Deriving a polynomial from nth differences

Let \(f(x) = ax^2 + bx + c.\) Then the first differences of \(f(x)\) can be written as \(g(x),\) as follows

$$\begin{align} g(x) =\,& f(x + 1) - f(x) \\ =\,& [a(x + 1)^2 + b(x + 1) + c] \\ -\,& [ax^2 + bx + c] \\ =\,& a(2x + 1) + b \\ =\,& (2a)x + (a + b) \end{align}$$

The first differences of \(g(x)\) will be the second differences of \(f(x).\) We write these as \(h(x),\) as follows

$$\begin{align} h(x) =\,& g(x + 1) - g(x) \\ =\,& (2a)(x + 1) + (a + b) \\ -\,& (2a)x + (a + b) \\ =\,& 2a \end{align}$$

Now, the pattern is clear

$$\begin{array}{} c & & a + b + c & & 4a + 2b + c & & 9a + 3b + c \\ & a + b & & 3a + b & & 5a + b & \\ & & 2a & & 2a & & \\ \end{array}$$

This pattern is useful, because it allows us to find a quadratic from its first 3 values. In fact, this method can be generalized to work for any polynomial. If you start with \(f\) being an \(n\)th degree polynomial, you will need to compute its nth differences to find the pattern. To use the pattern, you'll need the first \(n + 1\) values from the polynomial you're attempting to infer.