Fractals/Multibrot sets

Julia sets for polynomial functions

z^n+ z + c

Example

f (z) = z^3 + z + 0.6*I

z^n + c

Multibrot sets

How to compute powers of z :

$Z^2 = (x+iy)^2$ $\mathrm{Real} = x^2 - y^2$  $\mathrm{Imag} = 2  x  y$

$Z^3 = (x+iy)^3$ $\mathrm{Real} = x^3 - 3 y^2  x$  $\mathrm{Imag} = 3  x^2  y  - y^3$

$Z^4 = (x+iy)^4$ $\mathrm{Real} = x^4  - 6  x^2  y^2 + y^4$  $\mathrm{Imag} = 4x^3  y - 4  x  y^3$

$Z^5 = (x+iy)^5$ $\mathrm{Real} = x^5  - 10  x^3  y^2 + 5  x  y^4$  $\mathrm{Imag} = 5x^4  y - 10  x^2  y^3 + y^5$

$Z^6  = (x+iy)^6$ $\mathrm{Real} = x^6  - 15  x^4  y^2 +15  x^2  y^4 - y^6$  $\mathrm{Imag} = 6x^5  y - 20  x^4  y^2 + 6 x  y^5$

$Z^7   = (x+iy)^7$ $\mathrm{Real} = x^7  - 21  x^5  y^2 + 35  x^3  y^4 -  7  x  y^6$  $\mathrm{Imag} = 7x^6  y - 35  x^4  y^3 + 21 x^2  y^5 - y^7$

Fortan source code :

! Fortran program by P.M.J. Trevelyan
! http://philiptrevelyan.co.uk/
     PROGRAM FRACTAL
      IMPLICIT NONE
      INTEGER I,J,ITERATION,N,M
      PARAMETER(N=2000,M=50)
      REAL*8 U,V,X,Y,P,Q
      OPEN(99,FILE='Fractal_quad.dat')
 25   FORMAT(2F9.5,I3)
      DO 10, I=1,N
        DO 20, J=1,N
C Define first point z(n)=U+iV and k=X+iY
          U=0.D0
          V=0.D0
          X=I*3.2D0/(N-1.D0)-2.1D0
          Y=J*2.8D0/(N-1.D0)-1.4D0
          DO 30, ITERATION=1,M
C Calculate z(n+1) = z(n)**2 + k where z(n+1)=P+iQ
            P=U**2-V**2+X 
            Q=2.D0*U*V+Y
            U=P
            V=Q
C If |z|>2 stop iterating
            If (U**2+V**2.GT.4.D0) GOTO 100
 30       CONTINUE
 100      WRITE(99,25) X,Y,ITERATION
 20     CONTINUE
 10   CONTINUE
      STOP
      END

Parameter planes for positive integer powers :

• 

  z² + c

• 

  z³ + c

• 

  z⁴ + c

• 

  z⁵ + c

• 

  z⁶ + c

Negative powers

When d is negative the set surrounds but does not include the origin. There is interesting complex behaviour in the contours between the set and the origin, in a star-shaped area with (1 − d)-fold rotational symmetry. The sets appear to have a circular perimeter, however this is just an artifact of the fixed maximum radius allowed by the Escape Time algorithm, and is not a limit of the sets that actually extend in all directions to infinity.

• 

  z⁻² + c

• 

  z⁻³ + c

• 

  z⁻⁴ + c

• 

  z⁻⁵ + c

• 

  z⁻⁶ + c

z^2+c

See complex quadratic polynomials

z^3 + c

$z_{n+1} = z^3_n + c$

It can be computed by :

$X_{n+1} = X^3_n - 3X_n Y^2_n + C_x$

$Y_{n+1} = 3 X_n^2 Y_n - Y_n^3 + C_y$

z^n + m*z^(-d)

McMullen maps :

$z^n + \frac{m}{z^d}$

where : z and d are >=1

"These maps are known as `McMullen maps', since McMullen first studied these maps and pointed out that when (n;m) = (2; 3) and m is small, the Julia set is a Cantor set of circles."

z^n + m*z

Description

z^2+m*z

See complex quadratic polynomial

z^3+m*z

dynamic planes

z^3 + z

It can be found using Maxima CAS :

(%i2) z:zx+zy*%i;
(%o2) %i*zy+zx
(%i6) realpart(z+z^3);
(%o6) -3*zx*zy^2+zx^3+zx
(%i7) imagpart(z+z^3);
(%o7) -zy^3+3*zx^2*zy+zy

Finding roots and its multiplicity :

$z^3 + z = z$

$z^3 = 0$

so root z=0 has multiplicity 3.

