The Forthcoming IEEE Standard 1788 for Interval Arithmetic

Abstract

This is a slightly expanded form of the author’s talk of the same title at SCAN 2014, Würzburg. Angled towards people who use interval numerical methods little or not at all, it briefly describes how interval arithmetic works, the mindset required to use it effectively, why an interval arithmetic standard was needed, the setting up of IEEE Working Group P1788 for the purpose, the structure of the standard it has produced, some difficulties we encountered, and the current state of the P1788 project. During production of these Proceedings the 1788 standard has been published, but the talk’s original title has been kept.

A Proof of Interval Newton Properties

This appendix proves the properties stated in Sect. 2.2. It may be of interest because item (iv) of the Theorem does not seem to have appeared in the literature before. An interval extension of a real function f of real variables means an interval function ${\varvec{f}}$ of corresponding interval variables such that $y = f(x_1,\ldots ,x_n)$ is in ${\varvec{y}} = {\varvec{f}}({\varvec{x}}_1,\ldots ,{\varvec{x}}_n)$ whenever $x_i$ is in ${\varvec{x}}_i$ for each $i=1,\ldots ,n$.

Theorem 2

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be $C^1$ on an interval ${\varvec{x}}$, which may be unbounded. Let ${\varvec{f}}$ and ${\varvec{f}}\,'$ be interval extensions of f and its derivative $f'$, and let x be any point of ${\varvec{x}}$. Define the set

$$\begin{aligned} Y = x - {\varvec{f}}([x])\mathop {/\!\!/}{\varvec{f}}\,'({\varvec{x}}) \end{aligned}$$

where $\mathop {/\!\!/}$ denotes division in the sense of reverse multiplication. (Thus Y may be empty, an interval, or the union of two disjoint unbounded intervals.)

Then

• (i) Y contains all zeros of f in ${\varvec{x}}$.

• (ii) If $Y\cap {\varvec{x}}=\emptyset$, there are no zeros of f in ${\varvec{x}}$.

• (iii) If $0\notin {\varvec{f}}\,'({\varvec{x}})$, there is at most one zero of f in ${\varvec{x}}$.

• (iv) If Y is nonempty, bounded and $\subseteq {\varvec{x}}$, there is exactly one zero of f in ${\varvec{x}}$.

Proof

(i) Let $z\in {\varvec{x}}$ with $f(z)=0$. By the Mean Value Theorem

$$\begin{aligned} f(x) = f(x)-f(z) = (x-z)f'(\xi ). \end{aligned}$$

(5)

for some $\xi \in {\varvec{x}}$. By definition of interval extension, $f'(\xi )\in {\varvec{f}}'({\varvec{x}})$ and $f(x)\in {\varvec{f}}([x])$. Hence by the definition of reverse multiplication

$$ x-z \in {\varvec{f}}([x])\mathop {/\!\!/}{\varvec{f}}'({\varvec{x}}), $$

that is

$$ z \in x - {\varvec{f}}([x])\mathop {/\!\!/}{\varvec{f}}'({\varvec{x}}) $$

as required.

(ii) This is immediate from (i).

(iii) In (5) let both z and x be roots in ${\varvec{x}}$. Then we have $0=f(x)-f(z) = (x-z)f'(\xi )$. By hypothesis $0\notin {\varvec{f}}'({\varvec{x}})$ which implies $f'(\xi )\ne 0$. Hence $x-z=0$, $x=z$, so there is at most one root.

(iv) Write ${\varvec{b}}={\varvec{f}}'({\varvec{x}})$, ${\varvec{c}}={\varvec{f}}([x])$, both being nonempty by the definition of interval extension. By hypothesis $Y=x-{\varvec{c}}\mathop {/\!\!/}{\varvec{b}}$ is nonempty and bounded, so $Z={\varvec{c}}\mathop {/\!\!/}{\varvec{b}}$ is nonempty and bounded.

I claim $0\notin {\varvec{b}}$. For suppose $0\in {\varvec{b}}$. Then $0\notin {\varvec{c}}$, for if $0\in {\varvec{c}}$ then Z is the unbounded set $\mathbb {R}$, contrary to hypothesis. Now two subcases arise.

– Either ${\varvec{b}}$ is singleton [0], making Z empty, contrary to hypothesis.

– Or, ${\varvec{b}}$ contains 0 and another point, in which case it contains points arbitrarily close to 0. Since $0\notin {\varvec{c}}\ne \emptyset$, ${\varvec{c}}$ contains a nonzero point. Together these imply ${\varvec{c}}\mathop {/\!\!/}{\varvec{b}}$ is unbounded, again contrary to hypothesis.

Thus all the cases of $0\in {\varvec{b}}$ give a contradiction, proving $0\notin {\varvec{b}}$. Hence by part (iii) there is at most one root in ${\varvec{x}}$ and we must show there is at least one. If $f(x)=0$ there is nothing more to prove, so assume $f(x)\ne 0$.

Let $b^*$ be the bound of ${\varvec{b}}$ nearest 0, so it is finite, $\ne 0$ and in ${\varvec{b}}$. Let $z^*$ be the intercept on the x-axis of the line through (x, f(x)) with slope $b^*$, so

$$\begin{aligned} z^* = x - f(x)/b^*, \end{aligned}$$

(6)

equivalently

$$\begin{aligned} f(x)=(x-z^*)b^*. \end{aligned}$$

(7)

Since $f(x)\in {\varvec{c}}$ and $b^*\in {\varvec{b}}$, (6) shows $z^*\in Y\subseteq {\varvec{x}}$. Also $x\in {\varvec{x}}$ so by the Mean Value Theorem there is $\xi \in {\varvec{x}}$ with

$$\begin{aligned} f(x)-f(z^*) = (x-z^*)f'(\xi ). \end{aligned}$$

(8)

Subtracting this from (7) gives

$$\begin{aligned} f(z^*) = (x-z^*)(b^*-f'(\xi )) \end{aligned}$$

(9)

Now $f'(\xi )$ is in ${\varvec{b}}$ by the latter’s definition, so by the definition of $b^*$ it has the same sign as $b^*$ and at least as large absolute value, i.e. $f'(\xi )/b^* \ge 1$. Dividing (9) by (7) (recalling $f(x)\ne 0$) now gives

$$\begin{aligned} f(z^*)/f(x) = 1-f'(\xi )/b^* \le 0, \end{aligned}$$

so $f(z^*)$ has opposite (in the weak sense) sign to f(x). By the Intermediate Value Theorem f has a zero z between x and $z^*$. Since both the latter are in ${\varvec{x}}$ we have $z\in {\varvec{x}}$, and the result is proved.    $\square$