On the non-holonomic character of logarithms, powers, and the nth prime function
ON THE NON-HOLONOMIC CHARACTER OF LOGARITHMS,
POWERS,AND THE n TH PRIME FUNCTION
PHILIPPE FLAJOLET,STEFAN GERHOLD,AND BRUNO SALVY
Abstract.We establish that the sequences formed by logarithms and by
“fractional”powers of integers,as well as the sequence of prime numbers,
are non-holonomic,thereby answering three open problems of Gerhold [El.
https://www.wendangku.net/doc/3e11845616.html,b.11(2004),R87].Our proofs depend on basic complex analysis,
namely a conjunction of the Structure Theorem for singularities of solutions
to linear di?erential equations and of an Abelian theorem.A brief discussion is
o?ered regarding the scope of singularity-based methods and several naturally
occurring sequences are proved to be non-holonomic.
Es ist eine Tatsache,da?die genauere Kenntnis
des Verhaltens einer analytischen Funktion
in der N¨a he ihrer singul¨a ren Stellen
eine Quelle von arithmetischen S¨a tzen ist.1
—Erich Hecke [25,Kap.VIII]
Introduction
A sequence (f n )n ≥0of complex numbers is said to be holonomic (or P -recursive )if it satis?es a linear recurrence with coe?cients that are polynomial in the index n ,that is,
(1)p 0(n )f n +d +p 1(n )f n +d ?1+···+p d (n )f n =0,
n ≥0,for some polynomials p j (X )∈C [X ].A formal power series f (z )= n ≥0f n z n is holonomic (or ?-?nite )if it satis?es a linear di?erential equation with coe?cients that are polynomial in the variable z ,that is,(2)q 0(z )d e dz e f (z )+q 1(z )d e ?1
dz
e ?1
f (z )+···+q e (z )f (z )=0.for some polynomials q k (X )∈C [X ].It is well known that a sequence is holonomic if and only if its generatin
g series is holonomic.(See Stanley’s book [36]for the basic properties of these sequences and series.)By extension,a function analytic at 0is called holonomic if its power series representation is itself holonomic.
Holonomic sequences encapsulate many of the combinatorial sequences of com-mon interest,for instance,a wide class of sums involving binomial coe?cients.At the same time,they enjoy a varied set of closure properties and several formal mechanisms have been recognised to lead systematically to holonomic sequences—what we have in mind here includes ?nite state models and regular grammars leading to rational (hence,holonomic)functions,context-free speci?cations leading Date :January 21,2005.1“It is a fact that the precise knowledge of the behaviour of an analytic function in the vicinity of its singular points is a source of arithmetic properties.”
1
2PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
to algebraic (hence,holonomic)functions,a wide class of problems endowed with symmetry conducive to holonomic functions (via Gessel’s theory [22]).On these aspects,we may rely on general references like [19,36]as well as on many works of Zeilberger,who is to be held accountable for unearthing the power of the holonomic framework accross combinatorics;see [31,42].Thus,in a way,a non-holonomicity result represents some sort of a structural complexity lower bound.
This note answers three problems described as open in an article of Stefan Ger-hold [21]very recently published in the Electronic Journal of Combinatorics .Proposition 1.The sequence f n =log n is not holonomic.
(For de?niteness,we agree that log 0≡0here.)
For αan integer,the sequence
h n =n α
is clearly holonomic.(As a matter of fact,the generating function is rational if α∈Z ≥0and of polylogarithmic type if α∈Z <0.)Gerhold [21]proved that for any αthat is rational but not integral,h n fails to be holonomic.For instance,h n =√n fails to be holonomic because,in essence,Q (√2,√3,...)is not a ?nite extension of Q .
Proposition 2.For α∈C ,the sequence of powers h n =n αis holonomic if and only if α∈Z .
(We agree that h 0=0.)
Proposition 3.The sequence g n de?ned by the fact that g n is the n th prime is non-holonomic.
(We agree that g 0=1,g 1=2,g 2=3,g 3=5,and so on.)
Proposition 1,conjectured by Gerhold in [21]was only proved under the assump-tion that a di?cult conjecture of number theory (Schanuel’s conjecture)holds.The author of [21]describes the statement of our Proposition 2as a “natural conjec-ture”.Proposition 3answers an explicit question of Gerhold who writes:“we do not know of any proof that the sequence of primes is non-holonomic”.
Our proofs are plainly based on the combination of two facts.First,holonomic objects satisfy rich closure properties.In particular,we make use of closure of under sum,product and composition with an algebraic function.Second,the asymptotic behaviour of holonomic sequences,which is re?ected by the asymptotic behaviour at singularities of their generating functions,is rather strongly constrained.For instance,iterated logarithms or negative powers of logarithms are “forbidden”and estimates like
(3)a n ～n →+∞log log n,b (z )～z →1?1log(1?z )
,are su?cient to conclude that the sequence (a n )and the function b (z )are non-holonomic.(See details below.)A conjunction of the previous two ideas then per-fectly describes the strategy of this note:In order to prove that a sequence is non-holonomic,it su?ces to locate a “derived sequence”(produced by holonomicity-preserving transformations)that exhibits a suitable combination of kosher asymp-totic terms with a foreign non-holonomic element,like in (3).
NON-HOLONOMY3 We choose here to operate directly with generating functions.Under this sce-nario,one can rely on the well established classi?cation of singularities elaborated at the end of the nineteenth century by Fuchs[20],Fabry[13]and others.A summary of what is known is found in standard treatises,for instance the ones by Wasow[39] and Henrici[26].What we need of this theory is summarised by Theorem2of the next section.
The relation between asymptotic behaviour of sequences and local behaviour of the generating functions is provided by a classical Abelian theorem,stated as Theorem3below.
