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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 :
    exp0.jpg for 3x+sin[x]-exp[x]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.
    Worked out problems
     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

    Highlights of the scheme



     
     
     
    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