# Higher Order Differences

Recall from The Difference Operator that if $f$ is a real-valued function then $\Delta f(x) = f(x + 1) - f(x)$. We will now look at higher orders of the difference operator $\Delta$.

Definition: If $f$ is a real-valued function then the $p^{\mathrm{th}}$ Order Difference of $f$ denoted $\Delta^p f$ is defined to be $\Delta^p f(x) = \Delta^{p-1} f(x + 1) - \Delta^{p-1} f(x)$. |

For example, if $f(x) = x^2 + 2x - 1$ then the first, second, and third ordered differences of $f$ denoted $\Delta f$, $\Delta^2 f$, and $\Delta^3 f$ respectively are:

(1)As you can see, to determine $\Delta^p f$, it appears we need to compute $\Delta f$, $\Delta^2 f$, …, $\Delta^{p-1} f$. Interestingly enough, higher order differences seem to follow a pattern. Notice that:

(4)Notice the coefficients on $\Delta f$, $\Delta^2 f$, and $\Delta^3 f$. It appears as though the binomial coefficients are appear. Furthermore, we note that the terms are alternating from positive to negative. We generalize this observation in the following theorem.

Theorem 1: If $f$ is a real-valued function and $p \in \{1, 2, ... \}$ then $\displaystyle{\Delta^p f(x) = \binom{p}{0}(-1)^p f(x) + \binom{p}{1} (-1)^{p-1} f(x + 1) + ... + \binom{p}{p-1} (-1) f(x + p - 1) + \binom{p}{p} f(x + p) = \sum_{k=0}^{p} \binom{p}{k} (-1)^{p-k} f(x + k)}$. |

**Proof:**Let $S, T$ be operators where $S(f(x)) = f(x + 1)$ and $T(f(x)) = f(x)$. Then $S^p (f(x)) = f(x + p)$ and $T^p (f(x)) = f(x)$. Note that $\Delta = (S - T)$ and $\Delta^p = (S - T)^p$. Therefore:

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- By The Binomial Theorem we have that:

- Applying $\Delta^p$ to a real-valued function $f$ yields us: