Question

Consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{2x}{\sqrt{1 + 9x^2}}.$ If the composition of $f$, $(f \circ f \circ f \circ \dots \circ f)(x) \quad \text{(10 times)} = \frac{2^{10}x}{\sqrt{1 + 9\alpha x^2}},$ then the value of $\sqrt{3\alpha + 1}$ is equal to $\dots$.

Answer 1

To determine the value of $\alpha$, let’s analyze the repeated composition of $f(x)$.

1. Starting with $f(x) = \frac{2x}{\sqrt{1 + 9x^2}}$, we compute $f(f(x))$: $f(f(x)) = \frac{2f(x)}{\sqrt{1 + 9f(x)^2}} = \frac{4x}{\sqrt{1 + 9x^2 + 9 \cdot 2^2 x^2}} = \frac{2^2 x}{\sqrt{1 + 9(1 + 2)x^2}}.$ This gives us $\alpha_2 = 1 + 2$ for the second composition.
2. Repeating this process, we observe a pattern: after $n$ compositions, the denominator takes the form $\sqrt{1 + 9(1 + 2 + 2^2 + \cdots + 2^{n-1})x^2}.$
3. The series $1 + 2 + 2^2 + \cdots + 2^{n-1}$ is a geometric series that sums to $2^n - 1$. Therefore, after 10 compositions, we have: $\alpha = 2^{10} - 1 = 1023.$

Now, we calculate $\sqrt{3\alpha + 1}$:

$\sqrt{3\alpha + 1} = \sqrt{3 \cdot 1023 + 1} = \sqrt{3072 + 1} = \sqrt{3072} = 1024.$

Answer: 1024