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This report summarizes work done as part of Rice University's VIGRE program. VIGRE is a program of Vertically Integrated Grants for Research and Education in the Mathematical Sciences under the direction of the National Science Foundation. This module explores the topolgical properties of the sum of two central homogeneous Cantor sets and presents stronger sufficient conditions for these properties. This study was led by Dr. Danijela Damjanovic and Dr. David Damanik.

Introduction

The primary objective of this study is to give stronger characterizations for the arithmetic sum of two Cantor sets. We limited ourselves to the case of sums of two different mid- α Cantor sets. It is a straightforward question but still unsolved in the majority of cases. It was first posed by J. Palis and F. Takens in the context of Dynamical Systems [link] . It also has applications in Number Theory [link] , Physics [link] , and is interesting from a purely topological perspective.

In this module, we will give preliminary definitions, provide known results, and then present the results from our study.

Preliminaries

Mid- α Cantor sets

A Cantor Set C is a set with the following properties:

  • C is non-empty.
  • C is compact.
  • C is perfect.
  • C is totally disconnected.

These properties imply that C is uncountable. The canonical example is the Cantor ternary set T , constructed in the following way:

  1. Take T 0 = 0 , 1 to be the unit interval, and remove the "middle third" from T 0 to get T 1 = 0 , 1 3 2 3 , 1 ( [link] ).
  2. Remove again the "middle thirds" from the two remaining connected components of T 1 to get T 2 = 0 , 1 9 2 9 , 1 3 2 3 , 7 9 8 9 , 1 ( [link] ).
  3. Repeat this process. The desired Cantor ternary set is T = j = 0 T j ( [link] ).
The first stage in the construction of the Cantor ternary set.
The second stage in the construction of the Cantor ternary set.
This is only a visual approximation of the Cantor ternary set. The actual set contains no intervals, and all of the intervals seen here are broken up by gaps of size less than one pixel.

The construction of the Cantor ternary set may be generalized slightly by giving ourselves a varying parameter α . In this case, we had α = 1 3 . For example, if we take α = 1 2 , then we remove the "middle halves" of intervals at each stage.

These so-called mid- α Cantor sets are the building blocks for our study. However, it becomes more convenient to reference them in terms of λ = 1 2 1 - α , i.e. the lengths of the remaining intervals in the first stage of the construction. For a given λ , as done by Mendes and Oliveira, we denote the corresponding mid- 1 - 2 λ Cantor set as C λ [link] .

We may represent an arbitrary point x C λ in the form

x = n = 0 α n λ n where α n A λ = 0 , 1 - λ n .

With this notation in mind, we think of λ as a scaling factor and A λ as a set of offsets in the sense that C λ consists of two copies of itself, scaled by λ , and translated by the elements of A λ . That is,

C λ = λ · C λ λ · C λ + 1 - λ

where λ · C λ + 1 - λ is interpreted as

λ · x + 1 - λ x C λ .

Homogeneous cantor sets

We may further generalize the construction of the Cantor sets C λ to allow for more possibilities for the set of offsets A = a 0 , a 1 , , a k for some k 1 . In this case, we have a homogeneous Cantor set C λ , A , which can be represented (with a slight abuse of notation) as

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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