A 5 Equating coefficients of s on both sides yields two recurrence relations ( λ + ϕ ) f n = ϕ f n − 1 + λ f n + 1 A 6 and f n n − ( λ + ϕ ) b n = − ϕ b n − 1 − λ b n + 1.
A 4 Rearranging terms yields − s 2 b n n + s ( f n n − b n ( λ + ϕ ) ) + ( λ + ϕ ) f n = ϕ f n − 1 + λ f n + 1 − s ( ϕ b n − 1 + λ b n + 1 ).