TITLE: Where does $2\sqrt{d-1}$ come from in Ramanujan graphs?
QUESTION [8 upvotes]: Ramanujan graphs are the best spectral expanders: $\lambda_2 \le 2\sqrt{d-1}$. I'm looking for some intuition for this value $2\sqrt{d-1}$. 
Friedman showed that every random $d$-regular graph achieves $\lambda_2 = 2\sqrt{d-1}+\epsilon$ and Alon-Boppana shows that every sufficiently large $d$-regular graph must have $\lambda_2 \ge 2\sqrt{d-1} - o(1)$. Unfortunately both these results seem quite involved. Is there an easy way to see that the answer should be around $2\sqrt{d-1}$, or at very least $\Theta(\sqrt{d})$? Intuition for either direction would be appreciated.

REPLY [8 votes]: If you take an infinite regular tree of degree $d$ and fix one vertex $v$, then the number of closed walks of length $2k$ (there are none of odd length) starting at $v$ grows like $4^k(d-1)^k$ as $k\to\infty$.  To see this heuristically, note that you have to move towards $v$ half the time and away from $v$ half the time.  There are $2^{2k}$ choices for which steps you move away from $v$ and for each such step there are $d-1$ choices for which branch to take.   As David says, this is related to the question because Ramanujan graphs are locally tree-like.