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

*To*: mathgroup at smc.vnet.net*Subject*: [mg21041] Re: [mg21021] Re: [mg20923] Abs[a] Sin[Abs[a]]*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Sun, 12 Dec 1999 23:51:09 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

When I re-read my own messsage I saw with horror that I had written some really pretty awful nonsense. My excuse is that I must have been very tired (I usually write these messages after midnight). Anyway, when I wrote Simplify[Abs[Sin[a]], a < 0] I actually meant something like Simplify[Abs[Sinh[a]], a < 0] or in general Simplify[Abs[f[a]], a < 0] where f is some monotonically increasing non-polynomial odd function (which of course Sin is not, although in a temporary fit of madness I thought it was). But in fact actally this works exactly as it should provided one uses a suitable complexity function as pointed out in your reply: In[17]:= f[x_] := LeafCount[x] + 10 Count[x, _Abs, {0, Infinity}] In[19]:= Simplify[Abs[Sinh[a]], a < 0, ComplexityFunction -> f] Out[19]= -Sinh[a] To me this seems quite impressive: I had been assuming Simplify would have trouble with non polynomial functions in such situations. Andrzej Kozlowski -- > From: Adam Strzebonski <adams at wolfram.com> > Organization: Wolfram Reasearch, Inc > Date: Thu, 2 Dec 1999 21:41:32 -0500 (EST) > To: mathgroup at smc.vnet.net > Subject: [mg21041] [mg21021] Re: [mg20923] Abs[a] Sin[Abs[a]] > > 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 > 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: [mg21041] [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: Re: What kind of math problem is this?**

**Re: Q: Dickman function**

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

**PlotField3D and arbitrary coordinate systems**