lagrange polynomial

Given a set of $k\; +\; 1$ nodes $\backslash \{x\_0,\; x\_1,\; \backslash ldots,\; x\_k\backslash \}$, which must all be distinct, $x\_j\; \backslash neq\; x\_m$ for indices $j\; \backslash neq\; m$, the Lagrange basis for polynomials of degree $\backslash leq\; k$ for those nodes is the set of polynomials $\backslash \{\backslash ell\_0(x),\; \backslash ell\_1(x),\; \backslash ldots,\; \backslash ell\_k(x)\backslash \}$ each of degree $k$ which take values $\backslash ell\_j(x\_m)\; =\; 0$ if $m\; \backslash neq\; j$ and $\backslash ell\_j(x\_j)\; =\; 1$. Using the Kronecker delta this can be written $\backslash ell\_j(x\_m)\; =\; \backslash delta\_\{jm\}.$ Each basis polynomial can be explicitly described by the product:

$$\backslash begin\{aligned\}$$

\ell_j(x) &= \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)} \\[8mu]

&= \prod_{\begin{smallmatrix}0\le m\le k\\ m\neq j\end{smallmatrix}} \frac{x-x_m}{x_j-x_m} \vphantom\Bigg|.

\end{aligned}

Notice that the numerator $\backslash prod\_\{m\; \backslash neq\; j\}(x\; -\; x\_m)$ has $k$ roots at the nodes $\backslash \{x\_m\backslash \}\_\{m\; \backslash neq\; j\}$ while the denominator $\backslash prod\_\{m\; \backslash neq\; j\}(x\_j\; -\; x\_m)$ scales the resulting polynomial so that $\backslash ell\_j(x\_j)\; =\; 1.$

The Lagrange interpolating polynomial for those nodes through the corresponding values $\backslash \{y\_0,\; y\_1,\; \backslash ldots,\; y\_k\backslash \}$ is the linear combination:

$$L(x)\; =\; \backslash sum\_\{j=0\}^\{k\}\; y\_j\; \backslash ell\_j(x).$$

Each basis polynomial has degree $k$, so the sum $L(x)$ has degree $\backslash leq\; k$, and it interpolates the data because $L(x\_m)\; =\; \backslash sum\_\{j=0\}^\{k\}\; y\_j\; \backslash ell\_j(x\_m)\; =\; \backslash sum\_\{j=0\}^\{k\}\; y\_j\; \backslash delta\_\{mj\}\; =\; y\_m.$

The interpolating polynomial is unique. Proof: assume the polynomial $M(x)$ of degree $\backslash leq\; k$ interpolates the data. Then the difference $M(x)\; -\; L(x)$ is zero at $k\; +\; 1$ distinct nodes $\backslash \{x\_0,\; x\_1,\; \backslash ldots,\; x\_k\backslash \}.$ But the only polynomial of degree $\backslash leq\; k$ with more than $k$ roots is the constant zero function, so $M(x)\; -\; L(x)\; =\; 0,$ or $M(x)\; =\; L(x).$

Each Lagrange basis polynomial $\backslash ell\_j(x)$ can be rewritten as the product of three parts, a function $\backslash ell(x)\; =\; \backslash prod\_m\; (x\; -\; x\_m)$ common to every basis polynomial, a node-specific constant $w\_j\; =\; \backslash prod\_\{m\backslash neq\; j\}(x\_j\; -\; x\_m)^\{-1\}$ (called the barycentric weight), and a part representing the displacement from $x\_j$ to $x$:{{cite journal

| first1 = Jean-Paul | last1 = Berrut

| first2 = Lloyd N. | last2 = Trefethen |author-link = Lloyd N. Trefethen

| year = 2004

| title = Barycentric Lagrange Interpolation

| journal = SIAM Review

| volume = 46

| issue = 3

| pages = 501–517

| doi = 10.1137/S0036144502417715

| bibcode = 2004SIAMR..46..501B

| url = https://people.maths.ox.ac.uk/trefethen/barycentric.pdf

| doi-access = free

}}

$$\backslash ell\_j(x)\; =\; \backslash ell(x)\; \backslash dfrac\{w\_j\}\{x\; -\; x\_j\}$$

By factoring $\backslash ell(x)$ out from the sum, we can write the Lagrange polynomial in the so-called first barycentric form:

:$L(x)\; =\; \backslash ell(x)\; \backslash sum\_\{j=0\}^k\; \backslash frac\{w\_j\}\{x-x\_j\}y\_j.$

If the weights $w\_j$ have been pre-computed, this requires only $\backslash mathcal\; O(k)$ operations compared to $\backslash mathcal\; O(k^2)$ for evaluating each Lagrange basis polynomial $\backslash ell\_j(x)$ individually.

