Lagrange Interpolation

[NT123]

Homework Statement

I need to find a "Lagrange basis" corresponding to the function space spanned by the basis (1, x^2).

Homework Equations

I have been told the Lagrange polynomial is of the form

(x-x_1)...(x-x_(k-1))(x-x_(k+1))..(x-x_n) / (x_k-x_1)...(x_k-x_(k-1))(x_k-x_(k+1))..(x_k-x_n)

The Attempt at a Solution

Since a + bx^2 is symmetric about the y axis, I am guessing one of the basis elements should be of the form

(x-x_0)(x+x_0) / (x_1-x_0)(x_1+x_0) = (x^2 - x_0^2) / (x_1-x_0)(x_1+x_0). This has no terms in x so it seems to make sense. Not sure what to do about the other basis element.

 

[mighty2000]

Hello,

I can provide you with some guidance on finding the Lagrange basis corresponding to the function space spanned by the basis (1, x^2).

Firstly, it is important to understand the concept of a Lagrange basis. A Lagrange basis is a set of polynomials that form a basis for a given function space. These polynomials are constructed in such a way that they are orthogonal to each other, meaning that their inner product is equal to zero. This property is useful in many applications, including polynomial interpolation and numerical integration.

Now, let's move on to finding the Lagrange basis for the given function space. As you have correctly mentioned, the Lagrange polynomial for a set of n points is given by:

L(x) = (x-x_1)...(x-x_(k-1))(x-x_(k+1))..(x-x_n) / (x_k-x_1)...(x_k-x_(k-1))(x_k-x_(k+1))..(x_k-x_n)

In this case, we have two basis elements (1, x^2), which means we have two points (x_0, 1) and (x_1, x_1^2). Plugging these points into the Lagrange polynomial, we get:

L_1(x) = (x-x_1) / (x_0-x_1) = (x-x_1) / (x_1-x_0)

L_2(x) = (x-x_0) / (x_1-x_0)

Therefore, the Lagrange basis for the given function space is:

L_1(x) = (x-x_1) / (x_1-x_0)

L_2(x) = (x-x_0) / (x_1-x_0)

I hope this helps in your understanding of the Lagrange basis. If you have any further questions, please feel free to ask. Good luck with your studies!