Generating Functions ITT9131 Konkreetne Matemaatika Chapter Seven Domino Theory and Change Basic Maneuvers Solving Recurrences Special Generating Functions Convolutions Exponential Generating Functions Dirichlet Generating Functions Contents 1 Convolutions Fibonacci convolution m-fold convolution Catalan numbers 2 Exponential generating functions Next section 1 Convolutions Fibonacci convolution m-fold convolution Catalan numbers 2 Exponential generating functions Convolutions Given two sequences: hf0;f1;f2;:::i = hfni and hg0;g1;g2;:::i = hgni The convolution of hfni and hgni is the sequence * + * + f g f g f g f g f g f g f g f g h 0 0; 0 1 + 1 0; 0 2 + 1 1 + 2 0;:::i = ∑ k n−k = ∑ k ` : k k+`=n If F (z) and G(z) are generating functions on the sequences hfni and hgni, then their convolution has the generating function F (z) · G(z). Three or more sequences can be convolved analogously, for example: * + f g h f g h h nih nih ni = ∑ j k ` j+k+`=n and the generating function of this three-fold convolution is the product F (z) · G(z) · H(z). Convolutions Given two sequences: hf0;f1;f2;:::i = hfni and hg0;g1;g2;:::i = hgni The convolution of hfni and hgni is the sequence * + * + f g f g f g f g f g f g f g f g h 0 0; 0 1 + 1 0; 0 2 + 1 1 + 2 0;:::i = ∑ k n−k = ∑ k ` : k k+`=n If F (z) and G(z) are generating functions on the sequences hfni and hgni, then their convolution has the generating function F (z) · G(z). Three or more sequences can be convolved analogously, for example: * + f g h f g h h nih nih ni = ∑ j k ` j+k+`=n and the generating function of this three-fold convolution is the product F (z) · G(z) · H(z). Convolutions Given two sequences: hf0;f1;f2;:::i = hfni and hg0;g1;g2;:::i = hgni The convolution of hfni and hgni is the sequence * + * + f g f g f g f g f g f g f g f g h 0 0; 0 1 + 1 0; 0 2 + 1 1 + 2 0;:::i = ∑ k n−k = ∑ k ` : k k+`=n If F (z) and G(z) are generating functions on the sequences hfni and hgni, then their convolution has the generating function F (z) · G(z). Three or more sequences can be convolved analogously, for example: * + f g h f g h h nih nih ni = ∑ j k ` j+k+`=n and the generating function of this three-fold convolution is the product F (z) · G(z) · H(z). Next subsection 1 Convolutions Fibonacci convolution m-fold convolution Catalan numbers 2 Exponential generating functions Fibonacci convolution To compute ∑k fk fn−k use Fibonacci generating function (in the form given by Theorem 1 and considering that n 1 zn 1 ): ∑( + ) = (1−z)2 1 1 1 2 F 2 z p ( ) = 1 z − 5 − Φ 1 − Φbz 1 1 2 1 ! = − + 5 1 z 2 2 ( − Φ ) (1 − Φz)(1 − Φbz) (1 − Φbz) 1 2 1 n 1 nzn f zn n 1 nzn = 5 ∑ ( + )Φ − 5 ∑ n+1 + 5 ∑ ( + )Φb n>0 n>0 n>0 1 2 n 1 n n zn f zn = 5 ∑ ( + )(Φ + Φb ) − 5 ∑ n+1 n>0 n>0 1 2 n 1 2f f zn f zn = 5 ∑ ( + )( n+1 − n) − 5 ∑ n+1 n>0 n>0 1 2nf n 1 f zn = 5 ∑ ( n+1 − ( + ) n) n>0 Hence 2nf n 1 f f f n+1 − ( + ) n ∑ k n−k = 5 k Fibonacci convolution (2) On the previous slide the following was used: Property For any n 0: n n 2f f > Φ + Φb = n+1 − n Proof n n 1 n n 1 The equalities ∑n Φ z = 1 z , ∑n Φb z = , and Φ + Φb = 1 are used in the −Φ 1−Φbz following derivation: 1 1 1 z 1 z n n zn − Φb + − Φ ∑(Φ + Φb ) = 1 z + = = n − Φ 1 − Φbz (1 − Φz)(1 − Φbz) 2 − z 2 z z = = · − = 1 − z − z2 z 1 − z − z2 1 − z − z2 2 n n n−1 n = ∑fnz − ∑fnz = 2∑fnz − ∑fnz = z n n n n 2 f zn f zn = ∑ n+1 − ∑ n = n n 2f f zn = ∑( n+1 − n) n Q.E.D. Fibonacci convolution (2) On the previous slide the following was used: Property For any n 0: n n 2f f > Φ + Φb = n+1 − n Proof (alternative) We know from Exercise 6.28 that n n Φ + Φb = Ln = fn+1 + fn−1 ; with the convention f−1 = 1, is the nth Lucas number, which is the solution to the recurrence: L0 = 2; L1 = 1; L L L n 2 n = n−1 + n−2 8 > : By writing the recurrence relation for Fibonacci numbers in the form fn−1 = fn+1 − fn (which, incidentally, yields f−1 = 1), we get precisely Ln = 2fn+1 − fn. Q.E.D. Next subsection 1 Convolutions Fibonacci convolution m-fold convolution Catalan numbers 2 Exponential generating functions Spanning trees for fan Example: the fan of order 5: Spanning trees: Spanning trees for fan Example: the fan of order 5: Spanning trees: Spanning trees for fan (2) How many spanning trees can we make? We need to connect 0 to each of the four blocks: two ways to join 0 with f9;10g, one way to join 0 with f8g, four ways to join 0 with f4;5;6;7;g, three ways to join 0 with f1;2;3g, There is altogether 2 · 1 · 4 · 3 = 24 ways for that. In general: sn = ∑ ∑ k1k2 ···km m 0 > k1 + k2 + ··· + km = n k1;k2;:::;km > 0 Spanning trees for fan (2) How many spanning trees can we make? We need to connect 0 to each of the four blocks: two ways to join 0 with f9;10g, one way to join 0 with f8g, four ways to join 0 with f4;5;6;7;g, three ways to join 0 with f1;2;3g, There is altogether 2 · 1 · 4 · 3 = 24 ways for that. In general: sn = ∑ ∑ k1k2 ···km m 0 > k1 + k2 + ··· + km = n k1;k2;:::;km > 0 For example f4 = 4 + 3 · 1 + 2 · 2 + 1 · 3+2 · 1 · 1 + 1 · 2 · 1 + 1 · 1 · 2+1 · 1 · 1 · 1 = 21 | {z } | {z } Spanning trees for fan (2) How many spanning trees can we make? We need to connect 0 to each of the four blocks: two ways to join 0 with f9;10g, one way to join 0 with f8g, four ways to join 0 with f4;5;6;7;g, three ways to join 0 with f1;2;3g, There is altogether 2 · 1 · 4 · 3 = 24 ways for that. In general: sn = ∑ ∑ k1k2 ···km m 0 > k1 + k2 + ··· + km = n k1;k2;:::;km > 0 This is the sum of m-fold convolutions of the sequence h0;1;2;3;:::i. Spanning trees for fan (3) Generating function for the number of spanning trees: The sequence h0;1;2;3;:::i has the generating function z G(z) = : (1 − z)2 Hence the generating function forhfni is G(z) S(z) = G(z) + G 2(z) + G 3(z) + ··· = 1 − G(z) z = 1 z 2 1 z ( − ) ( − (1−z)2 ) z = (1 − z)2 − z z = : 1 − 3z + z2 Spanning trees for fan (3) Generating function for the number of spanning trees: The sequence h0;1;2;3;:::i has the generating function z G(z) = : (1 − z)2 Hence the generating function forhfni is G(z) S(z) = G(z) + G 2(z) + G 3(z) + ··· = 1 − G(z) z = 1 z 2 1 z ( − ) ( − (1−z)2 ) z = (1 − z)2 − z z = : 1 − 3z + z2 Consequently sn = f2n Next subsection 1 Convolutions Fibonacci convolution m-fold convolution Catalan numbers 2 Exponential generating functions Dyck language Denition The Dyck language is the language consisting of balanced strings of parentheses '[' and ']'. Another denition If X = fx;xg is the