Lindel¨of integral representations and asymptotic analysis...

Lindelof integral representations and asymptoticanalysis of a certain power series with

closed-form coefficients

Stefan Gerhold(joint work with P. Flajolet and B. Salvy)

January 20, 2009

Overview

I Contour integral representations of power series

∞∑n=1

φ(n)(−z)n

I Assumption: Taylor coefficients φ(n) are values of an analyticfunction φ(s)

I Analytic continuation

I Asymptotic analysis

I Focus on special case: φ(n) = ecnθ

I Applications: Finite differences, EXBERT distributions,non-holonomicity

Lindelof integrals

I Power seriesF (z) :=

∑n≥1

φ(n)(−z)n

I φ(s) analytic in a domain containing ]0,∞[

I Lindelof integral

Λ(z ; C) = − 1

2iπ

∫Cφ(s)zs π

sin(πs)ds

I contour C inside domain of analyticity of φ, andcontains 1, 2, . . .

Lindelof integrals

I Lindelof integral

Λ(z ; C) = − 1

2iπ

∫Cφ(s)zs π

sin(πs)ds

I Note: the residue of π/ sin πs at s = n equals (−1)n.

I Hence formally, by the residue theorem:∑n≥1

φ(n)(−z)n = Λ(z ; C).

Lindelof integrals: precise result

I Assume: φ(s) analytic for <(s) > 0

I Growth assumption: For some C > 0 and 0 ≤ A < π, we have

φ(s) < CeA|s| as s →∞ in <(s) ≥ 12 .

I Theorem. Then F (z) =∑

n≥1 φ(n)(−z)n is analytic in thesector −(π−A) < arg(z) < π−A, and has the representation

F (z) = − 1

2iπ

∫ 1/2+i∞

1/2−i∞φ(s)zs π

sin(πs)ds.

Proof idea

I Close the contour by a large half-circle on the right

I The contribution from the half-circle vanishes by the growthcondition

I Then use the residue theorem

1 2 3 4

-3i

-2i

-i

i

2i

3i

History and applications Lindelof integrals

I E. Lindelof, Le calcul des residus et ses applications a latheorie des fonctions, Gauthier- Villars, Paris, 1905.

I Analytic continuation of∑n≥1

nαzn,∑n≥1

(log n)zn

I Asymptotic analysis of these and similar functions(Flajolet 1999)

I Euler sums∑

Hn/nα (Flajolet, Salvy 1998)

I Also related to work by Barnes, Mellin, Ramanujan, Ford . . .

Our main example

I Define, for real c 6= 0 and 0 6= θ < 1, the function

E (z ; c , θ) :=∑n≥1

exp(cnθ)(−z)n, |z | < 1

I Lindelof integral

E (z) = − 1

2iπ

∫ 1/2+i∞

1/2−i∞exp(csθ)zs π

sin(πs)ds

I Analytic continuation to C \ ]−∞,−1]

I Singularities at −1 and ∞I Today’s topic: asymptotics of E (z ; c , θ)

Our main example

I Motivation:∑

n≥1 φ(n)(−z)n is well studied for

φ(n) = nα and φ(n) = (log n)m.

I The coefficient sequence

φ(n) = ecnθ

is a natural next step.

I Small application: asymptotics of finite differences, bysingularity analysis

Asymptotic results

I Asymptotics of

E (z) =∑n≥1

exp(cnθ)(−z)n

for c = ±1 and θ = −1, 12 (representative cases)

z →∞ z → −1

e1/n − e2√

log z

2√

π(log z)1/4

1

1 + z

e−1/n − 1√π(log z)−1/4 cos

(2√

log z − 14π

) 1

1 + z

e√

n −1− 1√π log z

+ . . .

√πe−1/8

(1 + z)3/2e

14(1+z)

e−√

n −1 +1√

π log z+ . . . E (1) + E ′(1)(1 + z) + . . .

Methods overview

I Asymptotics of

E (z) =∑n≥1

exp(cnθ)(−z)n

z →∞ z → −1

c > 0, θ < 0 saddle point series rearrangement

c < 0, θ < 0 2 saddle points series rearrangement

c > 0, 0 < θ < 1 Hankel contour Laplace

c < 0, 0 < θ < 1 Hankel contour Abel’s theorem

Saddle point method

I Parameters c > 0, θ < 0

I Guiding example:

E (z ; c = 1, θ = −1) =∑n≥1

e1/n(−z)n

I Lindelof integral

E (z) = − 1

2iπ

∫ 1/2+i∞

1/2−i∞e1/szs π

sin πsds

I Goal: asymptotics for z →∞

Saddle point method

I The function s 7→ e1/szs has an approximate saddle point at

s =1√log z

I New integration contour

s =1√log z

+ it, t ∈ R

I Main contribution for

|t| < (log z)−α, 23 < α < 3

4

Saddle point method

I The condition α > 23 ensures a uniform expansion

e1/szs

sin πs= 1

πL1/2 exp(2L1/2 − L3/2t2) · (1 + o(1)),

s = L−1/2 + it, |t| < L−α,

where L = log z .

