Consider the function $f:[0,1] \to \mathbb{R}$ defined by $$ f(x)= \begin{cases} 2^{- \bigg\{ 1+ \bigg( \log_2 \big( \frac{1}{x}\big) \bigg)^{\frac{1}{\beta}} \bigg\}^{\beta} }& \text{for }x \in (0,1] \\ 0 & \text{for }x=0, \\ \end{cases} $$ where $\beta \in (0, \infty)$ is a parameter.
Consider the iterations $$x_{k+1}=f(x_k), k=0,1, \dots; x_0>0.$$
Which of the following statements are true about the iteration ?
For $\beta=1,$ the sequence {$x_k$} converges to $0$ linearly with asymptotic rate of convergence $\log _{10}2$
For $\beta>1,$ the sequence {$x_k$} does not converges to $0$
For $\beta \in (0,1),$ the sequence {$x_k$} converges to $0$ sub-linearly
For $\beta \in (0,1),$ the sequence {$x_k$} converges to $0$ super-linearly
My attempt:
By putting $\beta=1,$ we get $f(x)=\frac{x}{2}$ for all $x \in [0,1].$ I am able to see that for this value of $\beta,$ the sequence $x_k$ converges to $0.$ But as for the convergence rate, I think it is with the convergence rate $\frac{1}{2}.$ So, option 1 is FALSE. But the key says that 1 is a correct option. If so, then, how to see that the convergence rate is $\log _{10}2$ ?
By putting $\beta=2$ and $x_0=1,$ I am able to see that the sequence $x_k$ is converging to $0.$ So, option 2 is FALSE, which is correct according to the key.
As for 3) and 4), I am confused on how to approach the problem. The key says 1 and 3 are correct options. Please help.
When the problem says "converges to $0$ linearly" this means that the number of decimal digits of precision increases at a linear rate. For example, the sequence $$\{0.9, 0.99, 0.999, 0.9999, \ldots \}$$ converges to $1$ at a linear rate of about $1$ digit per sequence term. In your case, $$\{0.5, 0.25, 0.125, 0.0625, \ldots \}$$ is also linear convergence, this time to $0$, and at a rate of $\log_{10} 2$ digits per iteration.
To study the behavior of the forward orbits of $f$, consider the auxiliary function $$g(x) = -\log_2 f(x) = (1 + (-\log_2 x)^{1/\beta})^\beta.$$ This suggests letting $y_n = -\log_2 x_n$, hence $$y_{n+1} = -\log_2 x_{n+1} = -\log_2 f(x_n) = g(x_n) = (1 + (-\log_2 x_n)^{1/\beta})^\beta = (1 + y_n^{1/\beta})^\beta.$$ And this in turn implies $$y_{n+1}^{1/\beta} = 1 + y_n^{1/\beta},$$ which suggests the same trick again, this time defining $$z_n = y_n^{1/\beta} = (-\log_2 x_n)^{1/\beta},$$ hence $$z_{n+1} = 1 + z_n.$$ This allows us to unfold the recurrence and obtain $$(-\log_2 x_n)^{1/\beta} = z_n = n + z_0 = n + (-\log_2 x_0)^{1/\beta}.$$ Now solving for $x_n$ yields $$\boxed{x_n = 2^{-(n + (-\log_2 x_0)^{1/\beta})^\beta}}.$$ For $x_0 \in (0,1]$, we note $-\log_2 x_0 \in [1, \infty)$, hence $$z_0 = (-\log_2 x_0)^{1/\beta} \in [1, \infty).$$ We have $$\lim_{n \to \infty} x_n = \lim_{n \to \infty} 2^{-(n + z_0)^\beta},$$ and since $\beta > 0$, $(n+z_0)^\beta \to \infty$ as $n \to \infty$. Hence $x_n \to 0$ for all initial choices of $x_0$ and $\beta$.
I leave the question of the rate of convergence is left as an exercise. How would you determine this based on the information provided above?