I have an $n \times n$ matrix for which, when I take its eigenvalues and plot them, I get a curve like this (for $n=3000$). Blue curve is the eigenvalues, large to small, largest value scaled to be 1, and horizontal axis scaled to be 1. As the size of the matrix increases, the shape of the curve stays the same.
The orange curve comes when I reverse the order of the eigenvalues and take the differences (a discrete differentiation), and then scale them to match maximum values. It appears to be an exact spot-on match. Discovered this while playing around. I slightly scaled them off from each other vertically to see the two curves.
So it seems that I am looking for an function that satisfies over the domain of $[0,1]$ $$ f(x)=- \alpha f'(1-x) $$ with $f(0)=1$ and $f(1)=0$.
Any thoughts? I plugged into trusty Mathematica and it burped on solving it. Seems like something people have tackled before.
hint
Successive differentiations give $$f'(x)=\alpha f''(1-x) $$ $$f''(x)=-\alpha f'''(1-x) $$
...........
thus for $x=1,0$
$$f'(0)=0$$ $$f'(1)=-\frac {1}{\alpha} $$ $$f''(0)=f'(1) $$ $$f''(1)=0$$
...... and so on.
A solution could be
$$f (x)=\sum_{n\ge 0} \frac {f^{(n)}(0)}{n!}x^n $$