       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:= F[x_, y_, a, b, c] := (a x + b x + c) ( {  {x},  {y}  } )
>
> In:= InverseFunction[[x_, y_, a, b, c]]
>
>>
> During evaluation of In:= Part::pspec: Part specification x_ is neither
> an integer nor a list of integers. >>
>
>
>
> Out= 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))}}

```

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