ProfesseurÉmérite,SorbonneUniversité,InstitutdeMathématiquesdeJussieu,Paris Interpolationofanalyticfunctionsandarithmeticapplications.MichelWaldschmidt December6-8,2019UniversityofDhaka,DepartmentofAppliedMathematicsBangladeshMathematicalSocietyThe21t

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December 6 - 8, 2019

University of Dhaka, Department of Applied Mathematics Bangladesh Mathematical Society

The 21th international Mathematics Conference 2019

Interpolation of analytic functions and arithmetic applications.

Michel Waldschmidt

Professeur Émérite, Sorbonne Université, Institut de Mathématiques de Jussieu, Paris http://www.imj-prg.fr/~michel.waldschmidt/

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Abstract

The very first interpolation formula for analytic functions is given by Taylor series. There are many other ways of interpolating analytic functions. Lagrange interpolation polynomials involve the values of the function at several points ; some derivatives may be included. We discuss other types of interpolation formulae, starting with Lidstone interpolation of a function of exponential type< π by its derivatives of even order at 0and 1. A new result is a lower bound for the exponential type of a transcendental entire function having derivatives of even order at two points taking integer values.

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The interpolation problem

An entire function is a holomorphic (=analytic) mapC→C. The graph{(z, f(z)) | z ∈C}has the power of continuum.

However, such a function is uniquely determined by a

countable set ; for instance by the sequence of coefficients of its Taylor series at a given pointz₀ :

f(z) =X

n≥0

f⁽ⁿ⁾(z₀)(z−z₀)ⁿ n! · Notation :

f⁽ⁿ⁾(z) = dⁿ dzⁿf(z).

There are other sequences of numbers which determine uniquely an entire function, at least if we restrict to some classes of entire functions.

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The interpolation problem

countable set ;for instance by the sequence of coefficients of its Taylor series at a given pointz₀ :

f(z) =X

n≥0

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The interpolation problem

f(z) =X

n≥0

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The interpolation problem

f(z) =X

n≥0

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References on interpolation :

Whittaker, J. M. (1935).

Interpolatory function theory, volume 33.

Cambridge University Press, Cambridge.

Gel’fond, A. O. (1952).

Calculus of finite differences.

Hindustan Publishing Corporation. VI (1971).

Boas, Jr., R. P. and Buck, R. C. (1964).

Polynomial expansions of analytic functions.

Ergebnisse der Mathematik und ihrer Grenzgebiete, 19, Springer-Verlag .

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Interpolation data

Given complex numbers{σ_i}i∈I, {a_i}i∈I and nonnegative integers{k_i}_i∈I, theinterpolation problem is to decide whether there exists an analytic functionf satisfying

f^(kⁱ⁾(σ_i) = a_i for all i∈I.

We will consider this question for f analytic everywhere in C (i.e.f an entire function) and I =N.

The unicity is given by the answer to the same question with a_i = 0 for all i∈I.

Taylor series: σ_n= 0 and k_n=n for all n ≥0. The solution, if it exists, is unique

f(z) = X

n≥0

a_nzⁿ

n!, f⁽ⁿ⁾(0) =a_n.

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Interpolation data

f(z) = X

n≥0

a_nzⁿ

n!, f⁽ⁿ⁾(0) =a_n.

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Interpolation data

f(z) = X

n≥0

a_nzⁿ

n!, f⁽ⁿ⁾(0) =a_n.

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Calculus of finite differences

Another classical interpolation problem is given by the data k_n = 0 and σ_n =n for all n≥0. Given complex numbersa_n, does there exist an entire functionf satisfying

f(n) =a_n for all n≥0?

The answer depends on the growth of the sequence (a_n)n≥0. The example of the functionsin(πz) shows that the solution is not unique in general. However we recover unicity by adding a condition on the growth of the solutionf.

For the existence, one uses interpolation formulae based on f(z) =f(0) +zf₁(z), f₁(z) =f₁(1) + (z−1)f₂(z),

fn(z) =fn(n) + (z−n)fn+1(z), . . .

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Calculus of finite differences

Another classical interpolation problem is given by the data k_n = 0 and σ_n =n for all n≥0. Given complex numbers a_n, does there exist an entire functionf satisfying

fn(z) =fn(n) + (z−n)fn+1(z), . . .

