The Antidifference Operator

The Antidifference Operator

We have already seen with The Difference Operator that if $f$ is a real-valued function then the difference operator applied to $f$ is $\Delta f = f(x + 1) - f(x)$ which is an analogue to the derivative $\frac{df}{dx}$ of $f$.

We will now look at another calculus analogue - this time with respect to the antiderivative of a real-valued function. Recall that if $f$ is a real-valued function that $F$ is said to be an antiderivative of $f$ is $\frac{dF}{dx} = f$. We now define the antidifference of a function.

Definition Let $f$ be a real-valued function. Then if $\Delta F(x) = f(x)$ then $F$ is said to be an Antidifference of $f$ using the notation $F = \Delta^{-1} f$ where $\Delta^{-1}$ is the Antidifference Operator.

We have already see that:

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

Therefore if $f(x) = 2x + 1$ then $F(x) = x^2$ is an antidifference of $f$.

Now recall that an antiderivative of a function is not unique. This is because if $F(x)$ is differentiable then so is $F(x) + C$ for any constant $C$ and so:

\begin{align} \quad \frac{d}{dx} (F(x) + C) = \frac{d}{dx} (F(x)) + \frac{d}{dx} (C) = \frac{d}{dx} (F(x)) \end{align}

The same holds for antidifferences. If $y = F(x)$ is an antidifference of $f$ then so is $y = F(x) + C$ since:

\begin{align} \quad \Delta (F(x) + C) = \Delta (F(x)) + \Delta (C) = \Delta (F(x)) + 0 = \Delta F(x) \end{align}

As a result, to denote the General antidifference of a function $f$ we usually tack on the constant $+ C$. Furthermore, the antidifference operator also satisfies the additivity property and homogeneity property and is hence linear. Right now we can formulate some antidifferences of some common functions from reversing the results we've noted on the More Properties of the Difference Operator page. They're summarized in the following theorem and can easily be verified.

Theorem 1: For $C$ as a constant, the following hold:
a) $\Delta^{-1} 0 = C$.
b) $\Delta^{-1} x^{\underline{k}} = \frac{1}{k+1} x^{\underline{k+1}} + C$.
c) $\Delta^{-1} 2^x = 2^x + C$
d) $\Delta^{-1} \binom{x}{k} = \binom{x}{k + 1} + C$.
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