Download February 7 Homework Solutions

February 7 Homework Solutions
(1)Produce the graph of the 4th iterate of the saw-tooth function and
determine the periodic points (intersection with diagonal)
All of the lines graphed above have the same slope and equidistant y-intercepts,
so they are parallel and the distance from one line to the next remains constant.
The unit is broken up into 16 equal parts so these points are:
1 2 3 4
14 15
, , , , ... ,
16 16 16 16
15 16
Written in base two, they are…
z1 = .0000
z5 = .0100
z9 = .1000
z13 = .1100
z2 = .0001
z6 = .0101
z10 = .1001
z14 = .1101
z3 = .0010
z7 = .0110
z11 = .1010
z15 = .1110
z4 = .0011
z8 = .0111
z12 = .1011
z16 = .1111
(2) Where exactly is .001101?
1
2
3
4
5
1
1
1
1
1
1
0.001101 = 0   0   1   1   0   1 
2
2
2
2
2
2
6
1 1
1
= 00  0
8 16
64
=
13
64
Another way of looking at this problem is to find this numbers place on the unit
interval the same way we would find it’s location in base 10. Look at each
decimal place and determine where that is on the number line.
Start with the unit interval.
The midpoint of this segment is 0.1.
There are two choices for the next place, 0 or 1. If it is a zero, the number will be
.01. If it is a one, the number will be .11. These two numbers are shown below.
This can be used to generate a rule. Find successive midpoints. If the next
place value is held by a 0, find the midpoint of the segment to the left (smaller). If
the next place value is held by a 1, find the midpoint of the segment to the right.
By continuing this process, we find the point .001101 on the unit interval.
(3) Given x = 0.1011010001101000011001010001000010.
|x – y| ≤1/64 and n so that |fn(x) – fn(y)| = ½.
Find y, so that
First we must find a number y so that the difference is less than or equal to 1/64.
Since 1/64 = 1/2^6, it would be the sixth place in our decimal, shown here in red:
x = 0.1011010001101000011001010001000010
For our number y, all digits will remain the same except for the one shown here
in red. The red one will become a 0.
y = 0.1011000001101000011001010001000010
When we find the difference, the only number that is any different is the one in
red so we get…
|x – y| = x = 0.0000010000000000000000000000000000 ;
which has a value base ten of 1/64. If we wanted the value to be less than 1/64,
we would do the same thing but just move one place to the right and make all
digits identical except for the seventh digit. Now when we find the difference it
will be 1/128 which is certainly less than 1/64. We could keep going forever in
this direction with the two values moving closer and closer together, but never
meeting.
For the second part of the question, we must find the number of iterations
needed so that the distance between the outputs of the saw tooth function are at
least ½ apart. We can use the base converter Dr. Peitgen gave us in class to
convert these numbers into base ten and then enter them in the Chaos Lab.
The numbers are 0.6890929379733279 and 0.7047179379733279
The values must go through six iterations before they are almost ½ apart. The
closest I got was when the outputs are 0.04385820048183924 and
0.5426094498569957 and the difference is approximately .498. These numbers
written in binary are
0.000010110011101001001010100000001011101100011100111110111
and 0.10001010111010000111001.
As you can see from the binary form of the numbers, the difference between
them is not ½, but a little less instead. If the difference were ½, the only digit that
would differ in the two numbers would be the first one to the right of the decimal
point. The rest of the digits would be identical.
(4)Let x be a point of period 5 of the Saw-Tooth Function. Assume by some
accident, the 60th digit in the binary representation of x changes from what
it was into the opposite. Describe the iteration of the Saw-Tooth function
of x and the changed initial value. How small is the difference initially (in
powers of ½, in powers of 1/10)? How large will the difference become
eventually and when?
Initially, the only difference between the two numbers is that one different value
in the 60th place. So in powers of ½, that would be (1/2)^60. To find what this
number would be in powers of 1/10, we need to use logs.
1
 
2
60
1
 
 10 
x
1
1
 x
60
2
10
2 60  10 x
 
 
log 2 60  log 10 x
60 log( 2)  x log( 10)
60 log 2
x
log 10
x
60 log 2
 60 log 2
1
x  18.06
(1/2)^60 is approximately (1/10)^18.06.
The saw tooth function chops off the first digit and then everything moves one
place over. This means that after one iteration, the difference would be (1/2)^59
and then after the second iteration it would be (1/2)58, and so on until we reach a
difference of (1/2)^1 after the 59th iteration. This value written as a power of 1/10
would be (1/10)^.301 approximately.
1
1
 x
2 10
2  10 x
log 2  log 10 x
log 2  x log 10
x
log 2 log 2

 .301
log 10
1
(5)Let x = 0.01000110110000010100111001011101110000…
Find the generating pattern of x. Continue it for at least another 30 digits.
Discuss how the iteration of the Saw-Tooth Function move around in the
unit interval [0,1].
The pattern is
.0 1 00 01 10 11 000 001 010 011 100 101 110 111 0000
Now continue the pattern by adding 1 to 0000 and keep adding 1 in the right
most place. Once we have reached 1111, we go on to 00000 and continue to do
the same.
0010 0011 0100 0101 0110 0111 1000 1001, 1010,…
In looking at the movement around the unit interval, we continually add 1/2^n
where n is the place (in the numbers directly above, n would be 4) and as the
place value moves farther out, n becomes increasingly large which makes 1/2^n
increasingly small.
(6)Create a GeoGebra implementation for the first 5 iterates of f(x) = ax(1-x),
where a is moveable by a slider between 1 and 4.
The 1st through 5th iterate is shown below.