Re: piecewise functions from logical relationships (ie. solving with constraints)

*To*: mathgroup at smc.vnet.net*Subject*: [mg81250] Re: piecewise functions from logical relationships (ie. solving with constraints)*From*: Chris Chiasson <chris.chiasson at gmail.com>*Date*: Tue, 18 Sep 2007 00:28:29 -0400 (EDT)*References*: <fclb12$fdk$1@smc.vnet.net>

On Sep 17, 2:40 am, "Ben Lewis" <benjamin.r.le... at gmail.com> wrote: > G'day all, > > Say I have an equation that defines some arbitrary relationship between > two variables x and y (or N equations and 2N variables), and furthermore > that I know a specific region for which the relationship is one-to-one > and onto. In other words, some function f (such that y=f[x]) and it's > inverse g (x=g[y]) are mathematically well-defined in that region by my > equation. Is there then any way to obtain both functions using Mathematica? > > I want the specific functions (f and g) so that, for particular points > in the region, I can change (coordinate) variables in either direction, > and also to compute various derivatives of these functions (particularly > the Jacobian matrix). > > For example: > > eq := y+1/2==Mod[x+0.1,1] > constr := -1/2<y<1/2 && 0<x<1 && y!=-0.4 && x!=0.9 > > Reduce produced output in a different form: > (x==2/5+y && -2/5<y<1/2) || (x==7/5+y && -1/2<y<-2/5) > > Something in a form like "g->Mod[#+2/5,1]&" would be ideal, or > similarly: "x->Piecewise[{{2/5+y,-2/5<y<1/2},{7/5+y,-1/2<y<-2/5}}]". > > -Ben I dunno how you made Reduce give you that answer (I didn't try very long), but, since you have it: In[1]:= (x==2/5+y&&-2/5<y<1/2)||(x==7/5+y&&-1/2<y<-2/5)/.{Or- >List,And[Equal[l_,r_],cond_]:>(l/;cond->r)} Out[1]= {x/;-(2/5)<y<1/2->2/5+y,x/;-(1/2)<y<-(2/5)->7/5+y}