Note1.A methodological remark is in order at this stage.A glance at(3)suggests two possible paths for proving non-holonomicity:one may a priori operate equally well with sequences or with generating functions.The latter is what we have opted to do here.The former approach with sequences seems workable,but it requires a strong structure theorem analogous to Theorem2below for recurrences,i.e., di?erence equations.An ambitious programme towards such a goal was undertaken by Birkho?and Trjitzinsky[5,6]in the1930’s,their works being later followed by Wimp and Zeilberger in[40].However,what is available in the classical literature is largely a set of formal solutions to di?erence equations and recurrences,and the relation of these to actual(analytic)solutions represents a di?cult problem evoked in[40,p.168]and[30,p.1138];see also[7]for recent results relying on multisummability. Note2.In this short paper,we do nothing but assemble some rather well-known facts of complex analysis,and logically organise them towards the goal of proving certain sequences to be non-holonomic.Our purpose is thus essentially pedagogical. As it should become transparent soon,a rough heuristic in this range of problem is the following:Almost anything is non-holonomic unless it is holonomic by design. (This na¨?ve remark cannot of course be universally true and there are surprises,e.g., some sequences may eventually admit algebraic or holonomic descriptions for rather deep reasons.Amongst such cases,we count the enumeration of k-regular graphs and various types of maps[22,23],the enumeration of permutations with bounded-length increasing subsequences,the Ap′e ry sequence[37]related to a continued fraction expansion ofζ(3),as well as the appearance of holonomic functions in the theory of modular forms,for which we refer to the beautiful exposition of Kontsevich-Zagier[27].)
1.Methods
From the most basic theorems regarding the existence of analytic solutions to di?erential equations(e.g.,[26,Th.9.1]),any function f(z)analytic at0that is holonomic can be continued analytically along any path that avoids the?nite setΣof points de?ned as roots of the equation p0(z)=0,where p0is the leading coe?cient in(1).Figuratively:
Theorem1(Finiteness of singularities).A holonomic function has only?nitely many singularities.
This theorem gives immediately as non-holonomic a number of sequences enu-merating classical combinatorial structures.
4PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
—Integer partitions,whose generating function is P(z)=
(1?z n)?1,as the
function admits the unit circle as a natural boundary.The same argument applies to integer partitions with summands restricted to any in?nite set
(e.g.,primes),partitions into distict summands,plane partitions,and so
on.More generally,combinatorial classes de?ned by an unlabelled set or multiset construction[19]are non-holonomic,unless a rather drastic com-binatorial simpli?cation occurs.
—Alternating(also known as zig-zag,up-and-down,cf2EIS A000111)per-mutations with exponential generating function tan z+sec z,as they have
the odd multiples ofπ
2as set of poles3.A similar argument applies to pref-
erential arrangements(also known as ordered set partitions or surjections, cf EIS A000670),Bernoulli numbers,and the like.
—Necklaces(equivalently Lyndon words,irreducible polynomials),whose gen-erating function admits the unit circle as a natural boundary.More gener-ally,“most”unlabelled cycles are non-holonomic.
—Unlabelled plane trees(EIS A000081),whose implicit speci?cation involves an unlabelled multiset construction.
In many cases,the criterion above is too brutal.For instance it does not preclude holonomicity for the Cayley tree function,
(4)T(z)=
n≥1n n?1
z n
n!
.
Indeed,the(multivalued)function T(z)has singularities at0,∞,e?1only.
A major theorem constrains the possible growth of a holonomic function near any of its singularities.Paraphrasing Theorem19.1of[39,p.111],we can state: Theorem2(Structure Theorem for singularities).Let there be given a di?erential equation of the form(2),a singular point z0,and a sector S with vertex at z0. Then,for z in a su?ciently narrow subsector S of S and for|z?z0|su?ciently small,there exists a basis of d linearly independent solutions to(2),such that any solution Y in that basis admits as z→z0in the subsector an asymptotic expansion of the form
(5)Y～exp
P(Z?1/r)
zα
∞
j=0
Q j(log Z)Z js,Z:=(z?z0),
where P is a polynomial,r an integer of Z≥0,αa complex,s a rational of Q>0, and the Q j are a family of polynomials of uniformly bounded degree.The quantities r,P,α,s,Q j depend on the particular solution and the formal asymptotic expansions of(5)are C-linearly independent.
(The argument is based on?rst constructing a formal basis of independent solu-tions,each of the form(5),and then applying to the possibly divergent expansions a summation mechanism that converts such formal solutions into actual analytic solutions.The restriction of the statement to a subsector is related to the Stokes phenomena associated to so-called“irregular”singularities.)
2In order to keep this note?nite,we refer to some of the combinatorial problems by means of their number in Sloane’s Encyclopedia of Integer Sequences(EIS),see[34].
3Stanley[35]describes an algebraic proof dependent on the fact that exp(z)is nonalgebraic (his Example4.5),then goes on to observe in his§4.a that sec z“has in?nitely many poles”.
NON-HOLONOMY5 This theorem implies that the sequence(n n?1/n!)(hence(n n))is non-holonomic. Indeed the Cayley tree function satis?es the functional equation
T(z)=ze T(z),
corresponding to the fact that it enumerates labelled nonplane trees.Set W(z)=?T(?z),which is otherwise known as the“Lambert W-function”.One has
W(x)=
x→+∞
log x?log log x+O(1),
as veri?ed by bootstrapping(see De Bruijn’s monograph[9,p.26]).This is enough to conclude that W,hence T,is non-holonomic as the log log(·)term is incompat-ible with Eq.(5).Observe that,conceptually,the proof involves considering the analytic continuation of T(z)and then extracting a clearly non-holonomic term in the expansion near a singularity.More of this in the next sections.
Amongst other applications,we may cite:
—Stanley’s children rounds[EIS A066166],with exponential generating func-tion(1?z)?z.The expansion as z→1,
(1?z)?z～
z→1
1
1?z
1+(1?z)log(1?z)+
(1?z)2log2(1?z)
2!
+···
,
contradicts the fact that logarithms can only appear with bounded degrees in holonomic functions.
—Bell numbers have OGF e e z?1.In this case,the double exponential be-haviour as z→+∞excludes them from the holonomic ring.
