Consider the function $$f : \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = \frac{2x}{\sqrt{1 + 9x^2}}$$. 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 _____

We want to determine $$\sqrt{3\alpha + 1}$$ where $$\underbrace{f \circ f \circ \cdots \circ f}_{10}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$$ and $$f(x) = \frac{2x}{\sqrt{1+9x^2}}\,. $$

To begin, we compute the second iterate by substituting $$f(x)=\frac{2x}{\sqrt{1+9x^2}}$$ into itself, yielding $$f(f(x)) = \frac{2 \cdot \frac{2x}{\sqrt{1+9x^2}}}{\sqrt{1 + 9\bigl(\frac{2x}{\sqrt{1+9x^2}}\bigr)^2}} = \frac{\frac{4x}{\sqrt{1+9x^2}}}{\sqrt{1+\frac{36x^2}{1+9x^2}}} = \frac{4x}{\sqrt{(1+9x^2)+36x^2}} = \frac{4x}{\sqrt{1+45x^2}}\,. $$ Hence $$f^{(2)}(x)=\frac{2^2x}{\sqrt{1+9\cdot 5x^2}}\,. $$

Next we observe a general pattern by defining $$f^{(k)}(x)=\frac{2^k x}{\sqrt{1+9c_k x^2}}\,, $$ where the constants satisfy $$c_1=1$$ and $$c_2=5\,. $$ Applying one more iteration gives $$f^{(k+1)}(x)=f\!\Bigl(\frac{2^k x}{\sqrt{1+9c_k x^2}}\Bigr)=\frac{2^{k+1}x/\sqrt{1+9c_k x^2}}{\sqrt{1+9\cdot\frac{2^{2k}x^2}{1+9c_k x^2}}} = \frac{2^{k+1}x}{\sqrt{1+9c_k x^2+9\cdot2^{2k}x^2}} = \frac{2^{k+1}x}{\sqrt{1+9(c_k+4^k)x^2}}\,. $$ Therefore the recurrence for the coefficients is $$c_{k+1}=c_k+4^k\,. $$

Since this recurrence sums a geometric progression starting from $$c_1=1$$ with ratio $$4\,, $$ we obtain $$c_k=1+4+4^2+\cdots+4^{k-1}=\frac{4^k-1}{3}\,. $$ Checking small values confirms that $$c_1=(4-1)/3=1$$ and $$c_2=(16-1)/3=5\,. $$ Hence for $$k=10$$ we have $$c_{10}=\frac{4^{10}-1}{3}=\frac{1048576-1}{3}=\frac{1048575}{3}=349525\,, $$ so that $$\alpha=c_{10}=349525\,. $$

Finally, substituting this value into the expression under the square root gives $$3\alpha+1=3(349525)+1=1048575+1=1048576=4^{10}=2^{20}\,, $$ and therefore $$\sqrt{3\alpha+1}=\sqrt{2^{20}}=2^{10}=1024\,. $$

Thus, the desired value is 1024.