In numerical analysis, the Lagrange interpolating polynomial is the unique polynomial of lowest

degree Degree may refer to: As a unit of measurement * Degree (angle), a unit of angle measurement ** Degree of geographical latitude ** Degree of geographical longitude * Degree symbol (°), a notation used in science, engineering, and mathematics ...

that

interpolates In the mathematical field of numerical analysis, interpolation is a type of estimation, a method of constructing (finding) new data points based on the range of a discrete set of known data points. In engineering and science, one often has a n ...

a given set of data. Given a data set of coordinate pairs

(x_j, y_j)

with

0 \leq j \leq k,

the

x_j

are called ''nodes'' and the

y_j

are called ''values''. The Lagrange polynomial

L(x)

has degree

\leq k

and assumes each value at the corresponding node,

L(x_j) = y_j.

Although named after

Joseph-Louis Lagrange Joseph-Louis Lagrange (born Giuseppe Luigi LagrangiaEdward Waring. It is also an easy consequence of a formula published in 1783 by Leonhard Euler. Uses of Lagrange polynomials include the Newton–Cotes method of numerical integration and Shamir's secret sharing scheme in cryptography. For equispaced nodes, Lagrange interpolation is susceptible to Runge's phenomenon of large oscillation.

Definition

Given a set of

k + 1

nodes

\

, which must all be distinct,

x_j \neq x_m

for indices

j \neq m

, the Lagrange basis for polynomials of degree

\leq k

for those nodes is the set of polynomials

\

each of degree

k

which take values

\ell_j(x_m) = 0

m \neq j

and

\ell_j(x_j) = 1

. Using the Kronecker delta this can be written

\ell_j(x_m) = \delta_.

Each basis polynomial can be explicitly described by the product:

&= \prod_ \frac. \end

Notice that the numerator

\prod_(x - x_m)

has

k

roots at the nodes

\_

while the denominator

\prod_(x_j - x_m)

scales the resulting polynomial so that

\ell_j(x_j) = 1.

The Lagrange interpolating polynomial for those nodes through the corresponding ''values''

\

is the linear combination:

L(x) = \sum_^ y_j \ell_j(x).

Each basis polynomial has degree

k

, so the sum

L(x)

has degree

\leq k

, and it interpolates the data because

L(x_m) = \sum_^ y_j \ell_j(x_m) = \sum_^ y_j \delta_ = y_m.

The interpolating polynomial is unique. Proof: assume the polynomial

M(x)

of degree

\leq k

interpolates the data. Then the difference

M(x) - L(x)

is zero at

k + 1

distinct nodes

\.

But the only polynomial of degree

\leq k

with more than

k

roots is the constant zero function, so

M(x) - L(x) = 0,

M(x) = L(x).

Barycentric form

Each Lagrange basis polynomial

\ell_j(x)

can be rewritten as the product of three parts, a function

\ell(x) = \prod_m (x - x_m)

common to every basis polynomial, a node-specific constant

w_j = \prod_(x_j - x_m)^

(called the ''barycentric weight''), and a part representing the displacement from

x_j

x

\ell_j(x) = \ell(x) \dfrac

By factoring

\ell(x)

out from the sum, we can write the Lagrange polynomial in the so-called ''first barycentric form'': :

L(x) = \ell(x) \sum_^k \fracy_j.

If the weights

w_j

have been pre-computed, this requires only

\mathcal O(k)

operations compared to

\mathcal O(k^2)

for evaluating each Lagrange basis polynomial

\ell_j(x)

individually. The barycentric interpolation formula can also easily be updated to incorporate a new node

x_

by dividing each of the

w_j

j=0 \dots k

(x_j - x_)

and constructing the new

w_

as above. For any

x,

\sum_^k \ell_j(x) = 1

because the constant function

g(x) = 1

is the unique polynomial of degree

\leq k

interpolating the data

\.

We can thus further simplify the barycentric formula by dividing

L(x) = L(x) / g(x)\colon

&= \sum_^k \fracy_j \Bigg/ \sum_^k \frac. \end

This is called the ''second form'' or ''true form'' of the barycentric interpolation formula. This second form has advantages in computation cost and accuracy: it avoids evaluation of

\ell(x)

; the work to compute each term in the denominator

w_j/(x-x_j)

has already been done in computing

\bigl(w_j/(x-x_j)\bigr)y_j

and so computing the sum in the denominator costs only

k-1

addition operations; for evaluation points

x

which are close to one of the nodes

x_j

catastrophic cancelation In numerical analysis, catastrophic cancellation is the phenomenon that subtracting good approximations to two nearby numbers may yield a very bad approximation to the difference of the original numbers. For example, if there are two studs, one L_ ...

would ordinarily be a problem for the value

(x-x_j)

, however this quantity appears in both numerator and denominator and the two cancel leaving good relative accuracy in the final result. Using this formula to evaluate

L(x)

at one of the nodes

x_j

will result in the

indeterminate Indeterminate may refer to: In mathematics * Indeterminate (variable), a symbol that is treated as a variable * Indeterminate system, a system of simultaneous equations that has more than one solution * Indeterminate equation, an equation that ha ...

