Lagrange polynomial

Given a set of k+1 data points

<math>(x_0, y_0),\ldots,(x_k, y_k)<math>

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

<math>L(x) := \sum_{j=0}^{k} y_j l_j(x)<math>

with the Lagrange basis polynomials defined as

<math>l_j(x) := \prod_{i=0, j\neq i}^{k} \frac{x-x_i}{x_j-x_i} = \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}}<math>

Proof

The function we are looking for has to be polynomial function L(x) of degree k with

<math>L(x_j) = y_j \qquad j=0,\ldots,k<math>

According to the Stone-Weierstrass theorem such a function exists and is unique. The Langrange polynomial is the solution to the interpolation problem.

As can be easily seen

1. <math>l_j(x)<math> is a polynomial and has degree k
2. <math>l_i(x_j) = \delta_{ij}, 0 \leq i,j \leq k <math>

Thus the function L(x) is a polynomial with degree k and

<math>L(x_i) = \sum_{j=0}^{k} y_j l_j(x_i) = y_i<math>.

Therefore L(x) is our unique interpolation polynomial.

Main idea

Solving an interpolation problem leads to a problem in linear algebra where we have to solve a matrix. Using a standard monomial basis for our interpolation polynomial we get the very complicated Vandermonde matrix. By choosing another basis, the Langrange basis we get the much simpler identity matrix = δ_i,j which we can solve instantly.

Usage

Example

We wish to interpolate <math>f(x)=\tan{x}<math> at the points

Now, the basis polynomials are:

<math>l_0(x)={x - x_1 \over x_0 - x_1}\cdot{x - x_2 \over x_0 - x_2}\cdot{x - x_3 \over x_0 - x_3}\cdot{x - x_4 \over x_0 - x_4}

            ={1\over 243} x (2x-3)(4x-3)(4x+3)<math>

<math>l_1(x)={x - x_0 \over x_1 - x_0}\cdot{x - x_2 \over x_1 - x_2}\cdot{x - x_3 \over x_1 - x_3}\cdot{x - x_4 \over x_1 - x_4}

            =-{8\over 243} x (2x-3)(2x+3)(4x-3)<math>

<math>l_2(x)={x - x_0 \over x_2 - x_0}\cdot{x - x_1 \over x_2 - x_1}\cdot{x - x_3 \over x_2 - x_3}\cdot{x - x_4 \over x_2 - x_4}

            ={1\over 243} (243-540x^2+192x^4)<math>

<math>l_3(x)={x - x_0 \over x_3 - x_0}\cdot{x - x_1 \over x_3 - x_1}\cdot{x - x_2 \over x_3 - x_2}\cdot{x - x_4 \over x_3 - x_4}

            =-{8\over 243} x (2x-3)(2x+3)(4x+3)<math>

<math>l_4(x)={x - x_0 \over x_4 - x_0}\cdot{x - x_1 \over x_4 - x_1}\cdot{x - x_2 \over x_4 - x_2}\cdot{x - x_3 \over x_4 - x_3}

            ={1\over 243} x (2x-3)(4x-3)(4x+3)<math>

Thus the interpolating polynomial then is

<math>{1\over 243}(f(x_0)x (2x-3)(4x-3)(4x+3)-8f(x_1)x (2x-3)(2x+3)(4x-3)<math>

<math>+f(x_2)(243-540x^2+192x^4)-8f(x_3)x (2x-3)(2x+3)(4x+3)+f(x_4)x (2x-3)(4x-3)(4x+3)\,<math>

<math>=-1.47748x+4.83456x^3\,<math>

Notes

The Langrange 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. But, as can be seen from the construction, each time a node x_k changes, all Langrange basis polynomials have to be recalculated. A better form of the interpolation polynomial for practical (or computational) purposes is the Newton polynomial. Using nested multiplication amounts to the same idea.

Lagrange and other interpolation at equally spaced points, as in the example above, yield a polynomial oscillating above and below the true function. This oscillation is lessened by choosing interpolation points at Chebyshev nodes.

The Lagrange basis polynomials are used in numerical integration to derive the Newton-Cotes formulas.

Lagrange interpolation is often used in digital signal processing of audio for the implementation of fractional delay FIR filters (e.g., to precisely tune digital waveguides in physical modelling synthesis).

See also

• Polynomial interpolation
• Newton form of the interpolation polynomial
• Bernstein form of the interpolation polynomial
• Newton-Cotes formulaspt:Polinômios de Lagrange