Finally,what is given is often a sequence rather than a function.Under such circumstances,it proves handy to be able to relate the asymptotic behaviour of f n as n→+∞to the asymptotic form of its generating function f(z),near a singularity. Such transfers exists and are widely known in the literature as Abelian theorems. We make use here of well-established principles in this theory,as found,e.g.,in the reference book by Bingham,Goldie,and Teugels[4].For convenience of exposition, we state explicitly one version used repeatedly here:
Theorem3(Basic Abelian theorem).Letφ(x)be any of the functions
(6)xα(log x)β(log log x)γ,α≥0,β,γ∈C.
Let(u n)be a sequence that satis?es the asymptotic estimate
u n～
n→∞
φ(n).
Then,the generating function,
u(z):=
n≥0
u n z n,
satis?es the asymptotic estimate
(7)u(z)～
z→1?Γ(α+1)
1
(1?z)
φ
1
1?z
.
This estimate remains valid when z tends to1in any sector with vertex at1, symmetric about the horizontal axis,and with opening angle<π.
6PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
Proof(sketch).We shall content ourselves here with brief indications since Corol-lary1.7.3p.40of[4]provides simultaneously the needed Abelian property and its real-analysis Tauberian converse4,at least in the case when z tends to1?along the real axis.
For simplicity,consider?rst the representative case whereφ(x)=log log x and one has exactly u n=φ(n)for n≥2,with u0=u1=0.Assume at this stage that z is real positive and set z=e?t,where t→0as z→1.We have
u(z)=
n≥2
φ(n)e?nt.
Take n1= t?1/log t?1 .Basic majorizations imply that the sum of the terms corresponding to n
(8)
n2
n=n1
φ(n)e?nt～
φ(1/t)
t
log t?1
1/log t?1
e?x dx～
log
log(1?z)?1
1?z
.
(Use the Euler-Maclaurin summation formula,then complete the tails.) The proof above applies when
z=e?t+iθ,with|θ|<θ0,
for someθ0<π
2.Once more only the central terms matter asymptotically;the in-
tegral is then to be taken along a line of angleθ,but it reduces to the corresponding integral along the positive real line,by virtue of the residue theorem.This su?ces to justify the extension of the estimate to sectors.The caseφ(x)=log log x is then settled.
The extension to u n=φ(n)whenφonly involves powers of logarithms and of iterated logarithms follows similar lines,as log n is also of slow variation.The inclusion of a power of n,in the form nαimplies that the integral in(8)should be modi?ed to include a factor tαleading to a real line integral that evaluates to the Gamma function,Γ(α+1).
Finally,simple modi?cations of the previous arguments show that if a sequence (v n)satis?es v n=o(φ(n)),forφ(n)any of the sequences of(6),then its generating function v(z)is a little-oh of the right hand side of(7).Decomposing u n=φ(n)+v n completes the proof of the statement. Note3.We have chosen to state the Abelian Theorem(Theorem3)for z varying in a cone of the complex plane,rather than the more customary real line.In this way we can avail ourselves of the comparatively simple Structure Theorem(as stated above in Theorem2)and avoid some of the possible hardships due to the Stokes phenomenon.
4The singularity analysis technology of Flajolet and Odlyzko[17,30]provides su?cient con-ditions for the converse complex-Tauberian implication.
NON-HOLONOMY 7
Here is a direct application of Theorem 3.Let π(x )be the number of primes less than or equal to x .By the Prime Number Theorem,one knows that
π(n )～n log n
.The Abelian Theorem permits us to conclude about the non-holonomic character of the sequence (π(n )),since
n ≥1
π(n )z n ～z →1?1(1?z )2log(1?z )?1
,which contradicts what the Structure Theorem permits.
Note 4.In this article,we concentrate on proofs of non-holonomicity based on analysis,that is,eventually,asymptotic approximations .On a di?erent register,powerful algebraic tools can be put to use in a number of situations.Consider-ations on power series have been used by Harris and Sibuya [24]to show:The reciprocal (1/f )of a holonomic function f is holonomic if and only if f /f is al-gebraic .For instance,this proves the non-holonomicity of the reciprocal of Gauss’2F 1hypergeometric,except in degenerate https://www.wendangku.net/doc/3e11845616.html,ing di?erential Galois theory,Singer generalized this result in [33].He characterized the possible polynomial re-lations between holonomic functions and also showed the following:A holonomic function f has to be algebraic if any of exp f or φ(f )is holonomic,with φan algebraic function of genus ≥1.An analogous result for sequences is given in [38,Chap.4]:If both f n and 1/f n are holonomic,then f n is an interlacing of hyperge-ometric sequences .
2.The logarithmic sequence
Let f n =log n (with log 0≡0)and let f (z )be its generating function,f (z )= n ≥1
(log n )z n .
We propose to show that a sequence derived from f n by means of holonomicity pre-serving transformations is non-holonomic.Consider a variant of the n th di?erence of the sequence f n ,namely
f n :=n k =1
n k (?1)k log k,whose ordinary generating function f (z ):= n ≥1 f n z n has positive radius of con-
vergence and satis?es (9) f (z )=11?z f ?z 1?z
.It is known that holonomic functions are closed under product and algebraic (hence
also,rational)substitutions.Thus,f and f
are such that either both of them are holonomic or none of them is holonomic.(See also Stanley’s paper [35,p.181]for a discussion of the fact that di?erencing preserves holonomicity.)
Next,Flajolet and Sedgewick proved in [18]that the sequence f n satis?es the
asymptotic estimate
(10) f n =log log n +O (1).
8PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
As a matter of fact,a full expansion is derived in [18,Th.4],based on the N¨o rlund-Rice integral representation [29]
(11) f n =(?1)n 2iπ H (log s )n !s (s ?1)···(s ?n )
ds,the use of a Hankel contour H ,and estimates akin to those used for the determi-nation of inverse Mellin transforms a?ected with an algebraic-logarithmic singular-ity [10].
The proof of Proposition 1can now be easily completed.By the Abelian estimate of Theorem 3applied to the asymptotic form (10)of f n ,we have f (z )～log log(1?z )?1 1?z
(z →1,z ∈S ),this in the whole of a sector S with vertex at 1and of opening angle <πextending towards the negative real axis symmetrically about the real axis.Assume a con-trario that f (z )is holonomic.Then f
(z ),which is associated to di?erences,is also holonomic.But then,given the Structure Theorem,a log-log asymptotic element valid in a subsector S can never result from a C -linear combination of elements,each having the form (5).A contradiction has thus been reached,and Proposition 1is established.
