In this section, we shall study the interpolation polynomial in the Lagrange form. Given a set of (n+1) data points and a function f, the aim is to determine a polynomial of degree n which interpolates f at the points in question.

## Interpolation

Given a set of $(n+1)$ data points

$(x_0,y_0),(x_1,y_1), \ldots ,(x_n,y_n)$

The points defined by $(x_i)_{0 \leq i\leq n}$ are called points of interpolation.

The points $(y_i)_{0 \leq i\leq n}$ are the values of interpolation. To interpolate a function $f$, we define the values of interpolation as follows: $$y_i=f(x_i), \quad \forall i=0, \ldots ,n$$

## Lagrange interpolation polynomial

The purpose here is to determine the unique polynomial of degree $n$, $P_n$ which verifies

$P_n(x_i)=f(x_i),\quad \forall i=0,\ldots,n.$

The polynomial which meets this equality is Lagrange interpolation polynomial

$P_n(x)= \sum_{k=0}^n l_k(x)f(x_k)$

where $l_k$ are polynomials of degree $n$ that form a basis of $\mathcal{P}_n$

$l_k(x)= \prod_{i=0,\, i\neq k}^{n} \dfrac{x-x_i}{x_k-x_i}=\dfrac{x-x_0}{x_k-x_0} \cdots \dfrac{x-x_{k-1}}{x_k-x_{k-1}} \dfrac{x-x_{k+1}}{x_k-x_{k+1}} \cdots \dfrac{x-x_{n}}{x_k-x_{n}}$

## Properties of Lagrange interpolation polynomial and Lagrange basis

• They are the $l_k$ polynomials which verify the following property:
$l_k(x_i) = \delta_{ki}=\left\{\begin{array}{ll} 1 & i=k \\ 0 & i\neq k \\ \end{array} \right.,\quad \forall i=0,\ldots,n.$
• They form a basis of the vector space $\mathcal{P}_n$ of polynomials of degree at most equal to $n$
$\sum_{k=0}^n \alpha_k l_k(x)=0$

By setting: $x=x_i$, we obtain:

$\sum_{k=0}^n \alpha_k l_k(x_i)=\sum_{k=0}^n \alpha_k \delta_{ki}=0\Longrightarrow\alpha_i=0$

The set $(l_k)_{0 \leq k\leq n}$ is linearly independent and consists of $n+1$ vectors. It is thus a basis of $\mathcal{P}_n$.

• Finally, we can easily see that:
$P_n(x_i)=\sum_{k=0}^n l_k(x_i)f(x_k)=\sum_{k=0}^n\delta_{ki} f(x_k)=f(x_i)$

## Existence and uniqueness of Lagrange interpolation polynomials

Existence. The proof is shown above. Actually, it corresponds to the construction of Lagrange interpolation polynomial with regard to the basis $(l_k)_{0 \leq k\leq n}$ of $\mathcal{P}_n$.

Uniqueness. Consider two elements $P_n$ and $Q_n$ of $\mathcal{P}_n$ which verify

$P_n(x_i)=Q_n(x_i)=f(x_i),\quad \forall i=0,\ldots,n.$

Let $R_n=(P_n-Q_n)\in \mathcal{P}_n$. The polynomial $R_n$ has $(n+1)$ roots which are exactly the $(x_i)_{0 \leq i\leq n}$ since

$R_n(x_i)=P_n(x_i)-Q_n(x_i)=f(x_i)-f(x_i)=0,\quad \forall i=0,\ldots,n.$

$R_n$ has then $(n+1)$ roots and $R_n\in \mathcal{P}_n$. Therefore,

$R_n=0 \Longrightarrow P_n=Q_n.$

We have thus shown the existence and uniqueness of Lagrange interpolation polynomial.

## Error in Lagrange interpolation

Assume that $f\in \mathcal{C}^{n+1}([a,b])$ and $x\in[a,b]$. Let $I$ be the closed set defined by $I=[\min(x,x_0),\max(x,x_n)]$ (the smallest closed set containing $x$ and $x_i,~\forall~i$).

