Divided Differences Table
The construction of a divided differences table is straightforward. Although we stop at third divided differences, higher differences can be obtained in the same way. First divided differences are derived from the definition. For example, to compute \(f[x_0,x_1,x_2]\), subtract the adjacent terms \(f[x_1,x_2] - f[x_0,x_1]\) and divide by \((x_2 - x_0)\). Similarly, \(f[x_0,x_1,x_2,x_3] = \frac{f[x_1,x_2,x_3] - f[x_0,x_1,x_2]}{x_3 - x_0}\). Newton's formula uses the main diagonal of this table.
🔹 Example: Divided Differences Table
Consider the points \( (0,1), (1,2), (2,9), (3,28) \). The divided differences table is:
| xi | f(xi) | f[xi,xi+1] | f[xi,...,xi+2] | f[xi,...,xi+3] |
|---|---|---|---|---|
| 0 | 1 | 1 | 1 | 1 |
| 1 | 2 | 3 | 2 | 6 |
| 2 | 9 | 7 | 3 | 6 |
| 3 | 28 | 19 | 1 | - |
Using Newton's formula with \(x_0=0\), the interpolation polynomial is:
\[ p_3(x) = 1 + 1(x-0) + 3(x-0)(x-1) + 1(x-0)(x-1)(x-2) = x^3 + 1 \]
This is the same polynomial (by uniqueness) as obtained with the Lagrange method. The quadratic polynomial through the first three points is:
\[ p_2(x) = 1 + 1(x-0) + 3(x-0)(x-1) \]
If we want to add a fourth point, e.g., \( (5,54) \), the table can be extended, and the new polynomial of degree 4 is simply:
\[ p_4(x) = p_3(x) - 3/5 (x-0)(x-1)(x-2)(x-3) \]
This demonstrates that higher-degree polynomials can be built recursively without starting over.