The Fundamental Theorem of the Calculus of Finite Differences

The Fundamental Theorem of the Calculus of Finite Differences

Recall from The Antidifference Operator that $F$ is said to be an antidifference of the real-valued function $f$ is $\Delta F = f$, or in using the antidifference operator $\Delta^{-1}$ we write $F = \Delta^{-1} f$.

We will now look at yet another theorem on antidifferences that is analogous to the Fundamental Theorem of Calculus.

 Theorem 1 (The Fundamental Theorem of the Calculus of Finite Differences): Let $f$ be a real-valued function and let $a$ and $b$ be integers such that $a \leq b$. If $F = \Delta^{-1} f$ then $\sum_{n=a}^{b} f(n) = F(b + 1) - F(a)$.
• Proof: Let $f$ be a real-valued function and let $a, b \in \mathbb{Z}$ where $a \leq b$.
• Since $F = \Delta^{-1} f$ we have that $\Delta F = f$, i.e., $\Delta F(x) = F(x + 1) - F(x) = f(x)$ and so:
(1)
\begin{align} \quad \sum_{n=a}^{b} f(n) = \sum_{n=a}^{b} (F(n + 1) - F(n)) \\ \quad \sum_{n=a}^{b} f(n) = \left ( F(a + 1) - F(a) \right ) + \left ( F(a + 2) - F(a + 1) \right ) + ... + \left ( F(b) - F(b - 1) \right ) + \left ( F(b + 1) - F(n) \right ) \\ \quad \sum_{n=a}^{b} f(n) = F(b + 1) - F(a) \quad \blacksquare \end{align}

Let's look at an example of applying the Fundamental Theorem of the Calculus of Finite Differences. Consider the function $f(x) = 2x + 1$. An antidifference of $f$ is the function $F(x) = x^2$. Suppose we want to evaluate the sum $\sum_{n=1}^{6} f(n) = \sum_{n=1}^{6} (2n + 1)$. Then:

(2)
\begin{align} \quad \sum_{n=1}^{6} (2n + 1) = F(6 + 1) - F(1) = F(7) - F(1) = (7)^2 - (1)^2 = 48 \end{align}

We will now look at a simply corollary to the Fundamental Theorem of the Calculus of Finite Differences.

 Corollary 1: For all $n \in \{0, 1, 2, ... \}$, $\displaystyle{\sum_{k=0}^{n} 2^k = 2^{n+1} - 1}$.
• Proof: Consider the function $f(x) = 2^k$. An antidifference of $f$ is $f$ itself, and so by the Fundamental Theorem of the Calculus of Finite Differences we have that:
(3)
\begin{align} \quad \sum_{k=0}^{n} f(k) = f(n+1) - f(0) = 2^{n+1} - 2^0 = 2^{n+1} - 1 \quad \blacksquare \end{align}