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Re: calling maximize recursively

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67644] Re: calling maximize recursively
  • From: dh <dh at metrohm.ch>
  • Date: Mon, 3 Jul 2006 06:38:03 -0400 (EDT)
  • References: <e82nm1$r2i$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Ho Garrio,
1) you must work with an numerical routine, as you can analytically 
compute the derivatives of function2. E.g.: 
NMaximize[function2[bestbound], {bestbound}]
2) your problem does not have a unique Maximum, every number >=0.5 
fullfills you conditions.
3) ensure that your function is only called with numerical arguments, 
e.g.: function2[bound_ /; NumberQ[bound]] := ...
4) If you know an approximate value of bestbound, you can help 
Mathematica by e.g.specifying this value: 
FindMaximum[function2[bestbound], {bestbound, 0.3}]

Daniel

garriott at gmail.com wrote:
> Hello all.
> 
> I am trying to call an instance of Maximize[] within Maximize[].
> 
> Essentially I'm trying to perform a traditional Bellman's equation
> optimization, where you maximize value[now] + value[later], and
> value[later] is itself another maximization, and so on.
> 
> Here is a very simplified example of a recursive Maximize[] call. If
> you can figure out a way to get this to work without deleting one of
> the two Mazimize[]s then you'll save the day for me! (I am aware this
> particular problem could be done far easier, but it's the setup here
> that matters, not the particular problem.)
> 
> 
> 
> function1[x_] := x - x^2;
> 
> function2[bound_] := Maximize[ {function1[x], x<bound} , x ];
> 
> Maximize [ function2[bestbound] , bestbound ]
> 
> 
> 
> (if you import this in to mathematica it gives an error, because it
> tries to pass bestbound as a pure variable to function2, rather than a
> candidate value of bestbound to function2)
> 
> Please help,
> Garrio
> 


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