I Thence the asymptotics of the central part of the integral:

−L1/2e2L1/2

2π

∫ L−α

−L−α

e−L3/2t2dt ∼ − e2L1/2

2√

πL1/4

Saddle point method

I The condition α < 34 allows to discard the tails

|t| > (log z)−α.

I Hence our result:∑n≥1

e1/n(−z)n ∼ − e2√

log z

2√

π(log z)1/4, z →∞

(in the sense of analytic continuation of the left side.)

I Holds in any sector with vertex at zero that does not containthe negative real axis.

I Straightforward generalization to arbitrary parameters c > 0,θ < 0.

Methods overview

I Asymptotics of

E (z) =∑n≥1

exp(cnθ)(−z)n

z →∞ z → −1





Two saddle points

I Parameters c < 0, θ < 0

I Guiding example:

E (z ; c = 1, θ = −1) =∑n≥1

e−1/n(−z)n

I Lindelof integral

E (z) = − 1

2iπ

∫ 1/2+i∞

1/2−i∞e−1/szs π

sin πsds

I Goal: asymptotics for z →∞

Two saddle points

I The function s 7→ e−1/szs has two approximate saddle pointsat

s = ± i√log z

I The new integration contour, passing through the two saddlepoints

-0.05

0

0.05

0.1Re-0.4

-0.2

0

0.2

0.4

Im0

1

2

3

g

-0.05

0

0.05

0.1Re-0.1 0 0.1 0.2

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Two saddle points

I Again, we take the central region |t| < (log z)−α with23 < α < 3

4 .

I The dominating factor of the integrand becomes

e±2i√

log z ,

producing oscillating behavior

I At each saddle point, we proceed as before

I Add the dominating parts of both saddle points

I Tail estimates: very tedious!

Two saddle points

I Result:∑n≥1

e−1/n(−z)n = − 1√π(log z)−1/4 cos

(2√

log z − 14π

)+ O((log z)−1/2+ε), z →∞.

I Again, this holds in any sector with vertex at zero that doesnot contain the negative real axis.

I Generalization to arbitrary parameters c < 0, θ < 0.

I An exponential factor exp(C (log z)θ

θ−1 ) appears for θ 6= −1.

Two saddle points

I For θ < −1, there are more than two saddle points

I Have to consider only the rightmost two

I Example: The surface |zs exp(−sθ)/ sin πs| for θ = −5 andz = 200 (six saddlepoints).

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

050

100150

200

Methods overview

I Asymptotics of

E (z) =∑n≥1

exp(cnθ)(−z)n

z →∞ z → −1





Hankel contour

I Parameters c , 0 < θ < 1

I Guiding example:

E (z ; c = 1, θ = 12) =

∑n≥1

e√

n(−z)n

I Lindelof integral

E (z ; 1, 12) = − 1

2iπ

∫ 1/2+i∞

1/2−i∞e√

szs π

sin πsds

I Near s = 0, we have

e√

s π

sin πs= s−1 + s−1/2 + O(1).

Hankel contour

I New integration contour (with L = log z and 0 < β < 1):

-L- Β

L-1

0

I The contributions of the two vertical lines are negligible.

I The U-shaped part is transformed into a Hankel contour H.

H

0

1

Hankel contour

I We want to evaluate the contour integrals∫zs

sλ, λ = 1, 1

2 .

I Substitutes log z = −w

I Use Hankel’s formula

1

Γ(λ)= − 1

2iπ

∫H

e−w

(−w)λdw .

I Hence ∫zs

sλ∼ 2iπ

Γ(λ)(log z)λ−1.

Hankel contour

I Result:∑n≥1

e√

n(−z)n = −1− 1√π log z

+ O

(1

log z

), z →∞.

I Again, this holds in any sector with vertex at zero that doesnot contain the negative real axis.

I Straightforward generalization to arbitrary parameters c ,0 < θ < 1.

Methods overview

I Asymptotics of

E (z) =∑n≥1

exp(cnθ)(−z)n

z →∞ z → −1





Behavior at the singularity z = −1

I Case 1: θ < 0. We can rewrite

E (z ; c , θ) =∑n≥1

∑k≥0

ck

k!nkθ(−z)n =

∑k≥0

ck

k!Li−kθ(−z), |z | < 1,

as a sum of polylogarithms

Liα(z) =∑n≥1

zn

nα.

I Asymptotics of Liα(z) are well known.

I For instance,∑n≥1

e1/n(−z)n =∑k≥0

1

k!Lik(−z) =

1

1 + z+ log

1

1 + z+ O(1).


I Case 2: c > 0, 0 < θ < 1.

