IB Mathematics Analysis and Approaches AHL 2.14—odd and Even Functions, Self-inverse, Inverse and Domain Restriction Questionbank

Suppose the iteration sequence converges to a limit $\alpha$. Explain why $\alpha$ must satisfy $\cos\alpha = \alpha$.

Use your graphic display calculator to find the value of $\alpha$, correct to three decimal places.

The behaviour of such iteration sequences can be visualised using a cobweb diagram. On the same axes, one draws $y = f(x)$ and $y = x$. Starting from $(x_0, 0)$, vertical and horizontal lines are drawn alternately between the two graphs, tracing the path of the iteration.

Now consider the family of functions $g(x) = kx(1 - x)$, where $k$ is a positive real constant.

Find, in terms of $k$, the two fixed points of $g$.

For $k = 2.5$ and $x_0 = 0.1$, calculate $x_1$, $x_2$ and $x_3$, and comment on the behaviour of the sequence in relation to the non-zero fixed point.

To understand why some fixed points attract nearby iterates, we study the behaviour of the error $e_n = x_n - L$, where $L$ is a fixed point.

For $x_n$ close to $L$, we may use the approximation $f(x) \approx f(L) + f'(L)(x - L)$.

Using this approximation, show that $e_{n+1} \approx f'(L) \cdot e_n$.

Hence explain why the fixed point $L$ attracts nearby iterates when $|f'(L)| < 1$, and repels them when $|f'(L)| > 1$.

Show that $g'(x^*) = 2 - k$, where $x^* = \dfrac{k-1}{k}$ is the non-zero fixed point of $g$.

Hence determine the range of values of $k$ for which the non-zero fixed point $x^*$ is an attracting fixed point in the interval $(0, 1)$.

Verify that your result from part 7 is consistent with the convergence observed in part 3 for $f(x) = \cos x$.

For $k = 3.2$, the non-zero fixed point is no longer attracting. Starting with $x_0 = 0.2$, use your GDC to iterate the function $g(x) = 3.2,x(1 - x)$ and describe the long-term behaviour of the sequence. State the values the sequence approaches, correct to three decimal places.