An Integral Formula for Taylor Coefficients of a Class of Analytic Functions

An Integral Formula for Taylor Coefficients of a Class of Analytic Functions

Journal of Mathematical Analysis and Applications 233, 266᎐275 Ž1999. Article ID jmaa.1999.6293, available online at http:rrwww.idealibrary.com on An...

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Journal of Mathematical Analysis and Applications 233, 266᎐275 Ž1999. Article ID jmaa.1999.6293, available online at http:rrwww.idealibrary.com on

An Integral Formula for Taylor Coefficients of a Class of Analytic Functions Titus Hilberdink Department of Mathematical Sciences, Goldsmiths College, New Cross, London SE14 6NW, England E-mail: [email protected] Submitted by Richard A. Duke Received September 8, 1998

In this paper, we prove that for a function f Ž z . analytic in ⺓ R  x g ⺢: x G 14 such that Ž1 y z . f Ž z . ª 0 as z ª 1 and for large < z <, < f Ž z .< s oŽ< z < k . for some k, the Taylor coefficients a n have the integral representation

an s

1 2␲ i



H1

⌬ f Žt. t nq1

dt

Žn G k.,

where ⌬ f Ž t . s lim ␧ ª 0 qŽ f Ž t q i ␧ . y f Ž t y i ␧ ... This formula is shown to have various applications particularly in determining the behaviour of a n as n ª ⬁. 䊚 1999 Academic Press

1. INTRODUCTION Many problems in probability and combinatorics involve generating functions from which we like to know as much as possible about the coefficients that are being studied. The standard way of deducing information about the coefficients is to find out where the generating function say f Ž z ., has its singularities. Then, from the behaviour of f near the singularities one can infer the behaviour of the coefficients a n Žsee, for example w3x, w6x, w8x, w9x.. If f Ž z . has only poles, then we have the formula

an s

1

f Ž z.

H 2␲ i ␥ z

nq1

dz s y Ý residues of

ž

266 0022-247Xr99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.

f Ž z. z nq1

/

at the poles of f ,

267

FORMULA FOR TAYLOR COEFFICIENTS

Žsubject to some growth condition on f Ž z . as < z < ª ⬁.. In many cases however, the functions involved do not have nicely behaved singularities and we cannot expect a simple formula as above. In the case where f can be analytically extended to the cut-plane ⺓ R  x: x G 14 , we show that we still have some formula for a n but the sum becomes an integral. This integral representation can then be used for various purposes including determining the asymptotic behaviour of the a n . In Section 3, we give some examples from probability and combinatorics where this formula can be applied. Notation. Throughout this paper we use the following conventions Žas x ª ⬁.: f Ž x . ; g Ž x . m lim f Ž x . rg Ž x . s 1, xª⬁

f Ž x . s O Ž g Ž x . . m f Ž x . F Ag Ž x . for some A, and all large x, f Ž x . s o Ž g Ž x . . m lim f Ž x . rg Ž x . s 0. xª⬁

Also, for f defined on ⺓ R  x: x G 14 , we write < f Ž z .< s oŽ< z < k . as z ª ⬁ to mean ᭙␧ ) 0, ᭚ z 0 such that ᭙ < z < ) z 0

Ž z not real and G 1 .

f Ž z . - ␧ < z < k.

We shall also need the notion of regular variation: A positive function f defined on some neighbourhood w A, ⬁. of infinity is regularly ¨ arying of index ␴ if it is measurable and for all ␭ ) 0, lim

xª⬁

f Ž ␭x. f Ž x.

s ␭␴ .

Let f be analytic in the region ⺓ R  x g ⺢: x G 14 . Thus f Ž z . s Ý⬁ns 0 a n z n which is valid for < z < - 1. We will assume that for x ) 1, the limit def

⌬ f Ž x . s limq Ž f Ž x q i ␧ . y f Ž x y i ␧ . . ␧ª0

exists. Also define for x ) 1, fq Ž x . and fy Ž x . to be the limits lim f Ž x q i ␧ . s fq Ž x .

