Re: Finding inverse of non-linear transformation

*To*: mathgroup at smc.vnet.net*Subject*: [mg116420] Re: Finding inverse of non-linear transformation*From*: "Christopher O. Young" <cy56 at comcast.net>*Date*: Mon, 14 Feb 2011 04:27:51 -0500 (EST)*References*: <ij8d29$396$1@smc.vnet.net>

On 2/13/11 5:50 AM, in article ij8d29$396$1 at smc.vnet.net, "Christopher O. Young" <cy56 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]] > I didn't have underscores for the a, b, and c the way I should have, but that wasn't the main problem. InverseFunction doesn't seem designed to find the inverse of vector transformations. But I think using "Solve" with added variables on the other sides of the equations gets me the result I'm looking for: Solve[{ (a x + b x + c) x == X, (a x + b x + c) y == Y }, {x, y} ] {{y -> (-((c Y)/(a + b)) - (Sqrt[c^2 + 4 a X + 4 b X] Y)/(a + b))/( 2 X), x -> (-c - Sqrt[c^2 + 4 a X + 4 b X])/( 2 (a + b))}, {y -> (-((c Y)/(a + b)) + ( Sqrt[c^2 + 4 a X + 4 b X] Y)/(a + b))/(2 X), x -> (-c + Sqrt[c^2 + 4 a X + 4 b X])/(2 (a + b))}}