Indexed Summation of Multiple of Mapping
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Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f: \closedint a b \to \mathbb A$ be a mapping.
Let $\lambda \in \mathbb A$.
Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.
Then we have the equality of indexed summations:
- $\ds \sum_{i \mathop = a}^b \map g i = \lambda \cdot \sum_{i \mathop = a}^b \map f i$
Proof
The proof goes by induction on $b$.
Basis for the Induction
Let $b < a$.
Then all indexed summations are zero.
Because $0 = \lambda \cdot 0$, the result follows.
This is our basis for the induction.
Induction Step
Let $b \geq a$.
We have:
\(\ds \sum_{i \mathop = a}^b \map g i\) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map g i + \map g b\) | Definition of Indexed Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^{b - 1} \map g i + \lambda \cdot \map f b\) | Definition of Product of Mapping with Scalar | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \cdot \sum_{i \mathop = a}^{b - 1} \map f i + \lambda \cdot \map fb\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \cdot \paren {\sum_{i \mathop = a}^{b - 1} \map f i + \map f b}\) | Distributive Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \cdot \sum_{i \mathop = a}^b \map f i\) | Definition of Indexed Summation |
By the Principle of Mathematical Induction, the proof is complete.
$\blacksquare$