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/

**References**:**Re: Abs function***From:*"David W. Cantrell" <DWCantrell@sigmaxi.org>