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# Lagrange polynomial

This image shows, for four points ((−9, 5), (−4, 2), (−1, −2), (7, 9)), the (cubic) interpolation polynomial L(x) (dashed, 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.

In numerical analysis, Lagrange polynomials are used for polynomial interpolation. For a given set of points ${\displaystyle (x_{j},y_{j})}$ with no two ${\displaystyle x_{j}}$ values equal, the Lagrange polynomial is the polynomial of lowest degree that assumes at each value ${\displaystyle x_{j}}$ the corresponding value ${\displaystyle y_{j}}$, so that the functions coincide at each point.

Although named after Joseph-Louis Lagrange, who published it in 1795, the method was first discovered in 1779 by Edward Waring.[1] It is also an easy consequence of a formula published in 1783 by Leonhard Euler.[2]

Uses of Lagrange polynomials include the Newton–Cotes method of numerical integration and Shamir's secret sharing scheme in cryptography.

Lagrange interpolation is susceptible to Runge's phenomenon of large oscillation. As changing the points ${\displaystyle x_{j}}$ requires recalculating the entire interpolant, it is often easier to use Newton polynomials instead.

## Definition

Here we plot the Lagrange basis functions of 1st, 2nd, and 3rd order on a bi-unit domain. Linear combinations of Lagrange basis functions are used to construct Lagrange interpolating polynomials. Lagrange basis functions are commonly used in finite element analysis as the bases for the element shape-functions. Furthermore, it is common to use a bi-unit domain as the natural space for the finite-element's definition.

Given a set of k + 1 data points

${\displaystyle (x_{0},y_{0}),\ldots ,(x_{j},y_{j}),\ldots ,(x_{k},y_{k})}$

where no two ${\displaystyle x_{j}}$ are the same, the interpolation polynomial in the Lagrange form is a linear combination

${\displaystyle L(x):=\sum _{j=0}^{k}y_{j}\ell _{j}(x)}$

of Lagrange basis polynomials

${\displaystyle \ell _{j}(x):=\prod _{\begin{smallmatrix}0\leq m\leq k\\m\neq j\end{smallmatrix}}{\frac {x-x_{m}}{x_{j}-x_{m}}}={\frac {(x-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x-x_{j-1})}{(x_{j}-x_{j-1})}}{\frac {(x-x_{j+1})}{(x_{j}-x_{j+1})}}\cdots {\frac {(x-x_{k})}{(x_{j}-x_{k})}},}$

where ${\displaystyle 0\leq j\leq k}$. Note how, given the initial assumption that no two ${\displaystyle x_{j}}$ are the same, then (when ${\displaystyle m\neq j}$) ${\displaystyle x_{j}-x_{m}\neq 0}$, so this expression is always well-defined. The reason pairs ${\displaystyle x_{i}=x_{j}}$ with ${\displaystyle y_{i}\neq y_{j}}$ are not allowed is that no interpolation function ${\displaystyle L}$ such that ${\displaystyle y_{i}=L(x_{i})}$ would exist; a function can only get one value for each argument ${\displaystyle x_{i}}$. On the other hand, if also ${\displaystyle y_{i}=y_{j}}$, then those two points would actually be one single point.

For all ${\displaystyle i\neq j}$, ${\displaystyle \ell _{j}(x)}$ includes the term ${\displaystyle (x-x_{i})}$ in the numerator, so the whole product will be zero at ${\displaystyle x=x_{i}}$:

${\displaystyle \forall ({j\neq i}):\ell _{j}(x_{i})=\prod _{m\neq j}{\frac {x_{i}-x_{m}}{x_{j}-x_{m}}}={\frac {(x_{i}-x_{0})}{(x_{j}-x_{0})}}\cdots {\frac {(x_{i}-x_{i})}{(x_{j}-x_{i})}}\cdots {\frac {(x_{i}-x_{k})}{(x_{j}-x_{k})}}=0.}$

On the other hand,

${\displaystyle \ell _{j}(x_{j}):=\prod _{m\neq j}{\frac {x_{j}-x_{m}}{x_{j}-x_{m}}}=1}$

In other words, all basis polynomials are zero at ${\displaystyle x=x_{j}}$, except ${\displaystyle \ell _{j}(x)}$, for which it holds that ${\displaystyle \ell _{j}(x_{j})=1}$, because it lacks the ${\displaystyle (x-x_{j})}$ term.

