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.
F(x) = x+2 (if x <= -1),
F(x) = -x (if -1 <= x <= 1),
F(x) = x-2 (if x >= 1.
To view the graphical analysis click
here.
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.
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.