How about some Proof?
Convergence and Bifurcations
Theorem 1 [Cauchy Convergence Theorem] Every Cauchy sequence is convergent;
and its corollary,
Corollary 2 [Contractive Sequence Theorem] If (xn) is a sequence, for which there is a number C < 1 such that |xn+2-xn+1| £ C ·|xn+1-xn|, then (xn) converges;
and its corollary,
Corollary 3its corollary,
Corollary 3 If f: [a,b] ® [a,b] is continuously differentiable, and sup{|f¢(x)| : x Î [a,b]} < 1, then, for any x1 Î [a,b], the sequence (x1, f(x1), f2(x1),¼, fn(x1) , ¼) is convergent.
Task 5. Calculate the derivative of f. Show that if
1 < a < 3, then |f¢(1-1/a)| < 1. Deduce that the
population sequence (xn) converges, provided the seed
x1 is `near' the limit (which is 1-1/a remember). (`Near'
means `for x sufficiently close to 1-1/a that |f¢(x)| is also
< 1.)
We know from our numerical experiments that for a1 < a < a2, the sequence (xn) does not converge but oscillates between two possible limits. These two numbers are the limits of the even subsequence and odd subsequence, respectively. In other words, they are limits of (x, f2(x), f4(x), ¼) for x = x1 and x = ) for x = x1 and x = f(x1). Thus, to find a2 we can do the following: (1) solve f2(x) = x (these are the potential limits of the sequence above, by the Algebra of Limits), (2) differentiate this expression, and then (3) a2 is the point in (0,1) where the absolute value of the derivative equals 1.
Task 6. Look at Figure 2. There are the graphs of f2(x) for various values of a. Which color shows the f2 with a closest to a2? Which color shows the f2 with a closest to a1? (Note: If the colors are not clear give the number of the curve, counting the highest curve `number 1'.)
Using a computer algebra program, I was able to carry out the algorithm for finding a2 above. The intermediate steps are horrible (including solving a quartic (power 4) equation), but the final answer is simple: a2 = 1 +Ö6. Does this match your value? The computer algebra package balked at finding a3!!