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Re: FindMinimum[Print[]]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86976] Re: FindMinimum[Print[]]
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Fri, 28 Mar 2008 03:13:28 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <fsg6o2$j61$1@smc.vnet.net>

Michaël Cadilhac wrote:

>   I'm really new to Mathematica (though I can already say wow), and,
> following the tutorial[1] (which might be quite outdated), one of the
> exercise got me in trouble.
> 
>   The author asks to reformulate the following actions
> 
> m = {{12,      1 + x,  4 - x,  x},
>      {4 - x,   11,     1 + x,  x},
>      {1 + x,   1 - x,  15,     x},
>      {x - 1,   x - 1,  x - 1,  x - 1}};
> expr = Max[Re[Eigenvalues[m]]];
> FindMinimum[expr, {x, 0, 1}]
> 
>   into a more optimized version.  In the course of doing that, I wanted
> to do something like
>     FindMinimum[Print[x]; x^2, {x, 1}],
> hoping to see how is this whole thing is expanded/parsed.  But, despite
> the fact that some articles on that newsgroup used the same form, this
> didn't print the iterations as expected.
> 
>   I wanted to understand how I should write
>     FindMinimum[Max[Eigenvalues[m]], {x, 0, 1}]
> so that the eigenvalues are computed on the fully numerical
> (non-symbolic) matrix.

The option StepMonitor is what you are looking for.

In[1]:= FindMinimum[Exp[x] + 1/x, {x, 1},
  StepMonitor :> Print["Step to x = ", x]]

During evaluation of In[1]:= Step to x = 0.788886

During evaluation of In[1]:= Step to x = 0.677316

During evaluation of In[1]:= Step to x = 0.706578

During evaluation of In[1]:= Step to x = 0.703587

During evaluation of In[1]:= Step to x = 0.703467

During evaluation of In[1]:= Step to x = 0.703467

Out[1]= {3.44228, {x -> 0.703467}}

In[2]:= m = {{12, 1 + x, 4 - x, x}, {4 - x, 11, 1 + x, x}, {1 + x,
     1 - x, 15, x}, {x - 1, x - 1, x - 1, x - 1}};
expr = Max[Re[Eigenvalues[m]]];
FindMinimum[expr, {x, 0, 1}, StepMonitor :> Print["Step to x = ", x]]

During evaluation of In[2]:= Step to x = 0.362425

During evaluation of In[2]:= Step to x = 0.362425

During evaluation of In[2]:= Step to x = 0.362159

During evaluation of In[2]:= Step to x = 0.362151

Out[4]= {16.8343, {x -> 0.362151}}


HTH,
-- 
Jean-Marc


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