It follows that ${\displaystyle y_{j}\ell _{j}(x_{j})=y_{j}}$, so at each point ${\displaystyle x_{j}}$, ${\displaystyle L(x_{j})=y_{j}+0+0+\dots +0=y_{j}}$, showing that ${\displaystyle L}$ interpolates the function exactly.

## Proof

The function L(x) being sought is a polynomial in x of the least degree that interpolates the given data set; that is, it assumes the value yj at the corresponding xj for all data points j:

${\displaystyle L(x_{j})=y_{j}\qquad j=0,\ldots ,k.}$

Observe that:

• In ${\displaystyle \ell _{j}(x)}$ 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 be a polynomial of degree at most k.
• ${\displaystyle \ell _{j}(x_{i})=\prod _{\begin{smallmatrix}m=0\\m\neq j\end{smallmatrix}}^{k}{\frac {x_{i}-x_{m}}{x_{j}-x_{m}}}.}$

Expand this product. Since the product omits the term where m = j, if i = j then all terms that appear are ${\displaystyle {\frac {x_{j}-x_{m}}{x_{j}-x_{m}}}=1}$. Also, if ij then one term in the product will be (for m = i), ${\displaystyle {\frac {x_{i}-x_{i}}{x_{j}-x_{i}}}=0}$, zeroing the entire product. So,

${\displaystyle \ell _{j}(x_{i})=\delta _{ji}={\begin{cases}1,&{\text{if }}j=i\\0,&{\text{if }}j\neq i\end{cases}},}$

where ${\displaystyle \delta _{ij}}$ is the Kronecker delta. So:

${\displaystyle L(x_{i})=\sum _{j=0}^{k}y_{j}\ell _{j}(x_{i})=\sum _{j=0}^{k}y_{j}\delta _{ji}=y_{i}.}$

Thus the function L(x) is a polynomial with degree at most k and where L(xi) = yi.

Additionally, the interpolating polynomial is unique, as shown by the unisolvence theorem at the polynomial interpolation article.

It's also true that:

${\displaystyle \sum _{j=0}^{k}\ell _{j}(x)=1\qquad \forall x}$

since it must be a polynomial of degree, at most, k and passes through all these k + 1 data points:

${\displaystyle (x_{0},1),\ldots ,(x_{j},1),\ldots ,(x_{k},1)}$

resulting in a horizontal line, since a straight line is the only polynomial of degree less than k + 1 that passes through k + 1 aligned points.

## 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 for our interpolation polynomial ${\textstyle L(x)=\sum _{j=0}^{k}x^{j}m_{j}}$, we must invert the Vandermonde matrix ${\displaystyle (x_{i})^{j}}$ to solve ${\displaystyle L(x_{i})=y_{i}}$ for the coefficients ${\displaystyle m_{j}}$ of ${\displaystyle L(x)}$. By choosing a better basis, the Lagrange basis, ${\textstyle L(x)=\sum _{j=0}^{k}l_{j}(x)y_{j}}$, we merely get the identity matrix, ${\displaystyle \delta _{ij}}$, 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.

