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# Lagrange polynomial

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### 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.[1]

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}

### Notes

Example of Lagrange polynomial interpolation divergence.

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 xk 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.[2]

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

## Barycentric interpolation

Using

\ell(x) = (x - x_0)(x - x_1) \cdots (x - x_k)
\ell'(x_j) = \frac{\mathrm{d} \ell(x)}{\mathrm{d} x}\Big|_{x=x_j} = \prod_{i=0,i \neq j}^k(x_j-x_i)

we can rewrite the Lagrange basis polynomials as

\ell_j(x) = \frac{\ell(x)}{\ell'(x_j)(x-x_j)}

or, by defining the barycentric weights[3]

w_j = \frac{1}{\ell'(x_j)}

we can simply write

\ell_j(x) = \ell(x)\frac{w_j}{x-x_j}

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

L(x) = \ell(x) \sum_{j=0}^k \frac{w_j}{x-x_j}y_j

which, if the weights w_j have been pre-computed, requires only \mathcal O(n) operations (evaluating \ell(x) and the weights w_j/(x-x_j)) as opposed to \mathcal O(n^2) for evaluating the Lagrange basis polynomials \ell_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 \dots 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)\equiv 1:

g(x) = \ell(x) \sum_{j=0}^k \frac{w_j}{x-x_j}.

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

L(x) = \frac{\sum_{j=0}^k \frac{w_j}{x-x_j}y_j}{\sum_{j=0}^k \frac{w_j}{x-x_j}}

which is referred to as the second form or true form of the barycentric interpolation formula. This second form has the advantage that \ell(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,...,xn we get the remainder R(x) = f(x) - L(x) which can be expressed as [4]

R(x) = f[x_0,\ldots,x_n,x] \ell(x) = \ell(x) \frac{f^{n+1}(\xi)}{(n+1)!}, \quad \quad x_0 < \xi < x_n,

where f[x_0,\ldots,x_n,x] and is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as

R(z) = \frac{\ell(z)}{2\pi i} \int_C \frac{f(t)}{(t-z)(t-z_0) \cdots (t-z_n)} dt.

The remainder can be bound as

|R(x)| \leq \frac{(x_n-x_0)^{n+1}}{(n+1)!}\max_{x_0 \leq \xi \leq x_n} |f^{(n+1)}(\xi)|.

## 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.