Theorem. $$\forall x\in [a,b],\quad \exists \xi \in I / f(x)-P_n(x)=\displaystyle \frac{f^{n+1}(\xi)}{(n+1)!} \displaystyle \prod_{i=0}^{n}(x-x_i)$$

Proof. There are two possible ways to prove this theorem: 1) the one encountered in [The polynomial interpolation in the form of Newton->article46] and, 2) the following proof.

If $x=x_i$, the problem is over:

$f(x)-P_n(x)=f(x_i)-P_n(x_i)=0$

Now, assume that $x\neq x_i$ and let us define the application $\Phi$ as follows: $$\Phi(x)=\displaystyle\frac{f(x)-P_n(x)}{\displaystyle\prod_{i=0}^{n}(x-x_i)}.$$

We also define the application $g$:

$g(x,t)=f(t)-P_n(t)-\displaystyle\prod_{i=0}^{n}(t-x_i)\Phi(x)$

$g(x,\cdot)$ is $(n+1)$ times differentiable and evaluates to zero at the $(n+2)$ points $x_0,x_1,\ldots,x_n,x$ of the interval $I$. By successively applying Rolle’s theorem, $g^{(n+1)}(x,\cdot)$ evaluates to zero at a point $\xi\in I$:

$g^{(n+1)}(x,\xi)=0$

The $(n+1)^{th}$ derivative of $g(x,\cdot)$ can be easily calculated:

$g^{(n+1)}(x,t)=f^{(n+1)}(t)-(n+1)!\Phi(x)$

By setting $t=\xi$, we have:

$g^{(n+1)}(x,\xi)=f^{(n+1)}(\xi)-(n+1)!\Phi(x)=0$

Therefore

$\Phi(x)=\displaystyle\frac{f^{(n+1)}(\xi)}{(n+1)!}$

We finally conclude that

$f(x)-P_n(x)=\displaystyle\frac{f^{n+1}(\xi)}{(n+1)!} \displaystyle\prod_{i=0}^{n}(x-x_i)$

Corollary. Assume that $f\in \mathcal{C}^{n+1}([a,b])$ and $x\in[a,b]$.

$\forall x\in [a,b],\quad |f(x)-P_n(x)|\leq \displaystyle\frac{\displaystyle|\prod_{i=0}^{n}(x-x_i)|}{(n+1)!}\sup_{x\in[a,b]}|f^{n+1} (x)|$

Alternatively stated:

$\forall x\in [a,b],\quad |f(x)-P_n(x)|\leq \displaystyle\frac{(b-a)^{n+1}}{(n+1)!}\sup_{x\in[a,b]}|f^{n+1} (x)|$

## Example: computing Lagrange interpolation polynomials

Given a set of three data points ${(0,1), (2,5),(4,17)}$, we shall determine the Lagrange interpolation polynomial of degree 2 which passes through these points.

First, we compute $l_0,l_1$ and $l_2$: $$l_0(x)=\displaystyle\frac{(x-2)(x-4)}{(0-2)(0-4)}=\displaystyle\frac{(x-2)(x-4)}{8}$$

$l_1(x)=\displaystyle\frac{x(x-4)}{(2-0)(2-4)}=-\displaystyle\frac{x(x-4)}{4}$ $l_2(x)=\displaystyle\frac{x(x-2)}{(4-0)(4-2)}=\displaystyle\frac{x(x-2)}{8}$

Lagrange interpolation polynomial is:

$\begin{array}{rcl} P_2(x)&=&1l_0(x)+5l_1(x)+17l_2(x)\\ &=&1+x^2 \end{array}$

## Scilab: computing Lagrange interpolation polynomial

The Scilab function lagrange.sci determines Lagrange interpolation polynomial. $X$ encompasses the points of interpolation and $Y$ the values of interpolation. $P$ is the Lagrange interpolation polynomial.

lagrange.sci

We thus have: