Untangling Complex Systems, page 80
alpha carbon. According to this procedure, it results that myoglobin has d ≈ 1 6
. 7 (Helman
et al. 1984). The Raman electron-spin-relaxation measurements indicate that the protein
vibrates like a compact structure (MacDonald and Jan 1986). What does this evidence
mean for d fr and dw?
11.16. Imagine having a fractal-like homodimeric reaction described by the differential equation
[11.20]. We carry out this chemical transformation in two reactors, maintained at the same
thermodynamic conditions of temperature, pressure, and so forth. The only difference is
the initial concentration of A. In the two reactors, we reach the same instantaneous con-
centration [ A] at two distinct delay times: in the first, at time t
i
= 15 s, whereas, in the sec-
ond, at time t = 5 s. Is the reaction proceeding with the same rate in the two reactors, both
having the same [ A] = [ A] ? If h
i
= 0.33, how much is the instantaneous rate coefficient?
11.19 SOLUTIONS TO THE EXERCISES
11.1. If you use MATLAB to solve this exercise and you plot the output disregarding the first 50
points, the script could be like that reported as follows.
function Henon_map(a, b)
N=1000000;
x=zeros(1,N);
y=zeros(1,N);
x(1)=0;
y(1)=0;
for i=1:N
x(i+1)=1+y(i)-a*(x(i))^2;
y(i+1)=b*x(i);
end
axis([-1,2,-1,1])
plot(x(50:N),y(50:N),’.’,’MarkerSize’,2);
fsize=15;
set (gca,’xtick’,[-1:1:1],’FontSize’,fsize)
set (gca,’ytick’,[-1:1:2],’FontSize’,fsize)
xlabel(‘itx’,’FontSize’,fsize)
ylabel(‘ity’,’FontSize’,fsize)
end
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Untangling Complex Systems
The command line will be Henon_map( 1.4 0.3), and the picture you obtain looks like
Figure 11.2.
11.2. If the length of C is 1
is (
+
=
) a u In ,
0
a.u. , the length of two segments of C 1
1 3 1 3 2 3 . .
C 2
we have four segments. The length of each segment is (1/9) a.u. , and the total length is,
of course, (4/9) a.u. In C , we have eight segments. The length of each segment is (1/27)
3
a.u. , and the total length is (8/27) a.u. After n iterations, the number of segments is 2 n, and the total length is (2 3) n. For n → ∞, the number of segments goes to infinite, and the full
length goes to zero.
11.3. If the length of K is 1
is ( ×
) a u
, we have 16 segments,
0
a.u. , the total length of K 1
4 1 3 . . In K 2
and the length of each one of them is (1/9) a.u. In K , we have ×
=
segments; the
3
4 16 64
length of each of them is (1/27) a.u. The total length will be (64/27) a.u. After n iterations,
the number of segments will be 4 n, and the full length will be (4 3) n. For n → ∞, the num-
ber of segments and the total length go to infinite.
11.4. (1) After one iteration, the number of triangles is 3; after two iterations, it is 32 = 9; after
three iterations, it is 33 = 27, and after n iterations, it is 3 n. In the end, the number of tri-
angles becomes infinite.
(2) If A is the initial full area of the triangle, after one iteration, the area of each triangle
is (1/4) A, and since the number of triangles is 3, the full area is (3/4) A. Analogously, after
two iterations, the area of each triangle is (1/4)2 A. Knowing the number of triangles, it
derives that the total area is (3/4)2 A. After three iterations, the area of the single triangle is
(1/4)3 A. The total area will be (3/4)3 A. After n iterations, the entire area will be (3/4) nA, i.e., it shrinks to zero for n → ∞.
(3) If L is the length of one side of the first triangle, the total perimeter is 3 L. The length
of one side after one iteration is (½) L, and the perimeter of just one triangle will be (3/2) L.
The total perimeter will be 3(3/2) L. After the second iteration, the length of one side of
a triangle is (1/4) L. The perimeter of one triangle is (3/4) L, and the total perimeter will
be 32(3/4) L. After the third iteration, the length of one side is (1/8) L; the perimeter of one
triangle is (3/8) L, and the total perimeter will be 33(3/8) L. Now, we can generalize and
state that, after n iterations, the length of one side is (1/2) nL, the perimeter of one triangle
is (3/2 n) L, and the total perimeter will be 3 n(3/2 n) L. In other words, the perimeter becomes longer and longer.