(%i1) z1:z^3+z;
(%o1) z^3+z
(%i2) solve(z1=z);
(%o2) [z=0]
(%i3) multiplicities;
(%o3) [3]

It means that there is a flower with 2 petals around fixed point z=0.

z^4 + mz

How to compute iteration :

(%i1) z:zx+zy*%i;
(%o1) %i*zy+zx
(%i2) m:mx+my*%i;
(%o2) %i*my+mx
(%i3) z1:z^4+m*z;
(%o3) (%i*zy+zx)^4+(%i*my+mx)*(%i*zy+zx)
(%i4) realpart(z1);
(%o4) zy^4-6*zx^2*zy^2-my*zy+zx^4+mx*zx
(%i5) imagpart(z1);
(%o5) -4*zx*zy^3+4*zx^3*zy+mx*zy+my*zx

See also f(z) = c(z^4-4z). It is a family 4.1 in program Mandel by Wolf Jung ( see main menu / New / 4. Quartic polynomials / 4.1 )

Here c = -m/4 and Mandelbrot set is rotated by 180 degrees.

Parameter plane

Parameter plane and Mandelbrot set for f(z) = z^4 + m*z

Period 1 components

Maxima CAS code :

(%i1) f:z^4+m*z;
(%o1) z^4+m*z
(%i2) e1:f=z;
(%o2) z^4+m*z=z
(%i3) d:diff(f,z,1);
(%o3) 4*z^3+m
(%i4) e2:d=w;
(%o4) 4*z^3+m=w
(%i5) s:eliminate ([e1,e2], [z]);
(%o5) [-(m-w)*(w+3*m-4)^3]
(%i6) s:solve([s[1]], [m]);
(%o6) [m=-(w-4)/3,m=w]

It means that there are 2 period 1 components :

• one with radius = 1 and center =0 ( m=w )
• second with radius 1/3 and center=4/3 ( m=-(w-4)/3 )

Dynamic planes

z^4+z

Julia set for f(z) = z^4 + z

Finding roots and its multiplicity :

$z^4 + z = z$

$z^4 = 0$

so root z=0 has multiplicity 4.

(%i1) z1:z^4+z;
(%o1) z^4+z
(%i2) solve(z1=z);
(%o2) [z=0]
(%i3) multiplicities;
(%o3) [4]

It means that there are 3 petals around fixed point z=0

How to compute iteration :

(%i17) z:x+y*%i;
(%o17) %i*y+x
(%i18) realpart(z+z^4);
(%o18) y^4-6*x^2*y^2+x^4+x
(%i19) imagpart(z+z^4);
(%o19) -4*x*y^3+4*x^3*y+y

z^4-iz

Critical orbits for f(z)=z^4-iz

First compute multiplier for internal angle=3/4 :

(%i1) m:exp(2*%pi*%i*3/4);
(%o1) -%i

Then find how to compute iteration :

(%i1) z:x+y*%i;
(%o1) %i*y+x
(%i2) z1:z^4-%i*z;
(%o2) (%i*y+x)^4-%i*(%i*y+x)
(%i3) realpart(z1);
(%o3) y^4-6*x^2*y^2+y+x^4
(%i4) imagpart(z1);
(%o4) -4*x*y^3+4*x^3*y-x

It is a parabolic Julia set with 12 petal flower

Critical points :

