Definition 8.1.1.
An interval on the real line is the set \(\boldsymbol{a}= [\underline{a},\overline{a}] = \{ x \; | \; \underline{a} \leq x \leq \overline{a} \}\text{.}\) If \(\underline{a}=\overline{a}\text{,}\) then the interval is called thin.
Section 8.1 Interval Arithmetic
We first start with arithmetic and applying functions to intervals. We start with the definition of an interval.
Note that we have defined an interval as a closed subset. That is, the endpoints of the interval are in the interval and that the interval can contain a single value.
Subsection 8.1.1 Basic Interval Operations
Since intervals are also sets, let’s define the following:
\begin{equation*}
\begin{aligned}\boldsymbol{a}\amp = \boldsymbol{b}\qquad \text{if $\underline{a}=\underline{b}$ and $\overline{a}= \overline{b}$}\\ \boldsymbol{a}\amp \subseteq \boldsymbol{b}\qquad \text{if $\underline{a}\geq \underline{b}$ and $\overline{a}\leq \overline{b}$}\\ \boldsymbol{a}\amp \subset \boldsymbol{b}\qquad \text{if $\boldsymbol{a}\subseteq \boldsymbol{b}$ and $\boldsymbol{a}\neq \boldsymbol{b}$}\\\end{aligned}
\end{equation*}
Also, the notion of intersection and union of sets can be extended to intervals if we are careful. Since in general, the union of two intervals is not necessarily an interval, then
\begin{equation*}
\begin{aligned}\boldsymbol{a}\cup \boldsymbol{b}\amp = [\min(\underline{a},\underline{b}),\max(\overline{a},\overline{b})]\end{aligned}
\end{equation*}
This is called the convex hull of the two intervals. Also
\begin{equation*}
\begin{aligned}\boldsymbol{a}\cap \boldsymbol{b}\amp = \begin{cases}\emptyset \amp \text{if $\overline{a}< \underline{b}$ and $\underline{a}>\overline{b}$} \\ [\max(\underline{a},\underline{b}),\min(\overline{a},\overline{b})] \amp \text{otherwise}\end{cases}\end{aligned}
\end{equation*}
If we define
\begin{equation*}
\begin{aligned}\rad(\boldsymbol{a})\amp = \frac{1}{2}(\underline{a}- \overline{a}) \\ \midpt(\boldsymbol{a})\amp =\frac{1}{2}(\underline{a}+ \overline{a}) \\ \diam(\boldsymbol{a})\amp = 2 \rad(\boldsymbol{a})\end{aligned}
\end{equation*}
to be the radius, midpoint and diameter of the interval. Additional, if we define
\begin{equation*}
\begin{aligned}\mig(\boldsymbol{a})\amp = \min_{a \in \boldsymbol{a}}a \\ \magn(\boldsymbol{a})\amp = \max_{a \in \boldsymbol{a}}a\end{aligned}
\end{equation*}
to be the “mignitude” and magnitude, which are the smallest (and largest respectively) distances between the origin and the interval. Another equivalent definition of these can be written:
\begin{equation*}
\begin{aligned}\mig(\boldsymbol{a})\amp = \begin{cases}0 \amp \text{if $0 \in \boldsymbol{a}$} \\ \min(|\underline{a}|,|\overline{a}|) \amp \text{otherwise} \\\end{cases}\\ \magn(\boldsymbol{a})\amp = \max(|\underline{a}|,|\overline{a}|)\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}|\boldsymbol{a}| = \abs(\boldsymbol{a}) = \{|a| \; | \; a \in \boldsymbol{a}\} = [\mig(\boldsymbol{a}), \magn(\boldsymbol{a})]\end{aligned}
\end{equation*}
Subsection 8.1.2 Operations on Intervals
Let \(A\) and \(B\) be intervals. We defined \(\star\) to be the four basic operations \(+,-,\cdot, \div\) on real numbers. Then
\begin{equation*}
\begin{aligned}A \star B\amp = \{ x \star y \; | \; x \in A, y \in B\}\end{aligned}
\end{equation*}
Example 8.1.2.
