1. The polynomials Λ

Michel Waldschmidt Interpolation, January 2021

Exercices: hints, solutions, comments Third course

1.Lets0, s1, s2 be three complex numbers. Give a necessary and sufficient condition for the following to hold.

There exist three sequences of polynomials(Λn,0(z))n≥0,(Λn,1(z))n≥0,(Λn,2(z))n≥0 such that any polynomial f∈C[z]can be written in a unique way as a finite sum

f(z) =X

n≥0

f⁽³ⁿ⁾(s0)Λn,0(z) +f⁽³ⁿ⁾(s1)Λn,1(z) +f⁽³ⁿ⁾(s2)Λn,2(z)

.

What is the degree ofΛn,j(z)? The leading term ? Write the six polynomials Λ0,0(z),Λ0,1(z),Λ0,2(z),Λ1,0(z),Λ1,1(z),Λ1,2(z).

1. The polynomials Λ

(n ≥ 0, i = 0, 1, 2) are defined by

Λ

(s

) = δ

δ

, n, k ≥ 0, i, j = 0, 1, 2.

By symmetry, it suffices to deal with i = 0.

• We start with n = 0 : a necessary and sufficient condition for the existence of a polynomial Λ

satisfying Λ

= 0 and

Λ

(s

) = 1, Λ

(s

) = Λ

(s

) = 0

is s

6= s

and s

6= s

. Assume from now on that s

, s

, s

are pairwise distinct. Then there is a unique such polynomial, it has degree 2 and is given by the Lagrange interpolation formula, namely

Λ

(z) = (z − s

)(z − s

) (s

− s

)(s

− s

) ·

• For n ≥ 1 the polynomial Λ

is the unique polynomial satisfying the differential equation Λ

= Λ

with the initial conditions

Λ

(s

) = Λ

(s

) = Λ

(s

) = 0.

It has degree 3n + 2 and leading term

z

.

• We explicit the solution for n = 1. The polynomial L(z) = 1

60

z

(s

− s

)(s

− s

) − 1 24

(s

+ s

)z

(s

− s

)(s

− s

) + 1 6

s

s

z

(s

− s

)(s

− s

) satisfies

L

(z) = z

(s

− s

)(s

− s

) − (s

+ s

)z

(s

− s

)(s

− s

) + s

s

(s

− s

)(s

− s

) = Λ

(z).

Hence the unique polynomial Λ

solution of the differential equation Λ

= Λ

with the initial conditions

Λ

(s

) = Λ

(s

) = Λ

(s

) = 0

is Λ

(z) = L(z) − (c

z

+ c

z + c

) where c

, c

, c

are the solutions of the system of equations







c

s

+ c

s

+ c

= L(s

), c

s

+ c

s

+ c

= L(s

), c

s

+ c

s

+ c

= L(s

).

2.Lets0, s1, s2 be three complex numbers. Give a necessary and sufficient condition for the following to hold.

There exist three sequences of polynomials(Mn,0(z))n≥0,(Mn,1(z))n≥0,(Mn,2(z))n≥0such that any polynomial f∈C[z]can be written in a unique way as a finite sum

f(z) =X

n≥0

f⁽³ⁿ⁾(s0)Mn,0(z) +f⁽³ⁿ⁺¹⁾(s1)Mn,1(z) +f⁽³ⁿ⁺²⁾(s2)Mn,2(z) .

What is the degree ofMn,j(z)? The leading term ? Write the six polynomials M0,0(z), M0,1(z), M0,2(z), M1,0(z), M1,1(z), M1,2(z).

2. The polynomials M

(n ≥ 0, i = 0, 1, 2) are defined by

M

(s

) = δ

δ

, n, k ≥ 0, i, j = 0, 1, 2.

As we will see, there is no condition for the existence and unicity of such polynomials.

• We start with n = 0.

The unique polynomial M

(z) satisfying M

= 0 and M

(s

) = 1, M

(s

) = M

(s

) = 0 is the constant polynomial M

(z) = 1.

The unique polynomial M

(z) satisfying M

= 0 and M

(s

) = M

(s

) = 0, M

(s

) = 1 is M

(z) = z − s

.

The unique polynomial M

(z) satisfying M

= 0 and M

(s

) = M

(s

) = 0, M

(s

) = 1 is

M

(z) = 1

2 z

− s

z + 1

2 s

(2s

− s

).