(%i12) s:GiveListOfCriticalPoints(f(z))
(%o12) [0.31498026247372*%i-0.54556181798586,-0.62996052494744*%i,0.31498026247372*%i+0.54556181798586]
(%i13) multiplicities
(%o13) [1,1,1]
(%i14) length(s)
(%o14) 3

with arguments in turns :

[0.41666666666667,0.75,0.083333333333334] = [5/12 , 9/12, 1/12]

Attracting vectors

Multiplier of fixed point −i is a fourth root of unity ( q=4), thus we examine 4-th iteration :

(%i1) z1:z^4-%i*z;
(%o1) z^4-%i*z
(%i2) z2:z1^4-%i*z1;
(%o2) (z^4-%i*z)^4-%i*(z^4-%i*z)
(%i3) z3:z2^4-%i*z2;
(%o3) ((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z))
(%i4) z4:z3^4-%i*z3;
(%o4) (((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z)))^4-%i*(((z^4-%i*z)^4-%i*(z^4-%i*z))^4-%i*((z^4-%i*z)^4-%i*(z^4-%i*z)))
(%i6) taylor(z4,z,0,20);
(%o6)/T/ z+(-76*%i-84)*z^13+(-36*%i+720)*z^16+(1812*%i-2556)*z^19+...

Next term after z is a :

(-76*%i-84)*z^13

so here :

• k=13 and n=m*q = k-1 = 12
• a = -76*%i-84

Attracting vectors satisfy :

$nav^n = -1$

so here :

$-12(76i + 84)v^{12} = -1$

$v^{12} = \frac{1}{912i + 1008}$

One can solve it in Maxima CAS :

(%i14) s:map('float,s);
(%o14) [1.007236559448514*%i+1.521106958434882,1.632845927320289*%i+0.81369898815363,
1.820935547602145*%i-0.11173896888541,1.521106958434882*%i-1.007236559448514,
0.81369898815363*%i-1.632845927320289,
-0.11173896888541*%i-1.820935547602145,-1.007236559448514*%i-1.521106958434882,
-1.632845927320289*%i-0.81369898815363,0.11173896888541-1.820935547602145*%i,
1.007236559448514-1.521106958434882*%i,
1.632845927320289-0.81369898815363*%i,0.11173896888541*%i+1.820935547602145]

With arguments in turns :

[0.093087406197659,0.17642073953099,0.25975407286433,0.34308740619766,0.42642073953099,0.50975407286433,
0.59308740619766,0.67642073953099,0.75975407286433,0.84308740619766,0.92642073953099,0.009754072864326]

different then arguments of critical points. Thus critical orbits form distorted 12-arms star

Find the fixed points :

(%i1) f:z^4-%i*z;
(%o1) z^4-%i*z
(%i2) s:solve(f=z);
(%o2) [z=((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,z=-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,z=(%i+1)^(1/3),z=0]
(%i4) multiplicities;
(%o4) [1,1,1,1]
(%i3) s:map(rhs,s);
(%o3) [((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,(%i+1)^(1/3),0]
(%i5) s:map('float,s);
(%o5) [0.5*(%i+1.0)^(1/3)*(1.732050807568877*%i-1.0),-0.5*(%i+1.0)^(1/3)*(1.732050807568877*%i+1.0),(%i+1.0)^(1/3),0.0]
(%i6) s:map(rectform,s);
(%o6) [0.7937005259841*%i-0.7937005259841,-1.084215081491351*%i-0.29051455550725,0.29051455550725*%i+1.084215081491351,0.0]

Compute the multiplier of fixed points :

(%i7) d:diff(f,z,1);
(%o7) 4*z^3-%i

Check the stability of fixed points :

(%i9) for z in s do disp(abs(ev(d)));
4.999999999999998
5.0
4.999999999999999
1
(%o9) done

Point z=0 is a parabolic point.

z^4 -z

Critical points, orbits for f(z) = z^4 - z

Julia set for f(z) = z^4 - z

Julia set for f(z) = z^4 - z with various number of star branches ( animation ) - repelling petals

It is a special case of polynomial from family :

$f_{\lambda}(z) = z^4 + \lambda z$

Here

$\lambda = -1 = e^{\pi i}$

so internal angle $\theta$ is :

$\theta = \frac{p}{q} = \frac{1}{2}$

(%i2) m:exp(2*%pi*%i/2);
(%o2) -1

Because :

$|\lambda| = 1$

it is a parabolic Julia set. Point $\lambda = -1 = e^{\pi i}$ is between two period one components ( root point ).

