# Download e-book for iPad: Algebraic Combinatorics by Ulrich Dempwolff

By Ulrich Dempwolff

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**Sample text**

By construction (Γ0 )0 = Γ and (∆0 )0 = ∆ and (1) follows. Set f0 = 1 and t0 = 0 and denote by f (X) and t(X) the EGF’s of the numbers fn and tn . ,Sk } hn n n≥0 n! X (2) with t|S1 | · · · t|Sk | . If [n] = S1 · · · Sk then t|S1 | · · · t|Sk | counts the number of rooted forests with components S1 , . . , Sk . We conclude hn = fn and (2) follows. On the other hand by (1) we obtain Xf (X) = n≥0 fn n+1 X = n! n≥0 tn X n+1 = t(X), (n + 1)! so that X exp(t(X)) = t(X). 16 tells us 1 1 nn−1 tn = [X n−1 ] exp(nX) = n!

As the polynomials lie dense in C[[X]] the identity holds in general. Definition Let R be a commutative ring with identity and R(X) the set of maps f : Z → R such that there exists a M = Mf ∈ Z with f (n) = 0 for n < M . Similar as in R[[X]] we define for f, g ∈ R(X) the sum by (f + ∞ g)(n) = f (n) + g(n) and the product by (f g)(n) = k=−∞ f (k)g(n − k) = n−N k=M f (k)g(n − k) if g(n) = 0 for n < N . Then R(X) is the ring of formal Laurent series which contains R[[X]] as a subring. The elements are denoted ∞ ∞ n n by f = f (X) = or by f (X) = if f (n) = 0 n=−∞ f (n)X n=M f (n)X for n < M .

Then H(X) = F (X)G(X). Proof. There exist precisely n k pairs (S, T ) with |S| = k. Thus: n n f (k)g(n − k). k h(n) = k=0 and n F (X)G(X) = n≥0 = n≥0 k=0 1 n! f (k) g(n − k) Xn k! (n − k)! n k=0 n f (k)g(n − k) X n = H(X) k follows. ✷ Remark Let f1 , . . ,Tk ) n≥0 fj (n) n n! X . f1 (|T1 |) · · · fk (|Tk |) where (T1 , . . , Tk ) ranges over the ordered partitions of size k of [n] and let n H(X) = n≥0 h(n) n! X be the associated EGF. Then F1 (X) · · · Fk (X) = H(X). 1 by an obvious induction.

### Algebraic Combinatorics by Ulrich Dempwolff

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