To proceed further, we substitute (A 3) into (A 2):Rearranging terms yieldsEquating coefficients ofon both sides yields two recurrence relationsandEquation (A 6) can be simplified aswhereSolution of (A 9) is given bywhere/λ.
To proceed further, we substitute (A 3) into (A 2): ( s n + λ + ϕ ) ( f n − s b n ) = ϕ ( f n − 1 − s b n − 1 ) + λ ( f n + 1 − s b n + 1 ).