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Calculus of finite differences

fn(z) =fn(n) + (z−n)fn+1(z), . . .

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Calculus of finite differences

fn(z) =fn(n) + (z−n)fn+1(z), . . .

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Further interpolation problems

We are going to consider the following interpolation problems :

I (Lidstone) :

f⁽²ⁿ⁾(0) =a_n, f⁽²ⁿ⁾(1) =b_n forn ≥0.

I (Whittaker) :

f⁽²ⁿ⁺¹⁾(0) =a_n, f⁽²ⁿ⁾(1) =b_n forn ≥0.

I (Poritsky) : Form ≥2 and σ₀, . . . , σ_m−1 in C,

f^(mn)(σ_j) = a_nj for n≥0 and j = 0,1, . . . , m−1.

I (Gontcharoff) : For(σ_n)n≥0 a sequence of complex numbers,

f⁽ⁿ⁾(σ_n) =a_n forn ≥0.

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Lidstone interpolation problem

The following interpolation problem was considered by G.J. Lidstone in 1930.

Given two sequences of complex numbers (an)n≥0 and (b_n)n≥0, does there exist an entire functionf satisfying

f⁽²ⁿ⁾(0) =a_n, f⁽²ⁿ⁾(1) =b_n forn ≥0 ? Is such a function f unique ?

The answer to unicity is plain : the functionsin(πz) satisfies these conditions witha_n =b_n= 0, hence there is no unicity, unless we restrict the question to entire functions satisfying some extra condition. Such a condition is a bound on the growth of f.

We start with unicity (a_n =b_n= 0) and polynomials.

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Lidstone interpolation problem

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Lidstone interpolation problem

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Lidstone interpolation problem

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Lidstone interpolation problem

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Lidstone interpolation problem

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Even derivatives at 0 and 1 : first proof

Lemma.Letf be a polynomial satisfying

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Thenf = 0.

First proof.

By induction on the degree of the polynomialf.

If f has degree ≤1, say f(z) = a₀z+a₁, the conditions f(0) =f(1) = 0imply a₀ =a₁ = 0, hence f = 0.

If f has degree ≤n with n ≥2 and satisfies the hypotheses, then f⁰⁰ also satisfies the hypotheses and has degree< n, hence by inductionf⁰⁰= 0 and thereforef has degree ≤1.

The result follows.

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Even derivatives at 0 and 1 : first proof

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Thenf = 0.

First proof.

The result follows.

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Even derivatives at 0 and 1 : first proof

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Thenf = 0.

First proof.

The result follows.

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Even derivatives at 0 and 1 : first proof

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Thenf = 0.

First proof.

The result follows.

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Even derivatives at 0 and 1 : first proof

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Thenf = 0.

First proof.

The result follows.

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Even derivatives at 0 and 1 : second proof

Second proof.

Letf be a polynomial satisfying

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

The assumptionf⁽²ⁿ⁾(0) = 0 for all n≥0 means that f is an odd function : f(−z) =−f(z).The assumption f⁽²ⁿ⁾(1) = 0 for all n≥0 means thatf(1−z) is an odd function :

f(1−z) =−f(1 +z). We deduce

f(z+ 2) =f(1 +z+ 1) =−f(1−z−1) =−f(−z) = f(z), hence the polynomialf is periodic, and therefore it is a

constant. Sincef(0) = 0, we conclude f = 0.

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Even derivatives at 0 and 1 : second proof

Second proof.

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

The assumptionf⁽²ⁿ⁾(0) = 0 for all n≥0 means that f is an odd function : f(−z) =−f(z). The assumption f⁽²ⁿ⁾(1) = 0 for all n≥0 means thatf(1−z) is an odd function :

f(1−z) =−f(1 +z).We deduce

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Even derivatives at 0 and 1 : second proof

Second proof.

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

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Even derivatives at 0 and 1 : second proof

Second proof.

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

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Even derivatives at 0 and 1 : second proof

Second proof.

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

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Even derivatives at 0 and 1 : third proof

Third proof.

Assume

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Write

f(z) = a₁z+a₃z³+a₅z⁵+a₇z⁷+ +· · ·+a_2n+1z²ⁿ⁺¹+· · ·

(finite sum).We have f(1) =f⁰⁰(1) =f^(ıv)(1) =· · ·= 0 :

a₁ +a₃ +a₅ +a₇ +· · · +a_2n+1 +· · ·= 0

6a3 +20a5 +42a7 +· · · +2n(2n+ 1)a2n+1 +· · ·= 0

120a5 +840a7 +· · · +^(2n+1)!_(2n−3)!a2n+1 +· · ·= 0

. .. ...

The matrix of this system is triangular with maximal rank.

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Even derivatives at 0 and 1 : third proof

Third proof.

Assume

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Write

f(z) = a₁z+a₃z³+a₅z⁵+a₇z⁷+ +· · ·+a_2n+1z²ⁿ⁺¹+· · ·

(finite sum). We have f(1) =f⁰⁰(1) =f^(ıv)(1) =· · ·= 0 :

a₁ +a₃ +a₅ +a₇ +· · · +a_2n+1 +· · ·= 0

6a3 +20a5 +42a7 +· · · +2n(2n+ 1)a2n+1 +· · ·= 0 120a5 +840a7 +· · · +^(2n+1)!_(2n−3)!a2n+1 +· · ·= 0

. .. ...

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Even derivatives at 0 and 1 : third proof

Third proof.

Assume

f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all n≥0.

Write

f(z) = a₁z+a₃z³+a₅z⁵+a₇z⁷+ +· · ·+a_2n+1z²ⁿ⁺¹+· · ·

(finite sum). We have f(1) =f⁰⁰(1) =f^(ıv)(1) =· · ·= 0 :

a₁ +a₃ +a₅ +a₇ +· · · +a_2n+1 +· · ·= 0

6a3 +20a5 +42a7 +· · · +2n(2n+ 1)a2n+1 +· · ·= 0 120a5 +840a7 +· · · +^(2n+1)!_(2n−3)!a2n+1 +· · ·= 0

. .. ...

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Even derivatives at 0 and 1

The fact that this matrix has maximal rank means that a polynomialf is uniquely determined by the numbers

f⁽²ⁿ⁾(0) and f⁽²ⁿ⁾(1) for n≥0.

Given numbers an and bn, all but finitely many of them are0, there is a unique polynomial f such that

f⁽²ⁿ⁾(0) =a_n and f⁽²ⁿ⁾(1) =b_n for all n ≥0.

Involution :z 7→1−z :

07→1, 17→0, 1−z 7→z.

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Even derivatives at 0 and 1

f⁽²ⁿ⁾(0) and f⁽²ⁿ⁾(1) for n≥0.

07→1, 17→0, 1−z 7→z.

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Even derivatives at 0 and 1

f⁽²ⁿ⁾(0) and f⁽²ⁿ⁾(1) for n≥0.

07→1, 17→0, 1−z 7→z.

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Even derivatives at 0 and 1

f⁽²ⁿ⁾(0) and f⁽²ⁿ⁾(1) for n≥0.

07→1, 17→0, 1−z 7→z.

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Lidstone expansion of a polynomial

G. J. Lidstone (1930). There exists a unique sequence of polynomialsΛ₀(z),Λ₁(z),Λ₂(z), . . . such that any polynomial f can be written as a finite sum

f(z) =X

n≥0

f⁽²ⁿ⁾(0)Λn(1−z) +X

n≥0

f⁽²ⁿ⁾(1)Λn(z).

This is equivalent to

Λ^(2k)_n (0) = 0 and Λ^(2k)_n (1) =δ_nk for n≥0 and k ≥0.

(Kronecker symbol).

A basis of the Q–space of polynomials in Q[z] of degree

≤2n+ 1 is given by the 2n+ 2 polynomials

Λ₀(z),Λ₁(z), . . . ,Λ_n(z), Λ₀(1−z),Λ₁(1−z), . . . ,Λ_n(1−z).

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Lidstone expansion of a polynomial

f(z) =X

n≥0

f⁽²ⁿ⁾(0)Λn(1−z) +X

n≥0

f⁽²ⁿ⁾(1)Λn(z).

(Kronecker symbol).

Λ₀(z),Λ₁(z), . . . ,Λ_n(z), Λ₀(1−z),Λ₁(1−z), . . . ,Λ_n(1−z).

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Lidstone expansion of a polynomial

f(z) =X

n≥0

f⁽²ⁿ⁾(0)Λn(1−z) +X

n≥0

f⁽²ⁿ⁾(1)Λn(z).