• Let n ≥ 1. For i = 0, 1, 2, the polynomial M

is the unique solution of the differential equation M

= M

with the initial condition

M

(s

) = M

(s

) = M

(s

) = 0.

For n ≥ 0 and i = 0 the solution is given by M

(z) =

z

.

The leading term of M

is

z

and the leading term of M

is

z

.

• We explicit the solution for n = 1.

The polynomial

A(z) = 1 24 z

− 1

6 s

z

satisfies

A

(z) = z − s

= M

(z).

Hence the unique polynomial M

solution of the differential equation M

= M

with the initial conditions

M

(s

) = M

(s

) = M

(s

) = 0,

is M

(z) = A(z) − (az

+ bz + c) where a, b, c are the solutions of the system of equations







as

+ bs

+ c = A(s

), 2as

+ b = A

(s

), 2a = A

(s

).

The polynomial

B(z) = 1

120 z

− s

24 z

+ s

(2s

− s

) 12 z

satisfies

B

(z) = 1

2 z

− s

z + 1

2 s

(2s

− s

) = M

(z).

Hence the unique polynomial M

solution of the differential equation M

= M

with the initial conditions

M

(s

) = M

(s

) = M

(s

) = 0

is M

(z) = B(z) − (az

+ bz + c) where a, b, c are the solutions of the system of equations







as

+ bs

+ c = B(s

), 2as

+ b = B

(s

), 2a = B

(s

).

3.Lets0, s1, s2 be three complex numbers. Give a necessary and sufficient condition for the following to hold.

There exist three sequences of polynomials(Nn,0(z))n≥0,(Nn,1(z))n≥0,(Nn,2(z))n≥0such that any polynomial f∈C[z]can be written in a unique way as a finite sum

f(z) = X

n≥0

f⁽³ⁿ⁾(s0)Nn,0(z) +f⁽³ⁿ⁾(s1)Nn,1(z) +f⁽³ⁿ⁺¹⁾(s2)Nn,2(z)

.

What is the degree ofNn,j(z)? The leading term ? Write the six polynomials N0,0(z), N0,1(z), N0,2(z), N1,0(z), N1,1(z), N1,2(z).

3. The polynomials N

(n ≥ 0, i = 0, 1, 2) are defined by

N

(s

) = N

(s

) = δ

, N

(s

) = δ

n, k ≥ 0.

The polynomial N

is deduced from N

by permuting s

and s

. So we need to deal only with N

and N

.

• Let n = 0. The conditions on N

(z) and N

(z) are

N

(s

) = 1, N

(s

) = N

(s

) = 0, N

(s

) = N

(s

) = 0, N

(s

) = 1.

Write

N

(z) = (z − s

)(az + b), N

(z) = c(z − s

)(z − s

),

so that N

(z) = a(2z −s

)+b. The conditions on the numbers a and b arise from the requirement N

(s

) = 1 and N

(s

) = 0 :

(as

+ b)(s

− s

) = 1,

a(2s

− s

) + b = 0,

while the condition on c come from N

(s

) = 1 :

c(2s

− s

− s

) = 1.

A necessary and sufficient condition for the existence and unicity of a solution to this system is s

6= s

, s

+ s

6= 2s

. Under this assumption, the solution is



 

 



 

 



N

(z) = (z − s

)(z − s

+ 2s

) (s

− s

)(s

+ s

− 2s

) , N

(z) = (z − s

)(z − s

)

(s

− s

)(2s

− s

− s

) ·

• For n ≥ 1 and i = 0, 1, 2, the polynomial N

is the unique solution of the differential equation N

= N

with the initial condition

N

(s

) = N

(s

) = N

(s

) = 0.

The degree of N

is 3n + 2, the leading coefficient of N

is 2

(3n + 2)!(s

− s

)(s

+ s

− 2s

) while the leading coefficient of N

is

2

(3n + 2)!(s

− s

)(2s

− s

− s

) ·

• We explicit the solution for n = 1. The polynomial N

is computed as follows : let A(z) be a primitive of N

. Then N

(z) = A(z) − (az

+ bz + c) where a, b, c are the solutions of







as

+ bs

+ c = A(s

), as

+ bs

+ c = A(s

), 2as

+ b = A

(s

).

In the same way, N

(z) = B(z) − (αz

+ βz + γ), where B(z) is a primitive of N

while α, β, γ are the solutions of







αs

+ βs

+ γ = B (s

), αs

+ βs

+ γ = B (s

), 2αs

+ β = B

(s

).

