Quote:
Originally Posted by Qfwfq
Plenty!
For every complex value there is a Julia set. I don't know if the map is injective but cardinality-wise my guess would be that it's  anyway. |
It is at least as you say injective, though I don't believe it is bijective.
Any one Julia Set so formed is likely to be in more than one algebraic map.
So 1-1 correspondence I think is out.
Quote:
Originally Posted by Qfwfq
The actual point though was that not all Julia sets are the quadratic ones, whereas the Mandelbrot set concerns only these.
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I am not sure of the exact definition of the Mandlebrot set being of only
quadratic equations. I had plotted many such maps base on nothing more
than the assumption that any Analytic Function over a piecewise defined
domain can form a Mandlebrot set.
For example
or
, etc.
maddog
