Generalized Lagrange Interpolation

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 $n+1$ points, resulting in a polynomial of degree $n$. Such a polynomial can be constructed from higher degree Lagrange interpolating polynomials

$$L^{(n)}_k(x) = \prod_{i=0,i\ne k}^n \frac{x - x_i}{x_k - x_i}.$$

They satisfy the special property,

$$L^{(n)}_k(x_j) = \delta_{j,k}.$$

The construction gives

$$P_{0,\ldots{},n}(x) = \sum_{k=0}^n L^{(n)}_k(x) y_k.$$

If we set $x = x_j$ for any $j$, we see that all the polynomials except $L^{(n)}_j$ vanish, leaving $P_{0,\ldots{},n}(x_j) = y_j$. Thus the polynomial passes through all the points, as required. Since there can be only one degree $n$ polynomial interpolating all $n+1$ points, our construction gives the unique result.

As an application of this result, let's write the polynomial interpolating the values at $x = 0.2, 0.3, 0.4$ in our table. With three points it must be a quadratic. The Lagrange construction gives

$$\begin{eqnarray*} P_{1,2,3} &=& -0.8218 \frac{(x - 0.3)(x - 0.4)}{(0.2 - 0.3)(0... ...\ &&+ 0.1048 \frac{(x - 0.2)(x - 0.3)}{(0.4 - 0.2)(0.4 - 0.3)} \end{eqnarray*}$$