\infty y_j/\infty

; computer implementations must replace such results by

L(x_j) = y_j

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 In mathematics the monomial basis of a polynomial ring is its basis (as a vector space or free module over the field or ring of coefficients) that consists of all monomials. The monomials form a basis because every polynomial may be uniquely writt ...

for our interpolation polynomial

L(x) = \sum_^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

L(x)

. By choosing a better basis, the Lagrange basis,

L(x) = \sum_^k l_j(x) y_j

, we merely get the

identity matrix In linear algebra, the identity matrix of size n is the n\times n square matrix with ones on the main diagonal and zeros elsewhere. Terminology and notation The identity matrix is often denoted by I_n, or simply by I if the size is immaterial o ...

\delta_

, 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 In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer ''n'' by several integers, then one can determine uniquely the remainder of the division of ''n'' by the product of thes ...

. Instead of checking for remainders of integers modulo prime numbers, we are checking for remainders of polynomials when divided by linears. Furthermore, when the order is large,

Fast Fourier transform A fast Fourier transform (FFT) is an algorithm that computes the discrete Fourier transform (DFT) of a sequence, or its inverse (IDFT). Fourier analysis converts a signal from its original domain (often time or space) to a representation in th ...

ation can be used to solve for the coefficients of the interpolated polynomial.

Example

We wish to interpolate

f(x) = x^2

over the domain

1 \leq x \leq 3

at the three nodes :

x_2 & = 3, & & & y_2 = f(x_2) & =9. \end

The node polynomial

\ell

is :

\ell(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6.

The barycentric weights are :

w_2 &= (3-1)^(3-2)^ = \tfrac12. \end

The Lagrange basis polynomials are :

\ell_2(x) &= \frac\cdot\frac = \tfrac12x^2 - \tfrac32x + 1. \end

The Lagrange interpolating polynomial is: :

&= x^2. \end

In (second) barycentric form, :

L(x) = \frac
  
  
 = \frac
  
  .

Notes

Runge's phenomenon in Lagrange polynomials

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 In numerical analysis, Chebyshev nodes are specific real algebraic numbers, namely the roots of the Chebyshev polynomials of the first kind. They are often used as nodes in polynomial interpolation because the resulting interpolation polynomial ...

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

Remainder in Lagrange interpolation formula

When interpolating a given function ''f'' by a polynomial of degree at the nodes

x_0,...,x_k

we get the remainder

R(x) = f(x) - L(x)

which can be expressed as :

\ell(x) = \ell(x) \frac, \quad \quad x_0 < \xi < x_k,

where

f_0,\ldots,x_k,x /math> is the notation for divided differences . Alternatively, the remainder can be expressed as a contour integral in complex domain as

: R(x) = \frac \int_C \frac dt = \frac \int_C \frac dt. The remainder can be bound as

:, R(x),  \leq \frac\max_ , f^(\xi), .

Derivation

Clearly,

R(x)

is zero at nodes. To find

R(x)

at a point

x_p

, define a new function

F(x)=R(x)-\tilde(x)=f(x)-L(x)-\tilde(x)

and choose

\tilde(x)=C\cdot\prod_^k(x-x_i)

where

C

is the constant we are required to determine for a given

x_p

. We choose

C

so that

F(x)

has

k+2

zeroes (at all nodes and

x_p

) between

x_0

and

x_k

(including endpoints). Assuming that

f(x)

k+1

-times differentiable, since

L(x)

and

\tilde(x)

are polynomials, and therefore, are infinitely differentiable,

F(x)

will be

k+1

-times differentiable. By

Rolle's theorem In calculus, Rolle's theorem or Rolle's lemma essentially states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them—that is, a point wher ...

F^(x)

has

k+1

zeroes,

F^(x)

has

k

zeroes...

F^

has 1 zero, say

\xi,\, x_0<\xi . Explicitly writing F^(\xi) :

: F^(\xi)=f^(\xi)-L^(\xi)-\tilde^(\xi) : L^=0,\tilde^=C\cdot(k+1)! (Because the highest power of x in \tilde(x) is k+1)

: 0=f^(\xi)-C\cdot(k+1)! The equation can be rearranged as

: C=\frac Since F(x_p) = 0 we have R(x_p)=\tilde(x_p) = \frac\prod_^k(x_p-x_i)

Derivatives

The

d

th derivatives of the Lagrange polynomial can be written as :

L^(x) := \sum_^ y_j \ell_j^(x)

. For the first derivative, the coefficients are given by :

\ell_j^(x) := \sum_^k \left \frac\prod_^k \frac \right /math>

and for the second derivative

: \ell^_j(x) := \sum_^ \frac \left \sum_^ \left( \frac\prod_^ \frac \right) \right .

Through recursion, one can compute formulas for higher derivatives.

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.

References

External links

*
ALGLIB
has an implementations in C++ / C# / VBA / Pascal.
GSL
has a polynomial interpolation code in C
SO
has a MATLAB example that demonstrates the algorithm and recreates the first image in this article

at ttp://numericalmethods.eng.usf.edu Holistic Numerical Methods Institutebr>Lagrange interpolation polynomial
on www.math-linux.com *
Dynamic Lagrange interpolation with JSXGraph
* Numerical computing with functions
The Chebfun Project

Excel Worksheet Function for Bicubic Lagrange Interpolation

Lagrange polynomials in Python
{{authority control Interpolation Polynomials Articles containing proofs