 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 | b-a | / 2N.

• C2. By testing the condition  | ci - c i-1| (where i are the iteration number) less than some tolerance limit, say epsilon, fixed a priori.

• C3. By testing the condition | f (ci ) | 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 E-3 (+ve) 4 0.34 0.393 0.367 -7.81 E-4 (-ve) 5 0.34 0.367 0.354 8.9 E-4 (+ve) 6 0.354 0.367 0.3605 -3.1 E-6 (-ve)
So one of the roots of 3x + sin(x) - exp(x) = 0 is approximately 0.3605.
 Exapmple 1 Find a root of cos(x) - x * exp(x) = 0 Solution Exapmple 2 Find a root of x4-x-10 = 0 Solution Exapmple 3 Find a root of x-exp(-x) = 0 Solution Exapmple 4 Find a root of exp(-x) * (x2-5x+2) + 1= 0 Solution Exapmple 5 Find a root of x-sin(x)-(1/2)= 0 Solution Exapmple 6 Find a root of exp(-x) = 3log(x) Solution Problems to workout

 Work out with the Bisection method 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^4-x-10  (iii) x-sin[x]-(1/2)  (iv) exp[(-x+2-1-2+1)]*(x^2+5x+2)+1  Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations