For large n, |
gn(ε) = 0 implies 0 = g(ε) = 2cos(√ε) |
So ε = ((2j+1)π/2)2. |
Then fn(cn,j) = 0 when |
cn,j = -2 + ((2j+1)2π2)/(4rn) = -2 + (6(2j+1)2π2)/(4n+1) |
So we have |
(cn - cn-1)/(cn+1 - cn) → 4 as n → ∞ |
for all j. |
So not only do the left-most |
Return to Hurwitz-Robucci scaling.