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Re: Re: Re: Problem with parametric minimization

• To: mathgroup at smc.vnet.net
• Subject: [mg90322] Re: [mg90279] Re: [mg90243] Re: [mg90186] Problem with parametric minimization
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sat, 5 Jul 2008 04:53:02 -0400 (EDT)
• References: <15947055.1215088335131.JavaMail.root@m08> <200807040758.DAA22279@smc.vnet.net>

Hmm... we seem to be using different logic because all this makes  
"perfect sense to me".

sol=Minimize[b*x + c, x]
{Piecewise[{{c, b == 0}}, -Infinity],
    {x -> Piecewise[{{0, b == 0}}, Indeterminate]}}

The answer consists of two parts (not two solutions).

In[9]:= sol[[1]]
Out[9]= Piecewise[{{c, b == 0}}, -Infinity]

gives the actual minima as a Piecewise function, which are c if b ==0  
and -Infinity if b!=0. So far so good.

In[10]:= sol[[2]]
Out[10]= {x -> Piecewise[{{0, b == 0}}, Indeterminate]}

gives the values of x, in terms of a rule involving a Piecwise  
function. This says that x is 0 when b =0  and there is no definite x  
when b != 0 (that's why x is Indeterminate). Note that a specific  
value x = 0 is given as a minimum point, although any other value  
could be given equally well. But this is completely in accordance with  
the way Minimize works in general, since it never attempts to find all  
the minima. It is also sensible to specify Indeterminate as a value of  
x for which the function is minimum when b!=0.

The Red Queen shook her head, `You may call it "nonsense" if you  
like,' she said, ` but I've heard nonsense, compared with which that  
would be as sensible as a dictionary!'

Lewis Carroll, "Through the Looking Glass".


Andrzej Kozlowski




On 4 Jul 2008, at 16:58, DrMajorBob wrote:

> Good idea, but the result is two solutions, the second of which is
> incorrect:
>
> Minimize[a*x^2 + b*x + c, x]
>
> {\[Piecewise]c	(b==0&&a==0)||(b==0&&a>0)
> ((-b^2+4 a c)/(4 a))	(b>0&&a>0)||(b<0&&a>0)
> -\[Infinity]	True
>
> ,{x->\[Piecewise]0	(b==0&&a==0)||(b==0&&a>0)
> -(b/(2 a))	(b>0&&a>0)||(b<0&&a>0)
> Indeterminate	True
>
> }}
>
> Here's Minimize failing in similar (spectacular) fashion for a LINEAR
> function:
>
> quadratic=a*x^2+b*x+c;
> linear=quadratic/.a->0
> Minimize[linear,x]
>
> c+b x
>
> {\[Piecewise]c	b==0
> -\[Infinity]	True
>
> ,{x->\[Piecewise]0	b==0
> Indeterminate	True
>
> }}
>
> The first solution is correct; the second is nonsense.
>
> (This is 64-bit 6.0.3 on the Mac.)
>
> Bobby
>
> On Thu, 03 Jul 2008 05:11:28 -0500, Bob Hanlon <hanlonr at cox.net>  
> wrote:
>
>> Use multiply or space to have your expression represent what you
>> intended.
>>
>> Minimize[a*x^2 + b*x + c, x]
>>
>>
>> Bob Hanlon
>>
>> ---- Stephan Heckmueller <stephan.heckmueller at gmx.net> wrote:
>>
>> =============
>> Hello,
>> I have a problem with with the examples for using the
>> Minimize function given in the documentation on wolfram.com:
>>
>> When trying:
>> Minimize[ax^2+bx+c,x]
>>
>> I receive the following error:
>>
>> Minimize::objv:
>> The objective function c+bx+ax^2 contains a nonconstant
>> expression c independent of variables (x).
>>
>> Is there something wrong with the command or the
>> documentation?
>>
>> Thanks.
>>
>>
>>
>>
>>
>
>
>
> -- 
> DrMajorBob at longhorns.com
>

• References:
  • Re: Re: Problem with parametric minimization
    • From: DrMajorBob <drmajorbob@att.net>