Problem. Consider the sequence a j for j N defined by the recurrence a j+1 = 2a j + j for j > 0

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1 GENERATING FUNCTIONS Give a ifiite sequece a 0,a,a,, its ordiary geeratig fuctio is A : a Geeratig fuctios are ofte useful for fidig a closed forula for the eleets of a sequece, fidig a recurrece forula, fidig averages or other statistical properties of a sequece, fidig asyptotic properties of a sequece, or provig idetities Proble Cosider the sequece a for N defied by the recurrece a + a + for > 0 ad the iitial coditio a 0 0 Fid a closed forula for a? Rear Oe could guess the forula ad the prove it by iductio Solutio Let A a The give recurrece relatio gives a + a + ad a + A + A a 0 a + a + A A A To fid a eplicit forula for a, we epad A as a power series; A so we coclude that a for N >0 >0, Proble Cosider the sequece a for N defied by the recurrece a + a + for > 0 ad the iitial coditio a 0 Solutio Let A a The give recurrece relatio gives A a 0 a + a + A A + A a + A To fid a eplicit forula for a, we first deterie a partial fractio decopositio A + α + β + γ Sice α + β + γ +, we have : α α : γ + γ 0: + β + β 0 MATH 40 : 07 page 46 of 5

2 Hece, epadig A as a power series produces A + + so we coclude that a + for N + + +, Propositio Fiboacci Geeratig Fuctio If F deotes the th Fiboacci uber, for N, ad F : F, the we have F Proof The Fiboacci recurrece relatio, with the iitial coditios F 0 0 ad F, give F F 0 F F + F + + F F F 0 + F F F F Proble Usig ordiary geeratig series, fid a closed forula for the Fiboacci ubers Solutio We first deterie a partial fractio decopositio for the Fiboacci geeratig fuctio By settig ϕ ± : ± 5, it follows that ϕ + ϕ ad F Sice α ϕ + β ϕ +, we have ϕ + : α ϕ ϕ+ ϕ : β α ϕ + + β ϕ ϕ + α ϕ + ϕ+ ϕ + ϕ 5, ϕ + ϕ ϕ β ϕ ϕ ϕ ϕ + 5 Hece, epadig A as a power series produces F 5 ϕ + 5 ϕ ϕ + ϕ 5 ϕ+ ϕ so we coclude that F for N Proble If F is the th Fiboacci uber, the show that F 0 + F + F + + F F + Solutio Sice F F, the left side of the equatio is F 0 + F + F + + F F, ad the right side is F + F F 0 F MATH 40 : 07 page 47 of 5

3 IDENTITIES FROM GENERATING SERIES Proble For N, prove that F + F 3 + F F + F + Solutio Recall that F F Sice we have F F F F F + + F +, the geeratig fuctio for the left side is F + F F F F Siilarly, we have F + F fuctio for the right side is F F + F F + F F, so the geeratig By etractig the coefficiets of or +, we establish the idetity o Fiboacci ubers Alterative Proof Sice we have F + F , etractig eve ad odd powers shows F Hece, the geeratig fuctio for the left side is 3 + ad F + F + F F , 3 + ad the geeratig fuctio for the right side is F Proble* For N, prove that F + F 4 + F F F + MATH 40 : 07 page 48 of 5

4 Solutio The geeratig fuctio for the left side is F 0 + F + + F 3 +, ad the geeratig fuctio for the right side is F Propositio Catala Geeratig Fuctio If C deotes the th Catala uber, for N, ad C : C, the we have C 4 Proof Sice C 0 ad the Catala Recurrece is C + C C 0 0 C C, we obtai C + C C C 0 C ± 4 C + 0 C Sice 4 +, we have , so C 4 Corollary We have C! +! +! + + Proof Sice the geeralized bioial theore ad the absorptio idetity establish that 4 4 / ! + + 4! we obtai C 4!! MATH 40 : 07 page 49 of 5

5 GENERATING FUNCTION WITH TWO VARIABLES Propositio Bioial Geeratig Fuctio For all N, we have + N Proof For N, set P : N The bioial coefficiets could be defied by the additive forula + for all > 0 ad all > 0, ad the iitial coditios 0 ad 0 0 for > 0 Hece, for > 0, we obtai [ ] P 0 >0 + P + P >0 0 P + P, ad P 0 It follows that P + for all N Proble Absorptio Forula Prove that usig ordiary geeratig fuctios Solutio Etractig the coefficiet of fro both sides of the equatio + + N + N N establishes the absorptio idetity Proble Trioial Revisio Reprove the idetity Solutio Etractig the coefficiet of y fro both sides of the equatio y N + y + + y + + y N + + y + y N y N N proves the trioial revisio idetity + Proble Upper Su For N, reprove the idetity Solutio Differetig the geoetric series ties gives +, ad ultiplyig are reideig the su yields + Hece, the geeratig fuctio for the left side is 0 + ad the geeratig fuctio for the right side is MATH 40 : 07 page 50 of 5

6 Proble Parallel Su Reprove the idetity Solutio Sice the equatio establishes the parallel su idetity , etractig the coefficiet of fro both sides of + + Propositio Multichoose Geeratig Fuctio For all N, we have , N Proof For N, set M : N The ultichoose coefficiets could be defied by the additive forula + for all > 0 ad all > 0, ad the iitial coditios 0 ad 0 0 for > 0 Hece, for > 0, we obtai M N + M + M N M M, ad M 0 It follows that M for all N + Proble Relatio to Bioial Coefficiets Reprove the idetity + Solutio Etract the coefficiets of fro + Proble Absorptio Reprove the idetity Solutio Etract the coefficiet of fro both sides of the equatio + N + N + Proble Parallel Su Reprove the idetity 0 Solutio Etract the coefficiet of fro both sides of the equatio + + N N + N 0 +, MATH 40 : 07 page 5 of 5