Proof
(related to Theorem: Fundamental Theorem of the Difference Calculus)
- By hypothesis, $F:D\to\mathbb C$ is an antidifference of a complex-valued function $f:C\to\mathbb C$, $D\subseteq\mathbb C.$
- Let $a,b\in\mathbb C$ and let $n\in\mathbb N$ be a natural number.
- By hypothesis, $F$ equals the indefinite sum $F(x)=\sum f(x).$
- Thus, its difference operator equals $\Delta F(x)=f(x).$
- By definition of the difference operator, we have $\Delta F(x)=F(x+1)-F(x).$
- Hence, $$\begin{array}{lcl}
F(a+1)-F(a)&=&f(a)\\
F(a+2)-F(a+1)&=&f(a+1)\\
\vdots&=&\vdots\\
F(a+n-1)-F(a+n-2)&=&f(a+n-2)\\
F(a+n)-F(a+n-1)&=&f(a+n-1)\\
\end{array}$$
- Adding up both sides of the above equations, we get
$$F(a+n)-F(a)=\sum_{x=a}^{a+n-1}f(x).$$
∎
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References
Bibliography
- Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960
