100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Exam (elaborations) TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton Gruiz(Solutions Manual) $15.49   Add to cart

Exam (elaborations)

Exam (elaborations) TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton Gruiz(Solutions Manual)

 8 views  0 purchase

Exam (elaborations) TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton Gruiz(Solutions Manual) Solutions to the problems1 Solution 2.2 Let the area of the largest regular triangle inscribed into the island be A0, which is also the area in the z...

[Show more]

Preview 3 out of 21  pages

  • November 14, 2021
  • 21
  • 2021/2022
  • Exam (elaborations)
  • Questions & answers
  • márton gruiz
book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:
All documents for this subject (1)
avatar-seller
Expert001
,CUUK255-Tell Et al June 9, 2006 16:45




Solutions to the problems1




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 × 4 9 = A0 /3 ∞
n−1 −n n
n=0 (4/9) . 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 = r12 , the only positive solution of the quadratic equation

r1D0 + r12D0 = 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
  −n+1
corners of the original square, ε = λn π n−1j=1 (1 − λ ) 2
j
; we therefore cover the
set with intervals of this size. In analogy with the solution of Problem 2.13, we find
that α = 2 ln λ/ln (λ/2). With λ = 0.6, α = 0.849.

Solution 2.16 See Fig. 1. The possible box probabilities are again pm = p1m p2n−m ,
 
m = 0, 1, ..., n. The number of boxes carrying pm at level n is Nm = 2m mn (the
total number of the boxes is 3n ). The logarithm of the total probability, Nm pm , is,



1 These are the solutions which do not appear in the book.


1

, CUUK255-Tell Et al June 9, 2006 16:45




2 Solutions to the problems




P(x)



15




10




5




x
Fig. 1. Distribution after the seventh step of construction. Similarly to Fig. 2.15
the typical intervals are shown under the graph, and the height of the columns
on these typical intervals is marked by a dashed line. (The continuous support is
not displayed, and only part of the central peak, of height 61.2, is visible.)

according to Stirling’s formula,

ln (Nm pm ) = n ln n − m ln m − (n − m) ln (n − m) + m ln 2 + m ln p1
+ (n − m) ln p2 .

The extremum belongs to a value m ∗ = 2np1 . The number of such boxes is
N ∗ ≡ Nm ∗ , leading to ln N ∗ = −n(2 p1 ln p1 + p2 ln p2 ). Since the resolution at
level n is ε = 3−n ,

2 p1 ln p1 + p2 ln p2
D1 = − .
ln 3

This is always less than unity for p1 = 1/3, and it is smaller, the smaller p1 is. The
same result follows from definition (2.18) by writing it as N ∗ pm ∗ ln pm ∗ = D1 n ln 3,
since N ∗ pm ∗ = 1.

Solution 3.1 The point mass moves tangentially. Since the tangent forms a slope
of inclination ϕ (see Fig. 3.1), the tangential acceleration is given by g sin ϕ, where
g is gravitational acceleration. This is, at the same time, the peripheral acceleration,
l ϕ̈; the Newtonian equation is therefore given by l ϕ̈ = g sin ϕ. For small ϕ, the sine
can be approximated by its argument, and the equation becomes ϕ̈ = (g/l)ϕ, which

is of the same type as equation (3.2), with repulsion parameter s0 = g/l.

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller Expert001. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $15.49. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

78998 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$15.49
  • (0)
  Add to cart