(Kronecker symbol).

Λ₀(z),Λ₁(z), . . . ,Λ_n(z), Λ₀(1−z),Λ₁(1−z), . . . ,Λ_n(1−z).

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Analogy with Taylor series

Given a sequence(a_n)n≥0 of complex numbers, the unique analytic solution (if it exists)f of the interpolation problem

f⁽ⁿ⁾(0) =a_n for all n≥0 is given by the Taylor expansion

f(z) =X

n≥0

a_nzⁿ n!·

The polynomials zⁿ/n! satisfy d^k

dz^k zⁿ

n!

z=0

=δ_nk for n≥0 and k ≥0.

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Lidstone polynomials

Λ₀(z) =z :

Λ₀(0) = 0, Λ₀(1) = 1, Λ⁽²ⁿ⁾₀ (0) = 0 forn ≥1.

Induction : the sequence of Lidstone polynomials is determined byΛ₀(z) =z and

Λ⁰⁰_n = Λn−1 for n≥1

with the initial conditionsΛ_n(0) = Λ_n(1) = 0 forn≥1.

LetL_n(z) be any solution of

L⁰⁰_n(z) = Λn−1(z).

Define

Λ_n(z) = −L_n(1)z+L_n(z).

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Lidstone polynomials

Λ₀(z) =z :

Λ₀(0) = 0, Λ₀(1) = 1, Λ⁽²ⁿ⁾₀ (0) = 0 forn ≥1.

L⁰⁰_n(z) = Λn−1(z).

Define

Λ_n(z) = −L_n(1)z+L_n(z).

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Lidstone polynomials

Λ₀(z) =z :

Λ₀(0) = 0, Λ₀(1) = 1, Λ⁽²ⁿ⁾₀ (0) = 0 forn ≥1.

L⁰⁰_n(z) = Λn−1(z).

Define

Λ_n(z) = −L_n(1)z+L_n(z).

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Lidstone polynomials

Λ₀(z) =z,

Λ⁰⁰_n = Λn−1, Λ_n(0) = Λ_n(1) = 0 forn ≥1.

Forn ≥0, the polynomial Λn is odd, it has degree 2n+ 1 and leading term _(2n+1)!¹ z²ⁿ⁺¹.

For instance

Λ₁(z) = 1

6(z³−z) and

Λ₂(z) = 1

120z⁵ − 1

36z³+ 7

360z = 1

360z(z² −1)(3z²−7).

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Lidstone polynomials

Λ₀(z) =z,

Λ⁰⁰_n = Λn−1, Λ_n(0) = Λ_n(1) = 0 forn ≥1.

For instance

Λ₁(z) = 1

6(z³−z) and

Λ₂(z) = 1

120z⁵ − 1

36z³+ 7

360z = 1

360z(z² −1)(3z²−7).

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Lidstone polynomials

Λ₀(z) =z,

Λ⁰⁰_n = Λn−1, Λ_n(0) = Λ_n(1) = 0 forn ≥1.

For instance

Λ₁(z) = 1

6(z³−z) and

Λ₂(z) = 1

120z⁵ − 1

36z³+ 7

360z = 1

360z(z² −1)(3z²−7).

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Lidstone polynomials

The polynomial f(z) = z²ⁿ⁺¹ satisfies

f^(2k)(0) = 0 fork ≥0, f^(2k)(1) =

( _(2n+1)!

(2n−2k+1)! for0≤k ≤n, 0 fork ≥n+ 1.

One deduces

z²ⁿ⁺¹ =

n−1

X

k=0

(2n+ 1)!

(2n−2k+ 1)!Λ_k(z) + (2n+ 1)!Λ_n(z), which yields the induction formula

Λ_n(z) = 1

(2n+ 1)!z²ⁿ⁺¹−

n−1

X

k=0

1

(2n−2k+ 1)!Λ_k(z).

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Lidstone polynomials

The polynomial f(z) = z²ⁿ⁺¹ satisfies

f^(2k)(0) = 0 fork ≥0, f^(2k)(1) =

( _(2n+1)!

(2n−2k+1)! for0≤k ≤n, 0 fork ≥n+ 1.

One deduces

z²ⁿ⁺¹ =

n−1

X

k=0

(2n+ 1)!

