great and typical tommy1729 idea :)

mike wrote :

On Mar 15, 8:54 pm, Mariano Suárez-Alvarez
<mariano.suarezalva...@xxxxxxxxx> wrote:
On Mar 15, 10:19 pm, mike3 <mike4...@xxxxxxxxx>
wrote:



On Mar 7, 5:35 am, "alainvergh...@xxxxxxxxx"
<alainvergh...@xxxxxxxxx>
wrote:

On 7 mar, 11:15, mike3 <mike4...@xxxxxxxxx>
wrote:
<snip>
Dear Mike,

I always wished getting out of these heavy
representations,

Regards,
Alain

How about if one used a simpler Riemann surface,
namely that of the
inverse
of z^2 + c, i.e. that of sqrt(z - c)? This is
simpler as it has only 2
leaves.
Note that sqrt(z - c) sends points on the Riemann
surface to points on
the
z-plane, and the map z^2 + c could then be
thought of as sending
points on
the z-plane to points on the Riemann surface. The
rub here would then
be:

1. finding a simple, easy representation for the
Riemann surface's
points

Let c be in C. Then the Riemann surface of f(z) =
sqrt(c - z)
is simply the set

  S = { (z, w) in C x C : w^2 + c = z }

The "complete function" corresponding to f (that
is,
the "extension" of f to its Riemann surface) is the
function F : S --> C such that F(z, w) = w.

The covering p : S --> C is given by p(z, w) = z,
and it is (outside of the ramification point 0)
a two-sheeted covering.

good math of mariano.

but not so helpfull ?




But this doesn't solve the problem... how to
represent the "dynamics"
of z^2 + c in such a way that the inverse map is
single valued as
well.
This involves iteration (repeated application) of
both z^2 + c and sqrt
(z - c).

indeed mike.


very recently however , i , the great tommy1729 :) , gave a new twist to iterations.

i considered extending the domain !!


w^2 + c = f(f(w))

f(w) maps complex w to an EXTENSION of the reals , different from C !

if also 2*w = f'(f(w)) * f'(w)

is satisfied , this might be what you look for !!


regards

the great :)

tommy1729
.

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