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]
>