Re: Interval arithmetic and Dependency problems

• To: mathgroup at smc.vnet.net
• Subject: [mg95484] Re: [mg95451] Interval arithmetic and Dependency problems
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Tue, 20 Jan 2009 05:43:50 -0500 (EST)

```Interval[{-1, 1}] Interval[{-1, 1}] is not equivalent to Interval[{-1, 1}]^2

In the product case there are two independent intervals and the product of these independent intervals can be anywhere on the Interval[{-1,1}]. With a single interval that is squared the result must be nonnegative.

Nonetheless, I agree that there are opportunities to improve the interval calculations.

Bob Hanlon

---- magma <maderri2 at gmail.com> wrote:

=============
I just started to have a look at Mathematica capabilities in Interval
arithmetic.

It seems that Mathematica does not consider the so called dependency problem
(see for instance wikipedia on Interval arithmetic for an
explanation).

For example evaluate

f[x_] := x^2 + x
f[Interval[{-1, 1}]]

and you get

Interval[{-1, 2}]

which is not really very good.
The correct or, if you prefer, best result should be

Interval[{-1/4,2}]

since -1/4 and 2 are really the min and max values of f over the given
interval.

In other words Mathematica is not capable of calculating f[x] when x is an
Interval and f is a general function.
Yet Mathematica, according to documentation, is capable of calculating f[x] on
an Interval, when f is restricted to a standard functions.
For example the "square" function

Interval[{-1, 1}]^2

Interval[{0, 1}]

But if the same  "square" is expressed as a product, the result is not
so good (tight):

evaluating
Interval[{-1, 1}] Interval[{-1, 1}]

gives
Interval[{-1, 1}]

It seems to me that a further function should be implemented in Mathematica.
Something called IntervalEvaluation or similar, with syntax like

IntervalEvaluation[f[x],x,Interval[...]]

Building this function should not be too difficult, using Maximize and
Minimize on the given interval.