Note 5.We have presented our proof in a way that seems to depend on the imported estimate (10)of the logarithmic di?erences.In this way,we could save a few analytic steps.A conceptually equivalent and self-contained proof would proceed from the asymptotic behaviour of the analytic continuation of f (z )as z →?∞.(See our earlier discussion of T (z )for a similar situation.)This can be achieved directly by means of a Lindel¨o f integral representation,
f (?z )=12iπ 1/2+i ∞1/2?i ∞(lo
g s )z s πsin πs
ds.(See Lindel¨o f’s monograph [28]for explanations from the mouth of the master and Flajolet’s paper [15]for related developments.)It can then be veri?ed,by deforming the line of integration into a Hankel contour,that non-holonomic elements crop up in the asymptotic expansion of f (z )at ?∞.
3.The sequence of powers
The proof of Proposition 2relies once more on the consideration of diagonal di?erences.It has been established in [18]that
w n :=n k =1
n k (?1)k k α,satis?es,for α∈C \Z ,
w n =(log n )?α
Γ(1?α)
1+O 1log n .For instance,
∞ k =1 n
k (?1)k √k =1√πlog n
+O (log n )?3/2 .
NON-HOLONOMY 9
By the Abelian theorem,this implies for instance that the generating function w (z )
of w n satis?es,for α=1
2as z →1
?w (z )～1π1 log(1?z )?111?z
,while,for general α∈Z ,the dominant asymptotics involve a factor of the form log(1?z )?1 ?α.
Such a factor is foreign to what the Structure Theorem provides as legal holonomic asymptotics,as soon as αis nonintegral.This completes the proof of Proposition 2.
Proposition 2implies in particular that the power sequences,
n √17,n i =cos log n +i sin log n,n π,...,
are non-holonomic.Note that some of these sequences do occur as valid asymptotic approximations of holonomic sequences,some of which even appear in natural com-binatorial problems.For instance,it is proved in [16]that the expected cost of a
partial match in a quadtree is holonomic and has the asymptotic form n
(√17?3)/2.4.The n th prime function
Our proof of Proposition 3will similarly involve detecting,in the generating func-tion g (z )associated to primes,some elements that are incompatible with holonomy and contradict the conclusions of the Structure Theorem.
The n th prime function n →g n (often also written p (n ))is a much researched function.In a way,this function is an inverse of the function π(x )that gives the number of primes in the interval [1,x ].By the Prime Number Theorem,the function π(x )satis?es
π(x )=Li(x )+R (x ),
where R (x )is of an order smaller than the main term and Li(x )is the logarithmic integral,Li(x )= x 2dt log t
～x log x 1+1!log x +2!(log x )2+··· .The precise description of the remainder term R (x )depends upon the Riemann hypothesis.However,it is known unconditionally that R is small enough that the relation π(x )=y can be inverted asymptotically by just inverting the main term Li(x ).In this way,one obtains (see [32]and references therein)an estimate due to Cipolla [8],
(12)g n =n log n +n log log n +O (n ).
The log-log term is once more a barrier to holonomicity.
To see this,we introduce now the function,
(z )=z (1?z )2log 11?z +z (1?z )2
,which is clearly holonomic.It satis?es,with H n =1+
12+···1n denoting the
harmonic number,
n =nH n =n log n +O (n ).
Then,by taking a di?erence,one gets
g n ? n =n log log n +O (n ).
10PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
Thus,the di?erence g (z )? (z )satis?es g (z )?h (z )～z →1?
log log(1?z )?1 (1?z )2.This last fact is incompatible with holonomicity,by the very same argument as in the previous section.
5.Conclusion
As we have strived to illustrate and as is otherwise seen in many areas of math-ematics (see the opening quotation),singularities are central to the understanding of properties of numeric sequences.It should be clear by now that a large number of sequences can be proved to be non-holonomic.Here is a brief recapitulation of methods and possible extensions.
1.Asymptotic discrepancies .In order to conclude that a sequence (u n )is non-holonomic,the following conditions are su?cient:(i )the generating function of the sequence (u n )(or one of its cognates)admits,near a singularity,an asymptotic expansion in a scale that involves logarithms and iterated logarithms;(ii )at least one term in that expansion is an iterated logarithm or a power of a logarithm with an exponent not in Z ≥0.It is then apparent from our earlier developments that sequences like n 7+1,1H n , log n H n ,log p (2n )p (n ),n √n +log n
,p (n 2),(H n the harmonic number and p (n )the n th prime function)fail to be holonomic.Also,techniques of this note extend easily to other slowly varying sequences,like
e √log n ,log log log n,...,
as well as to any sequence that involves any such term somewhere in its asymptotic expansion.
The singularity-based technology has otherwise been used to establish the non-algebraic character of sequences arising from combinatorics and the theory of for-mal languages in [14].Once more,such transcendence results imply that,struc-turally,the corresponding objects cannot be (unambiguously)encoded by words of a context-free language.For instance,two-dimensional walks on a regular lat-tice that are constrained to the ?rst quadrant cannot be described (via a length-preserving encoding)by means of an unambiguous context-free grammar.This property is neatly visible from a logarithmic component in the generating function of walks [19],which contradicts the Structure Theorem for algebraic functions (also known as Newton-Puiseux!).
2.In?nitude of singularities .A famous theorem of P′o lya and Carlson [3]
implies the following 5:A function analytic at the origin having integer coe?cients
and assumed to converge in the open unit disc is either a rational function or else it admits the unit circle as a natural boundary.Consider then the generating function g (z )of g n ≡p (n ),the n th prime function,which can be subjected to P′o lya-Carlson.Either it has a natural boundary,in which case it cannot be holonomic,since holonomic functions have isolated singularities.Else,it is rational;but this would be a clear contradiction,since no rational function can have coe?cients of
5Stefan Gerhold is grateful to Richard Stanley for pointing out the P′o lya-Carlson connection.
