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.

$\blacksquare$


Also see