Re: Inverse Function
- To: mathgroup at smc.vnet.net
- Subject: [mg123602] Re: Inverse Function
- From: Roger Bagula <roger.bagula at gmail.com>
- Date: Tue, 13 Dec 2011 05:38:53 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jc4plc$ckp$1@smc.vnet.net>
The M=F6bius bilinear transform has several kinds of inverse:
Inverse[{{a,b},{c,d}}
gives the inverse of the matrix usually taken as the Poincare disk
like:
a*d-b*c<=1
But if you take:
f[x_]=(a*x+b)/(c*x+d)
and
use
f[1/x]=(a*(1/x)+b)/(c*(1/x)+d)
you get:( Toral inverse)
g[x_]=(a+b*x)/(c+d*x)
The modular form based on the function f[x]:
f[(a*x+b)/(c*x+d)]=(c*x+d)^(2*n)*f[x]
leads to yet another much harder approach to the inverse.
I think you may want:
Solve[y==(a*x+b)/(c*x+d),x]
On Dec 12, 3:47 am, Harry Har <harryhar... at gmail.com> wrote:
> Hi, All,
>
> I'm newbie in Mathematica. I want to find the inverse function of y=(ax
> +b)/(cx+d). How to to this in Mathematica? Many thank's.
>
> Harry.