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For Newton's Method
 
g(x) = x -    f(x)
  f '(x)

 
  g'(x) = 1 -  f '2 - f f ''   f f '' 
  = 
     f '2    f '2 
if the multiplicity of the root m > 1 then f (x) and f '(x)  at x = s or zeros hence 
 
   g'(x)|x =s =  0 * f ''
    0

      Now by repeadtedly using L'Hospitals rule we get
 

  lim
x ® s 		    lim
  x ® s 		    f f ''    lim
  x ® s 		   f ' f '' + f f '''
 g'(x)    =
      = 
   f '2      2 f ' f ''

 
  =     lim
  x ® s 		  2 f ' f ''' + f ''2 + f f 'v 1
m		 )                           ( for m = 2 )
1
2
      =  ( 1 - 
      f ' f ''' + f ''2

 
  =     lim
  x ® s 		   3 f ' f 'v + 4f '' f ''' + f f v  
1
2
      3 f '' f ''' + f ' f 'v

 
 
  =     lim
  x ® s 		  7 f '' f 'v + 4 f ' f v + 4 f '''2 + f f v'  
1
2
         4 f '' f 'v + 3 f '''2 + f ' f v

 
=   1   *   4   =   2   = (1 -  1  )                                     (for m = 3)
 2  3  3  m

Similarly for m>3 also we can show that the coefficient of ei is not zero but equal to (1-1/m). In the same way the coefficient of ei2 can be obtained as (The derivation is left to the reader).
 

      1   f(m+1)s
m2(m+1)    f(m)s
 
Solution of Transcendental Equations | Solution of Linear System of Algebric Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differetial Equations
Numerical Solution of Partial Differential Equations