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Re: Symbolic use of numerical function FindRoot via ?NumericQ
*To*: mathgroup at smc.vnet.net
*Subject*: [mg48611] Re: Symbolic use of numerical function FindRoot via ?NumericQ
*From*: "Peter Pein" <petsie at arcor.de>
*Date*: Mon, 7 Jun 2004 05:33:51 -0400 (EDT)
*References*: <c9pddv$nuf$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
"Michael Beqq" <mbekkali at iastate.edu> schrieb im Newsbeitrag
news:c9pddv$nuf$1 at smc.vnet.net...
> I've use symbolical evalution of numerical functions all the time but
still
> have not mastered it. Now I got stuck on the following problem:
>
> I have a function g[x1,x2,y1,y2]. I need to solve for x1,x2 and y1,y2
that
> maximize g[x1,x2,y1,y2] in 2 steps,- in step 1 I need to find
x1*=x1[y1,y2]
> and x2*=x2[y1,y2]. Then I substitute solutions, *'s, into g[.] to get
> g[y1,y2] and then solve for solve for y1* and y2*. This is a classical
> 2-stage problem in Economics.
>
> Is there a way to do that in Mathematica 5 using FindRoot command. I
tried
> using SetDelayed and ?NumericQ options however get error messages that the
> function g[.]'s is not a list of numbers with dimension {2} at {2 values}.
>
> Here is more precise code:
>
> g[x1_x2_,y1_,y2_]=g[x1,x2,y1,y2]"g is some function of 4 variables";
>
{x1[y1_,y2_?Numeric],x2[y1_,y2_?Numeric]}:=Evaluate[{x1,x2}/.FindRoot[Evalua
> te[{D[g[x1,x2,y1,y2],x1]]==0,D[g[x1,x2,y1,y2],x2]]==0},{x1,x10},{x2,x20}]]
> (* another website showed different code, i.e. x[(y1_,y2_)?Number] but I
> assume it is just semantics*)
>
FindRoot[Evaluate[{D[g[y1,y2],y1]]==0,D[g[y1,y2],y2]]==0},{y1,y10},{y2,y20}]
> ]
>
> where {x10,x20,y10,y20}=some numbers.
>
> Thank you in advance.
>
>
Hi Michael,
calling a function of 4 variables with 2 variables will not work. I've got
difficulties to understand, why you do not solve all the 4 derivatives in
one step? And since x1* and x2* depend on y1 and y2 you have to use the
chain rule, when building the derivatives w.r.t. y1 and y2, or did I get
sth. wrong?
--
Peter Pein, Berlin
to write to me, start the subject with [
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