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MULLER METHOD

Muller's method is a generalization of the secant method. Instead of starting with two initial values and then joining them with a straight line in secant method, Mullers method starts with three initial approximations to the root and then join them with a second degree polynomial (a parabola), then the quadratic formula is used to find a root of the quadratic for the next approximation. That is if x0, x1 and x 2 are the initial approximations then x3 is obtained by solving the quadratic which is obtained by means of x0, x 1 and x2. Then two values among x0, x1 and x2 which are close to x3 are chosen for the next iteration


              x3 = x2 + z           ( * )
 
             where z =         -2c 
b±Ö( b2-4ac)

              a = D1/D2 ,  b = D 2 /D     and   c = f(x2)

              D = h0h1(h0-h1) ,      D1 = (f0-c)h1-(f1-c)h0 ,        D2 = (f1 -c)h02 - (f0-c)h12

              h0 = x0-x2 ,   h1 = x1-x2
 
 
 


Algorithm - Muller's Scheme

Given an equation f(x) = 0 
Let the initial guesses be x0,  x1 and x2
Let xi = x2,  xi-1 = x1 and xi-2 = x0
Compute f(xi-2), f(xi-1) and f(xi)

Do

           Compute 
                        h = x - xi,  hi = xi - xi-1  and  hi-1 = xi-1 - xi-2
                       li = hi / hi-1, di = 1 + li
                        gi = li2 fi-2 - di2 fi-1 +  ( li + di ) fi
                        ci = li  (li fi-2 - di fi-1 +   fi )    
                         li+1 = min (  -2di fi / (gi ± Ö(gi2- 4di fici)  ) )
                                                             (take minimum in absolute sense)

           Then xi+1 =  xi + ( xi  -  xi-1 ) li+1 ,       i = 2, 3, 4, . . . 

while (none of the convergence criterion C1 or C2 is met)
  • C1. Fixing apriori the total number of iterations N. 
  • C2. By testing the condition  | xi+1 - x i | (where i is the iteration number) less than some tolerance limit, say epsilon, fixed apriori. 
    • Numerical Example :
    Find the root of  3x+sin[x]-exp[x]=0                          [ Graph] 
    Let the initial guess be 0.0 and 1.0
    f(x) = 3x+sin[x]-exp[x]
    i
    0
    1
    2
    3
    4
    		 5 6
    xi
    0
    1
    0.471
    0.308
    0.363
    		 0.36 0.36
    So the iterative process converges to 0.36 in six iterations. 



    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

     
     
    Work out with theMuller  method here

    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