Re: -- g o g = f

On 2 fév, 01:19, quasi <qu...@xxxxxxxx> wrote:
On Mon, 02 Feb 2009 10:54:11 +1100, Gerry Myerson

<ge...@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
In article
<bd81c898-d9c4-4ed7-a71a-376fe62f4...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"alainvergh...@xxxxxxxxx" <alainvergh...@xxxxxxxxx> wrote:

Let us choose g(t)= 2t +1 , k(t)=-2t-3 ,
roots of (4x +3) ,
Using parametrized functions
we define h  | h(f(t)) = f(g(t))
          m  | m(f(t)) = f(k(t))
We obtain: h^(2](f(t)) = f(g^2(t)) = f(k^2(t)) = m^(2](f(t))
or h o h = m o m

Maybe so, but I wouldn't recommend telling someone
that his m o m is a h o h.

Hahaha.

quasi

Bonjour,

yes, strictly speaking Gerry is right.
I am sure we can define conditions (associativity,
intervals...) and make it mathematically correct,

Alain
.

Relevant Pages

  • Re: -- g o g = f
    ... roots of, ... Using parametrized functions ... but I wouldn't recommend telling someone ... Gerry Myerson ...
    (sci.math)
  • Re: -- g o g = f
    ... roots of, ... Using parametrized functions ... but I wouldn't recommend telling someone ...
    (sci.math)