Analytic transfer theorems (common cases) Rational functions. - PowerPoint PPT Presentation

Analytic transfer theorems (“common cases”) Rational functions. Meromorphic functions. Standard function scale. Supercritical sequences. Set schema (exp-log). Simple varieties of trees. Implicit tree-like classes. 1

Analytic transfer theorems (“common cases”) Q. Match each construction with the analytic transfer theorem best suited to solving it. Rational functions. R = SEQ ( SET > 1 ( Z )) Meromorphic functions. B = E + Z 0 + ( Z 1 + Z 0 × Z 1 ) × B Standard function scale. R = SET ( UCY C > 3 ( Z )) Supercritical sequences. B = Z × ( E + B ) × ( E + B ) Exp-log. B = E + Z × B × B Simple varieties of trees. Implicit tree-like classes. 2

AC SA Apps Q&A: a “simple variety of trees” Q. A “simple variety of trees” simple variety of trees B = z × (1 + B ) × (1 + B ) construction B ( z ) = z (1 + B ( z )) 2 OGF equation φ ( u ) = (1 + u ) 2 1 + 2 u + u 2 = 2 u + 2 u 2 characteristic 
 φ 0 ( u ) = 2 + 2 u equation φ 00 ( u ) = 2 λ = 1 solution 4 n binary trees with n internal nodes (Catalan) ∼ √ ~-approximation πn 3 3

AC SA Apps Q&A: 3 -ary trees Q. How many 3 -ary trees with n internal nodes? simple variety of trees B = Z × (1 + B ) 3 construction B ( z ) = z (1 + B ( z )) 3 OGF equation φ ( u ) = (1 + u ) 3 (1 + u ) 3 = 3 u (1 + u ) 2 characteristic 
 φ 0 ( u ) = 3(1 + u ) 2 equation φ 00 ( u ) = 6(1 + u ) λ = 1 / 2 solution (27 / 4) n ∼ ~-approximation p 8 πn 3 / 3 4

AC SA Apps Q&A: Motzkin trees with a restriction Def. A skinny Motzkin tree is an ordered, rooted, unlabelled tree whose node degrees are all 
 0, 1, or 2, with the restriction that the left child of every 2-node is either a leaf or a 1-node. ∼ cφ 2 N N − 3 / 2 Q. Show that the number of skinny Motzkin trees is for some constant c . A = Z + Z × A + Z × ( Z + Z × A ) × A construction A ( z ) = z + zA ( z ) + z 2 A ( z ) + z 2 A ( z ) 2 OGF equation implicit tree-like classes Φ( z, w ) = z + zw + z 2 w + z 2 w 2 = w characteristic 
 system Φ w ( z, w ) = z + z 2 + 2 z 2 w = 1 z = 1 / φ 2 w = φ solution ∼ cφ 2 N N − 3 / 2 ~-approximation 5