Notice that both systems have the same determinant

s

s

1 s

s

1 2s

1 0

= (s

− s

)(2s

− s

− s

).

4.On p. 11, check that if the determinantD(s)does not vanish, thenrj≤jfor allj= 0,1, . . . , m−1.

4.

Let z

, z

, . . . , z

be independent variables. Write z for (z

, z

, . . . , z

). Let K be the field Q (z

, z

, . . . , z

) and D(z) be the determinant

det k!

(k − r

)! z

0≤j,k≤m−1

∈ Q [z] ⊂ K.

Recall a!/(a − b)! = 0 for a < b.

For j = 0, 1, . . . , m − 1, the row vector v

=

k!

(k − r

)! z

k=0,1,...,m−1

=

0, 0, . . . , 0, r

!, (r

+ 1)!

1! z

, (r

+ 2)!

2! z

, . . . , (m − 1)!

(m − 1 − r

)! z

belongs to {0}

× K

. If r

> j for some j ∈ {0, 1, . . . , m − 1}, then the m − j vectors v

, v

, . . . , v

all belong to the subspace {0}

× K

of K

, the dimension of which is m − j − 1 ; hence the determinant D(z) vanishes.

This amounts to say that a triangular matrix with a zero on the diagonal has a zero deter- minant.

5.Prove the proposition p. 11.

5. Assume D(s) 6= 0. Then there exists a unique family of polynomials (Λ

(z))

satisfying

Λ

(s

) = δ

δ

, for n, k ≥ 0 and 0 ≤ j, ` ≤ m − 1.

For n ≥ 0 and 0 ≤ j ≤ m − 1 the polynomial Λ

has degree ≤ mn + m − 1.

The assumption D(s) 6= 0 means that the linear map C [z]

−→ C

L(z) 7−→ L

(s

)

0≤j≤m−1

is an isomorphism of C –vector spaces, C [z]

being the space of polynomials of degree ≤ m−1.

First proof. Assuming D(s) 6= 0, we prove by induction on n that the linear map ψ

: C [z]

−→ C

L(z) 7−→ L

(s

)

0≤`≤m−1,0≤k≤n

is an isomorphism of C –vector spaces. For n = 0 this is the assumption D(s) 6= 0. Assume ψ

is injective for some n ≥ 1. Let L ∈ ker ψ

. Then L

∈ ker ψ

, hence L

= 0, which means that L has degree < m. From the assumption D(s) 6= 0 we conclude L = 0.

The fact that ψ

is injective for all n implies that if a polynomial f ∈ C [z] satisfies f

(s

) = 0 for all k ≥ 0 and all ` with 0 ≤ ` ≤ m − 1, then f = 0. This shows the unicity of the solution Λ

of the system of equations

Λ

(s

) = δ

δ

, for n, k ≥ 0 and 0 ≤ j, ` ≤ m − 1.

Since ψ

is injective, it is an isomorphism, and hence surjective : for 0 ≤ j ≤ n − 1 there exists a unique polynomial Λ

∈ C [z]

such that Λ

(s

) = δ

δ

for 0 ≤ j, ` ≤ m − 1.

These conditions show that the set of polynomials Λ

for 0 ≤ k ≤ n, 0 ≤ j ≤ m − 1, is a basis of C [z]

: any polynomial f ∈ C [z] of degree ≤ m(n + 1) − 1 can be written in a unique way

f (z) =

m−1

X

j=0 n

X

k=0

a

Λ

(z),

and therefore the coefficients are given by a

= f

(s

).

Second proof. The conditions

Λ

(s

) = δ

δ

, for n, k ≥ 0 and 0 ≤ j, ` ≤ m − 1.

mean that any polynomial f ∈ C [z] has an expansion f (z) =

m−1

X

j=0

X

n≥0

f

(s

)Λ

(z),

where only finitely many terms on the right hand side are nonzero.

Assuming D(s) 6= 0, we first prove the unicity of such an expansion by induction on the degree of f . The assumption D(s) 6= 0 shows that there is no nonzero polynomial of degree

< m satisfying f

(s

) = 0 for all (n, j) with 0 ≤ n, j ≤ m − 1. Now if f is a polynomial satisfying f

(s

) = 0 for all (n, j) with n ≥ 0 and 0 ≤ j ≤ m − 1, then f

satisfies the same conditions and has a degree less than the degree of f . By the induction hypothesis we deduce f

= 0, which means that f has degree < m, hence f = 0. This proves the unicity.

