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Re: Interval arithmetic and Dependency problems

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  • Subject: [mg95484] Re: [mg95451] Interval arithmetic and Dependency problems
  • From: Bob Hanlon <hanlonr at>
  • Date: Tue, 20 Jan 2009 05:43:50 -0500 (EST)
  • Reply-to: hanlonr at

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> wrote: 

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

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

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


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

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

gives the correct answer

Interval[{0, 1}]

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

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

Interval[{-1, 1}]

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


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

Any comments?

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