Let \(A=[1,2]\text{,}\) \(B=[3,5]\text{,}\) and \(C=[-2,1]\text{,}\) find \(A+B\text{,}\)\(A-B\text{,}\) \(B-C\text{,}\) \(B \cdot C\text{,}\) \(A \div B\text{.}\)
Solution.
For each of these, we find all values in the resulting set, which typically means that we find the smallest possible value in the resulting set and the largest one.
-
\(\displaystyle A+B = [4,7]\)
-
\(\displaystyle A-B = [-4,-1]\)
-
\(\displaystyle B-C = [2,7]\)
-
\(\displaystyle B \cdot C = [-10,5]\)
-
\(\displaystyle A \div B = [1/5,2/3]\)
We can define these more specifically as
\begin{equation*}
\begin{aligned}\boldsymbol{a}+\boldsymbol{b}\amp = [\underline{a}+\underline{b},\overline{a}+\overline{b}] \\ \boldsymbol{a}- \boldsymbol{b}\amp = [\underline{a}-\overline{b},\overline{a}-\underline{b}] \\ \boldsymbol{a}\cdot \boldsymbol{b}\amp = [\min(\underline{a}\underline{b},\underline{a}\overline{b},\overline{a}\underline{b},\overline{a}\overline{b}),\max(\underline{a}\underline{b},\underline{a}\overline{b},\overline{a}\underline{b},\overline{a}\overline{b})] \\ \boldsymbol{a}\div \boldsymbol{b}\amp = \boldsymbol{a}\cdot [1/\overline{b},1/\underline{b}] \qquad \text{if $0 \not \in \boldsymbol{b}$}.\end{aligned}
\end{equation*}
Subsection 8.1.3 Computational Concerns
The formulas above are the beginning of the ability to computations with intervals.
-
For computer languages that allow new types, it is relatively straightforward to define a interval type, you simply need to store the min and max values of the interval.
-
Using the formulas above, the four basic operations can be computed quite easily and fast.
-
Except for when 0 is in the dividend interval in division. We will see that later.
Subsection 8.1.4 Expanding the idea of division
Let’s redefine the notion of division to be
\begin{equation*}
\begin{aligned}\boldsymbol{a}\div \boldsymbol{b}\amp = \{ c \; | \; a = cb, a \in \boldsymbol{a}, b \in \boldsymbol{b}\}\end{aligned}
\end{equation*}
Let’s look at an example of this: \(\boldsymbol{a}=[2,3]\) and \(\boldsymbol{b}=[-1,2]\text{.}\) Since \(0 \in \boldsymbol{b}\text{,}\) we will replace \(\boldsymbol{b}\) with
\begin{equation*}
\begin{aligned}\boldsymbol{b}' \cup \boldsymbol{b}''\amp = [-1,0)\cup(0,2)\end{aligned}
\end{equation*}
And note that we can define
\begin{equation*}
\begin{aligned}[2,3] \div [-1,0)\amp = \lim_{x \rightarrow 0^-}[2,3] \div [-1,x) \\\amp = \lim_{x \rightarrow 0^-}[2,3] \cdot (1/x,-1] \\\amp = \lim_{x \rightarrow 0^-}[3/x,-2] = (-\infty,-2]\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}[2,3] \div (0,2]\amp = \lim_{x \rightarrow 0^+}[2,3] \div (x,2] \\\amp = \lim_{x \rightarrow 0^+}[2,3] \cdot [1/2,1/x) \\\amp = \lim_{x \rightarrow 0^+}[1,1/x) = [1,\infty)\end{aligned}
\end{equation*}
So a way to define division containing 0 would be in this case is:
\begin{equation*}
\begin{aligned}[2,3] \div [-1,2] = [2,3] \div (\boldsymbol{b}' \cup \boldsymbol{b}) = (-\infty,-2] \cup [1,\infty)\end{aligned}
\end{equation*}
There are some issues with this. First, the result is not an interval. Secondly, if that wasn’t a problem, how to handle the infinite interval?