(2n−2k+ 1)!Λ_k(z) + (2n+ 1)!Λ_n(z), which yields the induction formula

Λ_n(z) = 1

(2n+ 1)!z²ⁿ⁺¹−

n−1

X

k=0

1

(2n−2k+ 1)!Λ_k(z).

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Order and exponential type

Orderof an entire function :

%(f) = lim sup

r→∞

log log|f|_r

logr where |f|_r = sup

|z|=r

|f(z)|.

Exponential typeof an entire function : τ(f) = lim sup

r→∞

log|f|_r r ·

If the exponential type is finite, thenf has order ≤1. If f has order<1, then the exponential type is 0.

Forτ ∈C\ {0}, the function e^{τ z} has order 1 and exponential type |τ|.

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Order and exponential type

%(f) = lim sup

r→∞

log log|f|_r

|z|=r

|f(z)|.

r→∞

log|f|_r r ·

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Order and exponential type

%(f) = lim sup

r→∞

log log|f|_r

|z|=r

|f(z)|.

r→∞

log|f|_r r ·

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Order and exponential type

%(f) = lim sup

r→∞

log log|f|_r

|z|=r

|f(z)|.

r→∞

log|f|_r r ·

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Order and exponential type

%(f) = lim sup

r→∞

log log|f|_r

|z|=r

|f(z)|.

r→∞

log|f|_r r ·

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Exponential type

An alternative definition of the exponential type is the following : f is of exponential type τ(f)if and only if, for all z₀ ∈C,

lim sup

n→∞

|f⁽ⁿ⁾(z₀)|^1/n =τ(f).

The equivalence between the two definitions follows from Cauchy’s inequalities and Stirling’s Formula.

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Exponential type

An alternative definition of the exponential type is the following : f is of exponential type τ(f)if and only if, for all z₀ ∈C,

lim sup

n→∞

|f⁽ⁿ⁾(z₀)|^1/n =τ(f).

The equivalence between the two definitions follows from Cauchy’s inequalities and Stirling’s Formula.

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Lidstone series : exponential type < π

Theorem (H. Poritsky, 1932).

Letf be an entire function of exponential type < π satisfying f⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0 for all sufficiently largen. Thenf is a polynomial.

This is best possible : the entire function sin(πz)has

exponential typeπ and satisfiesf⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0for all n≥0.

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Lidstone series : exponential type < π

Theorem (H. Poritsky, 1932).

This is best possible : the entire function sin(πz)has

exponential typeπ and satisfiesf⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(1) = 0for all n≥0.

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Lidstone series : exponential type < π

Proof.

Letf˜ =f −P, where P is the polynomial satisfying P⁽²ⁿ⁾(0) =f⁽²ⁿ⁾(0) and P⁽²ⁿ⁾(1) =f⁽²ⁿ⁾(1) forn≥0.

We havef˜⁽²ⁿ⁾(0) = ˜f⁽²ⁿ⁾(1) = 0 for all n≥0. The functions

˜f(z) and f(1˜ −z) are odd, hencef˜(z)is periodic of period 2.

Therefore there exists an entire functiong such that

˜f(z) =g(e^iπz). Since f˜(z) has exponential type < π, we deduce f˜= 0 and f =P.

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Lidstone series : exponential type < π

Proof.

We havef˜⁽²ⁿ⁾(0) = ˜f⁽²ⁿ⁾(1) = 0 for all n≥0.The functions

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Lidstone series : exponential type < π

Proof.

We havef˜⁽²ⁿ⁾(0) = ˜f⁽²ⁿ⁾(1) = 0 for all n≥0. The functions

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Exponential type < π : Poritsky’s expansion

Theorem (H. Poritsky, 1932).

The expansion

f(z) =

∞

X

n=0

f⁽²ⁿ⁾(0)Λ_n(1−z) +

∞

X

n=0

f⁽²ⁿ⁾(1)Λ_n(z)

holds for any entire functionf of exponential type < π.

We will check this formula for f_t(z) = e^tz with |t|< π, then deduce the general case.

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Solution of the Lidstone interpolation problem

Consequence of Poritsky’s expansion formula :

Let(a_n)_n≥0 and (b_n)_n≥0 be two sequences of complex numbers satisfying

lim sup

n→∞

|an|^1/n < π and lim sup

n→∞

|bn|^1/n < π.

