non-holonomic sequences and functions

Non-Holonomic Sequences and Functions

Stefan Gerhold

May 16, 2006

Holonomic Sequences and Functions

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

I Rather general way to specify concrete functions/sequences infinite terms

I Several algorithms are available for the symbolic manipulationof holonomic functions and sequences.

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

p0(n)an + · · ·+ pd(n)an+d = 0

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

p0(n)an + · · ·+ pd(n)an+d = 0

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

p0(n)an + · · ·+ pd(n)an+d = 0

Examples and Basic Properties

I Examples of holonomic functions: rational functions, exp, sin,cos, Bessel, etc.

I Examples of holonomic sequences: αn, n!,∑n

k=1 1/k,](involutions), etc.

I Sum and Product of two holonomic sequences (functions) areholonomic

I an is holonomic iff∑

n≥0 anzn is holonomic

I Definition can be extended to several variables(n1, . . . , nr , z1, . . . , zs)

Proving Non-Holonomicity

I Holonomic sequences ≈ algebraic numbers

I Annihilated by operators instead of polynomials

I If a sequence (function) is not obviously holonomic, it isusually not holonomic

I But how to come up with a rigorous proof?

I Lower bound on the algorithmic complexity of a sequence(function)

I Relations to many areas of mathematics:

I Analytic combinatorics, complex analysis, algebraic geometry,number theory

I Many ways to define sequences (functions) =⇒ ampleopportunities for research

Two Easy arguments

I For each holonomic sequence an there is α such thatan = O(n!α)

I Hence 22nand 2n2

are not holonomic (What about nn ?).

I Other argument: Singularities of a holonomic function f (z)are roots of pd(z).

I Hence tan z , z/(ez −1), and∏

(1− zn)−1 are not holonomic.

Two Easy arguments

I Hence 22nand 2n2

Two Easy arguments

I Hence 22nand 2n2

Two Easy arguments

I Hence 22nand 2n2

Infinitely Many Singularities

I We show that the sequence (ζ(n))n≥2 of values of theRiemann zeta function is not holonomic.

feven(z) :=∑n≥1

ζ(2n)zn, fodd(z) :=∑n≥1

ζ(2n + 1)zn

I Euler:feven(z) = 1

2 −π2 z cotπz

I Molteni (2001):

feven(z)/z − fodd(z) = γ − ψ(1 + z),

where ψ(z) = Γ′(z)/Γ(z) is the digamma function.

feven(z) :=∑n≥1

ζ(2n + 1)zn

2 −π2 z cotπz

I Molteni (2001):

feven(z) :=∑n≥1

ζ(2n + 1)zn

2 −π2 z cotπz

I Molteni (2001):

feven(z) :=∑n≥1

ζ(2n + 1)zn

2 −π2 z cotπz

I Molteni (2001):

Known Non-Trivial Results

I Harris, Sibuya (1985): The reciprocal 1/f of a holonomicfunction is holonomic iff f ′/f is algebraic.

I Singer (1986): Similar results about exponentiation andcomposition

I Van der Put, Singer (1997): The reciprocal 1/an of aholonomic sequence is holonomic iff an is an interlacement ofhypergeometric sequences.

I Polya-Carlson (1921): If f (z) has integer coefficients andradius of convergence 1, then f (z) is rational or has the unitcircle for its natural boundary.

Proofs by Number Theory (SG, 2004)

I W.l.o.g. the polynomials pk(n) have coefficients inQ(aj : j ≥ 0).

I an =√

n does not satisfy a recurrence

p0(n)an + · · ·+ pd(n)an+d = 0

with coefficients in Q(√

j : j ≥ 0)[n], since

[Q(√ρ1, . . . ,

√ρs) : Q] = 2s

for distinct primes ρ1, . . . , ρs .

I Other argument: transcendence of e implies non-holonomicityof nn.

I an =√

p0(n)an + · · ·+ pd(n)an+d = 0

[Q(√ρ1, . . . ,

√ρs) : Q] = 2s

I an =√

p0(n)an + · · ·+ pd(n)an+d = 0

[Q(√ρ1, . . . ,

√ρs) : Q] = 2s

Proofs by Asymptotics (P. Flajolet, SG, B. Salvy, 2005)

I Fuchs-Frobenius theory: Asymptotic expansion of holonomicfunctions as |z | → ∞ must be linear combination of series ofthe form

eP(z1/r )zα∑j≥0

Qj(log z)z−js ,

where P and Qj are polynomials, the Qj have boundeddegree, r ∈ N, α ∈ C, 0 < s ∈ Q.

I Hence log log z , eez−1 , and Lambert W are not holonomic.

Proofs by Asymptotics (P. Flajolet, SG, B. Salvy, 2005)

I Fuchs-Frobenius theory: Asymptotic expansion of holonomicfunctions as |z | → ∞ must be linear combination of series ofthe form

eP(z1/r )zα∑j≥0

Qj(log z)z−js ,

where P and Qj are polynomials, the Qj have boundeddegree, r ∈ N, α ∈ C, 0 < s ∈ Q.

