Play Video
1
Lagrange Interpolating Polynomials
Lagrange Interpolating Polynomials
::2013/12/15::
Play Video
2
Lagrangian Interpolation - Theory
Lagrangian Interpolation - Theory
::2010/03/09::
Play Video
3
Lagrangian Interpolation: Linear Interpolation: Example
Lagrangian Interpolation: Linear Interpolation: Example
::2010/03/09::
Play Video
4
Lagrange
Lagrange's Interpolation
::2013/04/10::
Play Video
5
Lagrange Polynomials
Lagrange Polynomials
::2011/04/10::
Play Video
6
Polynomial interpolation via Lagrange polynomials
Polynomial interpolation via Lagrange polynomials
::2014/09/10::
Play Video
7
5.2.4-Curve Fitting: Lagrange Interpolating Polynomials--Linear Interpolation
5.2.4-Curve Fitting: Lagrange Interpolating Polynomials--Linear Interpolation
::2013/09/20::
Play Video
8
LaGrange Interpolation
LaGrange Interpolation
::2012/11/28::
Play Video
9
Example of polynomial interpolation via Lagrange polynomials
Example of polynomial interpolation via Lagrange polynomials
::2014/09/10::
Play Video
10
Lagrange Interpolation Easily Explained on Casio fx-991ES Calculator!
Lagrange Interpolation Easily Explained on Casio fx-991ES Calculator!
::2014/02/22::
Play Video
11
Lagrange Error Bound
Lagrange Error Bound
::2012/05/27::
Play Video
12
Error or Remainder of a Taylor Polynomial Approximation
Error or Remainder of a Taylor Polynomial Approximation
::2011/09/15::
Play Video
13
Metodo de Interpolacion de Lagrange Matlab
Metodo de Interpolacion de Lagrange Matlab
::2013/03/18::
Play Video
14
Quadratic Lagrangian Interpolation: Example: Part 1 of  2
Quadratic Lagrangian Interpolation: Example: Part 1 of 2
::2010/03/17::
Play Video
15
Mathematics 128A - 2014-02-11: Interpolations and the Lagrange Polynomial;
Mathematics 128A - 2014-02-11: Interpolations and the Lagrange Polynomial;
::2014/02/11::
Play Video
16
Interpolation in MATLAB
Interpolation in MATLAB
::2013/08/21::
Play Video
17
Lagrange Polynomial
Lagrange Polynomial
::2011/12/09::
Play Video
18
Lagrange interpolation Modeling in Matlab::Hari
Lagrange interpolation Modeling in Matlab::Hari
::2014/04/22::
Play Video
19
Linear Algebra 12c: Applications Series - Polynomial Interpolation According to Lagrange
Linear Algebra 12c: Applications Series - Polynomial Interpolation According to Lagrange
::2014/12/13::
Play Video
20
Kezakoo - Polynomes interpolateurs de Lagrange
Kezakoo - Polynomes interpolateurs de Lagrange
::2013/07/21::
Play Video
21
11.02 Polynome angewendet, Näherung, Interpolation, Differentialgleichung
11.02 Polynome angewendet, Näherung, Interpolation, Differentialgleichung
::2010/11/15::
Play Video
22
Lagrangian Interpolation: Cubic Interpolation: Example: Part 1 of 2
Lagrangian Interpolation: Cubic Interpolation: Example: Part 1 of 2
::2010/03/24::
Play Video
23
Taylor
Taylor's Remainder Theorem - Finding the Remainder, Ex 1
::2011/07/02::
Play Video
24
Chapitre 1: INTERPOLATION-- Interpolation de Lagrange - cas n=2
Chapitre 1: INTERPOLATION-- Interpolation de Lagrange - cas n=2
::2014/08/24::
Play Video
25
WildLinAlg21: More bases of polynomial spaces
WildLinAlg21: More bases of polynomial spaces
::2011/03/21::
Play Video
26
Quadratic Lagrangian Interpolation: Example: Part 2 of  2
Quadratic Lagrangian Interpolation: Example: Part 2 of 2
::2010/03/17::
Play Video
27
Puppeteer interpolation methods: linear vs lagrange vs spline
Puppeteer interpolation methods: linear vs lagrange vs spline
::2013/10/22::
Play Video
28
5.2.5-Curve Fitting: Lagrange Interpolating Polynomials--Quadratic Interpolation
5.2.5-Curve Fitting: Lagrange Interpolating Polynomials--Quadratic Interpolation
::2013/09/20::
Play Video
29
Proof: Bounding the Error or Remainder of a Taylor Polynomial Approximation
Proof: Bounding the Error or Remainder of a Taylor Polynomial Approximation
::2011/09/15::
Play Video
30
Polynomial approximation of functions (part 1)
Polynomial approximation of functions (part 1)
::2008/04/28::
Play Video
31
Shifrin Math 3500 Day 55: Gram-Schmidt & Lagrange Interpolation
Shifrin Math 3500 Day 55: Gram-Schmidt & Lagrange Interpolation
::2014/12/05::
Play Video
32
MATLAB tutorial: Curve Fitting (quadratic, cubic, polynomial, etc)
MATLAB tutorial: Curve Fitting (quadratic, cubic, polynomial, etc)
::2010/04/08::
Play Video
33
