|
[Date Index]
[Thread Index]
[Author Index]
Re: Simplify and Abs
- To: mathgroup at smc.vnet.net
- Subject: [mg54641] Re: [mg54602] Simplify and Abs
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Fri, 25 Feb 2005 01:18:40 -0500 (EST)
- References: <200502240821.DAA13175@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 24 Feb 2005, at 09:21, Simon Anders 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]
>
> How do I make Matheamtica notice, that the assumptions constrain the
> argument of Abs[] to positive values?
>
> 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?
>
> TIA
> Simon
>
>
>
It seems to me that FullSimplify is indeed missing some rules for
Simplifying expressions involving Absolute. However, in the case when
you are dealing with real quantities there is a simple workaround;
FullSimplify[ComplexExpand[Abs[p - 1]], p < 1 && p > 1/2]
1 - p
In fact what ComplexExpand does here is:
ComplexExpand[Abs[x]]
Sqrt[x^2]
so when dealing only with reals you could use Sqrt[x^2] (for example by
defining your own function abs). Functions like FullSimplify are
generally better able to deal with expressions like Sqrt[x^2] than with
Abs.
Andrzej Kozlowski
Prev by Date:
Re: finding roots of 1 + 6*x - 8*x^3
Next by Date:
Re: Testing the 'type' of a root returned by Solve
Previous by thread:
Re: Simplify and Abs
Next by thread:
Re: Simplify and Abs
|