For large n,
gn(e) = 0 implies 0 = g(e) = 2cos(sqrt(e))
So e = ((2j+1)pi/2)2.
Then fn(cn,j) = 0 when
cn,j = -2 + ((2j+1)2pi2)/(4rn) = -2 + (6(2j+1)2pi2)/(4n+1)
So we have
(cn - cn-1)/(cn+1 - cn) -> 4 as n -> infinity
for all j.
So not only do the left-most
Return to Hurwitz-Robucci scaling.