␧ª0 q

lim f Ž x y i ␧ . s fy Ž x .

␧ ª0 q

whenever they exist. Thus, if these exist, we have ⌬ f Ž x . s fq Ž x . y fy Ž x .. Note that we do not assume their existence. We say f is bounded along Ž1, ⬁. if < f Ž z .< - A R, ␦ Žsome A independent of ␧ . for every z in the set  x q iy: 1 q ␦ F x F R, 0 - < y < F ␧ 4 , for any R, ␦ , ␧ ) 0.

268

TITUS HILBERDINK

2. RESULTS Our main result is the following: THEOREM 1. Let f be analytic in ⺓ R  x: x G 14 bounded along Ž1, ⬁., such that < f Ž z .< s oŽ< z < k . Ž for some k . and that Ž1 y z . f Ž z . ª 0 as z ª 1. Suppose also that for x ) 1, ⌬ f Ž x . exists. Then for n G k, we ha¨ e an s

1

⌬ f Ž t.



H 2␲ i 1

dt.

t nq1

Proof. By Cauchy’s Theorem, a n s Ž1r2␲ i .H␥ Ž f Ž z .rz nq 1 . dz, where ␥ can be taken to be the following positively-orientated contour: Choose R ) ␦ ) ␧ ) 0. Let ␥ s ⌫R j ⌫␧q j ⌫␧y j ⌫␦ with ⌫R s  Re i ␪ : < ␪ y ␲ < F ␲ y ␩ 4 ,

⌫␧" s  t " i ␧ : 1 q ␦ ⬘ F t F R⬘ 4 ,

⌫␦ s  1 q ␦ e i ␾ : < ␾ y ␲ < F ␲ y ␩ ⬘ 4 , where R⬘ q i ␧ s Re i␩ and ␦ ⬘ q i ␧ s ␦ e i␩ ⬘; that is, R⬘ s 'R 2 y ␧ 2 , ␦ ⬘ s '␦ 2 y ␧ 2 and sin ␩ s ␧rR, sin ␩ ⬘ s ␧r␦ . Thus an s

f Ž z.

1

H 2␲ i ␥ z

nq1

dz s

1

ž

2␲ i

H⌫ q H⌫

q ␧

R

q

H⌫

y ␧

q

H⌫



/

f Ž z. z nq1

dz.

We estimate each of these integrals in turn for large R, small ␦ and small ␧ . First we take ⌫␦ . Note that for z g ⌫␦ , <1 y z < s ␦ . 1

H 2␲ i ⌫



f Ž z. z

nq 1

dz F ␦ sup zg⌫␦

f Ž z. z

nq1

1

F

Ž1 y ␦ .

nq1

sup Ž 1 y z . f Ž z . zg⌫␦

which tends to zero as ␦ ª 0 independently of the way ␧ ª 0. Next f Ž z.

1

H 2␲ i ⌫

z

R

nq1

dz F

1 Rn

sup f Ž z . s o Ž R ky n . zg⌫R

which tends to zero as R ª ⬁ if n G k, independent of ␧ . Consider the remaining integrals over ⌫␧q and ⌫␧y as ␧ ª 0. f Ž z.

1

H 2␲ i ⌫

z

q ␧

s

nq 1

1

dz q R⬘

f Ž z.

1

H 2␲ i ⌫

y ␧

z nq1

f Ž t q i␧ .

H 2␲ i 1q ␦ ⬘ Ž t q i ␧ .

nq 1

y

dz f Ž t y i␧ .

Ž t y i␧ .

nq1

dt.

269

FORMULA FOR TAYLOR COEFFICIENTS

The integrand equals f Ž t q i␧ . y f Ž t y i␧ . t

q f Ž t y i␧ . ª

q f Ž t q i␧ .

nq 1

ž

⌬ f Ž t.

1 t

y

nq 1

ž

1

Ž t q i␧ .

