Re: Re: Uniform design

• To: mathgroup at smc.vnet.net
• Subject: [mg48117] Re: [mg48107] Re: Uniform design
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 14 May 2004 00:12:17 -0400 (EDT)
• References: <c7nnc7\$dm5\$1@smc.vnet.net> <200405130408.AAA26737@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```On 13 May 2004, at 13:08, Maxim wrote:

> Also, it's strange that Solve accepts intervals (Mathematica Help for
> Interval even gives such an example), but doesn't really support them:
>
> In[7]:=
> Solve[1/(x - 1) == Interval[{-1, 1}]]
>
> Out[7]=
> {{x -> Interval[{-Infinity, Infinity}]}}
>
> Not much point in treating this equation as Solve[1/(x-1)==a,x] and
> giving incorrect result.
>
I agree that it seems strange that this sort of thing was included in
the help browser, without additional comment,  for it can certainly
only be misleading. Interval arithmetic is strange and does not obey
usual rules:

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

Interval[{-Infinity, 0}, {2, Infinity}]

and

(1 + Interval[{-1, 1}])/Interval[{-1, 1}]

Interval[{-Infinity, Infinity}]

This means that the answer returned by Solve will depend on how you

Solve[1/(x - 1) == Interval[{-1, 1}]]

{{x -> Interval[{-Infinity, Infinity}]}}

Solve[x - 1 == 1/Interval[{-1, 1}]]

{{x -> Interval[{-Infinity, 0}, {2, Infinity}]}}

(What is actually weird is that

Solve[1/(x - 1) == Interval[{-1, 1}], x]

{}

while

Solve[x - 1 == 1/Interval[{-1, 1}], x]

{{x -> Interval[{-Infinity, 0}, {2, Infinity}]}})

Whether the original answer should be considered wrong or only
excessively "pessimistic" depends on the context. The usual context in
which  interval arithmetic is used is for error estimation, where it is
most important that it should  not return an interval smaller than the
correct one and at least in this case it does not.

Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/

```

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