## Sample Exam #1 Solutions

This sample exam is not intended to reflect the length of the actual exam. Rather, it is intended to show the types of questions that you should expect.

• 1. True/False.
• A. The function F(x) = x^2 + 1 has an attracting fixed point at the origin.
False, F(0)=1 so the origin in not fixed.

• B. Let (001001.....) and (1111....) be two sequences in the sequence space Sigma. Then the distance between these two points is 2/3.
False, d[s,t]=1/(2^0) + 1/(2^1) + 0/(2^2) + 1/(2^3) + 1/(2^4) + 0/(2^5) + . . .
=1 +1/2 +1/8 + 1/16 + 1/64 + ... > 1 > 2/3.

• C. The point with ternary expansion .01010101..... is an endpoint of the Cantor Middle-Thirds set.
False, enpoints end in all 0's or 1's.

• D. The point 1 lies on a cycle of period 2 for the function F(x) = x^2 - 1.
False, 1 is eventually cyclic. The orbit is <1,0,-1,0,-1,...>.

• E. If a continuous function F has a cycle of period 27, then it must have a cycle of period 28 too.
True, by Sarkovski's ordering since 27 is odd and 28 = 7*(2^2).

• 2. Quickies. Answers only -- no partial credit.
• A. The function F(x) = |x| has eventually fixed points at all x<0 since here F(x)=-x and F^2(x)=F(-x)=x.

• B. The function F(x) = A arctan x has an attracting fixed point at the origin when |A|<1. Notice F'(x)=A/(1 + x^2) so F'(0)=A and for x=0 to be an attracting fixed point |F'(0)|<1.

• C. The repeating sequence (1001...) with its orbit <1001 -> 0011 -> 0110 -> 1100 > represents one of the cycles of prime period 4 for the shift map on Sigma. The other two are given by (the orbits of) (0001....) and (0111...).

• D. The number whose ternary expansion is 121212.... = 1/3 + 2/(3^2) + 1/(3^3) + 2/(3^4) + . . .
=[1/3 + 1/(3^3) + 1/(3^5) + . . . ] + [2/(3^2) + 2/(3^4) + 2/(3^6) + . . .]
=1/3(1/9 + 1/81+ 1/243 + . . .)+ 2/9(1/9 + 1/81+ 1/243 + . . .)
=1/3*9/8 + 2/9*9/8 {since each grouping is a geometric series}
=5/8

• E. <1/3, 2/3> is a cycle of period 2 for the doubling map and <1/7, 2/7, 4/7> is a 3-cycle.

• 3. Use graphical analysis to give a complete orbit analysis of the function defined in three pieces:

F(x) = x+2 (if x <= -1),
F(x) = -x (if -1 <= x <= 1),
F(x) = x-2 (if x >= 1.

x=0 is an indifferent fixed point. {see blue point in graph analysis}
x in [-1,0) U (0,1] lies on a cycle of period 2. {see red orbit in graph analysis}
All even integers are eventually fixed. x in (1,2) U (2,4) is eventually periodic with period 2. {see green orbit in graph analysis}
x in (-4,-2) U (-2,-1) is eventually periodic with period 2. {see green orbit in graph analysis}
x=2 or -2 is eventually fixed. {see blue orbit in graph analysis}
All other orbits not mentioned are eventually periodic.

• 4. Definitions.
• A. Suppose p is a fixed point of F, then p is a repelling fixed point if |F'(p)|>1.
• B. A function F is called one-to-one if for any x and y in the domain of F with x not equal to y we have F(x) is not equal to F(y).
• C. The shift map 0^:SIGMA->SIGMA is defined by 0^(SoS1S2...) = (S1S2S3...)
That is, 0^ simply drops the first entry of any point in SIGMA.

• 5. Examples. Give an example of a continuous function F which has the following property (or prove that such a function does not exist). Note: a different function for each property.
• Exactly 3^n fixed points for F^n
Consider a tripling function similar to the doubling function.
F(x)=3x mod 1
In other words,
F(x)=3x (if 0<=x<1/3)
F(x)=3x-1 (if 1/3<=x<2/3)
F(x)=3x-2 (if 2/3<=x<1)
In general, the graph of F^n consists of 3^n linear pieces of slope 3^n each crossing the line y=x once creating 3^n fixed points for F^n.

• A repelling fixed point at the origin and an attracting fixed point at 1 (and not other fixed points or cycles).
Consider F(x)=sqrt(x):
F(0)=0 and F(1)=1 so both 0 and 1 are fixed points.
|F'(1)|=1/2<1 so 1 is atrracting.
|F'(0)|=+infinity>1 so 0 is repelling.

