CMT 시험대비2

1. 용어

f_n:seq from {0,1,2,...}->F

f(t):fps from {f_n}, :=G[f_n]

f(t,w):fps from {f_(n,k)}, bivariate formal power series from {f_(n,k)}

[t^n]f(t)

ord(f(t))

bar(f(t))

F[[t]], F_k[[t]]

L, laurent power series, 원소는 l(t)

(x)_k, falling factorial

[x]_k, rising factorial

(d(t),h(t))가 pRa

M(d(t),h(t)), riordan array

renewal array

R, riordan group

A, Appell subgroup

Lag, Lagrange subgroup

derivative subgroup, {M(h'(t),h(t)) s.t. h(t) in F_1[[t]]}

2. 정리

About F[[t]]

ord(f(t)+g(t)) >= min{ord(f(t)), ord(g(t)))

ord(f(t)g(t)) = ord(f(t)) + ord(g(t))

(F[[t]],+,*):ID, quotient field는 L

About F_0[[t]]

(F_0[[t]], *):abelian group((F_0[[t]], +, *)에서는 +에 대해 닫혀있지 않음)

About F_1[[t]]

(F_1[[t]], o):group

About L

[t^n]의 성질 중 f(t)g(t)와 f(g(t))와 bar(f(t))에 관해서 복습, 특히 마지막을 Lagrange Inversion formula([t^n]bar(f(t)), [t^n]{bar(f(t))}^k)

About riordan array, riordan group

R:group

M(d(t),h(t))*M(a(t),b(t))=M(d(t)a(h(t)),b(h(t))), riordan arrays are closed under the matrix product

multiplicative identity는 M(1,t)

M(d(t),h(t))의 multiplicative inverse는 M(1/d(bar(h(t)), bar(h(t)))

Every pRa can be seen as the product of an Appell by a Lagrange matrix

{f_(n,k)}:pRa iff te d(t) in F_0[[t]] and h(t) in F_1[[t]] s.t. f(t,w) = d(t)/{1-wh(t)}

pRa is characterized by d_(0,0), Z-seq, and A-seq

About functional equation w(t)=tΦ(w(t))

Φ in F_0[[t]] and F in F_0[[t]]일 때

[t^n]w(t) = 1/n * [t^(n-1)](Φ(t))^n, n>0

[t^n](w(t))^k = k/n * [t^(n-k)](Φ(t))^n, n>0

[t^n]F(w(t)) = 1/n * [t^(n-1)]F'(t)(Φ(t))^n = [t^n](F(t) * (Φ(t))^(n-1) * (Φ(t)-tΦ'(t)))

f(t), g(t) in L

[t^(-1)]f(t)g'(t) = - [t^(-1)]f'(t)g(t)

Φ in F_0[[t]] and F in F[[t]]일 때

(Diagonalization rule) {c_n}={[t^n]F(t)(Φ(t)^n)}의 generating function은 F(w)/(1-tΦ'(w)), w=tΦ(w)

F and Φ and φ in F_0[[t]]일 때

[t^(n-k)]F(t)(φ(t))^n(Φ(t))^k M(d(t),h(t)), where d(t) = F(w)φ(w)/(φ(w)-wφ'(w)), h(t)=wΦ(w), w=tφ(w)

About A-seq and B-seq and Z-seq of pRa, given pRa(=M(d(t),h(t))

About A-seq

An infinite lower triangular array D is pRa iff te A-seq of D(a_0 != 0)

(->방향은 M(d(t),h(t))에서 M(d(t),h(t))*M(A(t),M(t)) = M(d(t)h(t)/t, h(t))이용해서 M(t)=t보이고 좌우변 generic element비교)

(<-방향은 h(t)=tA(t) 이용해서 h(t) 구하고 d(t)는 D의 0-column읽어서 구함)

Given pRa, A-seq의 generating function바로 구하는 법, A(t)=t/bar(h(t)), A(y)=h(t)/t, y=h(t)

About B-seq

Given pRa, te B-seq(M(d(t),h(t))=M(d(t)h(t)/t, h(t))*M(A(t),t)^(-1)을 이용해서 구함, B(t)=(A(t))^(-1)임)

About Z-seq

Given pRa, te Z-seq

(z_0 = d_(1,0)/d_(0,0), d_(2,0)=z_0 * d_(1,0) + z_1 * d_(1,1)해서 z_1구하고 이런 방식으로 쭉 구함)

Given pRa, Z-seq의 generating function(implicit), d(t)=d_(0,0)/(1-tZ(h(t))), 따라서 Z(y)=(d(t)-d_(0,0))/(td(t)), y=h(t)

About renewal array

a pRa M(d(t),h(t)) is renewal array

iff d(t)=h(t)/t

iff A(y)=d_(0,0) + yZ(y) and d(0)=h(0) != 0

A:Appell subgroup

A:normal in R

Lag:Lagrange subgroup

L:subgroup in R

L giso (F_1[[t]], o)

3. 요약

P=M(1/(1-t), t/(1-t))의 경우 generic element가 계산 쉬움, binomial formula쓰면 됨

C=M(1/sqrt(1-4t), (1-sqrt(1-4t))/2)의 경우 generic element를 guess한 다음, [t^n]~~꼴로 써넣고 보니 n이 포함된 형태, 따라서 diagonalization rule사용하여 증명

4. 문제들