Nested functions Assuming that we have some function $L(x)$ such that $L(x) = x - \frac{x^2}{4}.$ Now, define $a_n$ as $$L \Bigl( L \Bigl( L \Bigl( \cdots L \Bigl( \frac{17}{n} \Bigr) \cdots \Bigr) \Bigr) \Bigr),$$ where we have $n$ iterations of $L.$ My question here is, what value does $n \cdot a_n$ approach as $n$ approaches infinity?
I tried to find some sort of pattern but it got nasty fast. I than tried to find a few small values and test them out, but they didn't quite work. How should I approach this problem? Thanks.

I'd like to post the solution I came up with.
Note that $0 < L(x) < x$ for $0 < x < 2.$ Assuming $n$ is sufficiently large, i.e. $n \ge 9,$ we have that $0 < a_n < \frac{17}{n} < 2.$
From $L(x) = x - \frac{x^2}{2},$ we can write
$$\frac{1}{L(x)} = \frac{1}{x - \frac{x^2}{2}} = \frac{2}{2x - x^2} = \frac{2}{x(2 - x)} = \frac{x + (2 - x)}{x(2 - x)} = \frac{1}{x} + \frac{1}{2 - x},$$ so
$$\frac{1}{L(x)} - \frac{1}{x} = \frac{1}{2 - x} \quad (*).$$ For a nonnegative integer $k,$ let $L^{(k)}(x)$ denote the $k$th iterate of $L(x).$ Then $0 < L^{(k)}(x) < x,$ so
$$0 < L^{(k)} \left( \frac{17}{n} \right) \le \frac{17}{n}.$$ Hence,
$$\frac{1}{2} < \frac{1}{2 - L^{(k)} (\frac{17}{n})} \le \frac{1}{2 - \frac{17}{n}} = \frac{n}{2n - 17}.$$ By equation $(*),$
$$\frac{1}{L^{(k + 1)} (\frac{17}{n})} - \frac{1}{L^{(k)} (\frac{17}{n})} = \frac{1}{2 - L^{(k)} (\frac{17}{n})},$$ so
$$\frac{1}{2} < \frac{1}{L^{(k + 1)} (\frac{17}{n})} - \frac{1}{L^{(k)} (\frac{17}{n})} \le \frac{n}{2n - 17}.$$ Summing over $0 \le k \le n - 1,$ we get
$$\frac{n}{2} < \frac{1}{L^{(n)} (\frac{17}{n})} - \frac{1}{\frac{17}{n}} \le \frac{n^2}{2n - 17}.$$ Since $a_n = L^{(n)} \left( \frac{17}{n} \right),$ this becomes
$$\frac{n}{2} < \frac{1}{a_n} - \frac{n}{17} \le \frac{n^2}{2n - 17}.$$ Dividing by $n,$ we get
$$\frac{1}{2} < \frac{1}{na_n} - \frac{1}{17} \le \frac{n}{2n - 17}.$$ As $n$ approaches infinity, $\frac{n}{2n - 17}$ approaches $\frac{1}{2},$ so if $L$ is the limit of $na_n,$ then
$$\frac{1}{L} - \frac{1}{17} = \frac{1}{2}.$$ Solving, we find $L = \boxed{\frac{34}{19}}.$
Thanks to the people who helped me with this problem.