I Guiding example:

E (z ; c = 1, θ = 12) =

∑n≥1

e√

n(−z)n

I Laplace method

I Summands have a peak near n = 14(log z)−2

I Second order expansion around this peak

I Evaluate resulting sum asymptotically

I Discard the tails


I Result of the Laplace method:

∑n≥1

e√

n(−z)n ∼√

πe−1/8

(1 + z)3/2exp(

1

4(1 + z)), z → −1.

I Straightforward generalization to arbitrary parameters c > 0,0 < θ < 1.


I Case 3: c < 0, 0 < θ < 1.

I Series defining E (z) and its derivatives converge at z = −1.

uk :=1

k!lim

z→−1+

dk

dzkE (z) = (−1)k

∑n≥1

(n

k

)exp(cnθ).

I Formal Taylor series at z = −1 is an asymptotic series for thefunction (extension of Abel’s theorem, see Ford 1916).

E (z ; c , θ) ∼z→−1+

u0 + u1(1 + z) + u2(1 + z)2 + . . .

I This formal series does not converge in any neighborhood ofz = −1.

Application: ecnθ

is not a holonomic sequence

I A function f (z) is holonomic if it satisfies an LODE

p0(z)f (0)(z) + · · ·+ pd(z)f (d)(z) = 0

with polynomial coefficients.

I A sequence (an) is holonomic if it satisfies an LORE

p0(n)an + · · ·+ pd(n)an+d = 0

with polynomial coefficients.

I Our results + classical LODE asymptotics imply: ecnθis not

holonomic (except for trivial cases).

I “Strong transcendence result” (algebraic power series areholonomic.)

Application: Finite differences

I Given a sequence fn, define

gn := Dn[f ] :=n∑

k=0

(n

k

)(−1)k fk

I Generating function:

g(z) =1

1− z

(f0 + F

(z

1− z

)),

whereF (z) =

∑n≥1

fn(−z)n.

Application: Finite differences

I Asymptotics of F (z) ⇒ asymptotics of g(z)

I If F (z) has “tame” asymptotics, we get the asymptoticsof gn = Dn[f ] by singularity analysis

I Example: For fn = e±√

n we have

g(z) ∼ −±1√π

1

1− z

(log

1

1− z

)−1/2

,

hence by singularity analysis

gn =n∑

k=0

(n

k

)(−1)ke±

√k ∼ − ±1√

π log n.

Application: Asymptotics of a certain EXBERT distribution

I EXBERT distributions (EXchangeable BERnoulli trials,Madsen 1993)

P[X = x ] =

(n

x

) n−x∑j=0

(n − x

j

)(−1)jπx+j , x ∈ {0, . . . , n},

where πk , 0 ≤ k ≤ n, is a probability distribution with alldifferences non-negative.

I “power model”: πk = exp((log p)ka) with 0 < a < 1 and0 < p < 1.

I Asymptotics for large n:

P[X = x ] ∼n→∞

{ − log pΓ(1−a)(log n)a x = 0

−a log pxΓ(1−a)(log n)a+1 x ≥ 1

.

Application: Finite differences (continued)

I Slowly varying functions and oscillating functions can besubjected to singularity analysis

I By our saddle point results, we find the amusing formulas

n∑k=1

(n

k

)(−1)ke1/k ∼ − e2

√log n

2√

π(log n)1/4,

n∑k=1

(n

k

)(−1)ke−1/k = −

cos(2√

log n − 14π

)√

π(log n)1/4+ o

((log n)−1/4

).

Heuristic partition asymptotics (Hardy)

I The function E (z) occurs in a heuristic argument to identify

the constant π√

2/3 in p(n) ≈ exp(π√

2n3 ).

I Generating function of the partition sequence

P(z) =∑n≥1

p(n)zn =∏k≥1

(1− zk)−1

I By an easy estimate

P(z) ≈ expπ2

6(1− z)=:

∑anz

n.

I What is the growth order of an?

I Polynomial growth ⇒∑

anzn too small

I Exponential growth ⇒∑

anzn too large


I Guess: an ≈ eBnbfor some constants B > 0 and 0 < b < 1.

I The growth order of∑eBnb

zn, z → 1,

is given by its maximum term

exp(C (1− z)−b/(1−b)), where C = B1/(1−b)bb/(1−b)(1− b).

I Matching this to P(z) ≈ exp π2

6(1−z) yields

p(n) ≈ exp(π

√2n

3).


I Relation to our work:

I Hardy’s argument needs the rough growth order of∑eBnb

zn, z → 1,

I We determine the asymptotics of∑eBnb

zn, z → 1,

exactly by Laplace’s method.

Conclusion

I Lindelof integrals provide analytic continuation of power seriesand allow to do asymptotics

I Asymptotics of E (z ; c , θ) depends strongly on the parameterregion (six different expansions)

I Extensions to related functions like∑n≥1

exp(cnθ(log n)α)(−z)n

should be relatively easy

Lindel¨of integral representations and asymptotic analysis...

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