Re: Simplify and Abs

• To: mathgroup at smc.vnet.net
• Subject: [mg54665] Re: [mg54602] Simplify and Abs
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 25 Feb 2005 01:19:27 -0500 (EST)
• References: <200502240821.DAA13175@smc.vnet.net> <b18de451a8a7765509c8ec460cea5330@mimuw.edu.pl>
• Sender: owner-wri-mathgroup at wolfram.com

```I have fallen yet again into the same old trap :-(

Use:

Refine[Abs[p - 1], p < 1 && p > 1/2]

1-p

The problem is that

LeafCount[Abs[p-1]]

4

while

LeafCount[1-p]

5

So Abs[p-1] is, according to the Default ComplexityFunction actually
simpler than 1-p

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/

On 24 Feb 2005, at 12:24, Andrzej Kozlowski wrote:

> 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
>
>

```

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