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Re: FindMinimum Problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg44675] Re: FindMinimum Problem
*From*: drbob at bigfoot.com (Bobby R. Treat)
*Date*: Thu, 20 Nov 2003 03:16:40 -0500 (EST)
*References*: <bpa1rv$19t$1@smc.vnet.net> <bpd1bi$c9c$1@smc.vnet.net> <bpffpl$lv6$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
This works for me using version 5.0:
foo[x_?NumericQ] := x^2 - a[0]*x /. First@
NDSolve[{a'[z] == x*z, a[1] == 1}, a, {z, -1, 1}]
FindMinimum[foo[x], {x, -1, 1}]
{-0.166667, {x -> 0.333333}}
Jens-Peer had left out "First@", so foo took on List values like {2.5}
rather than just the number 2.5. It's also important to make the range
{z,-1,1} encompass the range where a minimum might be found --
especially the two initial starting points, which were 1 and -1 in his
post.
The pattern x_?NumericQ matches x if it is numeric -- as it will be
when FindMinimum tries to iterate to a solution. If you don't use that
pattern, you get an error message when FindMinimum tries a symbolic x
value (which it does, for some reason I can't fathom).
Bobby
Jiang Xiao <jiang.xiao at physics.gatech.edu> wrote in message news:<bpffpl$lv6$1 at smc.vnet.net>...
> thank you for replying, I tried it, but it still doesn't work.
> what's the meaning of x_?NumericQ?
> if I let foo[x_?NumericQ]:=x^2-x, findminimum can't get output either,
> but no problem with foo[x_]:=x^2-x.
>
> Jiang
>
> Jens-Peer Kuska wrote:
> > Hi,
> >
> > you ust make a function
> >
> > foo[x_?NumericQ]:=x^2-a[0]*x/.NDSolve[{a'[z]==x*z,a[1]==1},a,{z,0,1}]
> >
> > FindMinimum[foo[x],{x,-1,1}]
> >
> > Regards
> > Jens
> >
> > Jiang Xiao wrote:
> >
> >>Hi, all,
> >> recently I am dealing with a problem as following, findminimum(over x)
> >>of a function f[a[0],x], where a[z] satisfies a differential equation
> >>a'[z]=x*z say. The code is like:
> >>FindMinimum[x^2-a[0]*x/.NDSolve[{a'[z]==x*z,a[1]==1},a,{z,0,1}],{x,-1,1}]
> >>
> >>the problem is that I can do it in mathematica 4.2, but can't in mathematica
> >>5.0 now. Do anybody where is the problem?
> >>
> >>thanks,
> >>
> >>Jiang
> >
> >
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