1. Solution of Nonlinear Equations (Root finding Problems

2. Bisection Method
When f(x) is continuous in the environment of a single root r, then f(x) changes sign through r. There is a small interval [a,b] including r such that f(a).f(b) < 0. Taking the midpoint m of [a,b], there are three possibilities.
f(m) = 0 ; then m is the root r.
f(m).f(a) < 0 ; then the root r is in [a,m]
f(m).f(b) < 0 ; then the root r is in [m,b]

3. Now we can restart the procedure with the smaller interval [a,m] or [m,b]. The interval becomes smaller and smaller, so we can find an approximation of the root r.
Example
t t = 0
has a positive root r in [0,1]. f(0)<0 and f(1)>0.

4. Since f(0.5) = 0.125 the root is in [0, 0.5].
and so we approach the root

6. Algorithm Input: Continuous function f(x) Interval ends a,b
Output: root of function f(x) in the interval [a,b], i.e. find x ∈ [a,b] such that f(x)=0
Assumptions:
f(x) is continuous on [a,b]
f(a) f(b) < 0

7. Steps:
Loop
1. Compute the midpoint c=(a+b)/2
2. Evaluate f(c ) if f(c)==0, stop c is the root
3. If f(a) f(c) < 0 then new interval [a, c]
If f(a) f( c) > 0 then new interval [c, b]
End loop

8. Example
Find the root of f(x)=x2 -7x+10
n the interval [1,4] with accuracy 0.01
1. Check the assumptions:
f(x) is continuous on [a,b] -> yes
f(1)f(4)=(12-7 X 1 +10)(42-7 X )=4 X -2=-8 < 0

9. Apply the Algorithm

11. Program
function [x,n]=bisection(f,a,b)
TOL=1e-2; %tolerance
NO=100; %maximum number of iterations
fx=2; n=0; % initial values
while (abs(fx)>TOL)&&(n<=NO)
n=n+1;
x=a; fa=eval(f);

12. x=(a+b)/2; fx=eval(f);
if (sign(fx)==sign(fa))
a=x;
else
b=x;
end

13. end
if n>NO
x=[];
n='No zeros in given interval';