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NEWTON'S METHOD

GRAPHICAL INTERPRETATION :Let the given equation be f(x) = 0 and the initial approximation for the root is x0. Draw a tangent to the curve y = f(x) at x0 and extend the tangent until x-axis. Then the point of intersection of the tangent and the x-axis is the next approximation for the root of f(x) = 0. Repeat the procedure with x0 = x1 until it converges. If m is the slope of the Tangent at the point x 0 and b is the angle between the tangent and x-axis then


m = tan b = f '(x0 ) =   f(x0)
 x 0-x1
(x0-x1) * f '(x0) = f(x0 )

 
x1 = x0   -
f(x0)
f '(x0)
This can be generalized to the iterative process as 
xi+1= xi  - 
f(xi)
i = 0, 1, 2, . . .
f '(xi)
FROM TAYLOR SERIES
The same also can be obtained from the Taylor series. Let x1= x0+ h be the root of f(x) = 0 .
f(x1) = f(x0+h) = f(x0 ) + hf '(x0) +   h2  f ''(x0) + . . .
 2

 

0 = f(x0) + hf '(x0 ) +   h2  f ''(x0) + . . .
 2
Þ

   h = -    f(x0)
 f '(x0 )
Þ

x1= x0 -    f(x0)
 f '(x0 )
or in general
xi+1= xi  - 
f(xi)
i = 0, 1, 2, . . .
f '(xi)

 

Algorithm - Newton's Scheme

Given an equation f(x) = 0 
Let the initial guess be x0 
Do 
 
 


xi+1= xi  - 
f(xi)
i = 0, 1, 2, . . .
f '(xi)

while (one 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 - xi | (where i is the iteration number) less than some tolerance limit, say epsilon, fixed apriori. 

    •  
    Numerical Example :
     
    Find a root of  3x+sin[x]-exp[x]=0

    Let the initial guess x0 be 2.0
    f(x) = 3x+sin[x]-exp[x]    f '(x) = 3+cos[x]-exp[x]
    i
    0
    1
    2
    3
    4
    xi
    2
    1.90016
    1.89013
    1.89003
    1.89003
    So the iterative process converges to 1.89003 in four iterations. 

    Example 2:Show that the intial approximation x0 for finding 1/N where N is a + ve integer, by the Newton's method must satisfy 0 < x0 < 2/N for convergence. Proof : Let f(x) = 1/x - N = 0

    Þ f '(x) = -1/x2

    and Newton's Method is 
     
     


    xi+1= xi  - 
    1/xi - N
    = 2xi - Nx2i 
    i = 0, 1, 2, . . .
    -1/x2i

    Now draw the curves y = x & y = 2x - Nx2ne2.jpgThe first curve is a straight line passing through origin and the second one is a parabola (x - 1/N)2 = -1/N(y - 1/N)The point of intersection of these two curves is the required value 1/N. From the figure, we find that any initial value outside the range 
    0 < x0 < 2/N diverges. If x0 = 0, the iterative does not converge to 1/N but remains zero always.


    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 the Newton's method here


    Note : 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 (v) (x+10) ^ (1/4) 
    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