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 [