In the end, the Sierpinski gasket is made of an infinite number of triangles having a
negligible area but an infinite perimeter.
11.5. For the determination of the dimension of the even-fifths Cantor set, we apply equation
[11.3]. The scale factor is r = 5. The number of copies is N = 3. Therefore, the dimension
is d = log(3) log( )
5 ≈ .
0 68.
11.6. If a square is scaled up by a factor of 3, its area increases by 32 = 9 because the square is
bi-dimensional. On the other hand, if the Sierpinski gasket is scaled up by a factor of 3, its
size increases by 31.58 ≈ 5.67.
11.7. The structure of the fractal after the first five iterations is shown in Figure 11.25.
After 11 iterations, it looks like a highly-ornamented letter C (see Figure 11.25).
LC0
LC1
LC2
LC3
LC4
LC5
LC11
FIGURE 11.25 The first five iterations, and the output of the eleventh one (LC ) for the generation of the
11
Lévy C curve fractal. The pictures have been obtained by using “Examples of Fractals” model, Complexity
Explorer project, http://complexityexplorer.org.
Chaos in Space
407
If
L is the length of the original segment, the scale factor is ( 2 ) L. The number of
copies is 2. Therefore, the dimension is d = log(2) log( 2) = 2. Surprise: d is an integer
number, which corresponds to the value of bi-dimensional objects. After n iterations, the
structure consists of 2 n segments, whose length is (1 2) nL. The total length grows at each
stage: after n iterations, it is (2/ 2) n L.
11.8. The results of the iterations for the different seeds are reported in the following tables. The
results listed in Tables 11.1 and 11.2 demonstrate that starting from either ( , 0 + i) or ( ,
0 − i),
the distance r oscillates between two values.
In the third case, starting from ( ,
0 2 − i), we see that the distance escapes to infinity
(Table 11.3).
TABLE 11.1
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0, + i)
Step ( n)
zn
r
0
0
0
1
+ i
1
2
−1+ i
2
3
− i
1
4
−1+ i
2
5
− i
1
TABLE 11.2
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0, − i)
Step ( n)
zn
r
0
0
0
1
− i
1
2
−1− i
2
3
+ i
1
4
−1− i
2
2
+ i
1
TABLE 11.3
Iterative Solutions of Equations [11.7],
Starting from ( z0, c) = (0, −
2 i)
Step ( n)
zn
r
1
2 − i
5
2
5 − 5 i
50
3
2 − 51 i
2605
4
−2595 − 205 i
2603
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Untangling Complex Systems
Finally, in the fourth case (see Table 11.4), it never escapes to infinity.
11.9. The outputs of the iterations are reported in the following tables (Tables 11.5 and 11.10).
When
c = 0, the seeds do not escape. On the other hand, when c = 1, all the three seeds
escape. The Julia sets for c = 0 and c = 1 are shown in Figure 11.26, obtained through the Julia set generator at the link http://www.shodor.org/interactivate/activities/JuliaSets/.
Note that for c = 1, there are no black areas, but only scattered black dots. The black
color denotes points that do not escape, but they approach the origin. In other words,
they are “prisoners.” The shades of gray colors represent the points that escape. Different
shades of gray are associated with different speeds of escape. For c = 0, the Julia set is a
circle of radius 1 containing “prisoners.”
TABLE 11.4
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0,0.01−0.5 i)
Step ( n)
zn
r
1
0.01− 0 5
. i
0 5
.
2
−0 2399
.
− 0 5
. 1 i
0 56
.
3
−0 19255
.
− 0 255
.
i
0 3195
.
4
−0 018
.
− 0 4017
.
i
0 402
.
5
−0.151− 0.4855 i
0 508
.
6
−0 203
.
− 0 3534
.
i
0 408
.
TABLE 11.5
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0,0)
Step ( n)
zn
0
0
1
0
2
0
TABLE 11.6
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0,1)
Step ( n)
zn
0
0
1
1
2
2
3
5
4
26
Chaos in Space
409
TABLE 11.7
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0.5,0)
Step ( n)
zn
0
0.5
1
0.25
2
0.0625
3
0.0039
TABLE 11.8
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0.5,1)
Step ( n)
zn
0
0.5
1
1.25
2
2.56
3
7.566
4
58.25
TABLE 11.9
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 1,0)
Step ( n)
zn
0
1
1
1
2
1
3
1
4
1
TABLE 11.10
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 1,1)
Step ( n)
zn
0
1
1
2
2
5
3
26
4
677
410
Untangling Complex Systems
c = 0
c = 1
FIGURE 11.26 Julia sets for c = 0 (on the left) and c = 1 (on the right).
