Birth, Growth, Death and Chaos. Part 2.You can answer by filling in the blank spaces. If there is not enough space attach other sheets. Problem 1. Explore the phenomenon of chaos. As you saw in the previous problem set, the behavior of the logistic map depends heavily on the choice of r. For some r (0 < r < 2), the population goes to a fixed point regardless of the initial population. For other choices of r (e.g., 2 < r < 2.5), the population ends up in a cycle with a fixed period that depends on r. Let's examine more closely what the behavior of the population is for various initial conditions P_{0} for various r. With c = 1, r = 1.7, and N = 50, set P_{0} = 0.01 and "Plot whole curve" on the software provided on the "Plotting NonLinear Models" webpage. Next, change P_{0} to 0.0101 and "Plot whole curve" again without clearing the previous plot. What do you observe? Are the population evolutions similar? Answer:
Next, with c = 1, r = 2.2, and N = 50, set P_{0} = 0.01 and "Plot whole curve". Now, change P_{0} to 0.0101 and "Plot whole curve" again without clearing the previous plot. Again, what do you observe? Are the population evolutions similar? Answer:
Now, with c = 1, r = 3, and N = 50, set P_{0} = 0.01 and "Plot whole curve". Next, change P_{0} to 0.0101 and "Plot whole curve" again without clearing the previous plot. Once more, what do you observe? Are the population evolutions similar? If not, at what iteration do the curves begin to noticably separate? Answer:
Now, with c = 1, r = 3, and N = 50, set P_{0} = 0.01 and "Plot whole curve". Next, change P_{0} to 0.0100001 and "Plot whole curve" again without clearing the previous plot. Once more, what do you observe? Are the population evolutions similar? If not, at what iteration do the curves begin to noticably separate? Answer:
Now, with c = 1, r = 3, and N = 50, set P_{0} = 0.01 and "Plot whole curve". Next, change P_{0} to 0.0100000001 and "Plot whole curve" again without clearing the previous plot. Once more, what do you observe? Are the population evolutions similar? If not, at what iteration do the curves begin to noticably separate? Answer:
For the last three examples, qualitatively describe the relationship between difference in initial population, P_{0}, and the iteration number when the curves separate. Answer:
After enough iterations, we can see that regardless of how close we make the initial conditions, eventually, the populations behave "independently". What is the name of this phenomenon of chaos? Answer:
Problem 2. A statistical analysis of population dynamics. Even though the long term behavior of the population may be chaotic, that does not mean we cannot say anything interesting about how large or small the population is likely to be at some point far in the future. The next two questions will look at population dynamics from a statistical perspective. That is to say, we will follow a population for a long time and see how often it falls within certain bounds (e.g., how often is it between 0 and 0.1? how often is it between 0.1 and 0.2? etc...). In this way we will build up a histogram (or probability distribution), showing which population ranges are visited most (and least) frequently. On the "Chaos and Probability" page, set c = 1, r = 1.7, N = 50, and P_{0} = 0.01, and press "Iterate" followed by "Draw Histogram". You should observe a single large peak in the histogram. Why is this the case? What is the percentage of time that the population is close to 1? Answer:
Increase N to 5000 and repeat (press "Iterate" followed by "Draw Histogram"). What are the differences of this histogram to the previous one? Why? Answer:
Now set c = 1, r = 2.2, N = 50, and P_{0} = 0.01 and press "Iterate" followed by "Draw Histogram". You should observe two large peaks in the histogram. Why is this the case? Answer:
Now set c = 1, r = 3, N = 50, and P_{0} = 0.01 and press "Iterate" followed by "Draw Histogram". What do you observe? Answer:
Increase N to 5000 and repeat? What are the differences of this histogram to the previous one? Answer:
Generate histograms for at least three other values of initial population (between 0 and 1). Are the histograms similar or different in each case? What does this say about the long term behavior of the population (what population values are most likely)? Answer:
Problem 3. Binary representations of rational numbers. We saw in class that any number (not just integers) can be represented in binary. For example. The decimal number 2.5 is written 10.1 in binary because 2.5 = 2^{1} + 2^{1}. Let's practice converting between decimal and binary. Convert the following decimal numbers to binary: Answer: 0.5 = 0.25 = 0.125 = 0.0625 = For the above examples what do you observe? Why? Answer:
Convert the following decimal numbers to binary: Answer: 0.625 = 1.625 = 2.875 = 7.375 = 5.18625 = Convert the following binary numbers to decimal: Answer: 1.01 = 11.11 = 101.001 = Problem 4. The Doubling Map. Consider the doubling map:
Apply the doubling map to these examples, where the numbers are written in binary. Please, give your answer in binary as well. Answer: 0.1101101 =>
0.1111 =>
What do you observe? Answer:
Problem 5. Fixed points. For the doubling map, we can easily see that the initial population .0000000000000000... (in binary) will always iterate to itself (we will use ... to notate that the obvious pattern (in this case zeros) is repeating forever). Thus, it is a fixed point (in this case corresponding to the zero population). What is the other fixed point (in binary) for the doubling map? Answer:
Problem 6. Orbits with period 2 and 3. The initial population .0101010101... (in binary) will, after two iterations through the doubling map, cycle back to where it started (.0101010101... => .1010101010... => .0101010101...). This means that .0101010101... and .1010101010... are points on an orbit with period 2. We can find out the corresponding fraction in decimal numbers by using the following trick: .01010101... (in binary) = 2^{2} + 2^{4} + 2^{6} + 2^{8} + etc... Let x = 2^{2} + 2^{4} + 2^{6} + 2^{8} + etc... Then 4x = 1 + 2^{2} + 2^{4} + 2^{6 }+ 2^{8} + ... Subtracting the first line from the second, we obtain 3x = 1 because all the terms cancel on the right hand side except the first. Thus x = 1/3 = 0.33333...(in decimal). We can verify that this leads to an orbit with period 2 by applying the doubling map (1/3 => 2/3 => 1/3). Note that we now also know that .1010101010... (in binary) = 2/3 = .66666... (in decimal). Now, let's look at orbits with period 3. The starting point .011011011... leads to an orbit with period 3. Verify that this is true by applying the doubling map to the binary representation. Answer:
Now we can use the trick to convert the starting point into a fraction.
.011011... = 2^{2 }+ 2^{3} + 2^{5} + 2^{6}
+ 2^{8} + 2^{9} + ... Let
Answer:
There is another orbit with period 3 (with different points from the one we just looked at). Find it and give its binary and decimal representation. Answer:
Problem 7. Orbits with period 4. There are three different orbits with period 4. One of them contains
the point, .000100010001... (in binary) which is .000100010001... = 2^{4}
+ 2^{8} + 2^{12} + ... Let
Find the two remaining orbits with period 4 and give their binary and rational representations. Answer:
Problem 8. Orbits with period 5. We have seen that there is one orbit with period 2, 2 different orbits with period 3, and 3 orbits with period 4. How many different orbits with period 5 are there? Explain how you obtained your answer. Answer:
Challenge Questions Problem C1. Irrational numbers and the doubling map. π is an irrational number, that is, there exist no integers a and b such that π = a/b. Prove that (π3) does not lead to a periodic orbit in the doubling map. Answer:
Problem C2. Tripling map. Consider the tripling map:
What are the fixed points for this map? Find one orbit with period 2. Answer:
Problem C3: Number of periodic orbits with period n.
Answer:
