Pollaczek contour integrals for the fixed-cycle traffic-light queue

0.27 MB·

by  Marko Boon

Pollaczek contour integrals for the
fixed-cycle traffic-light queue
M.A.A. Boon A.J.E.M. Janssen J.S.H. van Leeuwaarden
January 12, 2017
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Abstract
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The fixed-cycle traffic-light (FCTL) queue is the null model for intersections with static

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signaling,wherevehiclesarrive,formaqueueanddepartduringcyclescontrolledbyatraffic
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light. Classical analysis oftheFCTLqueuebasedontransformmethodsrequires acomputa-

] tionallychallengingstepoffindingthecomplex-valuedrootsofsomecharacteristicequation.
R We derive a novel contour-integral expression, reminiscent of Pollaczek integrals for bulk-
P service queues, fortheprobability generating function ofthesteady-state FCTL queue. This
h. representation will be the basis for effective algorithms. We show that it is straightforward
t to compute the queue-length distribution and all its moments using algorithms that rely on
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contourintegralsandavoidroot-findingproceduresaltogether.
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Keywords: fixed-cycletraffic-lightqueue;transformmethods;complexanalysis

1 AMS2010SubjectClassification. 60E10,60J10,60K25,68M20,90B20
v
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7 1 Introduction
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0 The fixed-cycle traffic-light (FCTL) queue is an intensivelystudied and widely applied stochastic
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1 modelintraffic engineering[2,5,6,10,11,15,17]. Vehiclesarrive toanintersectioncontrolled
0 by a traffic light and form a queue. The FCTL queue is traditionally modeled in discrete time.
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Time is divided into slots of unit length, and each slot corresponds to the time needed for a
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: delayed vehicle to depart from the queue. The green and red periods, of length g and r slots,
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i respectively, and thus the cycle length c = g+r, are assumed to be fixed multiples of one slot.
X
Those vehicles that arrive to the queue and are delayed, join the queue at the end of the slot in
r
a whichtheyarrive.
The key feature of the FCTL queue, which distinguishes it from traditional queues, is formu-
latedinthefollowingassumption:
FCTLassumptionForthosecyclesinwhichthequeueclearsbeforethegreenperiodterminates,
all vehicles that arrive during the residual green period pass through the system and experience
nodelaywhatsoever.
TheFCTLassumptionletsvehiclesthatarriveduringtheresidualgreenperiodpasstheinter-
section without slowing down, and therefore the discharge rate of these vehicles is larger than
∗Department ofMathematics andComputerScience, EindhovenUniversity ofTechnology, P.O. Box513, 5600MB
Eindhoven,TheNetherlands. Email: [email protected],[email protected]@tue.nl
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the discharge rate of the delayed vehicles (one per time unit). Because of the huge difference in
discharge rates of delayed vehicles (these vehicles have toaccelerate) and non-delayedvehicles,
the FCTL assumption is a sensible assumption. The FCTL assumption does, however, have some
severeconsequencesforthemathematicalanalysisofthequeuelength. LetX denotethequeue
k,n
lengthat time k in cycle n(time expressedin slots). Then, in cycle n, X is the queuelengthat
0,n
the beginning of the green period, and X the overflow defined as the queue length at the end
g,n
of the green period (and thus the beginning of the red period). Let A denote the total number
n
of vehicles that arrive at the intersection in between the two measurementsof the overflow X
g,n
and X . Thus A are the arrivals from X onwards in a consecutive red and green period.
g,n+1 n g,n
Further, A = Ad +Ap, where Ad denotes the number of delayed vehicles and Ap the number of
n n n n n
vehiclesthatpasswithoutdelayonbehalfoftheFCTLassumption. Theoverflowqueuecanthen
bedefinedas
X =max{X +Ad −g,0}. (1.1)
g,n+1 g,n n
The fact that Ad depends on both X and the exact specification of when the arrivals occur
n g,n
makes(1.1)hardtoanalyze. Tocapturethatlevelofdetail,letY denotethenumberofvehicles
k,n
that arrive to the intersection during slot k in cycle n. The random variables Y are assumed
k,n
independent and identically distributed, for all k and n. Notice that the above assumptions
together make that the queue lengths at the end of time slots can be modeled as a discrete-time
Markov chain. Using analytic techniques suitable for dealing with such Markov chains, Darroch
[5] obtained the probability generating function (PGF) of the steady-state overflow queue (the
numberof vehicles waiting in front of the traffic light at the end of a green period) and the PGF
of the steady-state delay was obtained in van Leeuwaarden [15]. Hence, all information about
the distribution ofthe steady-state overflowqueueandsteady-state delayin theFCTLqueuecan
be obtained from the results in [5, 15], including all moments of the steady-state queue length
anddelay, andthe distribution oftheoutputprocess(the wayvehiclesleavethe intersection).
