Interpolation
Interpolation is important concept in numerical analysis. Quite often functions may not be available explicitly but only the values of the function at a set of points, called nodes, tabular points or pivotal points. Then finding the value of the function at any  non-tabular point, is called interpolation.
 
Definition :

Suppose that the function f (x) is known at  (N+1)  points (x0, f0), (x1, f1), . . . , (xN, fNwhere the pivotal points xi spread out over the interval  [a,b]  satisfy  a  =  x<  x1 <   . . .   <  xN   =   b and  fi = f(xi) then finding the value of the function  at  any  non  tabular  point xs  (x0 < xs < xN)  is called interpolation.

Interpolation is done by approximating the required function using simpler functions such as, polynomials. Polynomial approximations assume the data as exact at the (N+1) tabular points and generate an Nth degree polynomial passing through these (N+1) points. However, if the given data has some errors then these errors also will reflect in the corresponding approximated function. More accurate approximations can be done using Splines and Chebicheve,  Legender and Hermite polynomials but polynomials of degree N or less passing through (N+1) points are easy to develop and useful in understanding numerical differenciation and numerical integral. Hence the present chapter is devoted to developing and using polynomial interpolation formulae to the required functions.
 

Linear Interpolation :

Consider the data ( x0, f0)(x1, f1). Problem is to find a function  f(x)  which passes through these two data points. Since there are only two data points available,   the maximum  degree  of  the  unique polynomial which passes through these points is one. Let us assume that f(x) = ax + b  is the straight line passing through the two points then

a x0  +  b  =  f0
a x1  +  b  =  f1
Solving for a and b gives
  a f1 - f0  ,    b =  f0 - f1 - f0 x0      =  f0x1 - f1x0



x1 - x0 x1 - x0
x1 - x0

 
  f(x)  
f1 - f0
  x   + 
f0x1 - f1x0


x1 - x0
x1 - x0
This is the required curve passing through the given two points.  Now one can substitute the value of   x Î (x0, x1in the equation to find f(x).   Similarly  the  same can be extended to find an unique Nth degree  polynomial  which  passes  through  (N+1)  data  points.   The  main  difficulty  in  the   above procedure is the lengthy calculations involved to arrive at the function f(x) which is a function of given data points for higher degree approximations.    For example let us consider the case with three data points. Let the points be
x0
x1
x2
f0
f1
f2
and the second degree polynomial approximation is  f (x) @ P2(x) = ax2 + bx + c   then we have
ax02 + bx0 + c = f0,      ax12 + bx1 + c = f1,            ax22 + bx2 + c = f2
Solving these three equations for a, b and c gives
(x2 - x1)f0 + (x0 - x2)f1 + (x1 - x0) f2
a =
x02(x1 - x2) + x12(x2 - x0) + x22(x0 - x1)

 
(x22 -x12)f0 + (x02 -x22)f1 + (x12 - x02) f2
b = 
x02(x1 - x2) + x12(x2 - x0) + x22(x0 - x1)

 
(x1 - x2)x1x2f0 + (x2 - x0)x0x2f1 + (x0 - x1)x0x1f2
c = 
x02(x1-x2) + x12(x2-x0) + x22(x0-x1)
The corresponding second degree polynomial approximation for the given data is
f (x)  = ax2 + bx + c

So  finding  the coefficients  of  the  polynomial  is not an easy task for polynomials of degree greater or equal to two if we follow the usual algebric means to solve the system of equations. The alternative is to simplify the calculations by making use of Vandermonde's determinants or by generating difference formulae. So in the  following sections  we  discuss  Newton's  difference  formulae  which  depend  on  difference  operator  and Lagrange's  method  which  depends  on Vandermonde's determinant.


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