Periodic points

Point z=0 is a root of multiplicity seven

$k = m*q + 1 = 3*2 +1 = 7$

for equation :

$f^2_{\lambda}(z) = z$

One can check it in Maxima CAS using numerical  :

(%i1) z1:z^4-z;
(%o1) z^4-z
(%i2) z2:z1^4-z1;
(%o2) (z^4-z)^4-z^4+z
(%i3) eq2:z2-z=0;
(%o3) (z^4-z)^4-z^4=0
(%i4) allroots(eq2);
(%o4) [z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=0.0,z=1.259921049894873,
z=0.7937005259841*%i-0.7937005259841,z=-0.7937005259841*%i-0.7937005259841,
z=1.084215081491351*%i-0.29051455550725,z=-1.084215081491351*%i-0.29051455550725,
z=0.29051455550725*%i+1.084215081491351,
z=1.084215081491351-0.29051455550725*%i,z=1.091123635971722*%i-0.62996052494744,
z=-1.091123635971722*%i-0.62996052494744]
(%i5) expand(eq2);
(%o5) z^16-4*z^13+6*z^10-4*z^7=0
(%i6) factor(eq2);
(%o6) z^7*(z^3-2)*(z^6-2*z^3+2)=0

and symbolic methods :

(%i1) z1:z^4-z;
(%o1) z^4-z
(%i2) solve(z1=z);
(%o2) [z=(2^(1/3)*sqrt(3)*%i-2^(1/3))/2,z=-(2^(1/3)*sqrt(3)*%i+2^(1/3))/2,z=2^(1/3),z=0]
(%i3) multiplicities;
(%o3) [1,1,1,1]
(%i4) z2:z1^4-z1;
(%o4) (z^4-z)^4-z^4+z
(%i5) solve(z2=z);
(%o5) [z=(2^(1/3)*sqrt(3)*%i-2^(1/3))/2,z=-(2^(1/3)*sqrt(3)*%i+2^(1/3))/2,z=2^(1/3),z=((%i+1)^(1/3)*(sqrt(3)*%i-1))/2,z=-((%i+1)^(1/3)*(sqrt(3)*%i+1))/2,z=(%i+1)^(1/3),z=(sqrt(3)*(1-%i)^(1/3)*%i-(1-%i)^(1/3))/2,z=-(sqrt(3)*(1-%i)^(1/3)*%i+(1-%i)^(1/3))/2,z=(1-%i)^(1/3),z=0]
(%i6) multiplicities;
(%o6) [1,1,1,1,1,1,1,1,1,7]

Number of petals = 6

$m*q = 3*2 = 6$

Atracting vectors Denominator of internal angle $\theta = \frac{p}{q} = \frac{1}{2}$ is $q = 2$ so one have to check second iteration of function :

(%i5) z1:z^4-z;
(%o5) z^4-z
(%i6) z2:z1^4-z1;
(%o6) (z^4-z)^4-z^4+z
(%i8) expand(z2);
(%o8) z^16-4*z^13+6*z^10-4*z^7+z

Next term after z is a -4z^7. Then :

• k = 7 and n=m*q = k-1 = 6
• a = -4

Attracting vectors satisfy :

$nav^n = -1$

so here :

$-24v^6 = -1$

$v^6 = \frac{1}{24}$

One can solve it using Maxima CAS :

