TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton

Solution 2.2 Let the area of the largest regular triangle inscribed into the island be
A0, which is also the area in the zeroth step of the Koch construction. In the first
step, three smaller triangles are added, with an area of A0/9 each. In the nth step,
the number of new triangles of area A0/9n is 3 × 4n−1. The total area of all the
small triangles is therefore A0
∞
n=1 3 × 4n−19−n = A0/3
∞
n=0(4/9)n . This is a
geometrical series of quotient 4/9. Thus, the area increment is finite:
A0/(3(1 − 4/9)). The total area of the Koch island is (8/5)A0.
Solution 2.4 D0 = ln 3/ ln 2 = 1.585.
Solution 2.5 (a) D0 = ln 4/ ln (1/r ); (b) D0 = ln 3/ ln 2 = 1.585;
(c) D0 = ln 5/ ln 3 = 1.465.
Solution 2.7 (a) D0 = ln 20/ ln 3 = 2.727; (b) D0 = 2.
Solution 2.8 With r2 = r 2
1 , the only positive solution of the quadratic equation
r D0
1
+ r 2D0
1
= 1 is D0 = ln[(
√
5 − 1)/2]/ ln r1, which yields, for r1 = 1/2,
D0 = 0.694.
Solution 2.9 The fractal can be decomposed into five similar parts. Four of these
are identical to the entire fractal reduced by a factor of r1 = 2/5, while the
reduction factor for the fifth part is r2 = 1/5. The equation for the dimension is
therefore 4(2/5)D0 + (1/5)D0 = 1, yielding D0 = 1.601.
Solution 2.14 The area (volume) of the preserved rectangles in the nth step is
Vn = nj
=1(1 − λ2 j ). (Its limiting value for λ = 0.6 is V = 0.517.) The smallest
distance occurring in this step is the size of the holes in the squares situated at the
corners of the original square, ε = λn
πn−1
j=1 (1 − λj )

2−n+1; we therefore cover the
set with intervals of this size. In analogy with the solution of Problem 2.13, we find
that α = 2 lnλ/ln (λ/2). With λ = 0.6, α = 0.849.
Solution 2.16 See Fig. 1. The possible box probabilities are again pm = pm
1 pn−m
2 ,
m = 0, 1, ..., n. The number of boxes carrying pm at level n is Nm = 2mn
m

(the
total number of the boxes is 3n ). The logarithm of the total probability, Nm pm, is,