For the existence, let us show that, under the assumption D(s) 6= 0, the recurrence relations Λ

= Λ

, Λ

(s

) = 0 for n ≥ 1, Λ

(s

) = δ

for 0 ≤ j, ` ≤ m − 1

have a unique solution given by polynomials Λ

(z), (n ≥ 0, j = 0, . . . , m − 1), where Λ

has degree ≤ mn + m − 1. Clearly, these polynomials will satisfy

Λ

(s

) = δ

δ

, for n, k ≥ 0 and 0 ≤ j, ` ≤ m − 1.

From the assumption D(s) 6= 0 we deduce that, for 0 ≤ j ≤ m−1, there is a unique polynomial Λ

of degree < m satisfying

Λ

(s

) = δ

for 0 ≤ ` ≤ m − 1.

By induction, given n ≥ 1 and j ∈ {0, 1, . . . , m − 1}, once we know Λ

(z), we choose a solution L of the differential equation L

= Λ

; using again the assumption D(s) 6= 0, we deduce that there is a unique polynomial L e of degree < m satisfying L e

(s

) = L

(s

) for 0 ≤ ` ≤ m − 1 ; then the solution is given by Λ

= L − L. e

6. Poritsky’s interpolation p. 31. Prove that the conditionD(s) 6= 0means thats0, s1, . . . , sm−1 are pairwise distinct.

Prove also that the function∆(t)has a zero at the origin of multiplicity at leastm(m−1)/2.

N.B.

The fact that the multiplicity is exactly

follows from the fact that the coefficient of

in the Taylor expansion at the origin of

is given by a product of two Vandermonde determinants

1

1!2!· · ·(m−1)!det

















1 1 · · · 1

1 ζ · · · ζ^m−1

1 ζ² · · · ζ^2(m−1)

. .

. . . . . . . . . .

1 ζ^m−1 · · · ζ^(m−1)2















 det















1 1 · · · 1

s0 s1 · · · sm−1

s²₀ s²₁ · · · s²m−1

. .

. . . . . . . . . .













 .

But this is not so easy to prove

6. Poritsky interpolation is the case

r

= r

= · · · = r

= 0.

The Vandermonde determinant

D(s) = det s

0≤j,k≤m−1

= det















1 s

s

· · · s

1 s

s

· · · s

1 s

s

· · · s

.. . .. . .. . . . . .. . 1 s

s

· · · s















= Y

0≤j<`≤m−1

(s

− s

)

does not vanish if and only if s

, s

, . . . , s

are pairwise distinct.

The determinant ∆(t) is the determinant of the following matrix















e

e

e

· · · e

e

e

e

· · · e

e

e

e

· · · e

.. . .. . .. . . . . .. . e

e

e

· · · e













 .

The value ∆(0) at t = 0 is 0. We use the multilinearity of the determinant : the derivative (with respect to t) is the sum of determinants where we derive the rows. The derivative of order k of the row

e

e

e

· · · e

is the row

(ζ

s

)

e

(ζ

s

)

e

(ζ

s

)

e

· · · (ζ

s

)

e

which takes the value

(ζ

s

)

(ζ

s

)

(ζ

s

)

· · · (ζ

s

)

at t = 0.

If we derive the same number of times two rows, the corresponding determinant vanishes at t = 0. Hence to get a nonzero derivative at 0 we need to take derivatives of order at least

0 + 1 + 2 + · · · + (m − 1) = m(m − 1)

2 ·

7.Letw= (wn)n≥0be a sequence of complex numbers. Prove that the sequence of polynomials(Ωw₀,w₁,...,wn−1(z))n≥0

defined byΩ_∅= 1and

Ωw0,w1,...,wn−1(z) = Z z

w₀

dt1

Z t1 w₁

dt2· · · Ztn−1

w_n−1

dtn

forn≥1satisfyΩw₀(z) =z−w0 and forn≥0,Ωw₀,w₁,w₂,...,wn(w0) = 0, Ω⁰_{w0,w1,w2,...,wn}(z) = Ωw₁,w₂,...,wn(z).

