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By Ulrich Dempwolff

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Take the logarithm of both sides of the equation. 2. Differentiate both sides and multiply through by X. 3. Clear the equation of fractions. 51 4. For each n, find the coefficient of X n on both sides of the equation and equate them. We apply this method in the case of the Bell numbers. 4 For all n ≥ 1 n−1 B(n) = k=0 Proof. The formula B(x) = n≥0 after applying the logarithm to: log n≥0 n−1 B(k). k 1 n n! B(n)X = exp(exp(X) − 1) transforms 1 B(n)X n = exp(X) − 1 n! We differentiate and multiply by X to get n n n≥1 n!

If f is invertible D(f −1 ) = −D(f )/f 2 . If ord(g) > 0 then D(f ◦ g) = D(f ) ◦ g · D(g). (b) D(f ) = 0 iff f = cX 0 , c ∈ K. (c) D(f ) = f iff f = c exp X, c ∈ K. (d) Assume f < 1. Then D(exp f ) = exp(f )D(f ). (e) Assume f < 1. Then D(log(1 + f )) = (1 + f )−1 D(f ). (f ) Assume f < 1 and a ∈ R. Then D((1 + f )a ) = a(1 + f )a−1 D(f ). (g) Assume f < 1. Then exp(log(1 + f )) = 1 + f and log(exp f ) = f . (h) Assume f , g < 1. Then exp(f + g) = exp(f ) exp(g). Proof. (Sketch) (a) follows from the definition of D.

X , G(X) = n≥0 g(n) n! X . Define further h : N → C by h(0) = g(0) and h(n) = k≥1 π∈Sym(n),c(π)=k f (|S1 |) · · · f (|Sk |)g(k), 52 n ≥ 1, where {S1 , . . , Sk } are the supports of the cycles of π with c(π) = k. Let n H(X) = n≥0 h(n) n! X be the associated EGF. Then H(X) = G( n≥1 f (n) n X ). n Proof. The number of k–cycles of a k–set is (k − 1)!. So if Π = {S1 , . . , Sk } is a partition of [n] we have precisely (|S1 | − 1)! · · · (|Sk | − 1)! permutations whose supports of the cycles induce the partition Π.

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