TheclassicalFCTLtreatmentin[5,15],however,comeswiththedisadvantagethatnumerical
evaluation is computationally cumbersome. This is because the probability generating function
(PGF)forthestationaryqueuelengthdistribution contains g−1boundarytermsthatneedtobe
found separately. The traditional way of determining these remaining unknowns consists of two
steps: findingthe g−1complex-valuedrootsandusingtheserootsasinputforasystemoflinear
equationswhosesolutiongivestheboundaryterms. Bothstepscanpresentnumericaldifficulties,
but were somehow assumed unavoidable in the mathematically rigorous treatment of the FCTL
queue[5,10,11,17,15]andofrelatedbulk-servicequeues[8,13,14].
Paperoutline. WepresentinSection2themainresultofthepaper,analternativerepresentation
for the transform solution in terms of a contour integral. We also explain how these contour
integrals lead to algorithms that are more efficient than existing algorithms. We present these
algorithmsinSection3andshowtheireffectiveness. Theproofofourmainresultispresentedin
Section 4 and uses several basic notions from complex analysis. We conclude in Section 5, and
discussthescopeofapplicationofourproofmethodandhowthecontourintegralsmightbeused
forasymptotic analysis.
2 Main result
We now briefly review the standard solution method for obtaining an exact transform solution
forthesteady-stateFCTLqueue. Wethenpresentthealternativecontourintegralrepresentation.
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2.1 Standard solution
Assume P(Y = 0) > 0, Y′(1) < 1, and Y(z) to be analytic in a region |z| < R with R > 1 and
R maximal. The key quantity in the mathematical analysis of the FCTL queue is the steady-state
overflow queue, defined as X = lim X . Clearly, to have stability, and for X to be well
g n→∞ g,n g
defined,
cE[Y]< g. (2.1)
Usingthekernelmethodandtransformtechniques,thePGFofX ,denotedbyX (z)=E[zXg]can
g g
be obtained using a by now classical line of reasoning. Introduce two more PGFs Y(z) = E[zY]
andA(z)=E[zA]=Ec[zY]. Thenitcanbeshown that[5,15]
(z−Y(z)) g−1q zkY(z)g−1−k
X (z)= k=0 k . (2.2)
g zg −A(z)
P
This expression still contains g unknowns q ,...,q , which can be found by exploiting the
0 g−1
analytic properties of PGFs. With Rouché’s theorem, it can be shown that the denominator of
(2.2) has g zeros on or within the unitcircle |z|=1. Becausea PGF is analytic and well-defined
in |z| ≤ 1, the numerator of X (z) should vanish at each of the zeros. This gives g equations.
g
One of the zeros equals 1, and leads to a trivial equation. However, the normalization condition
X (1) = 1 provides an additional equation. So that summarizes the highest level of general
g
developmentforFCTLqueueanalysis: transformtechniquesyieldanexpressionforX (z)thatin
g
ordertobeevaluateddemandsfinding g−1rootsinthecomplexplaneofthefunctionzg =A(z)
andsolvingasetof g linearequations.
2.2 Pollaczek integral
Wenowturntothealternative expression for X (z). Here isthe mainresultinthis paper:
g
Theorem 1 (Pollaczek integralfor FCTL) Thereisanε >0suchthatforallε∈(0,ε )
0 0
1 Y′(z)z−Y(z) w−Y(w) A(z)
X (w)=exp ln 1− dz , |w|<1+ε,
g 2πi z−Y(z) zY(w)−wY(z) zg
I|z|=1+ε (cid:18) (cid:19) !
(2.3)
withprincipalvalueofthelogarithm.