The barycentric interpolation formula can also easily be updated to incorporate a new node $x\_\{k+1\}$ by dividing each of the $w\_j$, $j=0\; \backslash dots\; k$ by $(x\_j\; -\; x\_\{k+1\})$ and constructing the new $w\_\{k+1\}$ as above.

For any $x,$ $\backslash sum\_\{j=0\}^k\; \backslash ell\_j(x)\; =\; 1$ because the constant function $g(x)\; =\; 1$ is the unique polynomial of degree $\backslash leq\; k$ interpolating the data $\backslash \{(x\_0,\; 1),\; (x\_1,\; 1),\; \backslash ldots,\; (x\_k,\; 1)\; \backslash \}.$ We can thus further simplify the barycentric formula by dividing $L(x)\; =\; L(x)\; /\; g(x)\backslash colon$

:$\backslash begin\{aligned\}$

L(x) &= \ell(x) \sum_{j=0}^k \frac{w_j}{x-x_j}y_j \Bigg/ \ell(x) \sum_{j=0}^k \frac{w_j}{x-x_j} \\[10mu]

&= \sum_{j=0}^k \frac{w_j}{x-x_j}y_j \Bigg/ \sum_{j=0}^k \frac{w_j}{x-x_j}.

\end{aligned}

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 $\backslash ell(x)$; the work to compute each term in the denominator $w\_j/(x-x\_j)$ has already been done in computing $\backslash bigl(w\_j/(x-x\_j)\backslash bigr)y\_j$ and so computing the sum in the denominator costs only $k$ addition operations; for evaluation points $x$ which are close to one of the nodes $x\_j$, catastrophic cancelation 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 $\backslash infty\; y\_j/\backslash infty$; computer implementations must replace such results by $L(x\_j)\; =\; y\_j.$

Each Lagrange basis polynomial can also be written in barycentric form:

:$$

\ell_j(x) = \frac{w_j}{x-x_j} \Bigg/ \sum_{m=0}^k \frac{w_m}{x-x_m}.

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 $L(x)\; =\; \backslash sum\_\{j=0\}^k\; x^j\; m\_j$, we must invert the Vandermonde matrix $(x\_i)^j$ to solve $L(x\_i)\; =\; y\_i$ for the coefficients $m\_j$ of $L(x)$. By choosing a better basis, the Lagrange basis, $L(x)\; =\; \backslash sum\_\{j=0\}^k\; l\_j(x)\; y\_j$, we merely get the identity matrix, $\backslash delta\_\{ij\}$, 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.

Furthermore, when the order is large, Fast Fourier transformation can be used to solve for the coefficients of the interpolated polynomial.

We wish to interpolate $f(x)\; =\; x^2$ over the domain $1\; \backslash leq\; x\; \backslash leq\; 3$ at the three nodes {{nobr|$\backslash \{1,\backslash ,\; 2,\backslash ,\; 3\backslash \}$:}}

: $$

\begin{align}

x_0 & = 1, & & & y_0 = f(x_0) & = 1, \\[3mu]

x_1 & = 2, & & & y_1 = f(x_1) & = 4, \\[3mu]

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

\end{align}

The node polynomial $\backslash ell$ is

:$\backslash ell(x)\; =\; (x-1)(x-2)(x-3)\; =\; x^3\; -\; 6x^2\; +\; 11x\; -\; 6.$

The barycentric weights are

:$\backslash begin\{align\}$

w_0 &= (1-2)^{-1}(1-3)^{-1} = \tfrac12, \\[3mu]

w_1 &= (2-1)^{-1}(2-3)^{-1} = -1, \\[3mu]

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

\end{align}

The Lagrange basis polynomials are

:$\backslash begin\{align\}$

\ell_0(x) &= \frac{x - 2}{1 - 2}\cdot\frac{x - 3}{1 - 3} = \tfrac12x^2 - \tfrac52x + 3, \\[5mu]

\ell_1(x) &= \frac{x - 1}{2 - 1}\cdot\frac{x - 3}{2 - 3} = -x^2 + 4x - 3, \\[5mu]

\ell_2(x) &= \frac{x - 1}{3 - 1}\cdot\frac{x - 2}{3 - 2} = \tfrac12x^2 - \tfrac32x + 1.

\end{align}

The Lagrange interpolating polynomial is:

:$\backslash begin\{align\}$

L(x) &= y_0 \cdot\ell_0(x)

+ y_1\cdot\ell_1(x)

+ y_2\cdot\ell_2(x)

= x^2.

