Indexed Summation of Multiple of Mapping

From ProofWiki
Jump to: navigation, search


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

Let $a, b$ be integers.

Let $\left[{a \,.\,.\, b}\right]$ denote the integer interval between $a$ and $b$.

Let $f: \left[{a \,.\,.\, b}\right] \to \mathbb A$ 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:

$\displaystyle \sum_{i \mathop = a}^b g \left({i}\right) = \lambda \cdot \sum_{i \mathop = a}^b f \left({i}\right)$


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:

\(\displaystyle \sum_{i \mathop = a}^b g \left({i}\right)\) \(=\) \(\displaystyle \sum_{i \mathop = a}^{b - 1} g \left({i}\right) + g \left({b}\right)\) $\quad$ Definition of Indexed Summation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = a}^{b - 1} g \left({i}\right) + \lambda \cdot f \left({b}\right)\) $\quad$ Definition of Product of Mapping with Scalar $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda \cdot \sum_{i \mathop = a}^{b - 1} f \left({i}\right) + \lambda \cdot f \left({b}\right)\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda \cdot \left({\sum_{i \mathop = a}^{b - 1} f \left({i}\right) + f \left({b}\right)}\right)\) $\quad$ Multiplication of Numbers Distributes over Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda \cdot \sum_{i \mathop = a}^b f \left({i}\right)\) $\quad$ Definition of Indexed Summation $\quad$

By the Principle of Mathematical Induction, the proof is complete.


Also see