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MathGroup Archive 2004

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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 
choose to write your equation:


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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