NON-HOLONOMY11 the asymptotic form n log n(by virtue of the Abelian Theorem,say).This proof6 seems to require the Prime Number Theorem but many weaker bounds that are elementary(e.g.,the ones due to Chebyshev)are su?cient for this purpose.It is pleasant to note that the P′o lya-Carlson Theorem was earlier employed in a similar fashion[2]in order to establish a structural lower bound in the theory of formal languages.
3.Arithmetic discrepancies7.In[14,p.294],a sketch was given of the solution to a conjecture of Stanley[35]to the e?ect that the generating function
S k(z)=
n≥0
2n
n
k
z n,
is transcendental for odd values of k,k=3,5,....(As already noted by Stanley,S k is clearly transcendental for k even because of the presence of logarithmic factors.) Longer algebraic proofs have since been published,see[41]and[1]for a discussion. It su?ces to remark,as shown by an Abelian argument,that the local expansion at the?nite singularity of S k for k=2 +1odd,satis?es
(13)d
dz
S2 +1
z
42 +1
～
z→1?
c π?
1
√
1?z
c ∈Q.
This asymptotic form is incompatible with algebraicity.Indeed,if S k were algebraic over C(z),it would be algebraic over Q(z),as it has rational Taylor coe?cients(this, by a famous lemma of Eisenstein[3]).But in that case,its Puiseux expansion could only involve algebraic numbers.Equation(13)contradicts this,by virtue of the transcendence ofπ.Et voila!In this last case,we have to play not only with the shape of an asymptotic expansion,but also with the arithmetic nature of its coe?cients.
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6Alternatively,Erd?o s,Maxsein,and Smith[11]have shown that an integer recurrent sequence [i.e.,one with a rational generating function]which consists only of primes is necessarily a periodic sequence,therefore involving only?nitely many di?erent values.We are indebted to Dr Alin Bostan for this remark.
7See the recent monograph of Everest al.[12]for a compendium of results relative to recurrence sequences.
12PHILIPPE FLAJOLET,STEF AN GERHOLD,AND BRUNO SAL VY
9.N.G.de Bruijn,Asymptotic methods in analysis,Dover,1981,A reprint of the third North
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puter Science49(1987),283–309.
15.,Singularity analysis and asymptotics of Bernoulli sums,Theoretical Computer Sci-
ence215(1999),no.1-2,371–381.
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quadtrees,Algorithmica10(1993),no.7,473–500.
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Philippe Flajolet:Algorithms Project,INRIA Rocquencourt,F-78153Le Chesnay (France),Philippe.Flajolet AT inria.fr
Stefan Gerhold:Research Institute for Symbolic Computation,Johannes Kepler University Linz(Austria),stefan.gerhold AT risc.uni-linz.ac.at
Bruno Salvy:Algorithms Project,INRIA Rocquencourt,F-78153Le Chesnay(France), Bruno.Salvy AT inria.fr
on the contrary的解析
On the contrary Onthecontrary, I have not yet begun. 正好相反，我还没有开始。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, the instructions have been damaged. 反之，则说明已经损坏。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I understand all too well. 恰恰相反，我很清楚 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I think this is good. ⑴我反而觉得这是好事。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I have tons of things to do 正相反，我有一大堆事要做 Provided by jukuu Is likely onthecontrary I in works for you 反倒像是我在为你们工作 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, or to buy the first good. 反之还是先买的好。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, it is typically american. 相反，这正是典型的美国风格。 222.35.143.196 Onthecontrary, very exciting.
恰恰相反，非常刺激。 https://www.wendangku.net/doc/3e11845616.html, But onthecontrary, lazy. 却恰恰相反，懒洋洋的。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I hate it! 恰恰相反，我不喜欢！ https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, the club gathers every month. 相反，俱乐部每个月都聚会。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I'm going to work harder. 我反而将更努力工作。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, his demeanor is easy and nonchalant. 相反，他的举止轻松而无动于衷。 https://www.wendangku.net/doc/3e11845616.html, Too much nutrition onthecontrary can not be absorbed through skin. 太过营养了反而皮肤吸收不了. https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, I would wish for it no other way. 正相反，我正希望这样 Provided by jukuu Onthecontrary most likely pathological. 反之很有可能是病理性的。 https://www.wendangku.net/doc/3e11845616.html, Onthecontrary, it will appear clumsy. 反之，就会显得粗笨。 https://www.wendangku.net/doc/3e11845616.html,
英语造句
一般过去式 时间状语：yesterday just now (刚刚) the day before three days ag0 a week ago in 1880 last month last year 1. I was in the classroom yesterday. I was not in the classroom yesterday. Were you in the classroom yesterday. 2. They went to see the film the day before. Did they go to see the film the day before. They did go to see the film the day before. 3. The man beat his wife yesterday. The man didn’t beat his wife yesterday. 4. I was a high student three years ago. 5. She became a teacher in 2009. 6. They began to study english a week ago 7. My mother brought a book from Canada last year. 8.My parents build a house to me four years ago . 9.He was husband ago. She was a cooker last mouth. My father was in the Xinjiang half a year ago. 10.My grandfather was a famer six years ago. 11.He burned in 1991
学生造句--Unit 1