11.10. The outputs of the first iterations for the two seeds proposed in this exercise are reported
in Tables 11.11 and 11.12. Both seeds belong to the Mandelbrot set. For the first seed ( c = −0.12 + 0. i
75 ), the distance from the origin oscillates periodically. For the second
seed ( c = 0.4 − .
0 i
3 ), the distance maintains small, but it changes irregularly.
By using the two seeds as c values of two distinct Julia sets, we know (remember
Figure 11.10) that the first gives rise to a “connected-type” and the second to a “Cantor-
type” Julia set, respectively (Tables 11.11 and 11.12).
11.11. The universe observed from the Earth looks like a Cantor-set. The stars are the points of
this “natural” Cantor set.
11.12. Norway is famous for its fjords that make its coast particularly irregular (see Figure 11.27).
If we apply the Box Counting method, we find that the dimension of Norway coast is
D = ( .
1 42 ± .
0 14). Feder in his book (1988) reports D ≈ 1.52. Since the Norway coast is
more jagged than that of Britain, its fractal dimension is larger.
11.13. By using glycerin as the viscous fluid, we obtain dense-branching morphologies (see
patterns a and c in Figure 11.16) because the surface tension at the interface between
the viscous and non-viscous fluids is quite strong. By using aqueous solutions of cross-
linked PVA as the viscous fluid, the surface tension at the interface is reduced, and it
is more likely to obtain dendrites. The dimensions of the fractal-like patterns shown in
TABLE 11.11
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0, −0.12+0.75 i)
Step ( n)
zn
r
1
−0 1
. 2 + 0 75
. i
0.76
2
−0 668
.
+ 0 5
. 7 i
0.878
3
0 00146
.
− 0 0116
.
i
0.0117
4
−0 1
. 2 + 0 75
. i
0.76
5
−0 668
.
+ 0 5
. 7 i
0.878
6
0 00156
.
− 0 0113
.
i
0.0114
Chaos in Space
411
TABLE 11.12
Iterative Solutions of Equations [11.7],
Starting from ( z0, c)=(
= 0, +0.4−0.3 i)
Step ( n)
zn
r
1
0 4
. − 0.3 i
0.5
2
0 47
.
− 0.54 i
0.716
3
0 329
.
− 0 808
.
i
0.873
4
−0 144
.
− 0 832
.
i
0.84
5
−0.271− 0.06 i
0.28
6
0 4698
.
− 0 2675
.
i
0.54
7
0.55 − 0.55 i
0.78
8
0 4
. − 0.9 i
0.98
9
−0 2
. 6 −1 02
. i
1.05
10
−0 575
.
+ 0 2
. 3 i
0.62
Norway
2.4
Slope: D = 1.42
2.2 Correlation coefficient r = 0.999
2.0
( l)) N 1.8
log( 1.6
1.4
1.2
−0.6
−0.4
−0.2
0.0
0.2
0.4
log(1/ l)
FIGURE 11.27 Determination of the dimension D of Norway coast by the Box Counting method.
Figure 11.16, determined by the Box Counting method, are reported in Figure 11.28. It is evident that the dense-branching morphologies have d = 1.72 as the dimension. The well-developed dendrite in (b) has d = 1.70. On the other hand, the other two dendrites in (d)
and (e) has smaller dimensions: d = 1.40 and 1.53, respectively.
11.14. The mass of the spherical silver nanoparticle having density d increases eight times
because M
( 4 3
π
, rf = 2 r, and hence Mf = 8 M . For the fractal nanoparticle, the
3
)
i = d
ri
i
i
growth of the mass is larger. In fact, M
3 5
.
f = 2
Mi ≈ 11 3
. Mi.
11.15.
Since myoglobin vibrates as a compact structure, d ≈ d fr . Thus, according to
equation [11.16], dw ≈
2.
412
Untangling Complex Systems
0.0
−0.1
l)
0.0
−0.2
log(1/
1.16.
−0.3
igure 1
−0.1
n Fn i
l)
d = 1.53 r = 0.9998
−0.4
−0.2
