Re: Finding inverse of non-linear transformation

*To*: mathgroup at smc.vnet.net*Subject*: [mg116413] Re: Finding inverse of non-linear transformation*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Mon, 14 Feb 2011 04:26:36 -0500 (EST)*References*: <ij8d29$396$1@smc.vnet.net>

Chris, A few remarks: 1. Your definition of the function F is a bit weird. Not using Blanks for a, b, and c, means the function will only be defined if you use precisely these variable names. Values as arguments can't be used. 2. InverseFunction is *not* used to *calculate* the inverse function of a given function, but is used by Mathematica to *represent* inverse functions that emerge in its calculations. Read the InverseFunction doc page for more details. 3. If you have a function f[x_] = ... then the inverse function can (sometimes and not always correctly because of non-unique inverses) be found by Solve[f[x]==fx,{x}] with fx a known result of f[x]. If you have a function of more variables f[x_,y_] = ..., then you have to solve two equations: Solve[{f[x,y]==fxy1, f[x,y]==fxy2},{x,y}] and hope for the best. Quite often you won't be able to find a solution. 4. You have to modify 3. to accommodate the 2x1 matrix you're using in the function definition, as the function now has two outputs instead of one. Cheers -- Sjoerd On Feb 13, 11:50 am, "Christopher O. Young" <c... at comcast.net> wrote: > I'm trying to find the inverse of the simple non-linear transformation > > (a x + b x + c) ({ > {x}, > {y}}) > > I'm trying to use this as a sort of counter-example to the usual projective > transformation, which has the polynomial in the denominator. Does > Mathematica have a way to do this? InverseFunction doesn't seem to work > here. > > In[1]:= F[x_, y_, a, b, c] := (a x + b x + c) ( { {x}, {y} }= ) > > In[2]:= InverseFunction[[x_, y_, a, b, c]] > > During evaluation of In[2]:= Part::pspec: Part specification x_ is neither > an integer nor a list of integers. >> > > Out[2]= InverseFunction[[x_, y_, a, b, c]] > > Any help very much appreciated. > > Chris Young > c... at comcast.net