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Estimating proportions of elements in finite symmetric and classical - - PowerPoint PPT Presentation

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Estimating proportions of elements in finite symmetric and classical groups Alice Niemeyer UWA, RWTH Aachen Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 1 / 64 Motivation Proportions of Elements Theorem Let G

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Estimating proportions of elements

in finite symmetric and classical groups Alice Niemeyer

UWA, RWTH Aachen

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 1 / 64

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SLIDE 2

Motivation

Proportions of Elements

Theorem Let G be a group in a family of groups. Then there exists some function c(N) of the size N of the input of G such that the proportion of elements in G with a particular property is at least c(N). Such a theorem is often hard to prove.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 2 / 64

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Motivation

Efficiency of algorithms

Question Will any lower bound do? Answer The lower bound affects two things: Number of searches until success on correct input. Number of searches until we “give up” on incorrect input.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 3 / 64

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Motivation

Efficiency of algorithms

Question Will any lower bound do? Answer The lower bound affects two things: Number of searches until success on correct input. Number of searches until we “give up” on incorrect input.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 3 / 64

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Introduction

Motivation

Proportions of elements in Sn: algorithmic applications theoretical interest applications to proportions in matrix groups

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 4 / 64

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Introduction

Notation

We write permutations in disjoint cycle notation. The number of cycles always refers to such a decomposition. n, m, k denote positive integers.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 5 / 64

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Previous Work

Euler (1707-1783)

Euler: Quaestio curiosa ex doctrina combinationis How many of the n! orderings of the numbers 1, . . . , n are such that no number remains in its natural place? How many derangements are there in Sn?

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 6 / 64

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Previous Work

Previous Results: k cycles

Let g(n, k) denote the proportion of elements in Sn with exactly k cycles. Sylvester (1861) g(n, k) = 1 n!

  • S⊆{1,...,n−1}

|S|=n−k

  • s∈S

s. n!g(n, k) is the Stirling number of the first kind.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 7 / 64

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Previous Work

Previous Results: k cycles

Let godd(n) denote the proportion of elements in Sn all of whose lengths are odd. Sylvester (1861) godd(n) = 1 n! ·

  

(1 · 3 · 5 · · · (n − 2))2 n n odd (1 · 3 · 5 · · · (n − 1))2 n even

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 8 / 64

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Previous Work

Previous Results: order of elements

Landau 1909 lim

n→∞

log (maxg∈Sn(o(g))) √n log(n) = 1.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 9 / 64

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Previous Work

Previous Results: order of elements

Erdös and Turán wrote a series

  • f papers on statistical group

theory. Erdös and Turán (1965) For ε, δ > 0 and n ≥ N0(ε, δ) |{g ∈ Sn | e(1/2−ε) log2(n) ≤ o(g) ≤ e(1/2+ε) log2(n)} n! ≥ 1 − δ.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 10 / 64

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Generating Functions

Generating Functions

Given a sequence of numbers, (an)n∈N, e.g., an the number of certain elements in Sn. Quote from Wilf’s Book A generating function is a clothesline on which we hang up a sequence of numbers for display. Suggested reading: Wilf’s book Generatingfunctionology [3] or Analytic Combinatorics by Flajolet and Sedgewick [4].

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 11 / 64

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Generating Functions

Ordinary Generating Functions

The Ordinary Generating Function for an is A(z) :=

  • n≥0

anzn. We denote the coefficient of zn by [zn]A(z).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 12 / 64

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Generating Functions

Exponential Generating Functions (egf)

We define the Exponential Generating Function for an is A(z) :=

  • n≥0

an n!zn. When do we use egf? When the coefficients grow very fast. E.g. in Sn the number of permutations is n! and we can hope that a proportion an/n! is manageably small.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 13 / 64

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Generating Functions

Generating Functions

We study generating functions as formal power series in the ring

  • f formal power series.

