# Enlarging a Rectangle Inside Another

$\vect{I}$ is a generalized rectangle in $\R^n$ with edges of length $l_i$. $\vect{J} \subseteq \vect{I}$ is the main character here. If the edges of $\vect{J}$ is allowed to coincide with those of $\vect{I}$, then the problem is trivial.

How about finding $\vect{J} \subseteq \interior \vect{I}$ such that $\volume \vect{I} - \volume \vect{J} < \varepsilon$?

Goal: Choose a big $L$ that depends on $\vect{I}$ and $\varepsilon$.1

$\begin{split} \volume \vect{J} :=& \prod_{i = 1}^n \left(l_i - \frac{\varepsilon}{L}\right) \\ \volume \vect{I} =& \prod_{i = 1}^n l_i \\ =& \prod_{i = 1}^n \left [\left(l_i - \frac{\varepsilon}{L}\right) + \frac{\varepsilon}{L} \right] \\ =& \left[ \prod_{i = 1}^n \left( l_i - \frac{\varepsilon}{L} \right) \right] + \sum_{i = 1}^n \left( l_i - \frac{\varepsilon}{L} \right) \left( \frac{\varepsilon}{L} \right)^{n - 1} \\ &+ \sum_{i = 2}^{n - 1} \sum_{\left\{ a_j \right\} \in S_i} \left[ \prod_{j = 1}^i \left( l_{a_j} - \frac{\varepsilon}{L} \right) \right] \left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n, \end{split}$

where $S_i$ is a set of $i$ numbers chosen from $\left\{ 1,\dots,n \right\}$, and $\left\{ a_j \right\}$ is an increasing sequence in $S_i$.

I assumed that $n > 2$. For $n = 2$, deleting the third term in the last step will do. For $n = 1$, the problem is trivial.

Then

$\begin{split} \volume \vect{I} =& \volume \vect{J} + \sum_{i = 1}^{n - 1} \sum_{\left\{a_j\right\} \in S_i} \left[ \prod_{j = 1}^i \left( l_{a_j} - \frac{\varepsilon}{L} \right) \right] \left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n \\ <& \volume \vect{J} + \sum_{i = 1}^{n - 1} \sum_{\left\{ a_j \right\} \in S_i} \left( \prod_{j = 1}^n l_{a_j} \right) \left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n \\ =& \volume \vect{J} + \sum_{i = 1}^{n - 1} \sum_{\left\{ a_j \right\} \in S_i} \volume \vect{I} \, \left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n \\ =& \volume \vect{J} + \sum_{i = 1}^{n - 1} {n \choose i} \volume \vect{I}\,\left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n \end{split}$

Rearanging the above inequality gives

$\volume \vect{I} - \volume \vect{J} < \sum_{i = 1}^{n - 1} {n \choose i} \volume \vect{I} \, \left( \frac{\varepsilon}{L} \right)^{n - i} + \left( \frac{\varepsilon}{L} \right)^n$

Assume $\varepsilon < 1, L \ge 1$, and let $M := \max\limits_{1 \le i \le n - 1} {n \choose i}$.

Then we have

$\begin{split} \volume \vect{I} - \volume \vect{J} <& \left( \frac{\varepsilon}{L} \right)^n + \sum_{i = 1}^{n - 1} {n \choose i} \volume \vect{I} \, \left( \frac{\varepsilon}{L} \right)^{n - i} \\ =& \frac{\varepsilon}{L} + M (n - 1) \volume \vect{I} \,\frac{\varepsilon}{L} \\ =& [1 + M (n - 1) \volume \vect{I}] \,\frac{\varepsilon}{L} \end{split}$

Hence, if $L \ge 1 + M (n - 1) \volume \vect{I}$, then $\volume \vect{I} - \volume \vect{J} < \varepsilon$.

1. $L$ is used to connote “large number”.