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Lagrange polynomial

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 Title: Lagrange polynomial Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

Lagrange polynomial

In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points x_j and numbers y_j, the Lagrange polynomial is the polynomial of the least degree that at each point x_j assumes the corresponding value y_j (i.e. the functions coincide at each point). The interpolating polynomial of the least degree is unique, however, and it is therefore more appropriate to speak of "the Lagrange form" of that unique polynomial rather than "the Lagrange interpolation polynomial", since the same polynomial can be arrived at through multiple methods. Although named after Joseph Louis Lagrange, who published it in 1795, it was first discovered in 1779 by Edward Waring and it is also an easy consequence of a formula published in 1783 by Leonhard Euler.

Lagrange interpolation is susceptible to Runge's phenomenon, and the fact that changing the interpolation points requires recalculating the entire interpolant can make Newton polynomials easier to use. Lagrange polynomials are used in the Newton–Cotes method of numerical integration and in Shamir's secret sharing scheme in cryptography. This image shows, for four points ((−9, 5), (−4, 2), (−1, −2), (7, 9)), the (cubic) interpolation polynomial L(x) (in black), which is the sum of the scaled basis polynomials y00(x), y11(x), y22(x) and y33(x). The interpolation polynomial passes through all four control points, and each scaled basis polynomial passes through its respective control point and is 0 where x corresponds to the other three control points.

Contents

• Definition 1
• Proof 2
• Main idea 3
• Examples 4
• Example 1 4.1
• Example 2 4.2
• Notes 4.3
• Barycentric interpolation 5
• Remainder in Lagrange interpolation formula 6
• Finite fields 7
• References 9

Definition

Given a set of k + 1 data points

(x_0, y_0),\ldots,(x_j, y_j),\ldots,(x_k, y_k)

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

L(x) := \sum_{j=0}^{k} y_j \ell_j(x)

of Lagrange basis polynomials

\ell_j(x) := \prod_{\begin{smallmatrix}0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m} = \frac{(x-x_0)}{(x_j-x_0)} \cdots \frac{(x-x_{j-1})}{(x_j-x_{j-1})} \frac{(x-x_{j+1})}{(x_j-x_{j+1})} \cdots \frac{(x-x_k)}{(x_j-x_k)},

where 0\le j\le k. Note how, given the initial assumption that no two x_i are the same, x_j - x_m \neq 0, so this expression is always well-defined. The reason pairs x_i = x_j with y_i\neq 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\neq j, \ell_j(x) includes the term (x-x_i) in the numerator, so the whole product will be zero at x=x_i:

\ell_{j\ne i}(x_i) = \prod_{m\neq j} \frac{x_i-x_m}{x_j-x_m} = \frac{(x_i-x_0)}{(x_j-x_0)} \cdots \frac{(x_i-x_i)}{(x_j-x_i)} \cdots \frac{(x_i-x_k)}{(x_j-x_k)} = 0.

On the other hand,

\ell_i(x_i) := \prod_{m\neq i} \frac{x_i-x_m}{x_i-x_m} = 1

In other words, all basis polynomials are zero at x=x_i, except \ell_i(x), for which it holds that \ell_i(x_i)=1, because it lacks the (x-x_i) term.

It follows that y_i \ell_i(x_i)=y_i, so at each point x_i, L(x_i)=y_i+0+0+\dots +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 \ell_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. \ell_j(x_i) = \prod_{m=0,\, m\neq j}^{k} \frac{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 \frac{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\neq j, i\neq j, contrary to i=j). Also if i \neq j then since m \neq j does not preclude it, one term in the product will be for m=i, i.e. \frac{x_i-x_i}{x_j-x_i} = 0, zeroing the entire product. So

1. \ell_j(x_i) = \delta_{ji} = \begin{cases} 1, & \text{if } j=i \\ 0, & \text{if } j \ne i \end{cases}

where \delta_{ij} is the Kronecker delta. So:

L(x_i) = \sum_{j=0}^k y_j \ell_j(x_i) = \sum_{j=0}^{k} y_j \delta_{ji} = y_i.

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 polynomial interpolation article.

Main idea

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=0k x j mj, we must invert the Vandermonde matrix (xi ) j to solve L(xi) = yi for the coefficients mj of L(x). By choosing a better basis, the Lagrange basis, L(x) = ∑j=0k lj(x) yj, we merely get the identity matrix, δij, 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.

Examples

Example 1

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

\begin{align} x_0 & = 1 & & & f(x_0) & = 1 \\ x_1 & = 2 & & & f(x_1) & = 4 \\ x_2 & = 3 & & & f(x_2) & =9. \end{align}

The interpolating polynomial is:

\begin{align} L(x) &= {1}\cdot{x - 2 \over 1 - 2}\cdot{x - 3 \over 1 - 3}+{4}\cdot{x - 1 \over 2 - 1}\cdot{x - 3 \over 2 - 3}+{9}\cdot{x - 1 \over 3 - 1}\cdot{x - 2 \over 3 - 2} \\[10pt] &= x^2. \end{align}

Example 2

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

 x_0=1\, f(x_0)=1\, x_1=2\, f(x_1)=8\, x_2=3\, f(x_2)=27\,

The interpolating polynomial is:

\begin{align} L(x) &= {1}\cdot{x - 2 \over 1 - 2}\cdot{x - 3 \over 1 - 3}+{8}\cdot{x - 1 \over 2 - 1}\cdot{x - 3 \over 2 - 3}+{27}\cdot{x - 1 \over 3 - 1}\cdot{x - 2 \over 3 - 2} \\[8pt] &= 6x^2 - 11x + 6. \end{align}