Santiago de Chile DEC 2006
SINGULAR COMBINATORICS
- A. Symbolic Methods
Philippe Flajolet, INRIA, Rocquencourt
http://algo.inria.fr/flajolet Based on Analytic Combinatorics, Flajolet & Sedgewick, C.U.P ., 2007+.
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A . Symbolic Methods Philippe Flajolet, INRIA, Rocquencourt - - PowerPoint PPT Presentation
Santiago de Chile DEC 2006 SINGULAR COMBINATORICS A . Symbolic Methods Philippe Flajolet, INRIA, Rocquencourt http://algo.inria.fr/flajolet ., 2007 + . Based on Analytic Combinatorics , Flajolet & Sedgewick, C.U.P 1 ANALYTIC COMBINATORICS
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∞
∞
n=0
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n−1
k=1
2(1−√1 − 4z) = 1 2− 1 2(1−4z)1/2.
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k≥1
k≥1
k≥1
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α z|α|.
γ = P α + P β. C(z) = A(z) + B(z)
γ = P α · P β. C(z) = A(z) · B(z)
α(1 + {α}), so that
α
n≥1
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z 1−z
1 n+1
n
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α∈A,β∈B
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n
k=0
k
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k≥0
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n
k
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∞
n=1
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n,k
n,k
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n!u(u + 1) · · · (u + n − 1); mean
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n≥0
h→0, h∈C
z=z0
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2z
4, +∞[.
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A f(z) dz does not depend on path. 6
B(z), with A, B analytic.
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−∞ dx 1+x4 =
π √ 2
|z|=1/2
|z|=2
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R, or
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1 1−2z ❀ Wn ⊲
1−z ❀ Dn n! ⊲
1 2
1 2z
3 and Un ⊲
1 1−f
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√n, etc.
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1 10(1 − z)−3 implies Dn ∼ n2/20.
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1−z
2: ρm ≈ 1 2(1 + 2−m−1).
2;error is exp. small.
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1 n!Dn = [zn] e−1 1−z + O(2−n) = e−1 +
n ∼ e−Hr,
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2(1 − √1 − 4z) ❀ c 4n √ πn3 .
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d dy y φ(y) = 0, i.e.,
τ φ(τ). All is computable.
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d dT (Te−T ) ≡ (1 − T)e−T = 0, that is, T = 1, z = e−1. Find:
z→e−1 1 −
en √ 2πn3 ; we rederive Stirling’s f. (since Tn = nn−1 by
2 U(z2)+···.
2U(z2) + · · · ).
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1 1−z(1+u) and smoothly
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n ∼ βnVar(B).
−∞
n
n
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nu−1 Γ(u) .
1 1−z/ρ.
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