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 P₀ for various r. 

With c = 1, r = 1.7, and N = 50, set P₀ = 0.01 and "Plot whole curve" on the software provided on the "Plotting Non-Linear Models" webpage. Next, change P₀ to 0.0101 and "Plot whole curve" again without clearing the previous plot. What do you observe? Are the population evolutions similar? 

Answer:
The population evolutions are very similar (while not identical) as they move toward the fixed point.
 

Next, with c = 1, r = 2.2, and N = 50, set P₀ = 0.01 and "Plot whole curve". Now, change P₀ to 0.0101 and "Plot whole curve" again without clearing the previous plot. Again, what do you observe? Are the population evolutions similar? 

Answer:
Again, while not identical, the populations in both cases evolve very similarly as they converge to a period-2 cycle. 
 

Now, with c = 1, r = 3, and N = 50, set P₀ = 0.01 and "Plot whole curve". Next, change P₀ 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:
In the beginning, the evolutions are again very similar until approximately n=6. After that (i.e. n>6) the graphs start to vary considerably, and they become totally different after just a few iterations.
 

Now, with c = 1, r = 3, and N = 50, set P₀ = 0.01 and "Plot whole curve". Next, change P₀ 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:
Again, in the beginning, the evolutions are again very similar, this time until approximately n=15. After that (i.e. n>15) the graphs start to vary considerably, and they become totally different after just a few iterations.
 

Now, with c = 1, r = 3, and N = 50, set P₀ = 0.01 and "Plot whole curve". Next, change P₀ 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:
Similar to the previous cases, in the beginning, the evolutions are very similar, this time until approximately n=27. After that (i.e. n>27) the graphs start to vary considerably, and they become totally different after just a few iterations.
 

For the last three examples, qualitatively describe the relationship between difference in initial population, P₀, and the iteration number when the curves separate. 

Answer:
The closer the initial populations to each other, the longer it takes for the curves to separate. But in each case, they do separate. 
 
 
 

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:
Sensitive dependence on initial conditions. 
 

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 P₀ = 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:
We observe a single large peak because the population P_n in this example converges to 1 (the fixed point). For N=50, the population is close to 84% of the time. 
 

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:
Since the population converges to 1, after some n P_n will be close to 1 for all n. This means that all the contibutions from the larger values of n will be to the single large peak, and this will increase the percentage of time the population is close to 1. For N=5000, we observe that 99.8% of the time, the population is close to 1. 
 

Now set c = 1, r = 2.2, N = 50, and P₀ = 0.01 and press "Iterate" followed by "Draw Histogram". You should observe two large peaks in the histogram. Why is this the case? 

Answer:
This is beacuse this time the population converges to a period-2 cycle, which means the population eventually oscillates between two values. Thus we observe in the histogram two peaks at the the bins corresponding to these two values. 
 

Now set c = 1, r = 3, N = 50, and P₀ = 0.01 and press "Iterate" followed by "Draw Histogram". What do you observe? 

Answer:
The histogram does not have distinguished peaks in this case. This means that P_n does not converge to any value, neither there is a periodic orbit. However, we see that the poulation seems to be more frequently close to the endpoints, i.e. in bin 0 or bin 19.
 

Increase N to 5000 and repeat? What are the differences of this histogram to the previous one? 

Answer:
We still observe that the orbits tend to visit points near the two ends much more often than near the middle. Also, unlike the case with N=50, the number of times the population visits consequtive bins is close to each other. 
 

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:
The histograms are virtually indistinguishable.  This shows that there is still some commonality between the orbits starting from different initial conditions. We can generalize our comments above, and say that, regardless of initial conditions, the orbits tend to visit points near the two ends much more often, and how much more time they spend near the ends than in the middle turns out to be independent of the initial population P₀.
 
 

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 = 2¹ + 2^-1. Let's practice converting between decimal and binary.

Convert the following decimal numbers to binary: 

Answer:

0.5 = 0.1

0.25 = 0.01

0.125 = 0.001

0.0625 = 0.0001

For the above examples what do you observe? Why? 

Answer:
Dividing a binary fraction by two corresponds to a left shift of the dot. (You can think of 0.1 as 00.1, so a left shift of the yields 0.01) This is similar to the case with decimal fractions, where dividing a number by 10 is done by shifting the dot one place left.
 

Convert the following decimal numbers to binary: 

Answer:

0.625 = 0.101

1.625 = 1.101

2.875 = 10.111

7.375 = 111.011

5.18625 = 101.001011110...

Convert the following binary numbers to decimal: 

Answer:

1.01 = 1.25

11.111 = 3.75

101.001 = 5.125

Problem 4. The Doubling Map.

Consider the doubling map:
x => 2x   (if x less than or equal to 1/2)
x => 2x-1 (if x is greater than 1/2 but less than or equal to 1)

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 => 0.101101
0.101101 => 0.01101
0.01101 => 0.1101 
0.1101 => 0.101
0.101 => 0.01

0.1111 => 0.111
0.111 => 0.11
0.11 => 0.1
0.1 => 1

What do you observe? 

Answer:
To apply the doubling map to a number x, we simply drop the first binary expansion bit.

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:
The other fixed point is 0.111111111... (which is equal to 1 -- see lecture notes about non-uniqueness of the binary representation of fractions). 
 

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:
.011011011...=> .110110110... =>.101101101...=>.011011011... (After applying the doubling map three times, we have the initial point back. This means that we have an orbit with period 3.)
 

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
  x =              2^-2 + 2^-3 + 2^-5 + 2^-6 + 2^-8 + 2^-9 + ... Then,
8x = 2 + 1 + 2^-2 + 2^-3 + 2^-5 + 2^-6 + 2^-8 + 2^-9 + ...
Subtracting the first line from the second, we obtain, 7x = 3 Thus, x = 3/7. Verify that this leads to an orbit with period 3 by applying the doubling map to 3/7. 

Answer:
3/7 =>  6/7 => 5/7 => 3/7.
 

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:
The starting point 0.001001001... leads to a different periodic orbit with period 3. (Applying the doubling map 3 times, we get our starting point back. Its decimal representation is:
x  = 0.001001001001
8x= 1.001001001001

Subtracting x from 8x, we obtain 7x=1, which yields x=1/7. We can verify that this point leads to a period-3 orbit again by applying the doubling map:
1/7 => 2/7 => 4/7 => 1/7.
 
 

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
    x =       2^-4 + 2^-8 + 2^-12 + ... Then,
16x = 1 + 2^-4 + 2^-8 + 2^-12 + ...
So, 15x = 1, and x = 1/15 = .06666... (in decimal) We can verify that this leads to an orbit with period 4 (in binary) .0001... => .0010... => .0100... => .1000... => .0001... (using the rational representation) 1/15 => 2/15 => 4/15 => 8/15 => 1/15. 

Find the two remaining orbits with period 4 and give their binary and rational representations. 

Answer:
The two remaining orbits are given by the starting points: 
1)   x=0.0011001100110011..., which is equal to 3/15.
2)   x=0.0111011101110111..., which is equal to 7/15
 
 
 

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:
There are 6 different orbits of period 5. The orbits are given by the following starting points: 
- .000010000100001...
- .000110001100011...
- .001010010100101...
- .001110011100111...
- .010110101101011...
- .011110111101111...