●I wonder if it’s because I have been at school for so long that I’ve grown so crazy about going home. ●It is because she wasn’t well that she fell far behind her classmates this semester. ●I can well remember that there was a time when I took it for granted that friends should do everything for me. ●In order to make a difference to society, they spent almost all of their spare time in raising money for the charity. ●It’s no pleasure eating at school any longer because the food is not so tasty as that at home. ●He happened to be hit by a new idea when he was walking along the riverbank. ●I wonder if I can cope with stressful situations in life independently. ●It is because I take things for granted that I make so many mistakes. ●The treasure is so rare that a growing number of people are looking for it. ●He picks on the weak mn in order that we may pay attention to him. ●It’s no pleasure being disturbed whena I settle down to my work. ●I can well remember that when I was a child, I always made mistakes on purpose for fun. ●It’s no pleasure accompany her hanging out on the street on such a rainy day. ●I can well remember that there was a time when I threw my whole self into study in order to live up to my parents’ expectation and enter my dream university. ●I can well remember that she stuck with me all the time and helped me regain my confidence during my tough time five years ago. ●It is because he makes it a priority to study that he always gets good grades. ●I wonder if we should abandon this idea because there is no point in doing so. ●I wonder if it was because I ate ice-cream that I had an upset student this morning. ●It is because she refused to die that she became incredibly successful. ●She is so considerate that many of us turn to her for comfort. ●I can well remember that once I underestimated the power of words and hurt my friend. ●He works extremely hard in order to live up to his expectations. ●I happened to see a butterfly settle on the beautiful flower. ●It’s no pleasure making fun of others. ●It was the first time in the new semester that I had burned the midnight oil to study. ●It’s no pleasure taking everything into account when you long to have the relaxing life. ●I wonder if it was because he abandoned himself to despair that he was killed in a car accident when he was driving. ●Jack is always picking on younger children in order to show off his power. ●It is because he always burns the midnight oil that he oversleeps sometimes. ●I happened to find some pictures to do with my grandfather when I was going through the drawer. ●It was because I didn’t dare look at the failure face to face that I failed again. ●I tell my friend that failure is not scary in order that she can rebound from failure. ●I throw my whole self to study in order to pass the final exam. ●It was the first time that I had made a speech in public and enjoyed the thunder of applause. ●Alice happened to be on the street when a UFO landed right in front of her. ●It was the first time that I had kept myself open and talked sincerely with my parents. ●It was a beautiful sunny day. The weather was so comfortable that I settled myself into the
英语句子结构和造句
高中英语~词性~句子成分~语法构成 第一章节：英语句子中的词性 1.名词：n. 名词是指事物的名称，在句子中主要作主语.宾语.表语.同位语。 2.形容词;adj. 形容词是指对名词进行修饰~限定~描述~的成份，主要作定语.表语.。形容词在汉语中是（的）.其标志是: ous. Al .ful .ive。. 3.动词：vt. 动词是指主语发出的一个动作，一般用来作谓语。 4.副词：adv. 副词是指表示动作发生的地点. 时间. 条件. 方式. 原因. 目的. 结果.伴随让步. 一般用来修饰动词. 形容词。副词在汉语中是（地）.其标志是：ly。 5.代词：pron. 代词是指用来代替名词的词，名词所能担任的作用，代词也同样.代词主要用来作主语. 宾语. 表语. 同位语。 6.介词：prep.介词是指表示动词和名次关系的词，例如：in on at of about with for to。其特征：
介词后的动词要用—ing形式。介词加代词时，代词要用宾格。例如：give up her（him）这种形式是正确的，而give up she（he）这种形式是错误的。 7.冠词：冠词是指修饰名词，表名词泛指或特指。冠词有a an the 。 8.叹词：叹词表示一种语气。例如：OH. Ya 等 9.连词：连词是指连接两个并列的成分，这两个并列的成分可以是两个词也可以是两个句子。例如：and but or so 。 10.数词：数词是指表示数量关系词，一般分为基数词和序数词 第二章节：英语句子成分 主语：动作的发出者，一般放在动词前或句首。由名词. 代词. 数词. 不定时. 动名词. 或从句充当。 谓语：指主语发出来的动作，只能由动词充当，一般紧跟在主语后面。 宾语：指动作的承受着，一般由代词. 名词. 数词. 不定时. 动名词. 或从句充当. 介词后面的成分也叫介词宾语。 定语：只对名词起限定修饰的成分，一般由形容
六级单词解析造句记忆MNO
M A: Has the case been closed yet? B: No, the magistrate still needs to decide the outcome. magistrate n.地方行政官，地方法官，治安官 A: I am unable to read the small print in the book. B: It seems you need to magnify it. magnify vt.１．放大，扩大；２．夸大，夸张 A: That was a terrible storm. B: Indeed, but it is too early to determine the magnitude of the damage. magnitude n.１．重要性，重大；２．巨大，广大 A: A young fair maiden like you shouldn’t be single. B: That is because I am a young fair independent maiden. maiden n.少女，年轻姑娘，未婚女子 a.首次的，初次的 A: You look majestic sitting on that high chair. B: Yes, I am pretending to be the king! majestic a.雄伟的，壮丽的，庄严的，高贵的 A: Please cook me dinner now. B: Yes, your majesty, I’m at your service. majesty n.１．［Ｍ－］陛下（对帝王，王后的尊称）；２．雄伟，壮丽，庄严 A: Doctor, I traveled to Africa and I think I caught malaria. B: Did you take any medicine as a precaution? malaria n.疟疾 A: I hate you! B: Why are you so full of malice? malice n.恶意，怨恨 A: I’m afraid that the test results have come back and your lump is malignant. B: That means it’s serious, doesn’t it, doctor? malignant a.１．恶性的，致命的；２．恶意的，恶毒的 A: I’m going shopping in the mall this afternoon, want to join me? B: No, thanks, I have plans already. mall n.（由许多商店组成的）购物中心 A: That child looks very unhealthy. B: Yes, he does not have enough to eat. He is suffering from malnutrition.
base on的例句
意见应以事实为根据. 3 来自辞典例句 192. The bombers swooped ( down ) onthe air base. 轰炸机 突袭 空军基地. 来自辞典例句 193. He mounted their engines on a rubber base. 他把他们的发动机装在一个橡胶垫座上. 14 来自辞典例句 194. The column stands on a narrow base. 柱子竖立在狭窄的地基上. 14 来自辞典例句 195. When one stretched it, it looked like grey flakes on the carvas base. 你要是把它摊直, 看上去就象好一些灰色的粉片落在帆布底子上. 18 来自辞典例句 196. Economic growth and human well - being depend on the natural resource base that supports all living systems. 经济增长和人类的福利依赖于支持所有生命系统的自然资源. 12 1 来自辞典例句 197. The base was just a smudge onthe untouched hundred - mile coast of Manila Bay. 那基地只是马尼拉湾一百英里长安然无恙的海岸线上一个硝烟滚滚的污点. 6 来自辞典例句 198. You can't base an operation on the presumption that miracles are going to happen. 你不能把行动计划建筑在可能出现奇迹的假想基础上.