Analytic questions, convergence etc. do not concern us just yet.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 14 / 64

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Generating Functions

Multiplication of Generating Functions

∞

  • n=0

anzn

  • ·

∞

  • n=0

bnzn

  • =

∞

  • n=0

n

  • k=0

akbn−k

  • zn.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 15 / 64

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Generating Functions

Example

Let b ≥ 1 be fixed integer and let an denote the number of permutations in Sn all of whose cycles have length at most b. List permutations by cycles of length d containing the point 1.

n−1

d−1

  • points for cycle of length d on 1

(d − 1)! different cycles on these an−d permutations on the remaining n − d points Then an = n! for n ≤ b and an n! = 1 n

min{b,n}

  • d=1

an−d (n − d)!.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 16 / 64

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Generating Functions

Example

Hence we get A(z) :=

∞

  • n=0

an n!zn = 1 +

∞

  • n=1

1 n

 

min{b,n}

  • d=1

an−d (n − d)!

  zn

= 1 +

b

  • d=1

∞

  • n=d

1 n an−d (n − d)!zn = 1 +

b

  • d=1

∞

  • n=0

1 n + d an n!zn+d

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 17 / 64

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Generating Functions

Example

Hence A′(z) =

b

  • d=1

∞

  • n=0

an n!zn+d−1 =

b

  • d=1

zd−1

∞

  • n=0

an n!zn =

b

  • d=1

zd−1A(z) Thus A′(z) A(z) =

b

  • d=1

zd−1

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 18 / 64

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Generating Functions

Example

A′(z) A(z) =

b

  • d=1

zd−1 and so log(A(z)) =

b

  • d=1

zd d . Therefore Generating Function A(z) = exp(

b

  • d=1

zd d ).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 19 / 64

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Generating Functions

Summary

Proportions of elements important for algorithms Generating functions useful description Generating functions can often be found

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 20 / 64

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Generating Functions

Coefficients

Question Can generating functions tell us about the limiting behaviour of the coefficients?

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 21 / 64

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Generating Functions

Saddlepoint Analysis

Based on results of W. Hayman. Theorem (See [4]) Let P(z) = n

j=1 ajzj have non-negative coefficients and

suppose gcd({j | aj = 0}) = 1. Let F(z) = exp(P(z)). Then [zn]F(z) ∼ 1 √ 2πλ exp(P(r)) r n , where r is defined as rP′(r) = n and λ =

  • r r

dr

2 P(r).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 22 / 64

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Generating Functions

Example Saddlepoint Analysis

Recall that A(z) = exp(

b

d=1 zd d ) is the generating function for the

number of elements all of whose cycles have length at most b. Let P(z) =

b

d=1 zd d . Then gcd({d | 1 d = 0}) = 1.

Find r n = rP′(r) = r b

d=1 r d−1 = b d=1 r d ≥ r b.

Find λ λ =

  • r r

dr

2 P(r) = r b

d=1 dr d−1 = b d=1 dr d ≤ b m d=1 r d = bn.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 23 / 64

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Generating Functions

Example Saddlepoint Analysis

Recall that A(z) = exp(

b

d=1 zd d ) is the generating function for the

number of elements all of whose cycles have length at most b. Let P(z) = b

d=1 zd d . Then gcd({d | 1 d = 0}) = 1.

Use r ≤ n1/b and λ ≥ bn and P(r) =

b

d=1 r d d ≥ 1 b

b

d=1 r d = n b

Hence [zn]A(z) ∼ 1 √ 2πλ exp(P(r)) r n ≥ 1 √ 2πbn

e

n

n/b

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 24 / 64

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Generating Functions

However ...

This can be difficult when cycle lengths grow with n: For m fixed let c(n, m) = 1 n!|{g ∈ Sn | gm = 1}|. Let Cm(z) =

∞

  • n=0

c(n, m)zn be the corresponding generating function.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 25 / 64

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Generating Functions

Previous Results: c(n, m)

Chowla, Herstein, and Scott (1952) Cm(z) = exp

 

1≤d|m

zd d

  .

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 26 / 64

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Generating Functions

Previous Results: c(n, m)

Warlimont (1978) 1 n + 2c n2 ≤ c(n, n) ≤ 1 n + 2c n2 + O

  • 1

n3−o(1)

  • ,

where c =

  

n odd 1 n even .