• All orbits escaping to infinity except for a repelling cycle of period two at 0 and 1.
Such a function does not exist.
In order to have a cycle of period two at 0 and 1 the function must have a fixed point inbetween.

• 6. State the Mean Value Theorem. Use it to prove that if F(p) = p and |F'(p)| < 1, then there is an interval about p in which all orbits tend to p.
Mean Value Theorem: Suppose F is a differentiable function on the interval a<=x<=b. Then there exists c between a and b for which the following quation is true: F'(c)=[F(b)-F(a)] / (b-a).
Proof: Since |F'(p)| < 1, there is a number L>0 such that |F'(p)| < L < 1. We may therefore choose a number D>0 so that |F'(p)| < L provided x belongs to the interval I=[p-D,p+D]. Now let y be any point in I. By the Mean Value Theorem
|F(y)-F(p)| / |y-p| < L
so that
|F(y)-F(p)| < L*|y-p|.
Since p is a fixed point, it follows that
|F(y)-p| < L*|y-p|.
This means that the distance from F(y) to p is smaller that the distance from y to p, since
0 < L < 1. In particular, F(y) also lies in the interval I. Therefore we may apply the same argument to F(y) and F(p), finding
|F^2(y)-p| = |F^2(y)-F^2(p)|
< L*|F(y)-F(p)|
< L^2*|y-p|.
Since L < 1, we have L^2 < L. This means that the points F^2(y) and p are even closer together that F(y) and p. Thus we may continue using this argument to find that,
for any n>0,
|F^n(y)-p| < L^n*|x-p|.
Now L^n->0 as n->infinity. Thus, F^n(y)->p as n->infinity.

• 7. Consider the family of functions F_A(x) = A(e^x -1) where A > 0. A bifurcation occurs for this family near 0 when A = 1.
• For which values of A is the fixed point at the origin attracting? For which values is it repelling?
Recall: F'(x)=A(e^x) and F'(0)=A
Therefore the fixed point at the origin is attracting for |A|<1 and repelling for |A|>1 it is considered indifferent when |A|=1.

• Describe the bifurcation that occurs at A = 1.
When 0 < A < 1 there are two fixed points one at the origin and one to the right of the origin. The fixed point at the origin is attracting while the other fixed point is repelling.
When A=1 the right most fixed point disappears and the fixed point at the origin becomes indifferent . It attracts from the left and repells from the right.
When A>1 another fixed point is born to the left of the origin. The fixed point at the origin becomes repelling while the new fixed point is attracting.

• Use graphical analysis on these functions at, before, and after the bifurcation.

• Sketch the phase portraits for these functions at, before, and after the bifurcation. To view the phase portraits click here.

• 8. In an essay, describe the sequence space Sigma. You should discuss the distance function on Sigma and give an example or two of how it is used. You should describe which points in Sigma are close together and why. In general, your essay should be a cohesive treatment of what Sigma is. You will be graded on the accuracy of your statements as well as the use of English in your essay (spelling, grammar, etc.)

The sequence space Sigma consists of all sequences of 0's and 1's, i.e. Sigma={(sos1s2...) | sj=0 or sj=1}
The distance function on Sigma is defind by
d[s,t]= the sum from i=0 to i=infinity of |si-ti| / 2^i
where s=s0s1s2... and t=t0t1t2.... If si=ti for i=0,1,...,n then |si-ti|=0 and so
d[s,t] <= 1/2^(n+1) + 1/2^(n+2) + 1/2^(n+3) + ...
= 1/2^(n+1)*[(1/2)^0 + (1/2)^1 + (1/2)^2 + (1/2)^3 + ...]
= 1/2^(n+1)*1/(1-1/2)=1/2^n
So if si=ti for i=0,1,...,n then s and t are close together. For example,
s=000... and t=111... are far apart since d[s,t]=(1/2)^0 + (1/2)^1 + (1/2)^2 + (1/2)^3 + ... =2
whereas, if s=000... and t=0...0111... {with n+1 0's}
then d[s,t]=1/2^(n+1) + 1/2^(n+2) + 1/2^(n+3) + ...=1/2^n.
so these sequences are close.
We may think of SIGMA as consisting of 2 halves: S0={(s) | s0=0} , S1={(s) | s0=1}. But the distance from S0 to S1 is at least 1 unit. But then So breaks into 2 halves: S00={(s) | s0=s1=0} and S01={(s) | s0=0, s1=1} and the distance between b>S00 and b>S01 is 1/2 and so forth.