The formula(2.3)for X (w) is essentiallyequivalentwith
g
1 wY(z)−zY(w) (zg −A(z))′
X (w)=exp ln dz , (2.4)
g 2πi Y(z)−z zg −A(z)
I|z|=1+ε (cid:18) (cid:19) !
except that the validity range is more delicate due to the more complicated argument of the
ln in (2.4). Formula (2.3) follows upon manipulating (2.4) using partial integration (details in
Section4).
Towards better numerical algorithms. The elegant contour-integral representation requires
model primitives as sole input. This representation will be the basis for effective algorithms. For
theempty-queueprobabilityandallmomentsweobtainexplicitcontourintegralsbyconsidering
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the derivatives of X (w) at w = 1. These contour integrals can be evaluated numerically in
g
a straightforward and fast manner. For calculating the entire steady-state distribution we use
numericaltransforminversionbasedonthesymboliccomputationofallderivativesofX (w)at0.
g
Theresultingalgorithmsarethefirsttonotusecomplex-valuedrootsandremarkablyelegantand
simplecomparedwithexistingalgorithms. Thecontourintegralsandnumericalimplementations
arediscussedinSection3. Therewealsodiscusshowthealgorithmsfortheoverflowqueuecarry
overtosimilaralgorithms forthe steady-state delaydistribution.
Historical notes. Integrals ofthis sort go a longway back in the history of queueing theory and
werefirstfoundintheground-breakingworkofPollaczekontheclassicalsingle-serverqueue(see
[1, 4, 7] for historical accounts). Let us point out the connection to the well known Pollaczek
type integral for the discrete bulk-service queue [8]. The analysis of the FCTL queue is greatly
simplified if all vehicles were delayed [3], so that all vehicles arrive while the queue lengthis at
least one and the complicated Ad random variable in (1.1) can be replaced by A . In that case
n n
(1.1) becomes a standard stochastic recursion driven by i.i.d. random variables and the FCTL
queuereducestotheclassicalbulk-servicequeue,aspecialcaseofthemoregeneralsingle-server
queueinvestigatedbyPollaczek. Let X denotethesteady-statequeuelengthinthatbulk-service
b
queue,definedas thesolutionofthestochastic equation
d
X =max{X +A−g,0}. (2.5)
b b
Pollaczek’sresultthensays that(see[8]foradirectderivation)
1 w−z (zg −A(z))′
X (w)=exp ln dz (2.6)
b 2πi 1−z zg −A(z)
I|z|=1+ε (cid:18) (cid:19) !
holds when |w| < 1+ε with ε positive and bounded by some constant. Observe the striking
similaritywith(2.4). WhiletheFCTLqueueishardertoanalyzethanthebulk-servicequeue,the
two contour-integral representations (2.4) and(2.6) onlydiffer in the logarithmic function. The
authorsfindthisquitesurprisingthemselves,particularlybecausethereseemsnowaytointerpret
the FCTLqueueas areflectedrandom walk(that is,arecursive structure with i.i.d. increments),
whileintheliteraturesofarthisseemstobeaprerequisiteforestablishingPollaczek-typecontour
integrals. Do observe that (2.6) is valid in an area that includes the unit disk while (2.4) is
guaranteed only in an open set containing [0,1], see Section 4. This objection does not hold
againsttherepresentation (2.3)of X (w).
g
Bernoulliarrivals. Thebulk-servicequeueservesasapopularapproximationoftheFCTLqueue
[3]. In fact,forBernoulliarrivals, sopertimeslotoneornoarrival, this approximation becomes
exact. Toseethis,substitute Y(z)=1−p+pzintothelogarithmicfunctionin(2.4),andobserve
that this gives the logarithmic function in (2.6). Obviously, when Y can take values larger than
one, the bulk-service queue is an approximation and yields an upper bound on the overflow
queue.
Sketch of the proof. The proof will find a way to go from representation (2.2) to contour
integrals. Asignificantstartinthis direction wasmade by[12],whorewrote (2.2)as
g−1−k
(z−Y(z))zg−1 g−1q Y(z)
k=0 k z
X (z)= . (2.7)
g zgP−A(z) (cid:16) (cid:17)
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Then denote the g roots of zg = A(z) on and within the unit circle by z = 1,z ,...,z . Now
0 1 g−1
here is where the authors in [12] took an eye-opening step: instead of using the g roots in the
traditional manner for finding the unknowns q and completing the transform (2.2), use these
k
rootsforfactorizingthenumeratorof(2.2). Noticethatthiscannotbedoneimmediately,because
interpreted as a function of z, the numerator is by no means a polynomial of degree g or less.