What are the degree and the leading term ofΩw₀,w₁,w₂,...,w_n(z)? Check Ω^(k)w₀,w₁,w₂,...,w_n(wk) =δkn

forn≥0andk≥0. Deduce that any polynomial is a finite sum

f(z) =X

n≥0

f⁽ⁿ⁾(wn)Ωw₀,w₁,w₂,...,w_n(z).

Check the formula for the Gontcharoff determinant p. 39.

Give a close formula for these polynomialsΩw₀,w₁,...,wn−1(z)when

•wn= 0for alln≥0.

•wn= 1for evenn≥0,wn= 0for oddn≥1.

•wn=nfor alln≥0.

7. The definition of these polynomials involving iterated integrals means that the sequence of polynomials (Ω

)

in C [z] is defined as follows : we set Ω

= 1, Ω

(z) = z − w

, and, for n ≥ 1, the polynomial Ω

(z) is the polynomial of degree n + 1 which is the primitive of Ω

vanishing at w

.

For n ≥ 0, we write Ω

for Ω

, a polynomial of degree n which depends only on the first n terms of the sequence w.

By induction we deduce that the leading term of Ω

is (1/n!)z

.

Starting from Ω

(w

) and using the differential equation, we deduce by induction Ω

(w

) = δ

for n ≥ 0 and k ≥ 0. It follows that the sequence (Ω

)

is the unique sequence of polynomials such that any polynomial P can be written as a finite sum

P (z) = X

n≥0

P

(w

)Ω

(z).

In particular, for N ≥ 0 we have z

N! =

N

X

n=0

1

(N − n)! w

Ω

(z).

This gives an inductive formula defining Ω

: for N ≥ 0,

Ω

(z) = z

N ! −

N−1

X

n=0

1

(N − n)! w

Ω

(z).

We also have

Ω

(z) = Ω

(z − w

).

With w

= 0, the first polynomials are given by

2!Ω

(z) = (z − w

)

− w

,

3!Ω

(z) = (z − w

)

− 3(w

− w

)

z + w

, 4!Ω

(z) = (z − w

)

− 6(w

− w

)

(z − w

)

− 4(w

− w

)

z + 6w

(w

− w

)

− w

.

Let us check that these polynomials are also given by the following determinant

Ω

(z) = (−1)

1 z

1!

z

2! · · · z

(n − 1)!

z

n!

1 w

1!

w

2! · · · w

(n − 1)!

w

n!

0 1 w

1! · · · w

(n − 2)!

w

(n − 1)!

0 0 1 · · · w

(n − 3)!

w

(n − 2)!

.. . .. . .. . . . . .. . .. . 0 0 0 · · · 1 w

1!

.

Indeed, the right hand side is a polynomial of degree n, vanishing at w

. Its derivative is obtained by replacing the first row with its derivative, namely

0 1 z 1!

z

2! · · · z

(n − 1)!

.

The determinant that we get reduces to a similar determinant as above but with w

, w

, . . . , w

replaced with w

, . . . , w

. Hence the sequence of determinants satisfies the differential equation characteristic of the sequence (Ω

)

.

• With the sequence w

= 0 for all n ≥ 0, we get Taylor polynomials Ω

(z) = z

n! ·

• With the sequence w = (1, 0, 1, 0, . . . , 0, 1, . . . ), that is w

= 1 for even n ≥ 0, w

= 0 for odd n ≥ 1, we recover the Whittaker polynomials

Ω

(z) = M

(z), Ω

(z) = M

(z − 1).

• With the arithmetic progression

(a, a + t, a + 2t, . . . , a + nt, . . . ),

w = (a + nt)

with a in C and t in C \ {0}, we get the sequence of Abel polynomials Ω

(z) = 1

n! (z − a)(z − a − nt)

for n ≥ 1. In particular for a = 0, t = 1, the sequence is w = (0, 1, 2, 3, . . . , n, . . . ), namely w

= n for all n ≥ 0, this is

Ω

(z) = 1

n! z(z − n)

for n ≥ 1.

Références

[Gontcharoff 1930] W. Gontcharoff, “Recherches sur les dérivées successives des fonctions ana- lytiques. Généralisation de la série d’Abel”, Ann. Sci. École Norm. Sup. (3) 47 (1930), 1–78.

MR Zbl

[Macintyre 1954] A. J. Macintyre, “Interpolation series for integral functions of exponential

type”, Trans. Amer. Math. Soc. 76 (1954), 1–13. MR Zbl