Iterated Function Sequences Accumulation: Help

  • Thread starter Thread starter aridneptune
  • Start date Start date
  • Tags Tags
    Function Sequences
Click For Summary
SUMMARY

The discussion focuses on the accumulation points of the iterated function sequence defined by the contractions A0, A1, and A2, which generate the Sierpinski right triangle. The user seeks to understand the behavior of the composite function (A1 ∘ A2)^n applied to an initial point . The user initially hypothesizes that iterating A2 after A1 converges to <0, 1>, but questions the validity of this assumption. The key conclusion is that the sequence accumulates at multiple points, necessitating a deeper analysis of the composite functions.

PREREQUISITES
  • Understanding of iterated function systems (IFS)
  • Familiarity with contraction mappings
  • Knowledge of the Sierpinski triangle and its properties
  • Basic concepts of composite functions in mathematics
NEXT STEPS
  • Explore the properties of contraction mappings in detail
  • Study the Sierpinski triangle's construction and its fractal nature
  • Learn about fixed points and their significance in iterative processes
  • Investigate the concept of accumulation points in sequences
USEFUL FOR

Mathematics students, particularly those studying fractals, iterative systems, and contraction mappings, as well as educators looking for examples of complex function behavior.

aridneptune
Messages
4
Reaction score
0

Homework Statement


We are given three contractions which generate the Sierpinski right triangle:

A0 = \frac{1}{2} <x , y>

A1 = \frac{1}{2} <x-1 , y> + <1 , 0>

A2 = \frac{1}{2} <x , y-1> + <0 , 1>

We are asked to find the points on which the sequence

(A2\circA1)n(<x0 , y0>) ) -- that's (A1 COMPOSITE A2)n

accumulates.

The Attempt at a Solution



Not quite sure how to approach this problem at all. I've figured that A1n takes any <x , y> to <1 , 0>, and that A2n takes any <x, y> to <0, 1>. So my first instinct was to say that iterating A2 n times on an n-iteration of A1 would just converge to <0 , 1>. However, I'm fairly sure this is incorrect. But how can this system accumulate on >1 point?

Any ideas/help would be greatly appreciated!
 
Last edited:
Physics news on Phys.org
my first thought is to write the n-th iterators in terms of <x0,y0>, where the n-th iteration of A1 with be the initial point for the n-th iteration of A2.
 
Sorry -- correction it's (A1 COMPOSITE A2)n (fixed above). Increasing n increases both applications of A1 and of A2.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
Replies
5
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K