However, by treating the function Y(z)/z as a variable itself, the summation in the numerator is
apolynomialofdegree g−1andcanbefactorizedas
g−1 Y(z) g−1−k g−1 Y(z) Y(z )
k
q =q − . (2.8)
k 0
z z z
k=0 (cid:18) (cid:19) k=1(cid:18) k (cid:19)
X Y
Afternormalizationusing X (1)=1thefactorization in(2.8)leadstotherepresentation
g
g−A′(1) z−Y(z) g−1Y(z)/z−Y(z )/z
X (z)= · ·zg−1 k k. (2.9)
g zg −A(z) 1−Y′(1) 1−Y(z )/z
k=1 k k
Y
Our proof then proceeds by interpreting (2.9) as the outcome of Cauchy’s residue theorem, the
classical tool from complex analysis to evaluate line integrals of analytic functions over closed
curves. Animportantstepis towrite
g−1 g−1
Y(z)/z−Y(z )/z zY(z )−z Y(z)
ln zg−1 k k = ln k k , (2.10)
1−Y(z )/z Y(z )−z
k=1 k k ! k=1 (cid:18) k k (cid:19)
Y X
and to regard (2.10) as the sum of residues at z = z . To construct an analytic function that,
k
in conjunction with Cauchy’s theorem and the closed curve |z| = 1+ε, returns (2.10) and has
singularities at z ,...,z , leads us to consider the integrand in (2.4). Here, the logarithmic
1 g−1
function
wY(z)−zY(w)
ln (2.11)
Y(z)−z
(cid:18) (cid:19)
follows from (2.10) and the singularities with appropriate residues are created through (zg −
A(z))′/(zg−A(z)). Aftercarefulconsiderationoftheanalyticpropertiesoftheintegrandin(2.4),
we then show that Cauchy’s theorem gives (2.9) from which (2.4) follows. As said, (2.3) is ob-
tainedbymanipulating(2.4),usingpartialintegration. TheformalproofofTheorem1presented
in Section4containsseveral challengingsteps,andrequires amongotherthings aproofthat the
function Y(z)/z is injective in a region that contains the unit disk, and a way to account for the
branchcutcausedbythelogarithm in(2.11)beingtakenovernegativevalues.
3 Novel FCTL algorithms
We now demonstrate how the integral representation in Theorem 1 yields efficient algorithms
for effectively all stationary performance metrics, including the stationary moments and queue
lengthdistribution. Thealgorithmsinthissectionarebasedontherepresentation(2.4)(butone
couldalsotake(2.3)). NotethatweonlyexpandX (w)atw=0andw=1,soinsidethevalidity
g
rangeof(2.4).
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3.1 Moments
The meanstationary overflowqueueEX is givenby X′(1)andtakesthe form
g g
1 Y(z)−zY′(1) (zg −A(z))′
EX = dz. (3.1)
g 2πi Y(z)−zY(1) zg −A(z)
I|z|=1+ε
This resultwasrecentlyobtainedin[12]usingadirectproofthatconvertedtheclassicalexpres-
sionforEX interms ofcomplexvaluedroots intotheintegral expression(3.1).
g
Fromthe PGF X (z) wecanin principledetermine allstationary moments. Define
g
1 (zg −A(z))′
f(w):= g(w,z) dz, (3.2)
2πi zg −A(z)
I|z|=1+ε
wY(z)−zY(w)
g(w,z):=ln , (3.3)
Y(z)−z
(cid:18) (cid:19)
1 k=0,
h (w):= (3.4)
k (hk−1(w)f′(w)+h′k−1(w) k=1,2,...
