Lagrange polynomial

Definition

Given a set of k + 1 data points

where no two x j are the same, the interpolation polynomial in the Lagrange form is a linear combination

of Lagrange basis polynomials

where 0 ≤ j ≤ k . Note how, given the initial assumption that no two x i are the same, x j − x m ≠ 0 , so this expression is always well-defined. The reason pairs x i = x j with y i ≠ y j are not allowed is that no interpolation function L such that y i = L ( x i ) would exist; a function can only get one value for each argument x i . On the other hand, if also y i = y j , then those two points would actually be one single point.

For all i ≠ j , ℓ j ( x ) includes the term ( x − x i ) in the numerator, so the whole product will be zero at x = x i :

On the other hand,

In other words, all basis polynomials are zero at x = x i , except ℓ i ( x ) , for which it holds that ℓ i ( x i ) = 1 , because it lacks the ( x − x i ) term.

It follows that y i ℓ i ( x i ) = y i , so at each point x i , L ( x i ) = y i + 0 + 0 + ⋯ + 0 = y i , showing that L interpolates the function exactly.

Proof

The function L(x) being sought is a polynomial in x of the least degree that interpolates the given data set; that is, assumes value y j at the corresponding x j for all data points j :

Observe that:

1. In ℓ j ( x ) there are k factors in the product and each factor contains one x, so L(x) (which is a sum of these k-degree polynomials) must also be a k-degree polynomial.
2. ℓ j ( x i ) = ∏ m = 0 , m ≠ j k x i − x m x j − x m

We consider what happens when this product is expanded. Because the product skips m = j , if i = j then all terms are x j − x m x j − x m = 1 (except where x j = x m , but that case is impossible, as pointed out in the definition section—in that term, m = j , and since m ≠ j , i ≠ j , contrary to i = j ). Also if i ≠ j then since m ≠ j does not preclude it, one term in the product will be for m = i , i.e. x i − x i x j − x i = 0 , zeroing the entire product. So

1. ℓ j ( x i ) = δ j i = { 1 , if  j = i 0 , if  j ≠ i

where δ i j is the Kronecker delta. So:

Thus the function L(x) is a polynomial with degree at most k and where L ( x i ) = y i .

Additionally, the interpolating polynomial is unique, as shown by the unisolvence theorem at the polynomial interpolation article.

A perspective from linear algebra

Solving an interpolation problem leads to a problem in linear algebra amounting to inversion of a matrix. Using a standard monomial basis for our interpolation polynomial L ( x ) = ∑ j = 0 k x j m j , we must invert the Vandermonde matrix ( x i ) j to solve L ( x i ) = y i for the coefficients m j of L ( x ) . By choosing a better basis, the Lagrange basis, L ( x ) = ∑ j = 0 k l j ( x ) y j , we merely get the identity matrix, δ i j , which is its own inverse: the Lagrange basis automatically inverts the analog of the Vandermonde matrix.

This construction is analogous to the Chinese Remainder Theorem. Instead of checking for remainders of integers modulo prime numbers, we are checking for remainders of polynomials when divided by linears.

Example 1

We wish to interpolate ƒ(x) = x² over the range 1 ≤ x ≤ 3, given these three points:

The interpolating polynomial is:

Example 2

We wish to interpolate ƒ(x) = x³ over the range 1 ≤ x ≤ 3, given these three points:

The interpolating polynomial is:

Notes

The Lagrange form of the interpolation polynomial shows the linear character of polynomial interpolation and the uniqueness of the interpolation polynomial. Therefore, it is preferred in proofs and theoretical arguments. Uniqueness can also be seen from the invertibility of the Vandermonde matrix, due to the non-vanishing of the Vandermonde determinant.

But, as can be seen from the construction, each time a node x_k changes, all Lagrange basis polynomials have to be recalculated. A better form of the interpolation polynomial for practical (or computational) purposes is the barycentric form of the Lagrange interpolation (see below) or Newton polynomials.

Lagrange and other interpolation at equally spaced points, as in the example above, yield a polynomial oscillating above and below the true function. This behaviour tends to grow with the number of points, leading to a divergence known as Runge's phenomenon; the problem may be eliminated by choosing interpolation points at Chebyshev nodes.

The Lagrange basis polynomials can be used in numerical integration to derive the Newton–Cotes formulas.

Barycentric form

Using

we can rewrite the Lagrange basis polynomials as

or, by defining the barycentric weights

we can simply write

which is commonly referred to as the first form of the barycentric interpolation formula.

The advantage of this representation is that the interpolation polynomial may now be evaluated as

which, if the weights w j have been pre-computed, requires only O ( n ) operations (evaluating ℓ ( x ) and the weights w j / ( x − x j ) ) as opposed to O ( n 2 ) for evaluating the Lagrange basis polynomials ℓ j ( x ) individually.

The barycentric interpolation formula can also easily be updated to incorporate a new node x k + 1 by dividing each of the w j , j = 0 … k by ( x j − x k + 1 ) and constructing the new w k + 1 as above.

We can further simplify the first form by first considering the barycentric interpolation of the constant function g ( x ) ≡ 1 :

Dividing L ( x ) by g ( x ) does not modify the interpolation, yet yields

which is referred to as the second form or true form of the barycentric interpolation formula. This second form has the advantage that ℓ ( x ) need not be evaluated for each evaluation of L ( x ) .

Remainder in Lagrange interpolation formula

When interpolating a given function f by a polynomial of degree n at the nodes x₀,...,x_n we get the remainder R ( x ) = f ( x ) − L ( x ) which can be expressed as

where f [ x 0 , … , x n , x ] is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as

The remainder can be bound as

First derivative

The first derivative of the lagrange polynomial is given by

where ℓ j ′ ( x ) = ℓ j ( x ) ⋅ ∑ m = 0 ,   m ≠ j k 1 x − x m .

Finite fields

The Lagrange polynomial can also be computed in finite fields. This has applications in cryptography, such as in Shamir's Secret Sharing scheme.