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Re: Re: Re: Re: Problem with
*To*: mathgroup at smc.vnet.net
*Subject*: [mg90328] Re: [mg90322] Re: [mg90279] Re: [mg90243] Re: [mg90186] Problem with
*From*: DrMajorBob <drmajorbob at att.net>
*Date*: Sun, 6 Jul 2008 07:18:08 -0400 (EDT)
*References*: <15947055.1215088335131.JavaMail.root@m08>
*Reply-to*: drmajorbob at longhorns.com
Yes, I know.
Bobby
On Sat, 05 Jul 2008 03:53:02 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl>
wrote:
> 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
>>
>
>
>
--
DrMajorBob at longhorns.com
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