- Q-rational
preperiodic points, i.e., F has only
finitely many
preperiodic points in P1(Q). The
uniform boundedness conjecture for
preperiodic points of...
-
invariant 1, and
there are no
other cycles.
Because all
numbers are
preperiodic points for F 2 , b {\displaystyle F_{2,b}} , all
numbers lead to 1 and...
-
topic of:
Fractals Preperiodic (Misiurewicz)
points in the
Mandelbrot set by
Evgeny Demidov M & J-sets
similarity for
preperiodic points. Lei's theorem...
- {\displaystyle f_{n}(x)=f_{m}(x)} then x is
called a
preperiodic point. All
periodic points are
preperiodic. If f is a
diffeomorphism of a
differentiable manifold...
- 2 {\displaystyle p=2} . All
natural numbers n {\displaystyle n} are
preperiodic points for F b {\displaystyle F_{b}} ,
regardless of the base. This is...
- {\displaystyle F_{b}(n)=n} . All
natural numbers n {\displaystyle n} are
preperiodic points for F b {\displaystyle F_{b}} ,
regardless of the base. This is...
- 2. {\displaystyle p=2.} All
natural numbers n {\displaystyle n} are
preperiodic points for F b {\displaystyle F_{b}} ,
regardless of the base. This is...
- (f)} of f ( z ) {\displaystyle f(z)} . If all the
critical points are
preperiodic, that is they are not
periodic but
eventually land on a
periodic cycle...
-
boundedness conjecture for
rational points Uniform boundedness conjecture for
preperiodic points Search for "Uniform
boundedness conjecture" on Wikipedia. Uniform...
- 2 {\displaystyle k=2} . All
natural numbers n {\displaystyle n} are
preperiodic points for SFD b {\displaystyle \operatorname {SFD} _{b}} , regardless...
