The general problem here is: given n points in the plane

find the polynomial

of degree less than or equal to n - 1 such that

Since a polynomial is a function of x the n numbers x₁, x₂, .. x_n have to be distinct.

The Case when n = 4 and x₁ = 1, x₂ = 2, x₃ = 3, x₄ = 4.

We shall first discuss how this is done when n = 4 and x₁ = 1, x₂ = 2, x₃ = 3, x₄ = 4. Once this is understood then the general case is easy to understand

So suppose we want to construct a polynomial passing through the four points where

is given. The basic idea is that the third degree polynomial

is equal to 0 for x = 2, x = 3, and x = 4. Furthermore if we define v₁(x) by

then a simple computation shows that

We continue with simple variations of this technique by setting

Then

Similarly, for

we have

To finish, we set

Then, by the properties of v₁(x), v₂(x),v₃(x), v₄(x), we have

as required. The polynomial p(x) assumes the values y₁, y₂, y₃, y₄ at x₁, x₂, x₃, x₄ , respectively. It is of degree at most 3 for each of the v_i(x) are cubic.

The polynomial p(x) may be of degree less than 3. To see this compute it when the four points are (1,1), (2,2), (3,3), (4,4), points which lie on the line y = x; p(x) will be linear. Similarly, if the points are (-1, 1), (0,0), (1,1) and (2,4), which lie on the parabola y = x² , p(x) will be a quadratic.

The General Case

In this case we define v_i(x) by

for i = 1, 2, .., n.. Then, by construction

Consequently if we define p(x) by

then

A tabulation of the v_i(x) for the n = 4 case is given below.