1

Ž t y i␧ .

nq1

nq1

y

1 t

nq1

/

/

pointwise, as ␧ ª 0 for 1 q ␦ F t F R

t nq 1

since f is uniformly bounded in the region. Also, the integrand is dominated by 2 Mrt nq 1 , where M s sup 1q ␦ F t F R < f Ž t q i ␧ .<. The contributions 1q ␦ R from H1q ␦ ⬘ and HR⬘ tend to zero as ␧ ª 0 by boundedness of f. So by Lebesque’s Dominated Convergence Theorem, limq

␧ª0

1 2␲ i

žH

⌫␧q

q

H⌫

y ␧

/

f Ž z. z

nq 1

dz s

1

⌬ f Ž t.

R

H 2␲ i 1q ␦

t nq1

dt.

Thus, for n G k, an s

1

lim

Rª⬁ , ␦ ª0 q

R

H 2␲ i 1q ␦

⌬ f Ž t. t

nq1

dt s

1



H 2␲ i 1

⌬ f Ž t. t nq1

dt

since the contribution from ⌫␦ and ⌫R tend to 0 as ␦ ª 0 and R ª ⬁ independently of ␧ , as desired. Remark. Theorem 1 can be extended to the case when f has other singularities outside the unit disc. Suppose that f is as before but has other singularities, the nearest Žleast in modulus. being at z s z 0 , where < z 0 < s ␤ ) 1. ŽThis time we don’t need a growth bound < f Ž z .< s oŽ< z < k ... Then the same methods give an s

1



H 2␲ i 1

⌬ f Ž t. t

nq1

dt q O

1

ž / ␩n

for any 1 - ␩ - ␤ .

In the proof we take R s ␩. We are often interested in finding the asymptotic behaviour of the coefficients a n as n ª ⬁ and from the above formula it can be seen that the main contribution of the integral formula comes from near t s 1. Thus the growth rate of a n depends on how ⌬ f Ž t . behaves near t s 1. The following lemma has been mentioned both in w1x and w5x but in both cases has the proof been omitted: LEMMA 2. Let f : Ž1, ⬁. ª Ž0, ⬁. be such that f Ž t . s O Ž t m . for t G 2 Ž some m. and let g Ž t . s f Ž1 q 1rt . for t ) 0. If g is regularly ¨ arying of

270

TITUS HILBERDINK

index y␴ with ␴ ) y1, then ⬁

H1

f Ž t. x

t

dt ;

␴ !gŽ x.

as x ª ⬁.

x

Proof. Let pŽ x . s H1⬁ f Ž t .rt x dt. We have pŽ x . s



H1

f Ž t. x

t

dt s

1

f Ž 1 q ␭rx .



H x 0

Ž 1 q ␭rx .

d␭ s

x

1

g Ž xr␭.



H x 0

Ž 1 q ␭rx .

x

d␭.

We find bounds for the RHS. Choose 0 - a - A - ⬁. Then 1

g Ž xr␭.



H x 0

Ž 1 q ␭rx . G s ;

1 x

Ha

A

gŽ x. x gŽ x. x

x

d␭

g Ž xr␭.

Ž 1 q ␭rx .

Ha Ha

A

A

d␭ ;

x

1 x

Ha

g Ž x . ␭␴

A

Ž 1 q ␭rx .

␭␴ eyx logŽ1q ␭ r x . d ␭ s

gŽ x. x

Ha

A

x

d␭

␭␴ ey ␭qOŽ1r x . d ␭

␭␴ ey ␭ d ␭ as x ª ⬁.

Hence lim inf xª⬁

xp Ž x . gŽ x.

G

Ha

A

␭␴ ey ␭ d ␭ for any A ) a ) 0.

Let a ª 0 and A ª ⬁ to obtain lim inf xª⬁

xp Ž x . gŽ x.

G ␴ !.

For an upper bound we proceed as follows: first, since f Ž t . s O Ž t m ., we have ⬁

H2

f Ž t. t

x

dt s O

ž



H2

t my x dt s O

1

so

/ ž / ž x2

x

gŽ x. x

/

.