Then the function f(z) =

∞

X

n=0

a_nΛ_n(1−z) +

∞

X

n=0

b_nΛ_n(z)

is the unique entire function of exponential type< π satisfying f⁽²ⁿ⁾(0) =a_n and f⁽²ⁿ⁾(1) =b_n for all n ≥0.

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Special case : e

^tz

for |t| < π

Consider Poritsky’s expansion formula f(z) =

∞

X

n=0

f⁽²ⁿ⁾(0)Λ_n(1−z) +

∞

X

n=0

f⁽²ⁿ⁾(1)Λ_n(z)

for the functionf_t(z) = e^tz where|t|< π.Since f⁽²ⁿ⁾_t (0) =t²ⁿ and f⁽²ⁿ⁾_t (1) =t²ⁿe^t it gives

e^tz =

∞

X

n=0

t²ⁿΛ_n(1−z) + e^t

∞

X

n=0

t²ⁿΛ_n(z).

Replacing t with −t yields e^−tz =

∞

X

n=0

t²ⁿΛ_n(1−z) + e^−t

∞

X

n=0

t²ⁿΛ_n(z).

Hence

e^tz −e^−tz = (e^t−e^−t)

∞

X

n=0

t²ⁿΛ_n(z).

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Special case : e

^tz

for |t| < π

Consider Poritsky’s expansion formula f(z) =

∞

X

n=0

f⁽²ⁿ⁾(0)Λ_n(1−z) +

∞

X

n=0

f⁽²ⁿ⁾(1)Λ_n(z)

for the functionf_t(z) = e^tz where|t|< π. Since f⁽²ⁿ⁾_t (0) =t²ⁿ and f⁽²ⁿ⁾_t (1) =t²ⁿe^t it gives

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z).

Replacing t with −t yields e^−tz =

∞

X

n=0

t²ⁿΛ_n(1−z) + e^−t

∞

X

n=0

t²ⁿΛ_n(z).

Hence

e^tz −e^−tz = (e^t−e^−t)

∞

X

n=0

t²ⁿΛ_n(z).

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Generating series

Lett∈C, t6∈iπZ. The entire function f(z) = sinh(tz)

sinh(t) = e^tz−e^−tz e^t−e^−t satisfies

f⁰⁰=t²f , f(0) = 0, f(1) = 1, hence f⁽²ⁿ⁾(0) = 0 and f⁽²ⁿ⁾(1) =t²ⁿ for all n ≥0.

For0<|t|< π and z ∈C, we deduce sinh(tz)

sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

Notice that

e^tz = sinh((1−z)t)

sinh(t) + e^t sinh(tz) sinh(t) ·

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Generating series

sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

Notice that

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Generating series

sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

Notice that

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Special case : e

^tz

From Poritsky’s expansion of an entire function of exponential type < π we deduced the formula

sinh(tz) sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

Let us prove this formula directly.

We will deduce

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z)

for|t|< π.

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Special case : e

^tz

sinh(tz) sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

We will deduce

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z)

for|t|< π.

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Special case : e

^tz

sinh(tz) sinh(t) =

∞

X

n=0

t²ⁿΛ_n(z).

We will deduce

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z)

for|t|< π.

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

Forz ∈C and |t|< π let

F(z, t) = sinh(tz) sinh(t)

with F(z,0) =z.

Fix z∈C. The function t7→F(z, t) is analytic in the disc

|t|< π and is an even function : F(z,−t) = F(z, t). Consider its Taylor series at the origin :

F(z, t) = X

n≥0

c_n(z)t²ⁿ

with c₀(z) = z.

We haveF(0, t) = 0 and F(1, t) = 1.

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

with F(z,0) =z.

F(z, t) = X

n≥0

c_n(z)t²ⁿ

with c₀(z) = z.

We haveF(0, t) = 0 and F(1, t) = 1.

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

with F(z,0) =z.

F(z, t) = X

n≥0

c_n(z)t²ⁿ

with c₀(z) = z.

We haveF(0, t) = 0 and F(1, t) = 1.

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

F(z, t) = e^tz −e^−tz

e^t−e^−t =X

n≥0

cn(z)t²ⁿ.

From

c_n(z) = 1 (2n)!

∂

∂t 2n

F(z,0)

it follows thatc_n(z)is a polynomial.

