BISECTION METHOD
Bisection method is the simplest among all the numerical schemes
to solve the transcendental equations. This scheme is based on the intermediate
value theorem for continuous functions
.
Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0.
Bisection scheme computes the zero, say c, by repeatedly halving the interval [a,b].
That is, starting with c = (a+b) / 2
the interval [a,b] is replaced either with [c,b] or with [a,c] depending
on the sign of f (a) * f (c) . This process is continued until the zero
is obtained. Since the zero is obtained numerically the value of c may
not exactly match with all the decimal places of the analytical solution of f (x) = 0 in the interval [a,b]. Hence any one of the following mechanisms can be used to stop the bisection iterations :
C1. Fixing a priori the total number of bisection iterations N i.e., the length of the interval or the maximum error after N iterations in this case is less than  ba  / 2^{N}.
C2. By testing the condition  c_{i}  c_{ i1}
(where i are the iteration number) less than some tolerance limit, say
epsilon, fixed a priori.
C3. By testing the condition  f (c_{i} )  less
than some tolerance limit alpha again fixed a priori.
Algorithm  Bisection Scheme
Given a function f (x) continuous on an interval [a,b] and f (a)
* f (b) < 0
Do
c = (a+b)/2
if f (a) * f (c) < 0 then
b = c
else a = c
while (none of the convergence criteria C1, C2 or C3 is satisfied) 
Numerical Example :
Find
a root of f (x) = 3x + sin(x)  exp(x) = 0.
The graph of this equation is given in the figure.
Its clear from the graph that there are two roots, one lies between 0 and 0.5 and the other lies between 1.5 and 2.0. Consider the function f (x) in the interval [0, 0.5] since f (0) * f (0.5) is less than zero.
Then the bisection iterations are given by
Iteration
No.

a

b

c

f(a) * f(c)

1

0

0.5

0.25

0.287 (+ve)

2

0.25

0.5

0.393

0.015 (ve)

3

0.65

0.393

0.34

9.69 E3 (+ve)

4

0.34

0.393

0.367

7.81 E4 (ve)

5

0.34

0.367

0.354

8.9 E4 (+ve)

6

0.354

0.367

0.3605

3.1 E6 (ve)

So one of the roots of 3x + sin(x)  exp(x) = 0 is approximately
0.3605.
Worked out problems
Exapmple 1 
Find a root of cos(x)  x * exp(x) = 0 
Solution 
Exapmple 2 
Find a root of x^{4}x10 = 0 
Solution 
Exapmple 3 
Find a root of xexp(x) = 0 
Solution 
Exapmple 4 
Find a root of exp(x) * (x^{2}5x+2)
+ 1= 0 
Solution 
Exapmple 5 
Find a root of xsin(x)(1/2)= 0 
Solution 
Exapmple 6 
Find a root of exp(x) = 3log(x) 
Solution 
Problems
to workout

Highlights
of the scheme
Work out with the here
Note : Please
enter equation like 3x+sin[x]exp[x]. Use "[ ]" brackets for transcendentals
and "( )" for others eg., 3x+sin[(x+2)]+(3/4). 'a' and 'b' are the limits
within which you are going to find the root. Few examples of how to enter
equations are given below ...
(i) exp[x]*(x^2+5x+2)+1
(ii) x^4x10
(iii) xsin[x](1/2) (iv)
exp[(x+212+1)]*(x^2+5x+2)+1

