Re: FullSimplify Question
- To: mathgroup at smc.vnet.net
- Subject: [mg48606] Re: FullSimplify Question
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Mon, 7 Jun 2004 05:33:41 -0400 (EDT)
- Organization: The University of Western Australia
- References: <c9tmvi$sek$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <c9tmvi$sek$1 at smc.vnet.net>,
"Dana DeLouis" <delouis at bellsouth.net> wrote:
> Hello. Hate to ask, but does anyone know what the issue is in the
> following?
>
> This equation basically is 5 when t is less than 0, and 1 when t>=0
>
> equ = 5 - 4*UnitStep[t]
>
> If I ask to "Simplify" this, by using the Sign function, I get the
> following, which is not even close
>
> FullSimplify[equ, TransformationFunctions -> {Sign}]
>
> 1 - UnitStep[t]
From the documentation:
TransformationFunctions is an option for Simplify and FullSimplify
which gives the list of functions to apply to try to t
So Sign is being _applied_ to equ. You can see this more clearly if you
try
FullSimplify[equ,
TransformationFunctions -> {Sign[Print[{#,Sign[#]}]; #] & }]
printing out the sub-expressions that FullSimplify is attempting to
simplify by application of Sign.
> The above is 1 when t<0, and 0 when t>=0.
>
> I have to use both functions "FourierTransform" and
> "InverseFourierTransform" to get the correct form.
>
> 3 - 2*Sign[t]
The transformation rule, UnitStep[x_] :> (Sign[x] + 1)/2 (valid
everywhere except at x==0) will do what you want:
Simplify[equ /. UnitStep[x_] :> (Sign[x] + 1)/2]
> My question is that I can't figure out why FullSimplify was so far off.
> Would Mathematica have been better to leave the expression unevaluated so as
> to use other methods?
You "told" Mathematica to apply Sign recursively in an attempt to
simplify the expression at hand ...
Cheers,
Paul
--
Paul Abbott Phone: +61 8 9380 2734
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