FORMULA FOR TAYLOR COEFFICIENTS

271

Hence we need only consider H12 f Ž t .rt x dt s 1rxH0x g Ž xr␭.rŽ1 q ␭rx . x d ␭. By regular variation of g, we know that g Ž x . s c Ž x . xy␴ exp

½

x

H1

␧ Ž t.

5

dt ,

t

where cŽ x . ª c ) 0 and ␧ Ž x . ª 0 as x ª ⬁. Let 0 - k F cŽ x . F K and < ␧ Ž x .< F ␧ for x G 1. Then g Ž xr␭. gŽ x.

F

K k

␭␴ exp ␧

Hx

xr ␭

dt

s Krk ␭␴q ␧

t

for 0 - ␭ F x.

Also for ␭ F x, x log Ž 1 q ␭rx . G x Ž ␭rx y ␭2r2 x 2 . s ␭ Ž 1 y ␭r2 x . G ␭r2 so that Ž1 q ␭rx .yx F ey ␭ r2 . Choose A ) 1 large. Then g Ž xr␭.

x

HA

g Ž x . Ž 1 q ␭rx .

x

d␭ F

K

x

H␭ k A

␴q ␧ y ␭ r2

e

d ␭ F ␦ Ž A. ,

where ␦ Ž A. ª 0 as A ª ⬁. Finally,

H0

A

g Ž xr␭ . g Ž x . Ž 1 q ␭rx .

x

d␭ ª

H0

A

␭␴ ey ␴ d ␭ as x ª ⬁.

Since A is arbitrary we may let A ª ⬁, so that lim sup xª⬁

xp Ž x . gŽ x.

F



H0

␭␴ ey ␴ d ␭ s ␴ !

from which the result follows.

3. APPLICATIONS A simple illustration of this integral formula comes from the power series for the function f Ž z. s

yz log Ž 1 y z .

s1y

z 2

y

s

z2 12

1 1 q zr2 q z 2r3 q ⭈⭈⭈

q ⭈⭈⭈ s 1 y



Ý ns1

an z n

272

TITUS HILBERDINK

for < z < - 1. Since the first few a n are positive, we may suspect that they are all positive. This is not trivial to show in an elementary sense. Theorem 1 gives an immediate answer to this and much more. The function f Ž z . has a singularity at z s 1 and can be analytically extended to the whole of ⺓ R  x: x G 14 , by defining logŽ re i ␪ . s log r q i ␪ for y␲ - ␪ - ␲ . Since lim ␧ ª 0q logŽ1 y t " i ␧ . s logŽ t y 1. " i␲ for t ) 1, we have f " Ž t . s ytrŽlogŽ t y 1. . i␲ .. Hence, for t ) 1, ⌬ f Ž t. s

yt

y

log Ž t y 1 . y i␲

yt log Ž t y 1 . q i␲

y2␲ it

s

2 Ž log Ž t y 1 . . q ␲ 2

.

Also Ž1 y z . f Ž z . ª 0 as z ª 1, f is bounded except near 1, and < f Ž z .< s oŽ< z <. as < z < ª ⬁. Thus Theorem 1 gives, for n G 1, an s y

1



H 2␲ i 1

⌬ f Ž t. t

dt s

nq1

1



H1

2

t n Ž Ž log Ž t y 1 . . q ␲ 2 .

dt.

Now it is immediatley obvious that a n ) 0 for each n G 1, and that a ny 1 ) a n . Moreover, the kth difference ⌬k a n Ž ⌬ a n s a n y a ny1 , ⌬kq 1 a n s ⌬Ž ⌬k a n .. satisfies ⌬ a n s Ž y1 . k

k

Ž t y 1.



H1

k 2

t n Ž Ž log Ž t y 1 . . q ␲ 2 .

dt,

so that Žy1. k ⌬k a n ) 0 for all n; i.e., a n is completely monotonic. Also, from Lemma 2 we can obtain the behaviour for large n thus an ;

1 n Ž log n .

2

as n ª ⬁.

