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

<p>$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$</p> <p>$$(f \circ f)(x)=\frac{2 f(x)}{\sqrt{1+9(f(x))^2}}=\frac{\frac{4 x}{\sqrt{1+9 x^2}}}{\sqrt{1+9 \times \frac{4 x^2}{1+9 x^2}}}=\frac{4 x}{\sqrt{1+45 x^2}}$$</p> <p>$$(f \circ f \circ f)(x)=\frac{4 \times \frac{2 x}{\sqrt{1+9 x^2}}}{\sqrt{1+45 \times \frac{4 x^2}{1+9 x^2}}}=\frac{8 x}{\sqrt{1+21 \times 9 x^2}}$$</p> <p>$(f \circ f \circ f \circ f)(x)=\frac{16 x}{\sqrt{1+85 \times 9 x^2}}$</p> <p>$\Rightarrow \alpha$ is $10^{\text {th }}$ term of $1,5,21,85, \ldots \alpha$ is $10^{\text {th }}$ term of</p> <p>$$\begin{aligned} & \frac{\left(2^1\right)^2-1}{3}, \frac{\left(2^2\right)^2-1}{3}, \frac{\left(2^3\right)^2-1}{3}, \frac{\left(2^4\right)^2-1}{3}, \ldots \\ \Rightarrow \quad & \alpha=\frac{\left(2^{10}\right)^2-1}{3} \\ \Rightarrow \quad & \sqrt{3 \alpha+1}=2^{10}=1024 \end{aligned}$$</p>