function;

$\begin{displaymath} f(x)=3x + sin(x) - e^x \end{displaymath}$

Look at to the plot of the function to learn where the function crosses the x-axis. MATLAB can do it for us:

>> fx = inline ( ' 3 *x + sin ( x) - exp ( x) ')
>> fplot ( fx, [ 0 2 ]) ; grid on

An algorithm for Halving the Interval (Bisection):
$\fbox{\parbox{10cm}{ To determine a root of $f(x)=0$\ that is accurate within a ... ...et $x_l=x_3$\ End If\\ Until $(\vert x_1 - x_2\vert) < 2*tolerance~value$\\ }}$
The MATLAB program for this algorithm is given.

function rtn=bisec(fx,xa,xb,n)
%bisec does n bisections to approximate
% a root of fx
x=xa; fa=eval(fx);
x=xb; fb=eval(fx);
for i=1:n;
  xc=(xa+xb)/2; x=xc; fc=eval(fx);
  X=[i,xa,xb,xc,fc];
  disp(X);
  if fc*fa<0
    xb=xc;
    else xa=xc;
  end
end

save with the name bisec.m. Then;

>> fx=' 3 *x + sin ( x) - exp ( x) '
>> bisec(fx,0,1,13)

Modify this MATLAB program for the bisection method for using a tolerance value of 1E-4.

Use the function used in the previous item, and write a MATLAB program for the method of false position (regula falsi):
An algorithm for the method of false position (regula falsi):
$\fbox{\parbox{10cm}{ To determine a root of $f(x) = 0$, given two values of $x_0... ...\\ Set $x_0= x_2$\\ End If\\ Until $\vert f(x_2)\vert<tolerance~value$.\\ }}$

To obtain the true value for the root

, which is needed to compute the actual error. MATLAB surely used a more advanced method than bisection.

>> solve('3*x + sin(x) - exp(x)')
ans=
.36042170296032440136932951583028

Tabulate the actual error values as the following;

Table 1: The Error Sequences

n	Bisection	Regula Falsi	Bisection	Regula Falsi
1
2
3
4
5
6
7
8
9
10
12
13
14
15

2004-10-15