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MathGroup Archive 2004

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Re: Re: Abs function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47351] Re: [mg47318] Re: Abs function
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 6 Apr 2004 06:36:39 -0400 (EDT)
  • References: <c4ge4p$5a$1@smc.vnet.net> <c4j8td$d00$1@smc.vnet.net> <200404050922.FAA21842@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 5 Apr 2004, at 18:22, David W. Cantrell wrote:

> Jens-Peer Kuska <kuska at informatik.uni-leipzig.de> wrote:
>> tha mean that the derivative of Abs[] can be different for
>> real and complex arguments. For real arguments you have
>>
>> D[Abs[x], x] // FullSimplify[#, Element[x, Reals]] &
>>
>> Sign[x]
>
> It's interesting that that works as desired, considering that
>
> In[1]:= Assuming[Element[x, Reals], Simplify[D[Abs[x], x] ]]
>
> Out[1]= Abs'[x]
>
> doesn't work as desired.


It's just the difference between Simplify and FullSImplify:


Assuming[Element[x, Reals], FullSimplify[D[Abs[x], x]]]


Sign[x]

Observe also that:

ComplexExpand[D[Abs[x], x]]

Sqrt[x^2]/x



>  Anyway, perhaps my prior suggestion of rewriting
> in terms of UnitStep is still of some use since
>
> In[2]:= Integrate[Abs[x], x] // FullSimplify[#, Element[x, Reals]] &
>
> Out[2]= Integrate[Abs[x], x]
>
> doesn't work as desired.
>
> BTW, I didn't mention it in my prior post, but perhaps the nicest form
> for that real antiderivative is just 1/2*x*Abs[x]. Could anyone manage
> (without going to lots of trouble) to get Mathematica to give that 
> form?

Well, almost:

Integrate[Abs[t], {t, 0, x},
    GenerateConditions -> False]

(x*Sqrt[x^2])/2



>
> Regards,
> David Cantrell
>
>
>> nikmatz wrote:
>>> i try to find the derivative of abs function
>>>
>>> D[Abs[x],x]
>>>
>>> and Mathematica answer
>>> Abs`[x]
>>>
>>> what mean that
>
>
>
Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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