This is because 1rŽŽlog t . 2 q ␲ 2 . is slowly varying Žregularly varying of index 0. as t ª ⬁. Without much difficulty we can obtain an asymptotic series Ž1rnŽlog n. 2 .Ý⬁ks 0 bk Žlog n.yk , but we omit the details. Application to the Luria-Delbruck ¨ distribution The probability generating function of the LDŽ m. distribution Žsee w7x, w10x, w11x. is given by gŽ z. s



Ý ns0

pn z n s Ž 1 y z .

m Ž1yz .rz

Ž < z < F 1. .

273

FORMULA FOR TAYLOR COEFFICIENTS

The aim here is to find the asymptotic behaviour of the probabilities pn . It was shown in w7x and w10x that pn ; mrn2 as n ª ⬁ and more accurate estimates were found in w11x using the singularity analysis of Flajolet and Odlyzko w6x. As before, logŽ1 y z . can be extended analytically to the region ⺓ R  x: x g w1, ⬁.4 and

Ž 1 y z . log Ž 1 y z .

s log Ž 1 y z . q o Ž 1 . s log < z < q O Ž 1 .

z

as < z < ª ⬁

we have g Ž z . s expŽ mŽ1 y z .logŽ1 y z .rz . is analytic in ⺓ R  x: x g w1, ⬁.4 and < g Ž z .< s O Ž< z < m . as < z < ª ⬁. Also g Ž z . ª 1 as z ª 1 and gq Ž x . and gy Ž x . both exist for x ) 1. Thus gq Ž x . s limq g Ž x q i ␧ . s limq exp ␧ª0

s exp

½

␧ ª0

mŽ 1 y x . x

s Ž x y 1.

m Ž1yx .rx

½

mŽ 1 y x y i ␧ . x q i␧

Ž log Ž x y 1 . y i␲ .

log Ž 1 y x y i ␧ .

5

5

e i␲ mŽ xy1.r x .

Similarly, we find gy Ž x . s Ž x y 1. mŽ1yx .r x eyi␲ mŽ xy1.r x. The conditions of Theorem 1 are satisfied and we obtain, for n ) m, pn s

1 2␲ i



H1

⌬ gŽ t. t nq1

dt s

1





H1

Ž t y 1 . m 1yt rt Ž

.

t nq1

sin  ␲ m Ž t y 1 . rt 4 dt.

From the integral representation and Lemma 2 it follows immediately that pn ; mrn2 as n ª ⬁ and it is straightforward Žalbeit laborious. to find an asymptotic expansion for pn by expanding the numerator of the integrand in powers of Ž t y 1. and logŽ t y 1.. Another application comes from combinatorics. In a recent paper w2x, Bona ´ showed that the total number sn of 1342-avoiding n-permutations is given by the generating function f Ž z. s



Ý ns0

sn z n s

32 z 1 q 20 z y 8 z 2 y Ž 1 y 8 z .

3r2

.

It was stated that this series gives the upper bound sn F 8 n and that n sn ª 8 as n ª ⬁. Much more accurate information can in fact be found by using this generating function. First we must investigate the zeros of the denominator; that is, when is 1 q 20 z y 8 z 2 y Ž1 y 8 z . 3r2 s 0? This function is analytic in the disc < z < - 1r84 and can be extended analytically

'

274

TITUS HILBERDINK

to the cut-plane ⺓ R w1r8, ⬁. by taking ␲ . Note that

're i␪ s 'r e i␪ r2 , where y␲ - ␪ -

Ž 1 q 20 z y 8 z 2 q Ž 1 y 8 z . 3r2 . . Ž 1 q 20 z y 8 z 2 y Ž 1 y 8 z . 3r2 . 2

3

s Ž 1 q 20 z y 8 z 2 . y Ž 1 y 8 z . s 64 z q 192 z 2 q 192 z 3 q 64 z 4 s 64 z Ž 1 q z .