## Examples

### Example 1

We wish to interpolate ƒ(x) = x2 over the range 1 ≤ x ≤ 3, given these three points:

{\displaystyle {\begin{aligned}x_{0}&=1&&&f(x_{0})&=1\\x_{1}&=2&&&f(x_{1})&=4\\x_{2}&=3&&&f(x_{2})&=9.\end{aligned}}}

The interpolating polynomial is:

{\displaystyle {\begin{aligned}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{aligned}}}

### Example 2

We wish to interpolate ƒ(x) = x3 over the range 1 ≤ x ≤ 4, given these four points:

 ${\displaystyle x_{0}=1}$ ${\displaystyle f(x_{0})=1}$ ${\displaystyle x_{1}=2}$ ${\displaystyle f(x_{1})=8}$ ${\displaystyle x_{2}=3}$ ${\displaystyle f(x_{2})=27}$ ${\displaystyle x_{3}=4}$ ${\displaystyle f(x_{3})=64}$

The interpolating polynomial is:

{\displaystyle {\begin{aligned}L(x)&={1}\cdot {x-2 \over 1-2}\cdot {x-3 \over 1-3}\cdot {x-4 \over 1-4}+{8}\cdot {x-1 \over 2-1}\cdot {x-3 \over 2-3}\cdot {x-4 \over 2-4}+{27}\cdot {x-1 \over 3-1}\cdot {x-2 \over 3-2}\cdot {x-4 \over 3-4}+{64}\cdot {x-1 \over 4-1}\cdot {x-2 \over 4-2}\cdot {x-3 \over 4-3}\\[8pt]&=x^{3}\end{aligned}}}

### Notes

Example of interpolation divergence for a set of 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 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.[3]

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

## Barycentric form

Using

${\displaystyle \ell (x)=(x-x_{0})(x-x_{1})\cdots (x-x_{k})}$
${\displaystyle \ell '(x_{j})={\frac {\mathrm {d} \ell (x)}{\mathrm {d} x}}{\Big |}_{x=x_{j}}=\prod _{i=0,i\neq j}^{k}(x_{j}-x_{i})}$

we can rewrite the Lagrange basis polynomials as

${\displaystyle \ell _{j}(x)={\frac {\ell (x)}{\ell '(x_{j})(x-x_{j})}}}$

or, by defining the barycentric weights[4]

${\displaystyle w_{j}={\frac {1}{\ell '(x_{j})}}}$

we can simply write

${\displaystyle \ell _{j}(x)=\ell (x){\frac {w_{j}}{x-x_{j}}}}$

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

${\displaystyle L(x)=\ell (x)\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}y_{j}}$

which, if the weights ${\displaystyle w_{j}}$ have been pre-computed, requires only ${\displaystyle {\mathcal {O}}(k)}$ operations (evaluating ${\displaystyle \ell (x)}$ and the weights ${\displaystyle w_{j}/(x-x_{j})}$) as opposed to ${\displaystyle {\mathcal {O}}(k^{2})}$ for evaluating the Lagrange basis polynomials ${\displaystyle \ell _{j}(x)}$ individually.

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

We can further simplify the first form by first considering the barycentric interpolation of the constant function ${\displaystyle g(x)\equiv 1}$:

${\displaystyle g(x)=\ell (x)\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}.}$

Dividing ${\displaystyle L(x)}$ by ${\displaystyle g(x)}$ does not modify the interpolation, yet yields

${\displaystyle L(x)={\frac {\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}y_{j}}{\sum _{j=0}^{k}{\frac {w_{j}}{x-x_{j}}}}}}$

which is referred to as the second form or true form of the barycentric interpolation formula. This second form has the advantage that ${\displaystyle \ell (x)}$ need not be evaluated for each evaluation of ${\displaystyle L(x)}$.

## Remainder in Lagrange interpolation formula

When interpolating a given function f by a polynomial of degree k at the nodes ${\displaystyle x_{0},...,x_{k}}$ we get the remainder ${\displaystyle R(x)=f(x)-L(x)}$ which can be expressed as[5]

${\displaystyle R(x)=f[x_{0},\ldots ,x_{k},x]\ell (x)=\ell (x){\frac {f^{(k+1)}(\xi )}{(k+1)!}},\quad \quad x_{0}<\xi

where ${\displaystyle f[x_{0},\ldots ,x_{k},x]}$ is the notation for divided differences. Alternatively, the remainder can be expressed as a contour integral in complex domain as

