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# Lagrange interpolation polynomial

Monday 25 June 2007,

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 data points , the points defined by are called points of interpolation. The points are the values of interpolation. To interpolate a function , we define the values of interpolation as follows:

### Lagrange interpolation polynomial

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

The polynomial which meets this equality is Lagrange interpolation polynomial

where are polynomials of degree that form a basis of

### Properties of Lagrange interpolation polynomial and Lagrange basis

They are the polynomials which verify the following property:

They form a basis of the vector space of polynomials of degree at most equal to

By setting: , we obtain:

The set is linearly independent and consists of vectors. It is thus a basis of .
Finally, we can easily see that:

### 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
of .

Uniqueness. Consider two elements and of which verify

Let . The polynomial has roots which are exactly the since

has then roots and . Therefore,

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

### Error in Lagrange interpolation

Assume that and . Let be the closed set defined by (the smallest closed set containing and ).

Theorem.

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

If , the problem is over:

Now, assume that and let us define the application as follows:

We also define the application :

is times differentiable and evaluates to zero at the points of the interval . By successively applying Rolle’s theorem, evaluates to zero at a point :

The derivative of can be easily calculated:

By setting , we have:

Therefore

We finally conclude that

Corollary.
Assume that and .

Alternatively stated:

### Example: computing Lagrange interpolation polynomials

Given a set of three data points , we shall determine the Lagrange interpolation polynomial of degree 2 which passes through these points.

First, we compute and :

Lagrange interpolation polynomial is:

### Scilab: computing Lagrange interpolation polynomial

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

lagrange.sci

We thus have:

## Forum posts

• Congratulations on the article! More could not run on your octave example it says that there is an error in the function call poly