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abs function on an interval?   Message List  
Reply | Forward Message #237 of 257 |
Re: [Frink] abs function on an interval?

Alan Stebbens wrote:
> I probably don't understand intervals well enough yet, but if
>
> x = new interval[-2,2]
> [-2,2]
>
> and
>
> x^2
> [0, 4]
>
> then why doesn't abs[x] yield [0,2]?
>
> Further, why doesn't abs[x+2] => [0, 4]?

As of the current release, the behavior of the abs[x] function is
changed for interval arguments (real or complex arguments are not
changed.) It now does what you expected. It returns an interval,
including both endpoints and a "main/middle" value if one exists.

This is after some lengthy discussions with practitioners in the
field, so I think that what Frink is doing now is quite reasonable. The
behavior matches the spirit of abs[x] for real numbers, and allows more
programs to accept interval arguments and produce expected results
without modification.

There are more details in the "what's new" file:
http://futureboy.us/frinkdocs/whatsnew.html

Thanks to Alan Stebbens for bringing up this issue.

--
Alan Eliasen | "Furious activity is no substitute
eliasen@... | for understanding."
http://futureboy.us/ | --H.H. Williams



Sun Apr 26, 2009 10:38 pm

aeliasen
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Message #237 of 257 |
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I probably don't understand intervals well enough yet, but if x = new interval[-2,2] [-2,2] and x^2 [0, 4] then why doesn't abs[x] yield [0,2]? Further, why...
Alan Stebbens
alan_stebbens
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Apr 14, 2009
7:18 am

... Good question. The short answer is that I'm following the convention for abs[x] that's defined in one of the seminal works on interval arithmetic, Ramon...
Alan Eliasen
aeliasen
Online Now Send Email
Apr 14, 2009
9:44 am

... As of the current release, the behavior of the abs[x] function is changed for interval arguments (real or complex arguments are not changed.) It now does...
Alan Eliasen
aeliasen
Online Now Send Email
Apr 26, 2009
10:37 pm
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