Consider the iterative function defined by: $$ x_{n+1} = f(x_n) $$ where $x_0\in [0, 1]$ and $$ f(x) = \sin\left(\pi \left(b^{rx(x-1)}\mod 1 \right)\right) $$ with $b, r > 0$. We aim to demonstrate that this function is sensitive to initial conditions. To do this, we will show that it is possible to find a value $r > 0$ such that $|f'(x)| > 1$ in a specific neighborhood within the interval $[0, 1]$.
First, let us examine the case where $b > 1$. In this case, the function can be written as:
$$ f(x) = \sin(\pi b^{r x (x-1)}) $$ The derivative of this function is given by: $$ f'(x) = r (2x - 1) \ln(b) b^{r x (x-1)} \cos(\pi b^{r x (x-1)}) $$ For values of $x$ close to 0 or 1, we have: $$ f'(x) \sim r \ln(b) $$
Thus, when $r > \frac{1}{\ln(b)}$, we get $|f'(x)| > 1$, which indicates that the function is sensitive to changes in the initial conditions in a neighborhood around these points.
Now, let us consider the case where $b < 1$. In this scenario, the function $f$ is not differentiable everywhere on the interval $[0, 1]$ because there exists a finite set of points where $f$ has discontinuities. This set is given by: $$ \Gamma = \{x \in [0, 1] : b^{r x (x-1)} \in \mathbb{Z}\} $$ that is, $\Gamma = \{x_0 = 0, x_1, \dots, x_{n-1}, x_n = 1\}$. For a point $x$ within an open interval $\left]x_i, x_{i+1}\right[$, the derivative of $f$ is: $$ f'(x) = r (2x - 1) \ln(b) b^{r x (x-1)} \cos(\pi b^{r x (x-1)}) $$ Furthermore, for $x$ close to 0 or 1, we have: $$ f'(x) \sim r |\ln(b)| $$ Thus, when $r > \frac{1}{|\ln(b)|}$, we again obtain $|f'(x)| > 1$, indicating sensitivity to initial conditions in certain regions of the interval $[0, 1]$.
In summary, we have shown that the function is sensitive to initial conditions by finding a condition on the parameter $r$ that ensures $|f'(x)| > 1$ in specific neighborhoods of the interval $[0, 1]$.
Question: Should I show that this problem is sensitive to initial conditions for any point in $[0, 1]$?${}$