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 27 / 64

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Generating Functions

Previous Results

Warlimont result tells us most g with gn = 1 are n-cycles second most g with gn = 1 have 2 cycles of length n/2, when n even

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 28 / 64

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Algorithmic Application

Algorithmic Application

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 29 / 64

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Algorithmic Application

A Presentation for Sn

Coxeter and Moser (1957) {r, s | r n = s2 = (rs)(n−1) = [s, r j]2 = 1 for 2 ≤ j ≤ n/2}. If r, s ∈ G with r 2 = 1 satisfy this presentation then r, s is isomorphic to Sn.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 30 / 64

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Algorithmic Application

Definition The transposition y matches the n-cycle x, if y moves two adjacent points in x. Lemma For n ≥ 5, an n-cycle and a matching transposition satisfy the presentation.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 31 / 64

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Algorithmic Application

A 1-sided Monte-Carlo algorithm

Recognise Sn as Black Box Groups (Beals et al. (see Seress’ book)): Input: G = X a black box group, n ≥ 5 Output: true and λ : G → Sn isomorphism false and most likely G ∼ = Sn Choose random elements in G to

1

find g ∈ G with gn = 1. λ(g) is n-cycle?

2

find a ∈ G with a2m = 1 where m ∈ {n − 2, n − 3} odd. λ(am) transposition?

3

find random conjugate h of am with [h, hg] = 1. λ(h) interchanges 2 points of λ(g)? Then define λ by λ(g) = (1, . . . , n) and λ(h) = (1, 2). We test, whether r, s ∼ = Sn via the presentation.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 32 / 64

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Algorithmic Application

Goal

Theorem Given Black Box Group G isomorphic to Sn, the probability that BBRECOGNISESN(G, n, ε) returns false is at most ε. Theorem The cost of the algorithm is O((nξ + n log(n)µ) log(ε−1)), where ξ is the cost of finding a random element in a Black Box Group and µ the cost of a Black Box Operation.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 33 / 64

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Algorithmic Application

Questions

1

What is the conditional probability that g ∈ Sn is an n-cycle, given that gn = 1?

2

What is the conditional probability that hm ∈ Sn is transposition, given that h2m = 1 (for m ∈ {n − 2, n − 3}

  • dd)?

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 34 / 64

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Estimates for Sn recognition

Our first goal

Work out the probability that g is an n-cycle given that gn = 1.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 35 / 64

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Estimates for Sn recognition

Conditional Probability

Let A ⊆ B then P(A | B) = P(A ∩ B) P(B) = P(A) P(B). P(A) proportion of n-cycles in Sn, namely 1/n. P(A | B) =

1 n

c(n, n). We need a lower bound for this probability.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 36 / 64

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Estimates for Sn recognition

Our goal

  • btain upper and lower bound for c(n, m)

correct first order term and hold for m very close to n practical bounds also for small n work out the conditional probability that g has a 2-cycle given that g2m = 1.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 37 / 64

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Estimates for Sn recognition

The Münchausen Method (Bootstrapping)

Produce a crude first estimate Insert the estimate into the next to produce better one

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 38 / 64

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Estimates for Sn recognition

The Münchausen Method (Bootstrapping)

Drawing by Theodor Hosemann (1807 - 1875)

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 39 / 64

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Estimates for Sn recognition

The Münchausen Method = Bootstrapping

First estimates in Beals et al. With Münchausen Method by CEP and N 2006.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 40 / 64

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Estimates for Sn recognition

The crude estimate

Define γ(m) :=

      

2 360 < n 2.5 60 < m ≤ 360 3.345 m ≤ 60 . Theorem 1 Let m, n ∈ N with m ≥ n − 1. Then c(n, m) ≤ 1 n + γ(m)m n2 .

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 41 / 64

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Estimates for Sn recognition

Proof-idea for crude estimate

Divide the problem into smaller ones by considering proportions in Sn (see Beals et al.)

