Change of Variables in Indexed Summation
Theorem
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $a, b, c, d$ 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 $g: \closedint c d \to \closedint a b$ be a bijection.
Then we have an equality of indexed summations:
- $\ds \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = c}^d \map f {\map g i}$
Outline of Proof
We use Indexed Summation over Translated Interval to translate $\closedint c d$ to $\closedint a b$.
This reduces the problem to Indexed Summation does not Change under Permutation.
Proof
Because $g : \closedint c d \to \closedint a b$ is a bijection, these sets are equivalent.
By Cardinality of Integer Interval, $\closedint a b$ has cardinality $b - a + 1$.
Thus:
- $b - a + 1 = d - c + 1$
Thus
- $c - a = d - b$
By Indexed Summation over Translated Interval:
- $\ds \sum_{i \mathop = c}^d \map f {\map g i} = \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }$
By Translation of Integer Interval is Bijection, the mapping $T : \closedint a b \to \closedint c d$ defined as:
- $\map T k = k + c - a$
is a bijection.
By Composite of Bijections is Bijection, $g \circ T$ is a permutation of $\closedint a b$.
We have:
| \(\ds \sum_{i \mathop = c}^d \map f {\map g i}\) | \(=\) | \(\ds \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }\) | Indexed Summation over Translated Interval | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^b \map f {\map g {\map T i} }\) | Definition of $T$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = a}^b \map f i\) | Indexed Summation does not Change under Permutation |
$\blacksquare$