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Re: Re: Re: solve and Abs

  • To: mathgroup at
  • Subject: [mg69517] Re: [mg69456] Re: [mg69398] Re: solve and Abs
  • From: Daniel Lichtblau <danl at>
  • Date: Thu, 14 Sep 2006 06:56:45 -0400 (EDT)
  • References: <>

Andrzej Kozlowski wrote:
> On 12 Sep 2006, at 20:40, Peter Pein wrote:
>>Andrzej Kozlowski schrieb:
>>>On 10 Sep 2006, at 20:20, Peter Pein wrote:
>>>>Richard J. Fateman schrieb:
>>>>>Is there a reason (in 5.1) for Solve[x-Abs[x]==0,x] to return {{}}?
>>>>>This supposedly means all variables can have all possible values.
>>>>>Reduce does better.
>>>>Oops, I should have read further...
>>>>Help says
>>>>"Solve gives {{}} if all variables can have all possible values."
>>>>I would consider this a bug, not a feature.
>>>>Sorry for overhasty posting,
>>>However, the documentation does not say "Solve gives {{}} if and  
>>>only if all variables can have all possible values." There are  
>>>other cases when Solve returns {{}}. For example, Solve also gives  
>>>{{}} when, for example, there is no "generic" solution. (This is  
>>>stated in the Help also). In addition, although this does not seem  
>>>to be stated, Solve never returns solutions that would have to be  
>>>given in conditional form (like the solution returned in this  
>>>example by Reduce). This does not appear to be stated in the help  
>>>but has been pointed out several times on this list.
>>>Andrzej Kozlowski
>>>Tokyo, Japan
>>all variables can have all possible values =:AVAPV
>>solve rteurns {{}} =:SRE
>>But if AVAPV=>SRE but not SRE=>AVAPV, what use has SRE? It would be  
>>better to admit "no solutions found" (by returning {}).
>>Or am I completely wrong?
>>Confused greetings,
>> Peter
> Oops, I think you are right. In fact, when Solve can find no generic  
> solutions it returns {} and not {{}}.
> Solve[{x + y == 1, x + y == a}, {x, y}]
> {}
> So it does seem that {} has a different meaning from {{}}, which  
> makes me now also confused ;-)
> Andrzej

An empty result, {}, indicates that either there are no generic 
solutions (as in the case above, because all solutions force a parameter 
  constraint a==1), or else that Solve found no solutions (can happen 
e.g. with transcendentals that get "algebraicized" but then fail to 
yield viable solutions using Solve's creaky technology).

An AVAPV result, in the terminology above, is a different fish. It means 
any value for the solve variable(s) will satisfy the "polynomialized" 
problem. So what does that mean? In the simple case that began this 
thread, Abs[x] is converted to Sqrt[x^2] (which many will realize is 
already a mistake off the real line. C'est la vie.) We make a new 
variable y which "encapsulates" Sqrt[x^2], and new equation 
y^2==(Sqrt[x^2])^2==x^2. So we are doing, in effect,

Solve[{x==y,y^2==x^2}, x, y]

(that is, eliminating y).

This correctly returns {{}}. Ordinarily when radicals are present Solve 
understands that parasite solutions may be found, and makes an attempt 
to check them. But in this case what can it check, when every possible 
value is a viable solution? So {{}} is the returned result.

Generally speaking, {{}} will be returned if there is some continuum on 
which all values are viable results. This is because a continuum of 
solutions implies (e.g. by the identity theorem of complex analysis) 
that, after polynomialization, all values are solutions. Even worse, 
there can be cases where all solutions will be parasites but 
algebraicization caused a continuum of apparent solutions to appear, 
etc. etc.

Daniel Lichtblau
Wolfram Research

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