英语造句大全
英语造句大全English sentence 在句子中，更好的记忆单词！ 1、(1)、able adj. 能 句子：We are able to live under the sea in the future. (2)、ability n. 能力 句子：Most school care for children of different abilities. (3)、enable v. 使。。。能句子：This pass enables me to travel half-price on trains. 2、(1)、accurate adj. 精确的句子：We must have the accurate calculation. (2)、accurately adv. 精确地 句子：His calculation is accurately. 3、(1)、act v. 扮演 句子：He act the interesting character. （2）、actor n. 演员 句子：He was a famous actor. (3)、actress n. 女演员 句子：She was a famous actress. (4)、active adj. 积极的 句子：He is an active boy. 4、add v. 加 句子：He adds a little sugar in the milk. 5、advantage n. 优势 句子：His advantage is fight. 6、age 年龄n. 句子：His age is 15. 7、amusing 娱人的adj. 句子：This story is amusing. 8、angry 生气的adj. 句子：He is angry. 9、America 美国n.
(完整版)主谓造句
主语+谓语 1. 理解主谓结构 1) The students arrived. The students arrived at the park. 2) They are listening. They are listening to the music. 3) The disaster happened. 2．体会状语的位置 1) Tom always works hard. 2) Sometimes I go to the park at weekends.. 3) The girl cries very often. 4) We seldom come here. The disaster happened to the poor family. 3. 多个状语的排列次序 1) He works. 2) He works hard. 3) He always works hard. 4) He always works hard in the company. 5) He always works hard in the company recently. 6) He always works hard in the company recently because he wants to get promoted. 4. 写作常用不及物动词 1. ache My head aches. I’m aching all over. 2. agree agree with sb. about sth. agree to do sth. 3. apologize to sb. for sth. 4. appear (at the meeting, on the screen) 5. arrive at / in 6. belong to 7. chat with sb. about sth. 8. come (to …) 9. cry 10. dance 11. depend on /upon 12. die 13. fall 14. go to … 15. graduate from 16. … happen 17. laugh 18. listen to... 19. live 20. rise 21. sit 22. smile 23. swim 24. stay (at home / in a hotel) 25. work 26. wait for 汉译英： 1.昨天我去了电影院。 2.我能用英语跟外国人自由交谈。 3.晚上7点我们到达了机场。 4.暑假就要到了。 5.现在很多老人独自居住。 6.老师同意了。 7.刚才发生了一场车祸。 8.课上我们应该认真听讲。9. 我们的态度很重要。 10. 能否成功取决于你的态度。 11. 能取得多大进步取决于你付出多少努力。 12. 这个木桶能盛多少水取决于最短的一块板子的长度。
初中英语造句
【it's time to和it's time for】 ——————这其实是一个句型,只不过后面要跟不同的东西. ——————It's time to跟的是不定式（to do）.也就是说,要跟一个动词,意思是“到做某事的时候了”.如： It's time to go home. It's time to tell him the truth. ——————It's time for 跟的是名词.也就是说,不能跟动词.如： It's time for lunch.(没必要说It's time to have lunch) It's time for class.(没必要说It's time to begin the class.) They can't wait to see you Please ask liming to study tonight. Please ask liming not to play computer games tonight. Don’t make/let me to smoke I can hear/see you dance at the stage You had better go to bed early. You had better not watch tv It’s better to go to bed early It’s best to run in the morning I am enjoy running with music. With 表伴随听音乐 I already finish studying You should keep working. You should keep on studying English Keep calm and carry on 保持冷静继续前行二战开始前英国皇家政府制造的海报名字 I have to go on studying I feel like I am flying I have to stop playing computer games and stop to go home now I forget/remember to finish my homework. I forget/remember cleaning the classroom We keep/percent/stop him from eating more chips I prefer orange to apple I prefer to walk rather than run I used to sing when I was young What’s wrong with you There have nothing to do with you I am so busy studying You are too young to na?ve I am so tired that I have to go to bed early
The Kite Runner-美句摘抄及造句
《The Kite Runner》追风筝的人--------------------------------美句摘抄 1.I can still see Hassan up on that tree, sunlight flickering through the leaves on his almost perfectly round face, a face like a Chinese doll chiseled from hardwood: his flat, broad nose and slanting, narrow eyes like bamboo leaves, eyes that looked, depending on the light, gold, green even sapphire 翻译：我依然能记得哈桑坐在树上的样子，阳光穿过叶子，照着他那浑圆的脸庞。他的脸很像木头刻成的中国娃娃，鼻子大而扁平，双眼眯斜如同竹叶，在不同光线下会显现出金色、绿色，甚至是宝石蓝。 E.g.: A shadow of disquiet flickering over his face. 2.Never told that the mirror, like shooting walnuts at the neighbor's dog, was always my idea. 翻译：从来不提镜子、用胡桃射狗其实都是我的鬼主意。E.g.:His secret died with him, for he never told anyone. 3.We would sit across from each other on a pair of high
翻译加造句
一、翻译 1. The idea of consciously seeking out a special title was new to me., but not without appeal. 让我自己挑选自己最喜欢的书籍这个有意思的想法真的对我具有吸引力。 2.I was plunged into the aching tragedy of the Holocaust, the extraordinary clash of good, represented by the one decent man, and evil. 我陷入到大屠杀悲剧的痛苦之中，一个体面的人所代表的善与恶的猛烈冲击之中。 3.I was astonished by the the great power a novel could contain. I lacked the vocabulary to translate my feelings into words. 我被这部小说所包含的巨大能量感到震惊。我无法用语言来表达我的感情（心情）。 4，make sth. long to short长话短说 5.I learned that summer that reading was not the innocent(简单的) pastime(消遣) I have assumed it to be., not a breezy, instantly forgettable escape in the hammock(吊床)，( though I’ ve enjoyed many of those too ). I discovered that a book, if it arrives at the right moment, in the proper season, will change the course of all that follows. 那年夏天，我懂得了读书不是我认为的简单的娱乐消遣，也不只是躺在吊床上，一阵风吹过就忘记的消遣。我发现如果在适宜的时间、合适的季节读一本书的话，他将能改变一个人以后的人生道路。 二、词组造句 1. on purpose 特意，故意 This is especially true here, and it was ~. (这一点在这里尤其准确，并且他是故意的) 2.think up 虚构，编造，想出 She has thought up a good idea. 她想出了一个好的主意。 His story was thought up. 他的故事是编出来的。 3. in the meantime 与此同时 助记：in advance 事前in the meantime 与此同时in place 适当地... In the meantime, what can you do? 在这期间您能做什么呢？ In the meantime, we may not know how it works, but we know that it works. 在此期间，我们不知道它是如何工作的，但我们知道，它的确在发挥作用。 4.as though 好像，仿佛 It sounds as though you enjoyed Great wall. 这听起来好像你喜欢长城。 5. plunge into 使陷入 He plunged the room into darkness by switching off the light. 他把灯一关，房