1

c1(n, m) those g which have 1, 2, 3 in same g-cycle

2

c2(n, m) those g which have 1, 2, 3 in 2 g-cycles

3

c3(n, m) those g which have 1, 2, 3 in 3 g-cycles Then c(n, m) = c1(n, m) + c2(n, m) + c3(n, m).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 42 / 64

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Estimates for Sn recognition

The pull

Enumerating g by g-cycle of length d on 1 yields: c(n, m) = 1 n

  • d|m

1≤d≤n

c(n − d, m) = 1 m + 1 n

  • d|m

1≤d≤m/2

c(n − d, m)

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 43 / 64

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Estimates for Sn recognition

The pull

Using the crude estimate: c(n, m) ≤ 1 m + 1 n

  • d|m

1≤d≤n

  • 1

n − d + γ(m)m (n − d)2

  • ≤

1 m + d(m)(2 + 4γ(m)) n2

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 44 / 64

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Estimates for Sn recognition

Corollary

The conditional probability that a random element g has an n-cycle given that it satisfies gn = 1 is at least 2/7. The conditional probability that a random element h has an m-cycle (m ∈ {n − 2, n − 3} and odd) given that it satisfies h2m = 1 is at least 1/4.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 45 / 64

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Proportions in classical groups

Proportion of elements in finite classical groups

Estimating Proportions Sommerschule 2011 46 / 64

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Proportions in classical groups

First Ideas

Lehrer (1992) tori and characters of Weyl group Theorem (Isaacs, Kantor & Spaltenstein, 1995) G finite simple group of Lie type, r a prime (not characteristic) dividing |G| with r > 3 h Coxeter number of corresponding simply connected simple algebraic group The proportion of r-singular elements in G is at least (1 − 1

r ) 1 h.

The connection between tori and F-conjugacy classes of Weyl group elements.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 47 / 64

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Proportions in classical groups

Aim:

Present a generalisation of their idea First used in collaboration with Lübeck to find elements that power up to special involutions [1] General the theory described in [2]

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 48 / 64

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Quokka Sets

Our Groups

connected reductive algebraic group G defined over the algebraic closure Fq of Fq of characteristic q0. F a Frobenius morphism of G GF = {g ∈ G | F(g) = g}, finite group of Lie type, e.g. GF = GL(n, q).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 49 / 64

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Quokka Sets

Our Groups

connected reductive algebraic group G defined over the algebraic closure Fq of Fq of characteristic q0. F a Frobenius morphism of G GF = {g ∈ G | F(g) = g}, finite group of Lie type, e.g. GF = GL(n, q).

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 49 / 64

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Quokka Sets

The Main Theorem

Quokka Theorem Let Q ⊆ GF be a quokka set. Then |Q| |GF| =

  • C∈CQ

|C| |W| · |T F

C ∩ Q|

|T F

C |

.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 50 / 64

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Quokka Sets

Jordan Decomposition

Every g ∈ GF has unique Jordan decomposition g = su = us, where

  • (s) co-prime to q0
  • (u) power of q0

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 51 / 64

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Quokka Sets

Quokka Sets

A quokka-set Q is a non-empty sub- set of GF such that a) If g ∈ GF has Jordan decomposition g = su then g ∈ Q ⇔ s ∈ Q. b) Q is a union of GF-conjugacy classes.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 52 / 64

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Quokka Sets

Tori

A torus is an algebraic group isomorphic to T ∼ = F

∗ q × · · · × F ∗ q.

