Permutation of Indices of Summation

From ProofWiki
(Redirected from Permutation of Range)
Jump to navigation Jump to search

Theorem

Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.

Let the fiber of truth of $R$ be finite.


Then:

$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$

where:

$\ds \sum_{\map R j} a_j$ denotes the summation over $a_j$ for all $j$ that satisfy the propositional function $\map R j$
$\pi$ is a permutation on the fiber of truth of $R$.


Infinite Series

Let the fiber of truth of $R$ be infinite.


Let $\ds \sum_{\map R i} a_i$ be absolutely convergent.


Then:

$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$


Proof

\(\ds \sum_{\map R {\map \pi j} } a_{\map \pi j}\) \(=\) \(\ds \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }\) Definition of Summation by Iverson's Convention
\(\ds \) \(=\) \(\ds \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sum_{j \mathop \in \Z} \sqbrk {i = \map \pi j}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i}\)
\(\ds \) \(=\) \(\ds \sum_{\map R i} a_i\)
\(\ds \) \(=\) \(\ds \sum_{\map R j} a_j\) Change of Index Variable of Summation

$\blacksquare$


Also known as

The operation of permutation of indices of a summation can be seen referred to as a permutation of the range.

However, as the term range is ambiguous in the literature, and as its use here is not strictly accurate (it is the fiber of truth of $R$, not its range, which is being permuted, its use on $\mathsf{Pr} \infty \mathsf{fWiki}$ is discouraged.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Permutation_of_Indices_of_Summation&oldid=536520"