|
[Date Index]
[Thread Index]
[Author Index]
Re: Simplify and Abs
- To: mathgroup at smc.vnet.net
- Subject: [mg54654] Re: Simplify and Abs
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Fri, 25 Feb 2005 01:18:59 -0500 (EST)
- References: <cvk3fu$d6j$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Simon Anders <simon.anders at uibk.ac.at> wrote:
> Hi,
>
> can it really be that this is already beyond Mathematica?
>
> In := FullSimplify[Abs[p - 1], p < 1 && p > 1/2]
>
> Out := Abs[-1 + p]
Strange.
> How do I make Matheamtica notice, that the assumptions constrain the
> argument of Abs[] to positive values?
Huh?
The assumptions constrain the argument of Abs[] to _negative_ values, and
thus the expression should simplify to 1 - p.
> Any suggestions how to treat these kinds of problems? Specifically, I
> have a list of products of absolute values of simple polynomials in p
> and I know that p is in the interval [0,1].
>
> I would like to know whether the polynomials have constant sign over the
> interval so that the Abs[] can be removed. Can this be done
> automatically?
How about using Refine instead? For example,
In[14]:= Refine[Abs[p - 1], 1/2 < p < 1]
Out[14]= 1 - p
But I don't understand why Simplify (or FullSimplify) doesn't give that
also. After all, the documentation states that
"Refine is one of the transformations tried by Simplify."
David Cantrell
Prev by Date:
Re: finding roots of 1 + 6*x - 8*x^3
Next by Date:
Re: Simplify and Abs
Previous by thread:
Re: Simplify and Abs
Next by thread:
Re: Simplify and Abs
|