${\displaystyle R(z)={\frac {\ell (z)}{2\pi i}}\int _{C}{\frac {f(t)}{(t-z)(t-z_{0})\cdots (t-z_{k})}}dt={\frac {\ell (z)}{2\pi i}}\int _{C}{\frac {f(t)}{(t-z)\ell (t)}}dt.}$

The remainder can be bound as

${\displaystyle |R(x)|\leq {\frac {(x_{k}-x_{0})^{k+1}}{(k+1)!}}\max _{x_{0}\leq \xi \leq x_{k}}|f^{(k+1)}(\xi )|.}$

### Derivation[6]

Clearly, ${\displaystyle R(x)}$ is zero at nodes. To find ${\displaystyle R(x)}$ at a point ${\displaystyle x_{p}}$, define a new function ${\displaystyle F(x)=f(x)-L(x)-R(x)}$ and choose ${\textstyle R(x)=C\cdot \prod _{i=0}^{k}(x-x_{i})}$ (This ensures ${\displaystyle R(x)=0}$ at nodes) where ${\displaystyle C}$ is the constant we are required to determine for a given ${\displaystyle x_{p}}$. Now ${\displaystyle F(x)}$ has ${\displaystyle k+2}$ zeroes (at all nodes and ${\displaystyle x_{p}}$) between ${\displaystyle x_{0}}$ and ${\displaystyle x_{k}}$ (including endpoints). Assuming that ${\displaystyle f(x)}$ is ${\displaystyle k+1}$-times differentiable, ${\displaystyle L(x)}$ and ${\displaystyle R(x)}$ are polynomials, and therefore, are infinitely differentiable. By Rolle's theorem, ${\displaystyle F^{(1)}(x)}$ has ${\displaystyle k+1}$ zeroes, ${\displaystyle F^{(2)}(x)}$ has ${\displaystyle k}$ zeroes... ${\displaystyle F^{(k+1)}}$ has 1 zero, say ${\displaystyle \xi ,\,x_{0}<\xi . Explicitly writing ${\displaystyle F^{(k+1)}(\xi )}$:

${\displaystyle F^{(k+1)}(\xi )=f^{(k+1)}(\xi )-L^{(k+1)}(\xi )-R^{(k+1)}(\xi )}$
${\displaystyle L^{(k+1)}=0,R^{(k+1)}=C\cdot (k+1)!}$ (Because the highest power of ${\displaystyle x}$ in ${\displaystyle R(x)}$ is ${\displaystyle k+1}$)
${\displaystyle 0=f^{(k+1)}(\xi )-C\cdot (k+1)!}$

The equation can be rearranged as

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

## Derivatives

The ${\displaystyle d}$th derivatives of the Lagrange polynomial can be written as

${\displaystyle L^{(d)}(x):=\sum _{j=0}^{k}y_{j}\ell _{j}^{(d)}(x)}$.

For the first derivative, the coefficients are given by

${\displaystyle \ell _{j}^{(1)}(x):=\sum _{\begin{smallmatrix}i=0\\i\not =j\end{smallmatrix}}^{k}\left[{\frac {1}{x_{j}-x_{i}}}\prod _{\begin{smallmatrix}m=0\\m\not =(i,j)\end{smallmatrix}}^{k}{\frac {x-x_{m}}{x_{j}-x_{m}}}\right]}$

and for the second derivative

${\displaystyle \ell _{j}^{(2)}(x):=\sum _{\begin{smallmatrix}i=0\\i\neq j\end{smallmatrix}}^{k}{\frac {1}{x_{j}-x_{i}}}\left[\sum _{\begin{smallmatrix}m=0\\m\neq (i,j)\end{smallmatrix}}^{k}\left({\frac {1}{x_{j}-x_{m}}}\prod _{\begin{smallmatrix}l=0\\l\neq (i,j,m)\end{smallmatrix}}^{k}{\frac {x-x_{l}}{x_{j}-x_{l}}}\right)\right]}$.

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.