(%i10) s:solve(z^6=1/24);
(%o10) [z=(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),z=(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),z=-1/(sqrt(2)*3^(1/6)),z=-(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),z=-(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),z=1/(sqrt(2)*3^(1/6))]
(%i11) s:map(rhs,s);
(%o11) [(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),-1/(sqrt(2)*3^(1/6)),-(sqrt(3)*%i+1)/(2^(3/2)*3^(1/6)),-(sqrt(3)*%i-1)/(2^(3/2)*3^(1/6)),1/(sqrt(2)*3^(1/6))]
(%i12) s:map('float,s);
(%o12) [0.29439796075012*(1.732050807568877*%i+1.0),0.29439796075012*(1.732050807568877*%i-1.0),-0.58879592150024,-0.29439796075012*(1.732050807568877*%i+1.0),-0.29439796075012*(1.732050807568877*%i-1.0),0.58879592150024]
(%i13) s:map(rectform,s);
(%o13) [0.50991222566388*%i+0.29439796075012,0.50991222566388*%i-0.29439796075012,-0.58879592150024,-0.50991222566388*%i-0.29439796075012,0.29439796075012-0.50991222566388*%i,0.58879592150024]
(%i14) s:map(carg_t,s);
(%o14) [0.5235987755983/%pi,1.047197551196598/%pi,1/2,1-1.047197551196598/%pi,1-0.5235987755983/%pi,0]
(%i15) s:map('float,s);
(%o15) [0.16666666666667,0.33333333333333,0.5,0.66666666666667,0.83333333333333,0.0]

So critical points lie on attracting vectors. Thus critical orbits tend straight to the origin under the iteration

How to compute $f_{\lambda}(z)$:

(%i2) z:x+y*%i;
(%o2) %i*y+x
(%i3) realpart(z^4-z);
(%o3) y^4-6*x^2*y^2+x^4-x
(%i4) imagpart(z^4-z);
(%o4) -4*x*y^3+4*x^3*y-y

Critical points :

s:GiveListOfCriticalPoints(f(z))
(%o8) [0.54556181798586*%i-0.31498026247372,-0.54556181798586*%i-0.31498026247372,0.62996052494744]

These points has arguments in turns : 1/3, 2/3, 0

z^5 + m*z

dynamic planes

z^5 + z

Julia set z+z^5. Image and src code

Finding roots and its multiplicity :

$z^5 + z = z$

$z^5 = 0$

so root z=0 has multiplicity 5. It means that there is a flower with 4 petals

around fixed point z=0.

It How to compute :

(%i23)  z:x+y*%i;
(%o23) %i*y+x
(%i24) realpart(z+z^5);
(%o24) 5*x*y^4-10*x^3*y^2+x^5+x
(%i25) imagpart(z+z^5);
(%o25) y^5-10*x^2*y^3+5*x^4*y+y

In c programs one must use temporary variable so it can be :

tempx = 5*x*y*y*y*y-10*x*x*x*y*y + x*x*x*x*x + x ; // temporary variable
y = y*y*y*y*y -10*x*x*y*y*y + 5*x*x*x*x*y + y ;
x=tempx;

It can be optimized

"...an escape time algorithm would take forever to generate that type of image, since the dynamics are so slow there. If you want resolution of 1/100, it would take roughly 2*10^8 iterates to move the point z0=0.01 to z=2 by iterating f(z)=z+z^5." ( Mark McClure

z^6+m*z

dynamical plane

z6+z on plane [-1.2;1.2]x[-1.2;1.2]. It has 5 petals

z^14 + m*z

dynamic plane

z^14 - z

Julia set for f(z) = z^14 - z . Image and c src code

Critical orbit for f(z) = z^14 - z . Image and Maxima CAS src code

How to compute iteration :

/* Maxima CAS session */
(%i1) z:x+y*%i;
(%o1) %i*y+x
(%i2) z1:z^14-z;
(%o2) (%i*y+x)^14-%i*y-x
(%i3) realpart(z1);
(%o3) -y^14+91*x^2*y^12-1001*x^4*y^10+3003*x^6*y^8-3003*x^8*y^6+1001*x^10*y^4-91*x^12*y^2+x^14-x
(%i4) imagpart(z1);
(%o4) 14*x*y^13-364*x^3*y^11+2002*x^5*y^9-3432*x^7*y^7+2002*x^9*y^5-364*x^11*y^3+14*x^13*y-y

f(z)=z^14-z, on [-1,2;1,2]x[-1,2;1,2] has 26 petals. Compare with image by Michael Becker.