The moments E[Xk] then follow from symbolically differentiating the PGF (2.4), and these
g
derivatives canbebe expressedas
dk dk
X(k)(w):= X (w)= exp f(w) =h (w)exp f(w) , (3.5)
g dwk g dwk k
(cid:0) (cid:1) (cid:0) (cid:1)
for k = 0,1,2,.... Using this recursive expression, X(k)(w) can be expressed in terms of f(w)
g
andthe first k derivatives of f(w),denotedby f(1)(w),...,f(k)(w) with
∂ j 1 (zg −A(z))′
f(j)(w):= g(w,z) dz
∂wj 2πi zg −A(z)
I|z|=1+ε
1 (zg −A(z))′
= g(j)(w,z) dz (3.6)
2πi zg −A(z)
I|z|=1+ε
and g(j)(w,z):= ∂j g(w,z),for j=1,2,...,k. Aftersubstitutingw=1,wecanexpressthefirstk
∂wj
momentsofX intermsofkcontourintegralsthatonlyinvolvethemodelprimitivesandthefirst
g
k momentsof Y. Using f(1)=0,thevariance of X givenbyVar(X )=h (1)+h (1)−(h (1))2
g g 2 1 1
takes theform
1 z2Var(Y)−zY(z)(1+E(Y2)−2EY) (zg −A(z))′
Var(X )= dz. (3.7)
g 2πi (z−Y(z))2 zg −A(z)
I|z|=1+ε
3.2 Overflow and queue-length distribution
Todeterminethe stationary distribution oftheoverflowqueueweusethat
1 dk 1
P(X =k)= X (w) = h (0)exp f(0) . (3.8)
g k!dwk g w=0 k! k
(cid:12)
(cid:12) (cid:0) (cid:1)
(cid:12)
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Firstobserve that
1 zP(Y =0) (zg −A(z))′
P(X =0)=exp f(0) =exp ln dz . (3.9)
g 2πi z−Y(z) zg −A(z)
I|z|=1+ε (cid:18) (cid:19) !
(cid:0) (cid:1)
Expressions for the other probabilities P(X = k) follow in a similar way, but require evaluating
g
theresultingfunctionat w=0insteadofw=1anddividingbyk!. Asaconsequence,P(X =k)
g
can be expressed in terms of f(0),f(1)(0),...,f(k)(0), again an expression that involves explicit
contourintegralsonly.
In principle, all other stationary performance metrics followfrom the distribution P(X = k)
g
(cf. [15]). However, someexplicit results from [15, 3] are given in terms ofq ,q ,...,q . These
0 1 g
are the boundary probabilities in the classical transform solution (2.2). In fact, q represents
j
the probability that the queue is empty at the end of slot j in the green period. To find these
probabilities, andthePGFsofthe queue-lengthdistributions attheendofallslots,wewrite
X (w)=X (w)Y(w)c−g, and q =exp f(0) Y(0)c−g. (3.10)
0 g 0
Wethenobtain therecursions (cid:0) (cid:1)
X (w)−q
k−1 k−1
X (w)=Y(w) +q , (3.11)
k k−1
w
X (w)−q
k−1 k−1
q =Y(0)lim +q , (3.12)
k k−1
w↓0 w
for k =1,2,...,g−1. The limit w ↓0 in (3.12) yields expressions in terms of the first k deriva-
tives of f(w) and Y(w), all of which can be evaluated symbolically, and then require numerical
evaluationofcontourintegrals.
3.3 Numerical examples
To illustrate the algorithms we now show some results for the FCTL queue with g = 20 and
c = 50. We consider Poisson arrivals with on average λ vehicles arriving per slot, and four
scenarios: λ = 0.2 (light traffic), λ = 0.3 (moderate traffic), λ = 0.36 (heavy traffic), and
λ=0.38 (extreme traffic). These arrival rates correspond to a volume/capacity ratio ρ =λc/g
rangingfrom0.5to0.95.
Figure 1(a) shows the mean queue lengths E[X ],...,E[X ] through one cycle. Observe
0 c
the strong cyclic behavior and the high sensitivity for ρ. Figure 1(b) shows the queue-length
distribution at cycle start, the moment that the traffic signal turns green and queue lengths are
expectedtopeak. Observethedifferencebetweenoperationgat75%or90%ofmaximalcapacity:
the probability that more than 20 vehicles are waiting is only 0.002 for λ = 0.3 and 0.32 for
λ=0.38. Figure1(c)depicts the distribution ofthe effective greentime G,definedin [3] as the
numberofslotsusedfordeparture ofdelayedvehicles. Wehave
q for k=0,
0
P(G=k)=qk−qk−1 for k=1,...,g−1, (3.13)
1−qg−1 for k= g.

