Indexed Summation over Translated Interval
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Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a$ and $b$ be integers.
Let $\closedint a b$ be the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Let $c\in\Z$ be an integer.
Then we have an equality of indexed summations:
- $\ds \sum_{i \mathop = a}^b f(i) = \sum_{i \mathop = a + c}^{b + c} \map f {i - c}$
Proof
The proof goes by induction on $b$.
Basis for the Induction
Let $b < a$.
Then:
- $b + c < a + c$
Thus both indexed summations are zero.
This is our basis for the induction.
Induction Step
Let $b \ge a$.
We have:
\(\ds \sum_{i \mathop = a + c}^{b + c} \map f {i - c}\) | \(=\) | \(\ds \sum_{i \mathop = a + c}^{b + c - 1} \map f {i - c} + \map f {b + c - c}\) | Definition of Indexed Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map f i + \map f b\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^b \map f i\) | Definition of Indexed Summation |
By the Principle of Mathematical Induction, the proof is complete.
$\blacksquare$