Summation of Multiple of Mapping on Finite Set

From ProofWiki
Jump to navigation Jump to search


Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $S$ be a finite set.

Let $f: S \to \mathbb A$ 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 summations on finite sets:

$\displaystyle \sum_{s \mathop \in S} g \left({s}\right) = \lambda \cdot \sum_{s \mathop \in S} f \left({s}\right)$

Outline of Proof

Using the definition of summation on a finite set, we reduce this to Indexed Summation of Multiple of Mapping.


Let $n$ be the cardinality of $S$.

Let $\sigma : \N_{< n} \to S$ be a bijection, where $\N_{< n}$ is an initial segment of the natural numbers.

By definition of summation, we have to prove the following equality of indexed summations:

$\displaystyle \sum_{i \mathop = 0}^{n - 1} g \left({\sigma \left({i}\right)}\right) = \lambda \cdot \sum_{i \mathop = 0}^{n - 1} f \left({\sigma \left({i}\right)}\right)$

By Multiple of Mapping Composed with Mapping, $g \circ \sigma = \lambda \cdot \left({f \circ \sigma}\right)$.

The above equality now follows from Indexed Summation of Multiple of Mapping.


Also see