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