Re: Abs[a] Sin[Abs[a]]

*To*: mathgroup at smc.vnet.net*Subject*: [mg21021] Re: [mg20923] Abs[a] Sin[Abs[a]]*From*: Adam Strzebonski <adams at wolfram.com>*Date*: Thu, 2 Dec 1999 21:41:32 -0500 (EST)*Organization*: Wolfram Reasearch, Inc*References*: <B46B1939.3546%andrzej@tuins.ac.jp>*Sender*: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote: > > Mathematica can only manage this under the assumption a>0. > In[1]:= > Simplify[ Abs[a] Sin[Abs[a]], a > 0] > Out[1]= > a Sin[a] > > One problem is that Simplify does not know that Abs[a] is -a for negative a. > > In[2]:= > Simplify[Abs[a], a < 0] > Out[2]= > Abs[a] This is a complexity function problem. Under the default complexity function Abs[a] is simpler than -a ( Times[-1, a] in FullForm, now it looks more complicated than Abs[a]...) If we define a complexity function assigning a heavier weight to Abs we can get the simplification. In[1]:= f[x_]:=LeafCount[x]+10 Count[x, _Abs, {0, Infinity}] In[2]:= Simplify[Abs[a], a<0, ComplexityFunction->f] Out[2]= -a In[3]:= Simplify[Abs[a] Sin[Abs[a]], a<0, ComplexityFunction->f] Out[3]= a Sin[a] > > In fact one would really like Simplify to manage more complicated cases > involving Abs, e.g.: > > In[3]:= > Simplify[Abs[Sin[a]], a < 0] > Out[3]= > Abs[Sin[a]] I presume you meant Sin[Abs[a]]. It works automatically, as soon as Abs[a]->-a works. In[4]:= Simplify[Sin[Abs[a]], a<0, ComplexityFunction->f] Out[4]= -Sin[a] > > Just for fun I have decided to write a quick hack that will give the right > answer for these cases. One has to modify Simplify in two ways. Firstly, we > add the above missing rule for Abs. This alone however is not enough. The > second change is a general rule for Simplify[expr, Element[something, > Reals]]. This is a more drastic change to the way Simplify works now but at > the moment I can't see anything that would be broken by it. Here is my > attempt: > > In[4]:= > Unprotect[Simplify]; > Simplify[expr_, x_ < 0] := > Simplify[expr /. {Abs[x] :> -x, Abs[Sin[x]] :> -Sin[x]}] > Simplify[expr_, Element[x_, Reals]] := > If[(Simplify[expr, x > 0] === > Simplify[expr, x < 0]) && ((Simplify[expr, x > 0] /. x :> 0) === > Simplify[expr, x == 0]), Simplify[expr, x > 0], Simplify[expr]] > Protect[Simplify]; > (Really we should also give rules for Simplify[expr_,{a___,x_<0,b___}] etc). > > Now indeed we get: > > In[5]:= > Simplify[Abs[a], a < 0] > Out[5]= > -a > > In[6]:= > Simplify[Abs[Sin[a]], a < 0] > Out[6]= > -Sin[a] > > and finally: > > In[7]:= > Simplify[ Abs[a] *Sin[Abs[a]], Element[a, Reals]] > Out[7]= > a Sin[a] The path of simplification of Abs[a]*Sin[Abs[a]] to a Sin[a] for a>0 is different than for a<0, so to get this simplification Simplify would need to do something similar to your second rule, i.e. simplify the whole expression under assumptions a>0, a==0, and a<0 separately and compare the results. This seems to have a too high complexity to be attempted automatically: an assumption that n variables are real would lead to calling Simplify with 3^n sets of assumptions (on each subexpression of the input expression...). A somewhat less costly version of this would be to just try replacing Abs[a] (for a assumed real) with a and with -a and comparing the results (without simplifying them). It would suffice in this example. I will experiment with adding a rule like this to (possibly Full)Simplify. Best regards, Adam Strzebonski Wolfram Research > > Of course there is a large number of other non-polynomial functions which > are odd and about which Simplify doesn't know so one would like to add a > rule for Simplify[Abs[f[a]],a<0] for all of these. Even this however would > not work in more complicated cases where f was some non-polynomial non > built-in function. The problem is that Simplify relies on the ability to > solve inequalities, which can really only be done in the case of polynomial > ones. > > > From: Edward Goldobin <gold at isitel1.isi.kfa-juelich.de> To: mathgroup at smc.vnet.net > > Organization: Research Center Juelich GmbH > > Date: Wed, 1 Dec 1999 01:49:54 -0500 (EST) > > To: mathgroup at smc.vnet.net > > Subject: [mg21021] [mg20923] Abs[a] Sin[Abs[a]] > > > > How to compell Mathematica to simplify > > Abs[a] Sin[Abs[a]] > > to > > a Sin[a] > > > > ? > > Why it does not do it automatically? I also tried > > assumptions like a :enum: Reals. > > > > Thanx > > Edward > >

**Re: calculus with Grassmann variables**

**Re: Adding interpolated values to a list**

**Re: Abs[a] Sin[Abs[a]]**

**Re: Re: Abs[a] Sin[Abs[a]]**