More Properties of the Difference Operator

Recall from The Difference Operator page that if $f$ is a real-valued function then the difference operator $\Delta$ on $f$ is defined to be:

(1)

\begin{align} \quad \Delta f = f(n + 1) - f(n) \end{align}

We proved some very basic properties of the difference operator. We showed that if $f$ and $g$ are real-valued functions that $\Delta$ satisfies the additivity property, that is $\Delta (f + g) = \Delta f + \Delta g$. We also showed that if $f$ is a real-valued function and $a \in \mathbb{R}$ then $\Delta (af) = a \Delta f$ and so $\Delta$ satisfies the homogeneity property. Furthermore, we showed that if $f$ is a constant function then $\Delta f = 0$.

So far, we've noted that $\Delta$ behaves very much like the differentiation operator $D$ on real-valued functions. Recall that for any real-valued function $f$ that provided the limit exists:

(2)

\begin{align} \quad D(f(x)) = \frac{df}{dx} = \lim_{h \to 0} \frac{f(x+ h) - f(x)}{h} \end{align}

By setting $h = 1$ and replacing the variable $x$ with $n$ we get the difference operator $\Delta$, so it's somewhat natural that $\Delta$ inherits many of the properties that the differentiation operator $D$ has. We will look at some more of these properties in the following theorems, the first of which gives us an analogue to the differentiation rule that $D(x^n) = nx^{n-1}$.

Theorem 1: If $f(x) = x^{\underline{n}} = x \cdot ( x - 1) \cdot ... \cdot (x - n + 1)$ then $\Delta f(x) = nx^{\underline{n-1}}$.

Recall that $x^{\underline{n}}$ is simply a Falling Factorial.

• Proof:

(3)

\begin{align} \quad \Delta f(x) = f(x + 1) - f(x) \\ \quad \Delta f(x) = (x + 1)^{\underline{n}} - x^{\underline{n}} \\ \quad \Delta f(x) = (x + 1) \cdot x \cdot ... \cdot (x + 1 - n + 1) - x \cdot (x - 1) \cdot ... \cdot (x - n + 1) \\ \quad \Delta f(x) = (x + 1) \cdot x \cdot ... \cdot (x - n + 2) - x \cdot (x - 1) \cdot ... \cdot (x - n + 1) \\ \quad \Delta f(x) = (\left [ x + 1 \right ] - \left [ x - n + 1 \right ]) \cdot (x \cdot x \cdot (x - 1) \cdot ... \cdot (x - n + 2)) \\ \quad \Delta f(x) = nx^{\underline{n-1}} \quad \blacksquare \end{align}

Recall from calculus that if $f(x) = e^x$ then $\frac{df}{dx} = e^x$. We would like to find a function $f$ such that $\Delta f = f$. Fortunately one such function exists as proven in Theorem 2.

Theorem 2: If $f(x) = 2^x$ then $\Delta f(x) = \Delta 2^x = 2^x$.

• Proof:

(4)

\begin{align} \quad \Delta f(x) = \Delta 2^x \\ \quad \Delta f(x) = 2^{x + 1} - 2^x \\ \quad \Delta f(x) = 2^x(2 - 1) \\ \quad \Delta f(x) = 2^x \quad \blacksquare \end{align}

Recall from The Generalized Binomial Coefficient Formula page that the generalized binomial coefficient for $n \in \mathbb{R}$ and $k \in \{0, 1, 2, ... \}$ is given by:

(5)

\begin{align} \quad \binom{n}{k} = \frac{n \cdot (\alpha - 1) \cdot ... \cdot (n - k + 1)}{k!} = \frac{n^{\underline{k}}}{k!} \end{align}

We will derive a rather interesting equality with regards to the difference of generalized binomial coefficients.

Theorem 3: If $f(x) = \binom{x}{k}$ then $\Delta f(x) = \binom{x}{k - 1}$.

• Proof: We use the homogeneity property of $\Delta$ as well as Theorem 1 to arrive at:

(6)

\begin{align} \quad \Delta f(x) = \Delta \binom{x}{k} \\ \quad \Delta f(x) = \Delta \frac{x^{\underline{k}}}{k!} \\ \quad \Delta f(x) = \frac{1}{k!} \Delta x^{\underline{k}} \\ \quad \Delta f(x) = \frac{1}{k!} \cdot k x^{\underline{k-1}} \\ \quad \Delta f(x) = \frac{x^{\underline{k-1}}}{(k - 1)!} \\ \quad \Delta f(x) = \binom{x}{k - 1} \quad \blacksquare \end{align}