Re: Evaluate, /., {{...}}, etc.

*To*: mathgroup at smc.vnet.net*Subject*: [mg70740] Re: Evaluate, /., {{...}}, etc.*From*: dh <dh at metrohm.ch>*Date*: Wed, 25 Oct 2006 01:40:07 -0400 (EDT)*References*: <ehkc9c$hqr$1@smc.vnet.net>

Hi Misha, I think you are fooling yourself. But first how braces and Replace work: a+b/.a->a1 is the same as a+b/.{a->a1} the braces make more sense if we have more than one rule, e.g.: a+b/.{a->a1,b->b1} a+b/.{{a->a1,b->b1},{a->a2,b->b2}} gives a list of 2 results, one for {a->a1,b->b1} and one for {a->a2,b->b2} r1[x1_]:=Evaluate[x1/.{{x1 -> 5 - x2/2}}] Because of the double barces, this defines a function that returns a list with 1 element, probably not what you want. r2[x1_]:=Evaluate[x1/.{x1 -> 10 - 2*x2}] defines exactly what you want. You must have made a mistake. Daniel misha wrote: > I'm having trouble figuring out how either Evaluate works or /. works > (with the logical understanding of "or"), as well as the asymmetric > appearance of curly brackets {} in what appears to me to be analogous > conditions. > > In the example below, I expect r1[x1_]:=Evaluate[x1/.solution1] to > produce a function r1 as a function of x1 from the equation provided by > solution1. Similar for r2. I notice that solution1 is {{x1 -> 5 - > x2/2}}, whereas solution2 is {x1 -> 10 - 2*x2}. Why "->" instead of "=" > or "=="? Why two curly brackets in the first case {{...}} and only one > in the second case {...}? > > More important, > > r1 produces what I expect, > In[...]:= ?r1 > Global`r1 > r1[x2_]:={5 - x2/2} > > while r2 does not. > In[...]:=?r2 > Global'r2 > r2[x1_]:=x2 > > I would expect the following > > r2[x1_]:=10-2*x2 > > (i.e., since this problem is symmetric, firm 1's "reaction function" > (r1) should be symmetric w.r.t. firm 2's "reaction function" (r2). > Again, why curly brackets in one case, but not in the other? > > Here is the full example (from > http://library.wolfram.com/infocenter/MathSource/551/): > > p1[x1_,x2_] := 10 - b1 x1 - c x2 > p2[x1_,x2_] := 10 - b2 x2 - c x1 > profit1[x1_,x2_] := Evaluate[p1[x1,x2]*x1] > profit2[x1_,x2_] := Evaluate[p2[x1,x2]*x2] > solution1=Solve[D[profit1[x1,x2],x1]==0,x1][[1]] > solution2=Solve[D[profit2[x1,x2],x2]==0,x2][[1]] > r1[x2_] := Evaluate[x1/.solution1] > r2[x1_] := Evaluate[x2/.solution2] > c=b1 > b1=b2=c=1 > isoprofitLines=ContourPlot[profit1[x1,x2],{x1,0,7},{x2,0,7},ContourShading->False] > reactionCurves=ParametricPlot[{{r1[t],t},{t,r2[t]}},{t,0,7}] > Show[isoprofitLines,reactionCurves] >