Sinceonlyonedelayedvehicledeparts perslot,this canalsobeconsideredtobethe distribution
of the platoon length consisting of delayed vehicles departing during one cycle. Observe that
7
(Xk) ℙX0k)
0.15
15
0.10
10
0.05
5
k k
10 20 30 40 50 10 20 30 40
(a) (b)
ℙ(Gk) ℙ(D[m]k)
0.08
0.07
0.5 0.06
0.2
0.02
0.1
0.01
k k
5 10 15 20 20 40 80 100
(c) (d)
Figure 1: Several performance measures for the FCTL queue in Section 3.3 with g =20, r =30
andPoissonarrivals. The colorsblue,orange,greenandredcorrespond tovolumeovercapacity
ratios of, respectively, 0.5, 0.75, 0.9 and 0.95. The subfigures show (a) mean queue lengths
during a cycle,(b) the queuelengthdistribution at the start ofgreen periods, (c) effective green
periods and(d)delaydistribution ofvehiclesarriving inslot10(forρ=0.9,0.95only).
P(G= g)is practicallyzerowhenρ=0.5,butas highas 0.84whenλ=0.36,whichmeansthat
onlyin16%ofthe cyclesthegreentimeis longenoughtoletthequeuevanish.
Finally, we consider the delay distribution of an arbitrary vehicle arriving in the 10-th slot,
which is during the green period. The stationary delay of a vehicle arriving in slot k, denoted
by D , is defined as the number of slots between arrival and departure, not including the slot
[k]
of arrival. Figure 1(d) shows D , which can be computed directly from X , i.e. the numberof
[10] 9
vehicles waiting at the start of the 10-th slot. If X = 0 we have that D = 0; otherwise the
9 [10]
delaycanbeexpressedasafunctionofthenumberofvehiclespresentatthearrivalofthetagged
vehicle. This function (studied in detail in [15]) should take into account interruptions due to
redperiods,which explainsthefragmentedhistograms inFigure 1(d).
4 Proof of the Pollaczek contour-integral representation
As explained briefly in Section 2, the proof of Theorem 1 exploits the factorized form (2.9) and
investigates in detail the logarithmic function (2.11). We present some useful properties of the
function Y(z)/z, visible in both (2.9) and (2.11). We then proceed to use Cauchy’s theorem to
8
obtain the contour-integral representation (2.4) for the case that 1 < w < 1+ε, and finally
manipulate(2.4)toobtain (2.3)onthe fullrange|w|<1+ε.
4.1 Auxiliary results
Before we prove Theorem 1 we present some auxiliary results for the function Y(z)/z. In [12,
Theorem 1] it was shown that the function Y(z)/z is injective on the disk |z| ≤ 1, so that all
Y(z )/z 6= Y(z )/z when z 6=z . For our proof we also need injectivity, but then for the larger
k k l l k l
diskwith radius t >1. Morespecifically,let
0
t :=sup{t ∈(0,R)| Y′(t)t−Y(t)≤0}. (4.1)
0
Lemma2 Thefunction Y(t)/t isstrictly decreasingin t ∈(0,t ].
