Periodic points of complex quadratic mappings explained

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.

Definitions

Let

fc(z)=z2+c

be the complex quadric mapping, where

z

and

c

are complex numbers.

Notationally,

f(k)c(z)

is the

k

-fold composition of

fc

with itself (not to be confused with the

k

th derivative of

fc

)—that is, the value after the k-th iteration of the function

fc.

Thus

f(k)c(z)=

(k-1)
f
c(f

c(z)).

p

are points

z

of the dynamical plane such that

f(p)c(z)=z,

where

p

is the smallest positive integer for which the equation holds at that z.

We can introduce a new function:

Fp(z,f)=f(p)c(z)-z,

so periodic points are zeros of function

Fp(z,f)

: points z satisfying

Fp(z,f)=0,

2p.

Number of periodic points

The degree of the polynomial

Fp(z,f)

describing periodic points is

d=2p

so it has exactly

d=2p

complex roots (= periodic points), counted with multiplicity.

Stability of periodic points (orbit) - multiplier

The multiplier (or eigenvalue, derivative)

p,z
m(f
0)
of a rational map

f

iterated

p

times at cyclic point

z0

is defined as:
p,z
m(f
0)

=λ=\begin{cases}fp(z0),&ifz0\neinfty\\

1
fp(z0)

,&ifz0=infty\end{cases}

where

fp\prime(z0)

is the first derivative of

fp

with respect to

z

at

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

is a root of unity;

abs(λ)=1

but multiplier is not a root of unity;

abs(λ)>1.

Periodic points

Period-1 points (fixed points)

Finite fixed points

Let us begin by finding all finite points left unchanged by one application of

f

. These are the points that satisfy

fc(z)=z

. That is, we wish to solve

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
} and

\alpha2=

1+\sqrt{1-4c
}.So for

c\inC\setminus\{1/4\}

we have two finite fixed points

\alpha1

and

\alpha2

.

Since

\alpha1=

1
2

-m

and

\alpha2=

1
2

+m

where

m=

\sqrt{1-4c
},

we have

\alpha1+\alpha2=1

.

Thus fixed points are symmetrical about

z=1/2

.

Complex dynamics

Here different notation is commonly used:[4]

\alphac=

1-\sqrt{1-4c
} with multiplier
λ
\alphac

=1-\sqrt{1-4c}

and

\betac=

1+\sqrt{1-4c
} with multiplier
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

can have at most one attractive fixed point.

These points are distinguished by the facts that:

\betac

is:

c\inM\setminus\left\{1/4\right\}

\alphac

is:

c

is in the main cardioid of the Mandelbrot set, in which case it is in the interior of a filled-in Julia set, and therefore belongs to the Fatou set (strictly to the basin of attraction of finite fixed point)

c

Special cases

An important case of the quadratic mapping is

c=0

. In this case, we get

\alpha1=0

and

\alpha2=1

. In this case, 0 is a superattractive fixed point, and 1 belongs to the Julia set.

Only one fixed point

We have

\alpha1=\alpha2

exactly when

1-4c=0.

This equation has one solution,

c=1/4,

in which case

\alpha1=\alpha2=1/2

. In fact

c=1/4

is the largest positive, purely real value for which a finite attractor exists.

Infinite fixed point

C

to the Riemann sphere (extended complex plane)

\hat{C

} by adding infinity:

\hat{C

} = \mathbb \cup \

and extend

fc

such that

fc(infty)=infty.

Then infinity is:

fc

:[6] f_c(\infty)=\infty=f^_c(\infty).

Period-2 cycles

Period-2 cycles are two distinct points

\beta1

and

\beta2

such that

fc(\beta1)=\beta2

and

fc(\beta2)=\beta1

, and hence

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

and

\alpha2

, computed above, since if these points are left unchanged by one application of

f

, then clearly they will be unchanged by more than one application of

f

.

Our 4th-order polynomial can therefore be factored in 2 ways:

First method of factorization

(z-\alpha1)(z-\alpha2)(z-\beta1)(z-\beta2)=0.

This expands directly as

x4-Ax3+Bx2-Cx+D=0

(note the alternating signs), where

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
} + \frac = \frac = 1

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

and

A=1+\beta1+\beta2

. Matching these against the coefficients from expanding

f

, we get

D=c\beta1\beta2=c2+c

and

A=1+\beta1+\beta2=0.

From this, we easily get

\beta1\beta2=c+1

and

\beta1+\beta2=-1

.

From here, we construct a quadratic equation with

A'=1,B=1,C=c+1

and apply the standard solution formula to get

\beta1=

-1-\sqrt{-3-4c
} and

\beta2=

-1+\sqrt{-3-4c
}.

Closer examination shows that:

fc(\beta1)=\beta2

and

fc(\beta2)=\beta1,

meaning these two points are the two points on a single period-2 cycle.

Second method of factorization

We can factor the quartic by using polynomial long division to divide out the factors

(z-\alpha1)

and

(z-\alpha2),

which account for the two fixed points

\alpha1

and

\alpha2

(whose values were given earlier and which still remain at the fixed point after two iterations):

(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]

Special cases

Again, let us look at

c=0

. Then

\beta1=

-1-i\sqrt{3
} and

\beta2=

-1+i\sqrt{3
},

both of which are complex numbers. We have

|\beta1|=|\beta2|=1

. Thus, both these points are "hiding" in the Julia set.Another special case is

c=-1

, which gives

\beta1=0

and

\beta2=-1

. This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.

Cycles for period greater than 2

The degree of the equation

f(n)(z)=z

is 2n; thus for example, to find the points on a 3-cycle we would need to solve an equation of degree 8. After factoring out the factors giving the two fixed points, we would have a sixth degree equation.

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
is equivalent to the logistic map case r = 4:

xn+1=4xn(1-xn).

Here the equivalence is given by

z=2-4x.

One of the k-cycles of the logistic variable x (all of which cycles are repelling) is
2\left(2\pi
2k-1
\sin

\right),

2\left(2 ⋅ 2\pi
2k-1
\sin

\right),\sin2\left(2

2 ⋅ 2\pi
2k-1

\right),\sin2\left(2

3 ⋅ 2\pi
2k-1

\right),...,\sin2\left(2k-1

2\pi
2k-1

\right).

Further reading

External links

Notes and References

  1. Alan F. Beardon, Iteration of Rational Functions, Springer 1991,, p. 41
  2. Alan F. Beardon, Iteration of Rational Functions, Springer 1991,, page 99
  3. http://www.ijon.de/mathe/julia/some_julia_sets_1_en.html Some Julia sets by Michael Becker
  4. http://www.math.nagoya-u.ac.jp/~kawahira/works/cauliflower.pdf On the regular leaf space of the cauliflower by Tomoki Kawahira Source: Kodai Math. J. Volume 26, Number 2 (2003), 167-178.
  5. http://www.ibiblio.org/e-notes/MSet/Attractor.htm Periodic attractor by Evgeny Demidov
  6. [Robert L. Devaney|R L Devaney]
  7. http://www.ibiblio.org/e-notes/MSet/Attractor.htm Period 2 orbit by Evgeny Demidov
  8. https://arxiv.org/abs/0802.2565 Gvozden Rukavina : Quadratic recurrence equations - exact explicit solution of period four fixed points functions in bifurcation diagram