LAGRANGE'S INTERPOLATION FORMULA
This is again an Nth degree polynomial approximation formula to the function f(x), which is known at discrete points xi, i = 0, 1, 2 . . . Nth. The formula can be derived from the Vandermonds determinant but a much simpler way of deriving this is from Newton's divided difference formula. If f(x) is approximated with an Nth degree polynomial then the Nth divided difference of f(x) constant and (N+1)th divided difference is zero. That is
f [x0, x1, . . . xn, x] = 0
From the second property of divided difference we can write
 f0 + fn fx =  0 + . . . + (x0 - x1) . . . (x0 - xn)(x0 - x) (xn - x0) . . . (xn - xn-1)(xn - x) (x - x0) . . . (x - xn)
or
 (x - x1) . . . (x - xn) (x - x0) .  .  . (x - xn-1) f(x)  = f0  + .   .   . + fn (x0 - x1) . . . (x0 - xn) (xn - x0) .  .  . (xn - xn-1)

 n ( n )fi S |  | x - xj j = 0 (xi - xj) i = 0 j ¹ 1
Since Lagrange's interpolation is also an Nth degree polynomial approximation to f(x) and the Nth degree polynomial passing through (N+1) points is unique hence the Lagrange's and Newton's divided difference approximations are one and the same. However, Lagrange's formula is more convinent to use in computer programming and Newton's divided difference formula is more suited for hand calculations.
Example : Compute f(0.3) for the data
 x 0 1 3 4 7 f 1 3 49 129 813
using Lagrange's interpolation formula (Analytic value is 1.831)

 (x - x1) (x - x2)(x- x3)(x - x4) (x - x0)(x - x1) (x - x2)(x - x3) f(x)  = f0+ . . . + f4 (x0 - x1) (x0 - x2)(x0 - x3)(x0 - x4) (x4 - x0)(x4 - x1)(x4 - x2)(x4 - x3)

 (0.3 - 1)(0.3 - 3)(0.3 - 4)(0.3 - 7) (0.3 - 0)(0.3 - 3)(0.3 - 4)(0.3 - 7) = 1+ 3 + (-1) (-3)(-4)(-7) 1 x (-2)(-3)(-6)

 (0.3 - 0)(0.3 - 1)(0.3 - 4)(0.3 - 7) (0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 7) 49 + 129 + 3 x 2 x (-1)(-4) 4 x 3 x 1 (-3)

 (0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 4) 813 7 x 6 x 4 x 3
= 1.831

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