In particular, T is abelian. Example: G = GL(n, ¯ Fq). T0 = diagonal matrices in G.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 53 / 64

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Quokka Sets

The Main Theorem

Quokka Theorem Let Q ⊆ GF be a quokka set. Then |Q| |GF| =

  • C∈CQ

|C| |W| · |T F

C ∩ Q|

|T F

C |

.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 54 / 64

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Quokka Sets

Maximal tori in GL(n, q)

GL(n, q) α = (a1, . . . , at) a partition of n. T F ∼ = Zqa1−1 × · · · × Zqat −1 W = Sn T F corresponds to the conjugacy class of Sn of permutations with cycle lengths a1, . . . , at.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 55 / 64

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Quokka Sets

The Main Theorem

Tc1

  • Q−

Torus

  • Q

T class Q−

Q

C C 1 C 2 in W −conj. class G

F

|Q| |GF| =

  • C∈CQ

|C| |W| · |T F

C ∩ Q|

|T F

C |

.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 56 / 64

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Quokka Sets

Lower Bounds for |Q|/|GF| Restrict to some C ∈ CQ Find a (uniform) lower bound for mC =

|T F

C ∩Q|

|T F

C |

those C. Problem in abelian groups. Find lower bounds for

C |C| |W| for those C.

For classical groups: Problem in Sn or related groups.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 57 / 64

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Quokka Sets

Example: GF = GL(n, q)

Let r be a prime not dividing q but dividing |GF|. Clearly the set Q of r-singular elements in GF is a quokka set: if g ∈ Q with g = su, then, since o(u) power of q0, we know r | o(g) if and only if r | o(s). g ∈ Q if and only if gh ∈ Q for any h ∈ GF.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 58 / 64

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Quokka Sets

Example: GF = GL(n, q)

We see r | |T F| if and only if r|qai − 1 for some ai. Fact from Number Theory: Let m denote the least positive integer with r | qm − 1. Then r|qai − 1 if and only if m | ai.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 59 / 64

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SLIDE 62

Quokka Sets

Example: GF = GL(n, q)

Formula |Q| |GF| =

  • C∈CQ

|C| |Sn| · |T F

C ∩ Q|

|T F

C |

. for GF = GL(n, q) W = Sn α = (a1, . . . , at) a partition of n. T F ∼ = Zqa1−1 × · · · × Zqat −1 T F corresponds to the conjugacy class C of Sn of permutations with cycle lengths a1, . . . , at.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 60 / 64

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SLIDE 63

Quokka Sets

Example: GF = GL(n, q)

If T F

C ∩ Q = ∅ then the proportion |T F

C ∩Q|

|T F

C |

≥ (1 − 1

r ).

Let c(n, m) denote the proportion of elements in Sn with a cycle

  • f length divisible by m. Hence

|Q| |GF| =

  • C∈CQ

|C| |Sn| · |T F

C ∩ Q|

|T F

C |

≥ (1 − 1 r )

  • C∈CQ

|C| |Sn| ≥ (1 − 1 r )c(n, m) Hence we reduced the problem to a problem of estimating a proportion in Sn.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 61 / 64

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SLIDE 64

Quokka Sets

Example: GF = GL(n, q)

Let c(n, m) denote the proportion of elements in Sn with a cycle

  • f length divisible by m. It can be seen that c(n, m) ≥ 1

m be an

inclusion-exclusion argument. Hence |Q| |GF| =

  • C∈CQ

|C| |Sn| · |T F

C ∩ Q|

|T F

C |

≥ (1 − 1 r )

  • C∈CQ

|C| |Sn| ≥ (1 − 1 r ) 1 m.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 62 / 64

slide-65
SLIDE 65

Anhang For Further Reading

For Further Reading I

Frank Lübeck, Alice C. Niemeyer, Cheryl E. Praeger. Finding involutions in finite Lie type groups of odd characteristic.

  • J. Algebra 321 (2009), no. 11, 3397–3417.

Alice C. Niemeyer, Cheryl E. Praeger. Estimating proportions of elements in finite groups of Lie type.

  • J. Algebra 324 (2010), no. 1, 122–145.

Herbert S. Wilf, generatingfunctionality (2nd edition), Academic Press, Inc., Boston, MA 1994.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 63 / 64

slide-66
SLIDE 66

Anhang For Further Reading

For Further Reading II

Philippe Flajolet and Robert Sedgewick, Analytic combinatorics, Cambridge University Press, 2009, also available online.

Alice Niemeyer (UWA, RWTH Aachen) Estimating Proportions Sommerschule 2011 64 / 64