I Hence log log z , eez−1 , and Lambert W are not holonomic.

Proofs by Asymptotics

I Basic Abelian theorem. Let φ(x) be any of the functions

xα(log x)β(log log x)γ , α ≥ 0, β, γ ∈ C. (1)

Let (an) be a sequence that satisfies the asymptotic estimate

an ∼n→∞

φ(n).

Then the generating function f (z) :=∑

n≥0 anzn satisfies

the asymptotic estimate

f (z) ∼z→1−

Γ(α+ 1)1

(1− z)φ

1− z

). (2)

I The sequence of prime numbers:

n-th prime = n log n + n log log n + O(n),

hence(n-th prime)/n −Hn ∼ log log n.

I The sequences of powers (α ∈ C \ Z):

n∑k=1

)(−1)kkα ∼ (log n)−α

Γ(1− α).

I The sequence of prime numbers:

n-th prime = n log n + n log log n + O(n),

hence(n-th prime)/n −Hn ∼ log log n.

I The sequences of powers (α ∈ C \ Z):

n∑k=1

)(−1)kkα ∼ (log n)−α

Γ(1− α).

e1/n by Asymptotics (P. Flajolet, SG, B. Salvy, 2005)

I Lindelof integral representation

∑n≥1

e1/n(−z)n = − 1

∫ 1/2+i∞

1/2−i∞

zse1/s

sinπsds

I Asymptotics (saddle point method)

∑n≥1

e1/n(−z)n ∼ − e2√

2√π(log z)1/4

as |z | → ∞.

I Hence e1/n is not holonomic.

I Work in progress: generalize asymptotics to αnβ.

∑n≥1

e1/n(−z)n = − 1

∫ 1/2+i∞

1/2−i∞

zse1/s

sinπsds

∑n≥1

e1/n(−z)n ∼ − e2√

2√π(log z)1/4

as |z | → ∞.

∑n≥1

e1/n(−z)n = − 1

∫ 1/2+i∞

1/2−i∞

zse1/s

sinπsds

∑n≥1

e1/n(−z)n ∼ − e2√

2√π(log z)1/4

as |z | → ∞.

∑n≥1

e1/n(−z)n = − 1

∫ 1/2+i∞

1/2−i∞

zse1/s

sinπsds

∑n≥1

e1/n(−z)n ∼ − e2√

2√π(log z)1/4

as |z | → ∞.

Closed-Form Sequences (J.P. Bell, SG, M. Klazar, F. Luca,2006)

I For sequences like nα, e1/n, there is a more direct way toexploit the additional structure

I an = f (n) for a smooth function f

I The recurrence means that

F (z ; p0, . . . , pd) :=d∑

pk(z)f (z + k)

vanishes at z = 0, 1, 2, . . . .

I Example: f (z) = zα. If F vanishes identically, the left handside of

f (z) = − 1

d∑k=1

pk(z)f (z + k)

is meromorphic at z = 0, hence α ∈ Z.

F (z ; p0, . . . , pd) :=d∑

pk(z)f (z + k)

vanishes at z = 0, 1, 2, . . . .

f (z) = − 1

d∑k=1

pk(z)f (z + k)

F (z ; p0, . . . , pd) :=d∑

pk(z)f (z + k)

vanishes at z = 0, 1, 2, . . . .

f (z) = − 1

d∑k=1

pk(z)f (z + k)

Zeroes of Analytic and of Elementary Functions

I Carlson’s theorem (1921). The function sinπz is the“smallest” function analytic for <(z) ≥ 0 that vanishes atz = 0, 1, 2, . . . .

I Shows non-holonomicity of nα, log n, e1/n, . . .

I Khovanskiı investigates the geometry of the zero set ofelementary functions in his book “Fewnomials”.

I Definition. Elementary functions are built by composingrational functions, exp(x), log(x), sin(x), cos(x), tan x ,arcsin(x), arccos(x), and arctan(x). The domain of definitionmust be such that arguments of sin and cos are bounded.

I Theorem (Khovanskiı). An elementary function has onlyfinitely many simple zeros in its domain of definition.

Results proved using Carlson or Khovanskiı

I For distinct complex u1, . . . , us , the sequenceΓ(n − u1)

α1 . . . Γ(n − us)αs is holonomic if and only if

α1, . . . , αs are integers.

I If a sequence from R(n, en) is holonomic, then thedenominator has just one summand.

I If (f (n))n≥1 is holonomic for an algebraic functionf : ]1,∞] → R, then f is a rational function.

Conclusion

I “Negative” results, but “positive” ones obtained along the way

I Opportunity to apply methods and results from various areas

I There is a good chance that holonomicity of a sequence canbe decided if it has (i) a closed form representation or (ii) aknown asymptotic expansion.

Conclusion

non-holonomic sequences and functions

Documents