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Lagrange Interpolation

Module by: Douglas L. Jones. E-mail the author

Summary: Lagrange's interpolation formula is a simple and clever method for finding the unique polynomial of order L that exactly passes through L+1 distinct samples of a signal.

Lagrange's interpolation method is a simple and clever way of finding the unique LLth-order polynomial that exactly passes through L+1 L 1 distinct samples of a signal. Once the polynomial is known, its value can easily be interpolated at any point using the polynomial equation. Lagrange interpolation is useful in many applications, including Parks-McClellan FIR Filter Design.

Lagrange interpolation formula

Given an LLth-order polynomial P⁢x= a 0 + a 1 ⁢x+...+ a L ⁢xL=∑ k =0L a k ⁢xk P x a 0 a 1 x ... a L x L k 0 L a k x k and L+1 L 1 values of P⁢ x k P x k at different x k x k , k∈01...L k 0 1 ... L , x i ≠ x j x i x j , i≠j i j , the polynomial can be written as P⁢x=∑ k =0LP⁢ x k ⁢(x− x 1 )⁢(x− x 2 )⁢...⁢(x− x k - 1 )⁢(x− x k + 1 )⁢...⁢(x− x L )( x k − x 1 )⁢( x k − x 2 )⁢...⁢( x k − x k - 1 )⁢( x k − x k + 1 )⁢...⁢( x k − x L ) P x k 0 L P x k x x 1 x x 2 ... x x k - 1 x x k + 1 ... x x L x k x 1 x k x 2 ... x k x k - 1 x k x k + 1 ... x k x L The value of this polynomial at other xx can be computed via substitution into this formula, or by expanding this formula to determine the polynomial coefficients a k a k in standard form.

Proof

Note that for each term in the Lagrange interpolation formula above, ∏ i =0,,i≠kLx− x i x k − x i ={1  if  x= x k 0  if  (x= x j )∧(j≠k) i 0, i k L x x i x k x i 1 x x k 0 x x j j k and that it is an L Lth-order polynomial in x x. The Lagrange interpolation formula is thus exactly equal to P⁢ x k P x k at all x k x k , and as a sum of LLth-order polynomials is itself an LLth-order polynomial.

It can be shown that the Vandermonde matrix ( 1 x 0 x 0 2... x 0 L 1 x 1 x 1 2... x 1 L 1 x 2 x 2 2... x 2 L ⋮⋮⋮⋱⋮ 1 x L x L 2... x L L )⁢ a 0 a 1 a 2 ⋮ a L =P⁢ x 0 P⁢ x 1 P⁢ x 2 ⋮P⁢ x L 1 x 0 x 0 2 ... x 0 L 1 x 1 x 1 2 ... x 1 L 1 x 2 x 2 2 ... x 2 L ⋮ ⋮ ⋮ ⋱ ⋮ 1 x L x L 2 ... x L L a 0 a 1 a 2 ⋮ a L P x 0 P x 1 P x 2 ⋮ P x L has a non-zero determinant and is thus invertible, so the L Lth-order polynomial passing through all L+1 L 1 sample points x j x j is unique. Thus the Lagrange polynomial expressions, as an L Lth-order polynomial passing through the L+1 L 1 sample points, must be the unique P⁢x P x .

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