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Re: Re: New user: Abs[] problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50157] Re: [mg50128] Re: New user: Abs[] problem
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Wed, 18 Aug 2004 01:20:06 -0400 (EDT)
*References*: <cfka7r$f7u$1@smc.vnet.net> <200408161045.GAA27783@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Janusz Kawczak wrote:
> Because you have infinite (uncountable) number of solutions. One way of
> approaching this kind of problems is to use Reduce in the following way:
>
> Reduce[x + 1 - Abs[2x - 4] + Abs[5 - x] == 0 && x \[Element] Reals, x]
>
> Of course, you can play with the domain of x. If left unspecified, it
> takes
> values from the complex domain.
>
> Also, to see some of the answers you may like to play with FindInstance,
> FindRoot and so on.
>
> Best, Janusz.
>
> Edson, Brusque wrote:
>
>
>>Hello,
>>
>> I'm starting Electrical Engineering and using Mathematica for
>>helping me in Calculus e Algebra studies.
>>
>> I'm trying to solve this equation on Mathematica:
>> x + 1 - |2x - 4| + |5 - x| = 0
>>
>> on the Math* notebook I'm typing:
>> Solve[x + 1 - Abs[2x - 4] + Abs[5 - x] == 0,x]
>>
>> but only got an empty output: {{}}
>>
>> Someone please can help me get Math* to solve this?
>>
>> Thank you very much,
>>
>> Edson Brusque
>>
>>---------------------------------------------------------------------
>>Edson Brusque C.I.Tronics Lighting Designers Ltda
>>Research and Development Blumenau - SC - Brazil
>>http://www.suporte.ind.br/ryan/netiqueta.htm www.citronics.com.br
>>---------------------------------------------------------------------
Let me add to this response. For nonpolynomial input Solve may make
transformations in an attempt to find some solutions. This can give rise
to parasite solutions which Solve will later attempt to remove.
The problem in this case is that the transformations may give rise to a
result with infinitely many solutions. Generally when this happens the
"polynomialized" version of the equation looks something like x==x
(quick example: Solve[x==Sqrt[x^2], x]).
The result is that any value of the variable suffices, so we get a
"full" solutions set {{}} (note that an empty solution set is denoted
{}). Unfortunately, at this point there is no way to weed out excess
solutions.
For such problems, as noted above, Reduce is likely to give a more
informative ("nuanced"?) result.
Daniel Lichtblau
Wolfram Research
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