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, it was first discovered in 1779 by Edward Waring and rediscovered 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.

Contents

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%2B1})}{(x_j-x_{j%2B1})} \cdots \frac{(x-x_k)}{(x_j-x_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 j\neq i, \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)=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%2B0%2B0%2B\dots %2B0=y_i, showing that L interpolates the function exactly.

Proof

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:

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

Observe that:

  1. In \ell_j(x) there are k terms in the product and each term 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}

Watch what happens if we expand this product. 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---if you tried to write out that term you'd find that m=j and since m\neq j, i\neq j, contrary to i=j). Also if i \neq j then since m \neq j doesn't 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.

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 Vandermonde matrix. By choosing another basis, the Lagrange basis, we get the much simpler identity matrix = δi,j which we can solve instantly: the Lagrange basis inverts the Vandermonde matrix.

This construction is the same as 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

Find an interpolation formula for ƒ(x) = tan(x) given this set of known values:

\begin{align} x_0 & = -1.5 & & & & & f(x_0) & = -14.1014 \\ x_1 & = -0.75 & & & & & f(x_1) & = -0.931596 \\ x_2 & = 0 & & & & & f(x_2) & = 0 \\ x_3 & = 0.75 & & & & & f(x_3) & = 0.931596 \\ x_4 & = 1.5 & & & & & f(x_4) & = 14.1014. \end{align}

The basis polynomials are:

\ell_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%2B3)
\ell_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%2B3)(4x-3)
\ell_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} ={3\over 243} (2x%2B3)(4x%2B3)(4x-3)(2x-3)
\ell_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%2B3)(4x%2B3)
\ell_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%2B3)(4x-3)(4x%2B3).

Thus the interpolating polynomial then is

\begin{align}L(x) &= {1\over 243}\Big(f(x_0)x (2x-3)(4x-3)(4x%2B3) \\ & {} \qquad {} - 8f(x_1)x (2x-3)(2x%2B3)(4x-3) \\ & {} \qquad {} %2B 3f(x_2)(2x%2B3)(4x%2B3)(4x-3)(2x-3) \\ & {} \qquad {} - 8f(x_3)x (2x-3)(2x%2B3)(4x%2B3) \\ & {} \qquad {} %2B f(x_4)x (2x%2B3)(4x-3)(4x%2B3)\Big)\\ & = 4.834848x^3 - 1.477474x. \end{align}

Example 2

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}%2B{4}\cdot{x - 1 \over 2 - 1}\cdot{x - 3 \over 2 - 3}%2B{9}\cdot{x - 1 \over 3 - 1}\cdot{x - 2 \over 3 - 2} \\[10pt] &= x^2. \end{align}

Example 3

We wish to interpolate ƒ(x) = x3 over the range 1 ≤ x ≤ 3, given these 3 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}%2B{8}\cdot{x - 1 \over 2 - 1}\cdot{x - 3 \over 2 - 3}%2B{27}\cdot{x - 1 \over 3 - 1}\cdot{x - 2 \over 3 - 2} \\[8pt] &= 6x^2 - 11x %2B 6. \end{align}

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 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.

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)

we can rewrite the Lagrange basis polynomials as

\ell_j(x) = \frac{\ell(x)}{x-x_j} \frac{1}{\prod_{i=0,i \neq j}^k(x_j-x_i)}

or, by defining the barycentric weights[1]

w_j = \frac{1}{\prod_{i=0,i \neq j}^k(x_j-x_i)}

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%2B1} by dividing each of the w_j, j=0 \dots k by (x_j - x_{k%2B1}) and constructing the new w_{k%2B1} 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).

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.

See also

References

  1. ^ Jean-Paul Berrut, Lloyd N. Trefethen (2004). "Barycentric Lagrange Interpolation". SIAM Review 46 (3): 501–517. doi:10.1137/S0036144502417715. 

External links