Task 5: Solutions
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 is `close' to 1-1/a.
The derivative of f is f¢(x) = a ·(1-2 x). Hence |f¢(1-1/a)| = |2-a|. For 1 < a < 3, clearly -1 < 2-a < +1, and so |f¢(1-1/a)¢(1-1/a)| < 1.
Clearly f is continuously differentiable. Let e = (1-|f¢(1-1/a)|) /2. From the above e > 0. By continuity
of f¢ at x = 1-1/a, there is a d > 0 so that |f¢(1-1/a)- f¢(x)| < e for x such that |(1-1/a)-x| £ d. Hence, for
such x, |f¢(x)| < |f¢(1-1/a)| + e < 1.
So by the Corollary to the Contractive Sequence Theorem, for any
x1 Î [(1-1/a)- d, (1-1/a)+ d], the sequence (xn) = (fn (x1) ) converges, as claimed.