How to find fixed points :

(%i1) z1:z^14-z;
(%o1) z^14-z
(%i2) solve(z1=z);
(%o2) [z=2^(1/13)*%e^((2*%i*%pi)/13),z=2^(1/13)*%e^((4*%i*%pi)/13),
     z=2^(1/13)*%e^((6*%i*%pi)/13),z=2^(1/13)*%e^((8*%i*%pi)/13),
     z=2^(1/13)*%e^((10*%i*%pi)/13),z=2^(1/13)*%e^((12*%i*%pi)/13),
     z=2^(1/13)*%e^(-(12*%i*%pi)/13),z=2^(1/13)*%e^(-(10*%i*%pi)/13),
     z=2^(1/13)*%e^(-(8*%i*%pi)/13),z=2^(1/13)*%e^(-(6*%i*%pi)/13),
     z=2^(1/13)*%e^(-(4*%i*%pi)/13),z=2^(1/13)*%e^(-(2*%i*%pi)/13),
     z=2^(1/13),z=0]
(%i3) multiplicities;
(%o3) [1,1,1,1,1,1,1,1,1,1,1,1,1,1]
(%i4) z2:z1^14-z1;
(%o4) (z^14-z)^14-z^14+z
(%i5) solve(z2=z);
(%o5) [z=2^(1/13)*%e^((2*%i*%pi)/13),z=2^(1/13)*%e^((4*%i*%pi)/13),
   z=2^(1/13)*%e^((6*%i*%pi)/13),z=2^(1/13)*%e^((8*%i*%pi)/13),
   z=2^(1/13)*%e^((10*%i*%pi)/13),z=2^(1/13)*%e^((12*%i*%pi)/13),
   z=2^(1/13)*%e^(-(12*%i*%pi)/13),z=2^(1/13)*%e^(-(10*%i*%pi)/13),
   z=2^(1/13)*%e^(-(8*%i*%pi)/13),z=2^(1/13)*%e^(-(6*%i*%pi)/13),
   z=2^(1/13)*%e^(-(4*%i*%pi)/13),z=2^(1/13)*%e^(-(2*%i*%pi)/13),
   z=2^(1/13),z=0,0=z^78-7*z^65+21*z^52-35*z^39+35*z^26-21*z^13+7,
   0=z^78-5*z^65+11*z^52-13*z^39+9*z^26-3*z^13+1]
(%i6) multiplicities;
(%o6) [1,1,1,1,1,1,1,1,1,1,1,1,1,27,1,1]

z^15 +m*z

dynamic plane

z^15-z

Julia set for f(z)= z^15-z

Critical points and orbits for f(z)= z^15-z

How to compute iterations :

/* Maxima CAS session */
(%i1) z:x+y*%i;
(%o1) %i*y+x
(%i2) z1:z^15-z;
(%o2) (%i*y+x)^15-%i*y-x
(%i3) realpart(z1);
(%o3) -15*x*y^14+455*x^3*y^12-3003*x^5*y^10+6435*x^7*y^8-5005*x^9*y^6+1365*x^11*y^4-105*x^13*y^2+x^15-x
(%i4) imagpart(z1);
(%o4) -y^15+105*x^2*y^13-1365*x^4*y^11+5005*x^6*y^9-6435*x^8*y^7+3003*x^10*y^5-455*x^12*y^3+15*x^14*y-y

Critical points :

