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NUMERICAL DIFFERENTIATION


  1. Methods based on interpolation: 


    2. Methods based on interpolation uses the ponomial approximation obtained by nterpolation to find the derivative of the function, which is known at discrete points in the interval [a, b]. 

    Consider the function f(x) known at a set of points, say, x0, x1, . . ., xn has function values f0, f1, . . ., fn respectively.

    Then by Newton's divided difference formula

    f(x) = pn(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2] 
                       +  . . .   + (x - x0) (x - x1) . . . (x - xn-1) f [x0, x1, . . ., xn]

    is an nth degree polynomial approximation to the function f(x).

    The r (r<n)th derivative of the function at any point x Î [a,b] can be obtained by finding  dr(pn(x))/dxr at x.

    Here, though f(x) and the approximated polynomial Pn(x) coinside at the nodal points xi, i = 0,1, . . . n, the values of derivatives of the function f(x) need not coinside with the value of derivatives of Pn(x) even at nodal points. The situation may be much worse at non-nodal points. 

    The error in the interpolation is given by 

    Er(x) = f(r)(x) - Pn(r)(x)





    Case(i): Interpolation with a linear polynomial

    If (x1, f1) and (x2, f2) be any two data points then from Lagrange's interpolation formula
     

     
    (x - x2)
    		  
    (x - x1)
    		  
    f(x)  =  P1(x)  =
    f1  + 
     f2
     
    (x1 - x2) 
    		  
    (x2 - x1)
    		  

     
     
    f1
    		  
     f2
    		  
    f '(x)  =  P1'(x)  =
      + 
    		 
     
    (x1 - x2) 
    		  
    (x2 - x1)
    		  

     
         
     f2 - f1
    		  
       
    = 
     
     
     
     
    (x2 - x1)
     

    Error in the approximation
     

    E1(x) = f '(x) - P1'(x)

     
     
    [
    		 (x - x1) (x - x2)
     
    		 ] /
    = 
    		 f(2)(x)
     
      2!
    		  
    x0 < x < xn
      2x - x1 - x2
     + 
    		 (x - x1) (x - x2)
     
     = 
    		 f(2)(x)
    		 d/dx   f(2)(x)
     
    2
      2
    		  

    since x is unknown and is a function of x, finding d/dx  f(2)(x) at x =x is not possible. However, at the given datapoints since the coefficient of this term is zero, the total error can be written as
     

      2x - x1 - x2
     E1(x)  = 
    		 f(2)(x)
     
    2

    at the nodal points.

    Case(ii): Interpolation with a second degree polynomial

    Non-uniform grid points :

    Again by Lagrange's formula
     

     
    (x - x2)(x - x3)
    		   (x - x1)(x - x3)
    (x - x1)(x - x2)
    		  
    f(x) = P2(x) = 
     f1 + 
    		 f2+ 
     f3
     
    (x1 - x2)(x1 - x3)
    		   (x2 - x1)(x2 - x3)
    (x3 - x1)(x3 - x2)
    		  

     
     
    (2x - x2 - x3)
    		   (2x - x1 - x3)
    (2x - x1 - x2)
    		  
    f '(x) = P2'(x) = 
     f1 + 
    		 f2+ 
     f3
     
    (x1 - x2)(x1 - x3)
    		   (x2 - x1)(x2 - x3)
    (x3 - x1)(x3 - x2)
    		  

    Uniform grid points

    The error in the approximation

    E'2(x) = [(x - x1)(x - x2)(x - x3)]  1/2!  f(3)(x)               x1 < x < x2
     

    Case(iii): Interpolation with a polynomial of degree n.

    Non-uniform grid points :

    The nth degree Lagranges interpolation formula for f(x)
     

    n
    f(x) = Pn(x) =
    S
    		 li(x)fi
    i=0

     
    li(x) =        p(x) 
    (x - xi)p'(xk)

    with p(x)  = (x - x0) (x - x1) . . . (x - xn)
     

    n
    f '(x) = Pn'(x) =
    S
    		 li'(x)fi
    i=0

    The error in the approximation is
     

      p'(x)
     + 
    		 p(x)
     
     E'n(x)  = 
    		 f(n+1)(x)
    		 d/dx   f(n+1)(x)
     
    (n + 1)!
      (n + 1)!
    		  
    where   x0 < x < xn
    Again, since x is an unknown in (x0, xn) finding  E'n(x) at x = xis not feasible. However at nodal points xi, i = 0,1, . . ., n since p(xi) = 0 we get
     
      p'(x)
     
     
     E'n(x)  = 
    		 f(n+1)(x)  
     where   x0 < x < xn
     
    (n + 1)!
     
    		  

    provided  d/dx (f(n+1)(x)) is finite

    These formulae can be extended to find any rth derivative at a point of a function f(x) in the interval 
    (x0, xn) by taking the rth derivative of Pn(x). For example, for any  1 <r <n we get 
     

    n
    f (r)(x) = Pn(r)(x) =
    S
    		 li(r)(x)fi
    i=0
    at any point x. 
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