0
Proof. Wehave that
Y(t) y
0
= + y + y t+... , 0< t <R, (4.2)
1 2
t t
is strictlyconvexsince y >0,withderivative
0
Y(t) ′ Y′(t)t−Y(t)
= , 0< t <R. (4.3)
t t2
(cid:18) (cid:19)
Since Y′(1) < Y(1) = 1, we have that t > 1. Now consider the following cases: (i) y =0
0 k
for k =2,3,..., (ii) there is a k =2,3,... such that y 6=0. For case (i) Y(t)/t = y t−1+ y is
k 0 1
strictlydecreasingin t >0since y >0. Forcase (ii), y >0forsome k≥2,andso
0 k
∞
Y′(t)t−Y(t)=−y + (k−1)y tk (4.4)
0 k
k=2
X
is strictlyincreasingin t ∈(0,R). From thedefinition of t ,we thengetthat
0
Y′(t)t−Y(t)<0, t ∈(0,t ), (4.5)
0
andso Y(t)/t isstrictlydecreasing in t ∈(0,t )by(4.3). ƒ
0
Lemma3 Thefunction Y(z)/z isinjectiveontheopendisk|z|< t ,so thatfor |z|< t ,|w|< t
0 0 0
Y(z) Y(w)
= ⇒z=w. (4.6)
z w
Proof. In case (i), y =0 for k =2,3,..., we have Y(z)/z = y z−1+ y and the result is trivial
k 0 1
since y >0. Forcase(ii), thereis a k=2,3,... suchthat y 6=0,welet|z|< t ,|w|< t . Then
0 k 0 0
Y(z) Y(w) w−z ∞
− = y + y (zk−1−wk−1)
0 k
z w zw
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12) Xk=2 (cid:12)
(cid:12) (cid:12) (cid:12) y ∞ zk−1−wk(cid:12)−1
(cid:12) (cid:12)=|(cid:12)z−w| − 0 + y (cid:12) . (4.7)
zw k z−w
(cid:12) Xk=2 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
9
Let t :=max{|z|,|w|}< t . Then|y /(zw)|≥ y /t2 while
0 0 0
zk−1−wk−1
= zk−2+wzk−3+···+zwk−3+wk−2 ≤(k−1)tk−2. (4.8)
z−w
(cid:12) (cid:12)
(cid:12) (cid:12)
Therefore,wh(cid:12)en z6=w, (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
Y(z) Y(w) y ∞
− ≥|z−w| 0 − (k−1)y tk−2 >0 (4.9)
z w t2 k
(cid:12) (cid:12) k=2
(cid:12) (cid:12) (cid:16) X (cid:17)
by(4.4)and(4.5). T(cid:12)(cid:12)his proves (4.6(cid:12)(cid:12)). ƒ
Lemma4 Letε>0besuch that1+ε< t ,andtake w∈(1,1+ε). For|z|< t ,
0 0
wY(z)−zY(w)
∈(−∞,0] ⇔ 1≤z≤w. (4.10)
Y(z)−z
Proof. For a≤0,
wY(z)−zY(w) Y(z) Y(w)−a
=a ⇔ = . (4.11)
Y(z)−z z w−a
Since 1 < Y(w) < w, the function (Y(w)−a)/(w−a) increases from Y(w)/w at a = 0 to 1 at
a =−∞ when a decreases from 0 to −∞. Since Y(v)/v decreases strictly in v ∈[1,w], there is
forany a≤0aunique v =v(a)∈[1,w] suchthat
Y(v) Y(w)−a
= . (4.12)
v w−a
SincebyLemma3 Y(z)/z is injective in|z|< t weget(4.10). ƒ
0
As a consequenceof Lemma 4, taking the principal value logarithm in (2.11) when 1< w <
1+ε < t , we obtain a function of z that is analytic in the open disk |z| < t , with branch cut
0 0
[1,w].
4.2 Contour integral for (2.4)
We next consider the function zg −A(z) that has its zeros in |z| ≤ 1 at z = z = 1,z ,...,z ,
0 1 g−1
whileitsotherzeroshavemodulusgreaterthanone. LetR bethezerooutside|z|≤1ofsmallest
0
modulus; we have R is real and larger than one. Take ε>0 such that 1+ε<min{t ,R } and
0 0 0
considertheintegral
1 wY(z)−zY(w) (zg −A(z))′
I(w)= ln dz. (4.13)
2πi Y(z)−z zg −A(z)
I|z|=1+ε (cid:18) (cid:19)
Chooseδ>0suchthatδ< 1(w−1)andδ<1+ε−wwhile|z −1|>δ,k=1,...,g−1. Now
2 k
letC bethepositivelyorientedcontourconsistingofthecirclesC (δ)andC (δ)ofradiiδaround
1 w
1 and w, respectively, togetherwith the linesegments L (δ)={z = t±i0|1+δ≤ t ≤w−δ}.
±
See Fig. 2 for the positioning of the contour C with its four components in the disk |z| < 1+ε
andrelative tothe zerosofzg −A(z). Then,byCauchy’stheorem,
g−1 wY(z )−z Y(w) 1 wY(z)−zY(w) (zg −A(z))′
k k
I(w)= ln + ln dz. (4.14)
Y(z )−z 2πi Y(z)−z zg −A(z)
k=1 (cid:18) k k (cid:19) IC (cid:18) (cid:19)
X
10