Returning to our table, if we examine the change in the function
between 0.1 and 0.2 and between 0.2 and 0.3 we see that there is a
dramatic change in slope. So clearly our result would be improved if
we selected an interpolating polynomial with curvature. It takes
three points to fix a parabola. Why stop there? We could include all
points, resulting in a polynomial of degree . Such a
polynomial can be constructed from higher degree Lagrange
interpolating polynomials

They satisfy the special property,

The construction gives

If we set for any , we see that all the polynomials except vanish, leaving . Thus the polynomial passes through all the points, as required. Since there can be only one degree polynomial interpolating all points, our construction gives the unique result.

As an application of this result, let's write the polynomial interpolating the values at in our table. With three points it must be a quadratic. The Lagrange construction gives