3

so the zeros can only occur when z s 0 or z s y1. When z s y1, we get 1 y 20 y 8 y 9 3r2 s y27 y 27 s y54 / 0. Also the simple root at 0 is cancelled in the numerator by the 32 z. Hence, our original function f Ž z . has no singularities other than the branch point at z s 1r8 inherited from Ž1 y 8 z . 3r2 and is uniformly bounded outside neighbourhoods of 1 and ⬁. Let g Ž z . s f Ž zr8. s 32 zrŽ8 q 20 z y z 2 y 8Ž1 y z . 3r2 . s Ý sn 8yn z n which has the branch point at z s 1. Note also that for t ) 1, both gq Ž t . and gy Ž t . exist and are given by g " Ž t . s limq g Ž t " i ␧ . s ␧ª0

32 t 8 q 20 t y t 2 . 8i Ž t y 1 .

3r2

because '1 y t " i ␧ ª "i't y 1 as ␧ ª 0 through positive values. Hence ⌬ g Ž t . s gq Ž t . y gy Ž t . s

s

512 i Ž t y 1 .

Ž t q 8.

32 t Ž 2.8i Ž t y 1 .

3r2

.

2 2

3 Ž 8 q 20 t y t . q 64 Ž t y 1 .

3r2

3

.

As < z < ª ⬁, < f Ž z .< ª 0 and Theorem 1 can be applied for n G 0. Thus we have sn s

8n 2␲ i



H1

512 i Ž t y 1 .

3r2

t nq1 Ž t q 8 .

3

dt s

8n 2␲



H1

512 Ž t y 1 .

3r2

t nq1 Ž t q 8 .

3

dt.

To find the asymptotic behaviour, let hŽ t . s 512Ž t y 1. 3r2rt Ž t q 8. 3 and consider hŽ1 q 1rt . as t ª ⬁. h Ž 1 q 1rt . s

512 ty3 r2

Ž 1 q 1rt . Ž 9 q 1rt .

3

;

512 729

ty3 r2

275

FORMULA FOR TAYLOR COEFFICIENTS

which is regularly varying of index y␴ s y3r2. Thus ␴ s 3r2 ) y1 and Lemma 2 may be applied. We obtain sn ;

8n 2␲

⭈ Ž 3r2 . !

512 729

ny3r2y1 ;

64

8n

243'␲ n5r2

as n ª ⬁.

It is also straightforward to obtain an asymptotic series for sn using this integral representation. We omit the details.

REFERENCES 1. N. H. Bingham, C. M. Goldie, and J. L. Teugels, ‘‘Regular variation,’’ Cambridge University Press, 1987. 2. M. Bona, ´ Exact enumeration of 1342-avoiding permutations; A close link with Labelled trees and planar maps, J. Combin. Theory Ser. A 80 Ž1997., 257᎐272. 3. B. L. J. Braaksma and D. Stark, A Darboux-type Theorem for Slowly Varying Functions, J. Combin. Theory Ser. A 77 Ž1997., 51᎐66. 4. N. G. de Bruijn, ‘‘Asymptotic Methods in Analysis,’’ North-Holland Publishing Co., 1970 Ž3rd edition.. 5. W. Feller, ‘‘An introduction to probability theory and its applications,’’ Vol. I and II, Wiley, New York, 1968 Ž3rd edition.. 6. P. Flajolet and A. M. Odlyzko, Singularity analysis of generating functions, SIAM J. Discrete Math. 3 Ž1990., 216᎐240. 7. C. M. Goldie, Asymptotics of the Luria-Delbruck ¨ Distribution, J. Appl. Probab. 32 Ž1995., 840᎐841. 8. T. W. Hilberdink, On the Taylor coefficients of the composition of two analytic functions, Ann. Acad. Sci. Fenn. Ser. A I 21 Ž1996., 189᎐204. 9. A. M. Odlyzko, ‘‘Asymptotic enumeration methods,’’ Handbook of Combinatorics Vol. 2, Elsevier, 1995. 10. A. G. Pakes, Remarks on the Luria-Delbruck ¨ distribution, J. Appl. Probab. 30 Ž1993., 991᎐994. 11. H. Prodinger, Asymptotics of the Luria-Delbruck ¨ distribution via singularity analysis, J. Appl. Probab. 33 Ž1996., 282᎐283.