Polynômes de Lagrange
Polynômes de Lagrange
::2013/07/20::
Play Video
34
Python Interpolation 1 of 4: 1d interpolation with interp1d
Python Interpolation 1 of 4: 1d interpolation with interp1d
::2009/12/07::
Play Video
35
Lagrange polynomials linearly independent
Lagrange polynomials linearly independent
::2014/09/10::
Play Video
36
Cálculo Numérico: Método de Lagrange para Interpolação
Cálculo Numérico: Método de Lagrange para Interpolação
::2014/04/17::
Play Video
37
LAGRANGE ERROR BOUND
LAGRANGE ERROR BOUND
::2013/01/13::
Play Video
38
30. Shamir
30. Shamir's Secret Sharing / Lagrange interpolation
::2014/10/11::
Play Video
39
How to Solve Polynomial Equations in MATLAB. [HD]
How to Solve Polynomial Equations in MATLAB. [HD]
::2014/09/01::
Play Video
40
Mod-01 Lec-14 Example for Lagrange interpolation
Mod-01 Lec-14 Example for Lagrange interpolation
::2011/11/24::
Play Video
41
Vidéo 3sur54  Interpolation de Lagrange   cas n=2  Analyse Numérique pour Ingénieurs
Vidéo 3sur54 Interpolation de Lagrange cas n=2 Analyse Numérique pour Ingénieurs
::2014/01/24::
Play Video
42
Newtoninterpolation , Teil 1
Newtoninterpolation , Teil 1
::2010/03/19::
Play Video
43
A137435 - Lagrange Interpolation
A137435 - Lagrange Interpolation
::2014/12/02::
Play Video
44
Mod-01 Lec-15 Lagrange interpolation contd...
Mod-01 Lec-15 Lagrange interpolation contd...
::2011/11/24::
Play Video
45
Taylor
Taylor's Theorem with Remainder
::2010/10/09::
Play Video
46
Lecture 10 - Error In Interpolation Polynomial
Lecture 10 - Error In Interpolation Polynomial
::2008/01/19::
Play Video
47
LaGrange Error for Taylor Polynomials
LaGrange Error for Taylor Polynomials
::2014/03/18::
Play Video
48
Unit 10-4: Taylor Polynomials and LaGrange Error
Unit 10-4: Taylor Polynomials and LaGrange Error
::2013/04/16::
Play Video
49
DiffGeom4: The differential calculus for curves, via Lagrange!
DiffGeom4: The differential calculus for curves, via Lagrange!
::2013/08/14::
Play Video
50
Error of approximation by polynomials
Error of approximation by polynomials
::2013/08/31::
NEXT >>
RESULTS [51 .. 101]
From Wikipedia, the free encyclopedia
Jump to: navigation, search

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.

Definition[edit]

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 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), 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[edit]

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:

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

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[edit]

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[edit]

Example 1[edit]

The tangent function and its interpolant

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 Lagrange 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+3)
\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+3)(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+3)(4x+3)(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+3)(4x+3)
\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+3)(4x-3)(4x+3).

Thus the interpolating polynomial then is

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

Example 2[edit]

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 3[edit]

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[edit]

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[edit]

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[3]

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+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).

Finite fields[edit]

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[edit]

References[edit]

  1. ^ Meijering, Erik (2002), "A chronology of interpolation: from ancient astronomy to modern signal and image processing", Proceedings of the IEEE 90 (3): 319–342, doi:10.1109/5.993400 .
  2. ^ Quarteroni, Alfio; Saleri, Fausto (2003), Scientific Computing with MATLAB, Texts in computational science and engineering 2, Springer, p. 66, ISBN 9783540443636 .
  3. ^ Jean-Paul Berrut & Lloyd N. Trefethen (2004). "Barycentric Lagrange Interpolation". SIAM Review 46 (3): 501–517. doi:10.1137/S0036144502417715. 

External links[edit]

Wikipedia content is licensed under the GFDL License
Powered by YouTube
MASHPEDIA
LEGAL
  • Mashpedia © 2014