This article describes periodic points of some complex quadratic maps. A map is a formula for computing a value of a variable based on its own previous value or values; a quadratic map is one that involves the previous value raised to the powers one and two; and a complex map is one in which the variable and the parameters are complex numbers. A periodic point of a map is a value of the variable that occurs repeatedly after intervals of a fixed length.
These periodic points play a role in the theories of Fatou and Julia sets.
Let
fc(z)=z2+c
be the complex quadric mapping, where
z
c
Notationally,
f(k)c(z)
k
fc
k
fc
fc.
f(k)c(z)=
(k-1) | |
f | |
c(f |
c(z)).
p
z
f(p)c(z)=z,
where
p
We can introduce a new function:
Fp(z,f)=f(p)c(z)-z,
so periodic points are zeros of function
Fp(z,f)
Fp(z,f)=0,
2p.
The degree of the polynomial
Fp(z,f)
d=2p
d=2p
The multiplier (or eigenvalue, derivative)
p,z | |
m(f | |
0)=λ |
f
p
z0
p,z | |
m(f | |
0) |
=λ=\begin{cases}fp(z0),&ifz0\neinfty\\
1 | |
fp(z0) |
,&ifz0=infty\end{cases}
where
fp\prime(z0)
fp
z
z0
Because the multiplier is the same at all periodic points on a given orbit, it is called a multiplier of the periodic orbit.
The multiplier is:
abs(λ).
A periodic point is[2]
abs(λ)<1;
abs(λ)=0;
0<abs(λ)<1;
abs(λ)=1;
λ
abs(λ)=1
abs(λ)>1.
Periodic points
Let us begin by finding all finite points left unchanged by one application of
f
fc(z)=z
z2+c=z,
which can be rewritten as
z2-z+c=0.
Since this is an ordinary quadratic equation in one unknown, we can apply the standard quadratic solution formula:
\alpha1=
1-\sqrt{1-4c | |
\alpha2=
1+\sqrt{1-4c | |
c\inC\setminus\{1/4\}
\alpha1
\alpha2
Since
\alpha1=
1 | |
2 |
-m
\alpha2=
1 | |
2 |
+m
m=
\sqrt{1-4c | |
we have
\alpha1+\alpha2=1
Thus fixed points are symmetrical about
z=1/2
Here different notation is commonly used:[4]
\alphac=
1-\sqrt{1-4c | |
λ | |
\alphac |
=1-\sqrt{1-4c}
and
\betac=
1+\sqrt{1-4c | |
λ | |
\betac |
=1+\sqrt{1-4c}.
Again we have
\alphac+\betac=1.
Since the derivative with respect to z is
Pc'(z)=
d | |
dz |
Pc(z)=2z,
we have
Pc'(\alphac)+Pc'(\betac)=2\alphac+2\betac=2(\alphac+\betac)=2.
This implies that
Pc
These points are distinguished by the facts that:
\betac
c\inM\setminus\left\{1/4\right\}
\alphac
c
c
An important case of the quadratic mapping is
c=0
\alpha1=0
\alpha2=1
We have
\alpha1=\alpha2
1-4c=0.
c=1/4,
\alpha1=\alpha2=1/2
c=1/4
C
\hat{C
\hat{C
and extend
fc
fc(infty)=infty.
Then infinity is:
fc
Period-2 cycles are two distinct points
\beta1
\beta2
fc(\beta1)=\beta2
fc(\beta2)=\beta1
fc(fc(\betan))=\betan
for
n\in\{1,2\}
fc(fc(z))=(z2+c)2+c=z4+2cz2+c2+c.
Equating this to z, we obtain
z4+2cz2-z+c2+c=0.
This equation is a polynomial of degree 4, and so has four (possibly non-distinct) solutions. However, we already know two of the solutions. They are
\alpha1
\alpha2
f
f
Our 4th-order polynomial can therefore be factored in 2 ways:
(z-\alpha1)(z-\alpha2)(z-\beta1)(z-\beta2)=0.
This expands directly as
x4-Ax3+Bx2-Cx+D=0
D=\alpha1\alpha2\beta1\beta2,
C=\alpha1\alpha2\beta1+\alpha1\alpha2\beta2+\alpha1\beta1\beta2+\alpha2\beta1\beta2,
B=\alpha1\alpha2+\alpha1\beta1+\alpha1\beta2+\alpha2\beta1+\alpha2\beta2+\beta1\beta2,
A=\alpha1+\alpha2+\beta1+\beta2.
We already have two solutions, and only need the other two. Hence the problem is equivalent to solving a quadratic polynomial. In particular, note that
\alpha1+\alpha2=
1-\sqrt{1-4c | |
and
\alpha1\alpha2=
(1-\sqrt{1-4c | |
)(1+\sqrt{1-4c})}{4} |
=
12-(\sqrt{1-4c | |
) |
2}{4}=
1-1+4c | |
4 |
=
4c | |
4 |
=c.
Adding these to the above, we get
D=c\beta1\beta2
A=1+\beta1+\beta2
f
D=c\beta1\beta2=c2+c
A=1+\beta1+\beta2=0.
From this, we easily get
\beta1\beta2=c+1
\beta1+\beta2=-1
From here, we construct a quadratic equation with
A'=1,B=1,C=c+1
\beta1=
-1-\sqrt{-3-4c | |
\beta2=
-1+\sqrt{-3-4c | |
Closer examination shows that:
fc(\beta1)=\beta2
fc(\beta2)=\beta1,
meaning these two points are the two points on a single period-2 cycle.
We can factor the quartic by using polynomial long division to divide out the factors
(z-\alpha1)
(z-\alpha2),
\alpha1
\alpha2
(z2+c)2+c-z=(z2+c-z)(z2+z+c+1).
The roots of the first factor are the two fixed points. They are repelling outside the main cardioid.
The second factor has the two roots
-1\pm\sqrt{-3-4c | |
These two roots, which are the same as those found by the first method, form the period-2 orbit.[7]
Again, let us look at
c=0
\beta1=
-1-i\sqrt{3 | |
\beta2=
-1+i\sqrt{3 | |
both of which are complex numbers. We have
|\beta1|=|\beta2|=1
c=-1
\beta1=0
\beta2=-1
The degree of the equation
f(n)(z)=z
There is no general solution in radicals to polynomial equations of degree five or higher, so the points on a cycle of period greater than 2 must in general be computed using numerical methods. However, in the specific case of period 4 the cyclical points have lengthy expressions in radicals.[8]
In the case c = –2, trigonometric solutions exist for the periodic points of all periods. The case
zn+1
2-2 | |
=z | |
n |
xn+1=4xn(1-xn).
z=2-4x.
| ||||
\sin |
\right),
| ||||
\sin |
\right),\sin2\left(2
| ||||
\right),\sin2\left(2
| ||||
\right),...,\sin2\left(2k-1
2\pi | |
2k-1 |
\right).