\end{align}

In (second) barycentric form,

:$$

L(x) = \frac

{\displaystyle \sum_{j=0}^2 \frac{w_j}{x-x_j}y_j}

{\displaystyle \sum_{j=0}^2 \frac{w_j}{x-x_j}}

= \frac

{\displaystyle \frac{\tfrac12}{x - 1} + \frac{-4}{x - 2} + \frac{\tfrac92}{x - 3}}

{\displaystyle \frac{\tfrac12}{x - 1} + \frac{-1}{x - 2} + \frac{\tfrac12}{x - 3}}.

File:Runge's phenomenon in Lagrange polynomials.svg

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.{{cite book|title=Scientific Computing with MATLAB|volume=2|series=Texts in computational science and engineering |author-link1= Alfio Quarteroni |first1=Alfio|last1=Quarteroni|first2=Fausto|last2=Saleri|publisher=Springer|year=2003|isbn=978-3-540-44363-6|page=66|url=https://books.google.com/books?id=fE1W5jsU4zoC&pg=PA66}}.

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

When interpolating a given function f by a polynomial of degree {{mvar|k}} at the nodes $x\_0,...,x\_k$ we get the remainder $R(x)\; =\; f(x)\; -\; L(x)$ which can be expressed as{{AS ref|25, eqn 25.2.3|878}}

:

where $f[x\_0,\backslash ldots,x\_k,x]$ is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as

:$R(x)\; =\; \backslash frac\{\backslash ell(x)\}\{2\backslash pi\; i\}\; \backslash int\_C\; \backslash frac\{f(t)\}\{(t-x)(t-x\_0)\; \backslash cdots\; (t-x\_k)\}\; dt\; =\; \backslash frac\{\backslash ell(x)\}\{2\backslash pi\; i\}\; \backslash int\_C\; \backslash frac\{f(t)\}\{(t-x)\backslash ell(t)\}\; dt.$

The remainder can be bound as

:$|R(x)|\; \backslash leq\; \backslash frac\{(x\_k-x\_0)^\{k+1\}\}\{(k+1)!\}\backslash max\_\{x\_0\; \backslash leq\; \backslash xi\; \backslash leq\; x\_k\}\; |f^\{(k+1)\}(\backslash xi)|.$

Clearly, $R(x)$ is zero at nodes. To find $R(x)$ at a point $x\_p$, define a new function $F(x)=R(x)-\backslash tilde\{R\}(x)=f(x)-L(x)-\backslash tilde\{R\}(x)$ and choose $\backslash tilde\{R\}(x)=C\backslash cdot\backslash prod\_\{i=0\}^k(x-x\_i)$ 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 $f(x)$ is $k+1$-times differentiable, since $L(x)$ and $\backslash tilde\{R\}(x)$ are polynomials, and therefore, are infinitely differentiable, $F(x)$ will be $k+1$-times differentiable. By Rolle's theorem, $F^\{(1)\}(x)$ has $k+1$ zeroes, $F^\{(2)\}(x)$ has $k$ zeroes... $F^\{(k+1)\}$ has 1 zero, say

:$F^\{(k+1)\}(\backslash xi)=f^\{(k+1)\}(\backslash xi)-L^\{(k+1)\}(\backslash xi)-\backslash tilde\{R\}^\{(k+1)\}(\backslash xi)$

:$L^\{(k+1)\}=0,\backslash tilde\{R\}^\{(k+1)\}=C\backslash cdot(k+1)!$ (Because the highest power of $x$ in $\backslash tilde\{R\}(x)$ is $k+1$)

:$0=f^\{(k+1)\}(\backslash xi)-C\backslash cdot(k+1)!$

The equation can be rearranged as{{Cite web|url=https://sam.nitk.ac.in/sites/default/Numerical_Methods/Interpolation/interpolation.pdf|title=Interpolation|pages=12–15|archive-url=https://web.archive.org/web/20170215114507/http://www.sam.nitk.ac.in/sites/default/Numerical_Methods/Interpolation/interpolation.pdf|archive-date=2017-02-15}}

:$C=\backslash frac\{f^\{(k+1)\}(\backslash xi)\}\{(k+1)!\}$

Since $F(x\_p)\; =\; 0$ we have $R(x\_p)=\backslash tilde\{R\}(x\_p)\; =\; \backslash frac\{f^\{k+1\}(\backslash xi)\}\{(k+1)!\}\backslash prod\_\{i=0\}^k(x\_p-x\_i)$

The {{mvar|d}}th derivative of a Lagrange interpolating polynomial can be written in terms of the derivatives of the basis polynomials,

:$L^\{(d)\}(x)\; :=\; \backslash sum\_\{j=0\}^\{k\}\; y\_j\; \backslash ell\_j^\{(d)\}(x).$

Recall (see {{slink|#Definition}} above) that each Lagrange basis polynomial is

$$\backslash begin\{aligned\}$$

\ell_j(x) &= \prod_{\begin{smallmatrix}m = 0\\ m\neq j\end{smallmatrix}}^k \frac{x-x_m}{x_j-x_m}.