改写句子练习2标准答案
The effective sentences:(improve the sentences!) 1.She hopes to spend this holiday either in Shanghai or in Suzhou. 2.Showing/to show sincerity and to keep/keeping promises are the basic requirements of a real friend. 3.I want to know the space of this house and when it was built. I want to know how big this house is and when it was built. I want to know the space of this house and the building time of the house. 4.In the past ten years,Mr.Smith has been a waiter,a tour guide,and taught English. In the past ten years,Mr.Smith has been a waiter,a tour guide,and an English teacher. 5.They are sweeping the floor wearing masks. They are sweeping the floor by wearing masks. wearing masks,They are sweeping the floor. 6.the drivers are told to drive carefully on the radio. the drivers are told on the radio to drive carefully 7.I almost spent two hours on this exercises. I spent almost two hours on this exercises. 8.Checking carefully,a serious mistake was found in the design. Checking carefully,I found a serious mistake in the design.
用以下短语造句
M1 U1 一. 把下列短语填入每个句子的空白处（注意所填短语的形式变化）： add up (to) be concerned about go through set down a series of on purpose in order to according to get along with fall in love (with) join in have got to hide away face to face 1 We’ve chatted online for some time but we have never met ___________. 2 It is nearly 11 o’clock yet he is not back. His mother ____________ him. 3 The Lius ___________ hard times before liberation. 4 ____________ get a good mark I worked very hard before the exam. 5 I think the window was broken ___________ by someone. 6 You should ___________ the language points on the blackboard. They are useful. 7 They met at Tom’s party and later on ____________ with each other. 8 You can find ____________ English reading materials in the school library. 9 I am easy to be with and _____________my classmates pretty well. 10 They __________ in a small village so that they might not be found. 11 Which of the following statements is not right ____________ the above passage? 12 It’s getting dark. I ___________ be off now. 13 More than 1,000 workers ___________ the general strike last week. 14 All her earnings _____________ about 3,000 yuan per month. 二.用以下短语造句： 1.go through 2. no longer/ not… any longer 3. on purpose 4. calm… down 5. happen to 6. set down 7. wonder if 三. 翻译： 1.曾经有段时间，我对学习丧失了兴趣。（there was a time when…） 2. 这是我第一次和她交流。（It is/was the first time that …注意时态） 3.他昨天公园里遇到的是他的一个老朋友。（强调句） 4. 他是在知道真相之后才意识到错怪女儿了。（强调句） M 1 U 2 一. 把下列短语填入每个句子的空白处（注意所填短语的形式变化）： play a …role (in) because of come up such as even if play a …part (in) 1 Dujiangyan(都江堰) is still ___________in irrigation(灌溉) today. 2 That question ___________ at yesterday’s meeting. 3 Karl Marx could speak a few foreign languages, _________Russian and English. 4 You must ask for leave first __________ you have something very important. 5 The media _________ major ________ in influencing people’s opinion s. 6 _________ years of hard work she looked like a woman in her fifties. 二.用以下短语造句： 1.make (good/full) use of 2. play a(n) important role in 3. even if 4. believe it or not 5. such as 6. because of
英语造句
English sentence 1、(1)、able adj. 能 句子：We are able to live under the sea in the future. (2)、ability n. 能力 句子：Most school care for children of different abilities. (3)、enable v. 使。。。能 句子：This pass enables me to travel half-price on trains. 2、(1)、accurate adj. 精确的 句子：We must have the accurate calculation. (2)、accurately adv. 精确地 句子：His calculation is accurately. 3、(1)、act v. 扮演 句子：He act the interesting character.（2）、actor n. 演员 句子：He was a famous actor. (3)、actress n. 女演员 句子：She was a famous actress. (4)、active adj. 积极的 句子：He is an active boy. 4、add v. 加 句子：He adds a little sugar in the milk. 5、advantage n. 优势 句子：His advantage is fight. 6、age 年龄n. 句子：His age is 15. 7、amusing 娱人的adj. 句子：This story is amusing. 8、angry 生气的adj. 句子：He is angry. 9、America 美国n. 句子：He is in America. 10、appear 出现v. He appears in this place. 11. artist 艺术家n. He is an artist. 12. attract 吸引 He attracts the dog. 13. Australia 澳大利亚 He is in Australia. 14.base 基地 She is in the base now. 15.basket 篮子 His basket is nice. 16.beautiful 美丽的 She is very beautiful. 17.begin 开始 He begins writing. 18.black 黑色的 He is black. 19.bright 明亮的 His eyes are bright. 20.good 好的 He is good at basketball. 21.British 英国人 He is British. 22.building 建造物 The building is highest in this city 23.busy 忙的 He is busy now. 24.calculate 计算 He calculates this test well. 25.Canada 加拿大 He borns in Canada. 26.care 照顾 He cared she yesterday. 27.certain 无疑的 They are certain to succeed. 28.change 改变 He changes the system. 29.chemical 化学药品