Sum of Summations over Overlapping Domains/Example
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Example of Sum of Summations over Overlapping Domains
- $\ds \sum_{1 \mathop \le j \mathop \le m} a_j + \sum_{m \mathop \le j \mathop \le n} a_j = \paren {\sum_{1 \mathop \le j \mathop \le n} a_j} + a_m$
Proof
Let $\map R j$ be the propositional function $1 \mathop \le j \mathop \le m$.
Let $\map S j$ be the propositional function $m \mathop \le j \mathop \le n$.
Then we have:
\(\ds \map R j \lor \map S j\) | \(=\) | \(\ds \paren {1 \mathop \le j \mathop \le m} \lor \paren {m \mathop \le j \mathop \le n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 \mathop \le j \mathop \le n}\) |
and:
\(\ds \map R j \land \map S j\) | \(=\) | \(\ds \paren {1 \mathop \le j \mathop \le m} \land \paren {m \mathop \le j \mathop \le n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {j = m}\) |
The result follows from Sum of Summations over Overlapping Domains.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: $(12)$