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Re: Interval[{a,b}]-Interval[{a,b}] = 0?

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  • Subject: [mg66644] Re: [mg66613] Interval[{a,b}]-Interval[{a,b}] = 0?
  • From: Daniel Lichtblau <danl at>
  • Date: Thu, 25 May 2006 02:58:37 -0400 (EDT)
  • References: <>
  • Sender: owner-wri-mathgroup at

Richard Fateman wrote:
> Is this a bug or a feature?
> Notice that Interval[{-1,1}]-Interval[{-1,1}] is Interval[{-2,2}].
> (I'm using Mathematica 5.1)

It is a very important feature in that this is exactly how intervals are 
supposed to work (read this as though it were ALL IN CAPS). Intervals 
are to be treated as independent values, and interval arithmetic (as 
presented in the literature) should indeed do exactly as you see above. 
For ranges of values that are meant to correlate e.g. obtaining zero from

With[{x=Interval[...]}, x-x]

the fact is that interval arithmetic is not the correct tool. Likewise 
for finding max and min values of a function of x over some interval on 
the real line, though sometimes interval arithmetic is used to get fast 
overshooting of such bounds.

Here is a related example I ran across yesterday, from a bug report sent 
in by a user. We want to get a sensible result from

Limit[ArcTan[Sin[x]], x->Infinity]

Here is a sensible result from my development version (I am ignoring 
some of the things stated in the recent sci.math.symbolic thread:

In[4]:= InputForm[Limit[ArcTan[Sin[x]], x->Infinity]]
Out[4]//InputForm= Interval[{-Pi/4, Pi/4}]

Here is a result that is not so good. It is what prompted the user to 
send in a bug report.

In[1]:= Limit[ArcTan[Sin[x]], x->Infinity]
Out[1]= 0

Why did we get 0? Well, we found some limits of summands from a series 
representation. The series used was not optimal, but that's beside the 
point. Here is an intermediate form of the limit sum.

Log[(2 - (1 + I)*Interval[{-1, 1}] + (1 + I)*Interval[{-1, 1}])/2] -
   Log[(2 - (1 + I)*Interval[{-1, 1}] + (1 + I)*Interval[{-1, 1}])/2]

Notice that this is of the form Log[ii]-Log[ii] where ii contains 
Interval objects. Mathematica's eager Plus handling cancels the 
summands, giving 0. We are (today) experimenting with weakening this 
handler to not coalesce terms containing Interval. If this proves to 
give acceptable results on relevant parts of our test suite then we will 
be able to give the better interval result shown above.

Daniel Lichtblau
Wolfram Research

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