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
>>>
>