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.



Where does $2\sqrt{d-1}$ come from in Ramanujan graphs?
Yes, see [this paper by Ram Murty.](http://www.mast.queensu.ca/~murty/ramanujan.pdf) The basic point is that the sum of squares of the eigenvalues is the trace of the square of the adjacency matrix, which is equal to $d n.$


