If $f(x) = \frac{x}{\sqrt{x^2 + 1}}$, what is $\underbrace{f(f(f( \dots f}_{2013}(x) \dots )))$?

Question

Given the function

$$f(x) = \frac{x}{\sqrt{x^2 + 1}},$$

I need to evaluate the iterated (nested) function

$$\underbrace{f(f(f( \dots f}_{2013}(x) \dots ))).$$

I believe the alternative notation for this is $f^{2013}(x)$. I've tried looking at the problem as a geometric progression, but I haven't gotten very far. Any help would be appreciated.

Answer 1

Let $g_k(x) = \frac{x}{\sqrt{kx^2 + 1}}$, such that $f = g_1$ $$\begin{align*} g_1(g_k(x)) & = \frac{\frac{x}{\sqrt{kx^2 + 1}}}{\sqrt{\left(\frac{x}{\sqrt{kx^2 + 1}}\right)^2+1}} = \frac{\frac{x}{\sqrt{kx^2 + 1}}}{\sqrt{\frac{x^2}{\sqrt{kx^2 + 1}^2} + 1}} \\ & = \frac{\frac{x}{\sqrt{x^2 + 1}}}{\sqrt{\frac{x^2 + kx^2 + 1}{x^2 + 1}}} = \frac{x}{\sqrt{(k+1)x^2 + 1}} \\ & = g_{k+1}(x) \end{align*}$$ We have now proven, that $f^k(x) = g_k(x)$, so $$f^{2013}(x) = g_{2013}(x) = \frac{x}{\sqrt{2013x^2 + 1}}$$

Answer 2

Hint: $$f(f(x)) = \frac{x}{\sqrt{2x^2 + 1}}, \quad f(f(f(x))) = \frac{x}{\sqrt{3x^2 + 1}}, \quad \ldots$$