Define $f : \mathbb R\rightarrow \mathbb R$ by $f(x) = (3x^2+1)/(x^2+3).$ Let $f^{\circ1} = f,$ and let $f^{\circ n} = f^{\circ(n-1)} \circ f$ for all integers $n \geq 2.$ Which of the following statements is correct ?
- $\lim_{n\to \infty}f^{\circ n}(1/2) = 1,$ but $\lim_{n\to \infty}f^{\circ n}(2) $ does not exist;
- $\lim_{n\to \infty}f^{\circ n}(1/2) = 1,$ and $\lim_{n\to \infty}f^{\circ n}(2) = 1 $;
- $\lim_{n\to \infty}f^{\circ n}(1/2)$ does not exist but $\lim_{n\to \infty}f^{\circ n}(2) = 1 $;
- Neither $\lim_{n\to \infty}f^{\circ n}(1/2)$ nor $\lim_{n\to \infty}f^{\circ n}(2)$ exist.
I know that, $f(1) =1$ f has unique fixed point. since, Range of $f$ is inside $[0, 3]$, we can consider $f$ as a map from [0, 3] to [0, 3]. If $c= \sup |f'(x) |<1$ on [a b] $x_{n+1} = f(x_n) $ converges to the fixed point for any initial choice.
Actually here $|f'(x) |\leq 1$, so I am unable to conclude the correct option.
First notice that
$$ x < f(x) \iff (x-1)^3 < 0 \iff x< 1$$
$$ x > f(x) \iff (x-1)^3 > 0 \iff x > 1$$
Hence there is only one fixed point which is $x = 1$.The derivative is given by
$$ f'(x) = \frac{16x}{(x^2+3)^2}$$ so the function $f$ is increasing on $(0,+\infty).$
Let $x_0 \in (0,1)$ and define the sequence $$ x_{n+1} = f(x_n).$$
Since $f$ is increasing it follows $$ x_n < 1\implies f(x_n)< f(1) = 1$$ for all $n$. In particular the sequence $x_{n+1}$ is increasing and bounded above so it converges (to the fixed point $x = 1$).
Repeating a similar argument for $x_0 \in (1,+\infty)$ will show that the sequence is bounded below and decreasing and hence converges to the fixed point.
The only valid statement is therefore the second statement.