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:


 * $\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = c}^d \map f {\map g i}$

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:
 * $\displaystyle \sum_{i \mathop = c}^d \map f {\map g i} = \sum_{i \mathop = a}^b \map f {\map g {i + c - a} }$

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:

Also see

 * Change of Variables in Summation over Finite Set