(%i1) m:-1;
f:z^15+ m*z;
d:diff(f,z,1);
s:solve(d=0,z)$
s:map(rhs,s)$
s:map(rectform,s)$
s:map('float,s);
multiplicities;
(%o1) -1
(%o2) z^15-z
(%o3) 15*z^14-1
(%o7) 
[0.35757475986465*%i+0.74251163973317,
0.64432745317147*%i+0.51383399763062,
0.80346319222004*%i+0.1833852305369,
0.80346319222004*%i-0.1833852305369,
0.64432745317147*%i-0.51383399763062,
0.35757475986465*%i-0.74251163973317,
-0.8241257452789,
-0.35757475986465*%i-0.74251163973317,
-0.64432745317147*%i-0.51383399763062,
-0.80346319222004*%i-0.1833852305369,
0.1833852305369-0.80346319222004*%i,
0.51383399763062-0.64432745317147*%i,
0.74251163973317-0.35757475986465*%i,
0.8241257452789]
(%o8) [1,1,1,1,1,1,1,1,1,1,1,1,1,1]

It means that here are 14 critical points and 14 critical orbits.

Fixed points :

kill(all);
remvalue(all);


/*------------- functions definitions ---------*/

/* function */
f(z):=z^15 -z;

/* find fixed points  returns a list */
GiveFixedPoints():= block
(
  [s],
  s:solve(f(z)=z),
  /* remove "z="  from list s */
  s:map('rhs,s),
  s:map('rectform,s),
  s:map('float,s),
  return(s)
)$

compile(all);


ff:GiveFixedPoints();
multiplicities;
length(s);


for i:1 thru length(ff) step 1 do
  (z:ff[i],
   disp("z= ",z, " abs(d(z))= ",abs(15*z^14-1)));

Result is :

(%i12) ff:GiveFixedPoints()
(%o12) [0.45590621928146*%i+0.94669901916834,0.82151462051137*%i+0.65513604843564,1.024411975933374*%i+0.23381534859391,
1.024411975933374*%i-0.23381534859391,0.82151462051137*%i-0.65513604843564,0.45590621928146*%i-0.94669901916834,-1.050756638653219,-
0.45590621928146*%i-0.94669901916834,-0.82151462051137*%i-0.65513604843564,-1.024411975933374*%i-0.23381534859391,0.23381534859391-
1.024411975933374*%i,0.65513604843564-0.82151462051137*%i,0.94669901916834-0.45590621928146*%i,1.050756638653219,0.0]
(%i13) multiplicities
(%o13) [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
(%i14) length(s)
(%o14) 14
(%i15) for i thru length(ff) do (z:ff[i],disp("z= ",z," abs(d(z))= ",abs(15*z^14-1)))
z= 0.45590621928146*%i+0.94669901916834 ; abs(d(z))= 28.99999999999996
z= 0.82151462051137*%i+0.65513604843564 ; abs(d(z))= 28.99999999999998
z= 1.024411975933374*%i+0.23381534859391 ; abs(d(z))= 28.99999999999999
z= 1.024411975933374*%i-0.23381534859391 ; abs(d(z))=  28.99999999999997
z= 0.82151462051137*%i-0.65513604843564 ; abs(d(z))= 29.00000000000001
z= 0.45590621928146*%i-0.94669901916834 ; abs(d(z))=  28.99999999999995
z= -1.050756638653219 ;                   abs(d(z))= 29.00000000000003
z=  -0.45590621928146*%i-0.94669901916834; abs(d(z))= 28.99999999999995
z= -0.82151462051137*%i-0.65513604843564; abs(d(z))=  29.00000000000001
z= -1.024411975933374*%i-0.23381534859391 ; abs(d(z))= 28.99999999999997
z= 0.23381534859391-1.024411975933374*%i abs(d(z))= 28.99999999999999
z= 0.65513604843564-0.82151462051137*%i ; abs(d(z))= 28.99999999999998
z= 0.94669901916834-0.45590621928146*%i ; abs(d(z))=  28.99999999999996
z= 1.050756638653219 ;                    abs(d(z))= 29.00000000000003
z= 0.0 ;                                   abs(d(z))= 1.0

So only z=0 is parabolic fixed points, the rest of them are repelling

Inside dynamics under generalized power setting

The orbit dynamics of the set can become more complex when the power is something other than 2.0. The iterated function can become multivalued and the structure of the set is then affected by the 'arbitrary' choice of which value is chosen.