From

∂

∂z 2

F(z, t) = t²F(z, t)

we deduce

c⁰⁰_n=c_n−1 for n≥1.

Sincec_n(0) =c_n(1) = 0 forn≥1we concludec_n(z) = Λ_n(z).

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

e^t−e^−t =X

n≥0

cn(z)t²ⁿ.

From

c_n(z) = 1 (2n)!

∂

∂t 2n

F(z,0)

From

∂

∂z 2

F(z, t) = t²F(z, t)

we deduce

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Expansion of F (z, t) = sinh(tz)/ sinh(t)

e^t−e^−t =X

n≥0

cn(z)t²ⁿ.

From

c_n(z) = 1 (2n)!

∂

∂t 2n

F(z,0)

From

∂

∂z 2

F(z, t) = t²F(z, t)

we deduce

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From e

^tz

to exponential type < π

Hence a special case of the Poritsky’s expansion formula

f(z) =

∞

X

n=0

f⁽²ⁿ⁾(0)Λ_n(1−z) +

∞

X

n=0

f⁽²ⁿ⁾(1)Λ_n(z),

which holds for any entire function f of exponential type < π, is

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z)

for|t|< π.

Conversely, from this special case (that we proved directly) we are going to deduce the general case by means of Laplace transform (R.C. Buck, 1955, kernel expansion method).

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From e

^tz

to exponential type < π

Hence a special case of the Poritsky’s expansion formula

f(z) =

∞

X

n=0

f⁽²ⁿ⁾(0)Λ_n(1−z) +

∞

X

n=0

f⁽²ⁿ⁾(1)Λ_n(z),

which holds for any entire function f of exponential type < π, is

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z)

for|t|< π.

Conversely, from this special case (that we proved directly) we are going to deduce the general case by means of Laplace transform (R.C. Buck, 1955, kernel expansion method).

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Laplace transform

Let

f(z) = X

n≥0

a_n n!zⁿ

be an entire function of exponential typeτ(f). The Laplace transform off, viz.

F(t) = X

n≥0

a_nt⁻ⁿ⁻¹,

is analytic in the domain|t|> τ(f). From Cauchy’s residue Theorem, it follows that forr > τ(f) we have

f(z) = 1 2πi

Z

|t|=r

e^tzF(t)dt.

Hence

f⁽²ⁿ⁾(z) = 1 2πi

Z

|t|=r

t²ⁿe^tzF(t)dt.

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Laplace transform

Let

f(z) = X

n≥0

a_n n!zⁿ

F(t) = X

n≥0

a_nt⁻ⁿ⁻¹,

f(z) = 1 2πi

Z

|t|=r

e^tzF(t)dt.

Hence

f⁽²ⁿ⁾(z) = 1 2πi

Z

|t|=r

t²ⁿe^tzF(t)dt.

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Laplace transform

Let

f(z) = X

n≥0

a_n n!zⁿ

F(t) = X

n≥0

a_nt⁻ⁿ⁻¹,

f(z) = 1 2πi

Z

|t|=r

e^tzF(t)dt.

Hence

f⁽²ⁿ⁾(z) = 1 2πi

Z

|t|=r

t²ⁿe^tzF(t)dt.

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Laplace transform

Let

f(z) = X

n≥0

a_n n!zⁿ

F(t) = X

n≥0

a_nt⁻ⁿ⁻¹,

f(z) = 1 2πi

Z

|t|=r

e^tzF(t)dt.

Hence

f⁽²ⁿ⁾(z) = 1 2πi

Z

|t|=r

t²ⁿe^tzF(t)dt.

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Laplace transform

Assumeτ(f)< π. Let r satisfyτ(f)< r < π. For |t|=r we have

e^tz =

∞

X

n=0

∞

X

n=0

t²ⁿΛ_n(z).

We deduce f(z) = X

n≥0

Λ_n(1−z) 1

2πi Z

|t|=r

t²ⁿF(t)dt

+

X

n≥0

Λ_n(z) 1

2πi Z

|t|=r

t²ⁿe^tF(t)dt

and therefore f(z) =X

n≥0

f⁽²ⁿ⁾(0)Λ_n(1−z) +X

n≥0

f⁽²ⁿ⁾(1)Λ_n(z),

where the last series are absolutely and uniformly convergent forz on any compact in C.