\end{aligned}

The first derivative can be found using the product rule:

:$\backslash begin\{align\}$

\ell_j'(x) &= \sum_{\begin{smallmatrix}i=0 \\ i\not=j\end{smallmatrix}}^k \Biggl[ \frac{1}{x_j-x_i}\prod_{\begin{smallmatrix}m=0 \\ m\not = (i , j)\end{smallmatrix}}^k \frac{x-x_m}{x_j-x_m} \Biggr]

\\[5mu]

&= \ell_j(x)\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{x-x_i}.

\end{align}

The second derivative is

:$\backslash begin\{align\}$

\ell_j''(x)

&= \sum_{\begin{smallmatrix}i=0 \\ i\ne j\end{smallmatrix}}^{k} \frac{1}{x_j-x_i} \Biggl[ \sum_{\begin{smallmatrix}m=0 \\ m\ne(i,j)\end{smallmatrix}}^{k} \Biggl( \frac{1}{x_j-x_m}\prod_{\begin{smallmatrix}n=0 \\ n\ne(i,j,m)\end{smallmatrix}}^{k} \frac{x-x_n}{x_j-x_n} \Biggr) \Biggr]

\\[10mu]

&= \ell_j(x)

\sum_{0 \leq i < m \leq k}

\frac{2}{(x-x_i)(x - x_m)}

\\[10mu]

&= \ell_j(x)\Biggl[\Biggl(\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{x-x_i}\Biggr)^2-\sum_{\begin{smallmatrix}i=0 \\i\not=j\end{smallmatrix}}^k \frac{1}{(x-x_i)^2}\Biggr].

\end{align}

The third derivative is

:$\backslash begin\{align\}$

\ell_j'''(x)

&= \ell_j(x)

\sum_{0 \leq i < m < n \leq k}

\frac{3!}{(x-x_i)(x - x_m)(x - x_n)}

\end{align}

and likewise for higher derivatives.

Note that all of these formulas for derivatives are invalid at or near a node. A method of evaluating all orders of derivatives of a Lagrange polynomial efficiently at all points of the domain, including the nodes, is converting the Lagrange polynomial to power basis form and then evaluating the derivatives.

The Lagrange polynomial can also be computed in finite fields. This has applications in cryptography, such as in Shamir's Secret Sharing scheme.

• Neville's algorithm
• Newton form of the interpolation polynomial
• Bernstein polynomial
• Carlson's theorem
• Lebesgue constant
• The Chebfun system
• Table of Newtonian series
• Frobenius covariant
• Sylvester's formula
• Finite difference coefficient
• Hermite interpolation

{{reflist}}

{{wikibooks|Algorithm Implementation|Mathematics/Polynomial interpolation|Polynomial interpolation}}

• {{springer|title=Lagrange interpolation formula|id=p/l057170}}
• [https://www.alglib.net/interpolation/polynomial.php ALGLIB] has an implementations in C++ / C# / VBA / Pascal.
• [https://www.gnu.org/software/gsl/ GSL] has a polynomial interpolation code in C
• [https://stackoverflow.com/questions/11029615/lagrange-interpolation-method/11552763 SO] has a MATLAB example that demonstrates the algorithm and recreates the first image in this article
• [https://nm.mathforcollege.com/chapter-05-04-lagrange-method/ Lagrange Method of Interpolation — Notes, PPT, Mathcad, Mathematica, MATLAB, Maple]
• [http://www.math-linux.com/spip.php?article71 Lagrange interpolation polynomial] on www.math-linux.com
• {{MathWorld|urlname=LagrangeInterpolatingPolynomial|title=Lagrange Interpolating Polynomial}}
• [http://mathformeremortals.wordpress.com/2013/01/15/bicubic-interpolation-excel-worksheet-function/ Excel Worksheet Function for Bicubic Lagrange Interpolation]
• [https://learn.64bitdragon.com/articles/computer-science/numerical-analysis/lagrange-interpolation Lagrange polynomials in Python]

{{Joseph-Louis Lagrange}}

{{authority control}}

Category:Interpolation

Category:Polynomials

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