Re: Simplify and Abs

*To*: mathgroup at smc.vnet.net*Subject*: [mg54668] Re: [mg54602] Simplify and Abs*From*: DrBob <drbob at bigfoot.com>*Date*: Fri, 25 Feb 2005 01:19:34 -0500 (EST)*References*: <200502240821.DAA13175@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

FullSimplify is giving the simplest result already. >> the assumptions constrain the argument of Abs[] to positive values That's just not true. 2/3 is less than one, greater than 1/2, but 2/3 - 1 is not positive. This works, on the other hand: FullSimplify[Abs[p-1],p>1] -1+p Bobby On Thu, 24 Feb 2005 03:21:06 -0500 (EST), 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] > > 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 > > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**Simplify and Abs***From:*Simon Anders <simon.anders@uibk.ac.at>