alphabet, then the Dyck language is the subset D of words u of X ∗ which satisfy 1 ujx = jujx , where jujx is the number of letters x in the word u, and ∗ 2 if u is factored as vw, where v and w are words of X , then jvjx > jvjx . Dyck language (2) Let Cn be the number of words in the Dyck language D having exactly n pairs of parentheses. If u = vw for u 2 D, then the prex v 2 D i the sux w 2 D Then every word u 2 D of length > 2 has a unique writing u = [v]w such that v;w 2 D (possibly empty) but [p 62 D for every prex p of u (including u itself). Hence, for any n > 0 Cn = C0Cn−1 + C1Cn−2 + ··· + Cn−1C0 The number series hCni is called Catalan numbers, from the Belgian mathematician Eugène Catalan. Let us derive the closed formula for Cn in the following slides. Catalan numbers Step 1 The recurrent equation of Catalan numbers for all integers C C C n 0 n = ∑ k n−1−k + [ = ]: k n Step 2 Write down C(z) = ∑n Cnz : C z C zn C C zn n 0 zn ( ) = ∑ n = ∑ k n−1−k + ∑[ = ] n k;n n C zk z C zn−1−k 1 = ∑ k ∑ n−1−k + k n k n = ∑Ck z z ∑Cnz + 1 k n = zC 2(z) + 1 Catalan numbers Step 1 The recurrent equation of Catalan numbers for all integers C C C n 0 n = ∑ k n−1−k + [ = ]: k n Step 2 Write down C(z) = ∑n Cnz : C z C zn C C zn n 0 zn ( ) = ∑ n = ∑ k n−1−k + ∑[ = ] n k;n n C zk z C zn−1−k 1 = ∑ k ∑ n−1−k + k n k n = ∑Ck z z ∑Cnz + 1 k n = zC 2(z) + 1 Catalan numbers (2) Step 3 Solving the quadratic equation zC 2(z) − C(z) + 1 = 0 for C(z) yields directly: p 1 1 4z C z ± − ( ) = 2z : (Solution with "+" isn't proper as it leads to C0 = C(0) = ¥.) Step 4 From the binomial theorem we get: p 1=2 k 1 −1=2 k 1 − 4z = (−4z) = 1 + (−4z) ∑ k ∑ 2k k − 1 k>0 k>1 −1=2 n2n Using the equality for binomials n = (−1=4) n we nally get p 1 − 1 − 4z 1 −1=2 k 1 C(z) = = (−4z) − 2z ∑ k k − 1 k>1 n −1=2 (−4z) = ∑ n n + 1 n>0 2n zn = ∑ n n + 1 n>0 Catalan numbers (2) Step 3 Solving the quadratic equation zC 2(z) − C(z) + 1 = 0 for C(z) yields directly: p 1 1 4z C z ± − ( ) = 2z : (Solution with "+" isn't proper as it leads to C0 = C(0) = ¥.) Step 4 From the binomial theorem we get: p 1=2 k 1 −1=2 k 1 − 4z = (−4z) = 1 + (−4z) ∑ k ∑ 2k k − 1 k>0 k>1 −1=2 n2n Using the equality for binomials n = (−1=4) n we nally get p 1 − 1 − 4z 1 −1=2 k 1 C(z) = = (−4z) − 2z ∑ k k − 1 k>1 n −1=2 (−4z) = ∑ n n + 1 n>0 2n zn = ∑ n n + 1 n>0 −1=2 n2n Proof that n = (−1=4) n We prove a bit more: for every r 2 R and k > 0, k 2k 1 (2r) r k · r − = 2 22k −1=2 n2n Proof that n = (−1=4) n We prove a bit more: for every r 2 R and k > 0, k 2k 1 (2r) r k · r − = 2 22k Indeed, 1 k 1 3 1 r k r r r r 1 r r k 1 r k 1 · − 2 = · − 2 · ( − ) · − 2 ···( − − ) · − 2 − + 2r 2r − 1 2r − 2 2r − 3 2r − 2k − 2 2r − 2k + 1 = 2 · 2 · 2 · 2 ··· 2 · 2 2k (2r) = 22k −1=2 n2n Proof that n = (−1=4) n We prove a bit more: for every r 2 R and k > 0, k 2k 1 (2r) r k · r − = 2 22k Indeed, 1 k 1 3 1 r k r r r r 1 r r k 1 r k 1 · − 2 = · − 2 · ( − ) · − 2 ···( − − ) · − 2 − + 2r 2r − 1 2r − 2 2r − 3 2r − 2k − 2 2r − 2k + 1 = 2 · 2 · 2 · 2 ··· 2 · 2 2k (2r) = 22k Then for r = k = n, dividing by (n!)2 and using nn = n!, n n − 1=2 1 2n n = 4 n : k k and as r k = (−1)k (−r) = (−1)k (−r + k − 1) , −1=2 n − (n − 1=2) − 1 (−1)n 2n n = n = 4n n Q.E.D.