Hermite interpolation sequences over fields

Linear Algebra and its Applications 439 (2013) 66–77

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Hermite interpolation sequences over fields L.L. Stachó, R. Vajda ∗ Bolyai Institute, University of Szeged, Aradi Vértanúk tere 1, H-6720 Szeged, Hungary

ARTICLE INFO

ABSTRACT

Article history: Received 20 July 2012 Accepted 27 February 2013 Available online 1 April 2013

By a Hermite interpolation sequence we mean a sequence of Hermite interpolation polynomials of degree 0, 1, . . . such that consecutive terms satisfy the differentiation conditions of the previous ones. We extend this concept to arbitrary fields from the reals by purely algebraic means based on the possibility of formal Taylor expansions of rational fractions around any point of the underlying field. As an application we obtain recursion-free explicit formulas for the entries of triangular decompositions of generalized Hermite–Vandermonde matrices. © 2013 Elsevier Inc. All rights reserved.

Submitted by P. Sˇ emrl AMS classification: 65D05 Keywords: Hermite interpolation Taylor expansion Vandermonde–Hermite matrix Triangular factorization

1. Introduction Lagrange and Hermite interpolations are taught, using recursion methods, to a wide range of students as a chapter of complex numerical analysis, see e. g., [3]. Recently interest has arisen on recursion free closed formulas concerning them in the setting of triangular decompositions of Vandermonde matrices over generic fields [4,5]. The aim of this note is to show that the basic ideas of Spitzbart’s paper [1] which provide a natural generalization of the Lagrangian approach to Hermite interpolation with higher derivatives (actually the generalized basic polynomials are Ajk there) can be realized by purely algebraic means based on the possibility of formal Taylor expansions of rational fractions over fields. We continue these arguments to achieve a generalization of the Newtonian construction as well in a more flexible formulation which may be of interest in education. We conclude the paper with the application of the results to get a recursion-free explicit triangular decomposition of generalized Hermite–Vandermonde matrices.

∗ Corresponding author. E-mail addresses: [email protected] (L.L. Stachó), [email protected] (R. Vajda). 0024-3795/$ - see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.02.032

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

67

2. Basic concepts Throughout the paper we work in the setting of the polynomial ring K[x] of all formal expressions  p(x) = nk=0 ck xk (c0 , c1 , . . . , cn ∈ K; n ∈ Z+ ) where K is an arbitrary field. Let us emphasize that n n k k k=0 ck x is not identified with its functional representation K  ξ  → k=0 ck ξ as it is usual in the  n ( m) classical real case. Though the formal derivatives p (x) = k=m k(k − 1) · · · (k − m + 1)ck xk−m are  (m) well defined, in the case of χ := char(K)  = 0, we have inconveniently xk = 0 for m  χ since k(k − 1) · · · (k − m + 1) = modχ (k(k − 1) · · · (k − m + 1)) in K. Instead, we reformulate the defining constraints of Hermite interpolation in terms of the Taylor coefficients  p|ka

:= coefficient of x for p(x + a) := k



n 

=0

c (x + a)

=

n 

=k



 k

c a−k

corresponding to the terms p(k) (a)/k! in the classical case. Indeed we have the Taylor expansion  p = p(x) = nk=0 p|ka (x − a)k for any point a ∈ K. Let X := (x0 , x1 , x2 , . . .) be an arbitrary sequence in K indexed over Z+ . Define

ωnX :=

j: j


(x − xj ),

ν X (n, i) := # j : j < n, xj = xi

with the convention ω0X := n, i = 0, 1, . . . we have 1

ωnX = (x − xi )ν

X

(n,i)

  x − xj    j:j
(x−xi )+(xi −xj )

xj  =xi

X

(2.1)

= 1 = x0 and with # standing for cardinality. Observe that for any

∅

⎡

= (x − xi )ν



⎤

⎣ (xi

(n,i) ⎢ ⎢

j:j
⎥

− xj )⎥ ⎦ [1 + (x − xi ) pol(x)] .

In particular, xi is a root of ωnX with multiplicity ν X (n, i) and therefore ωnX is the unique polynomial of degree  n such that ν X (i,i) 

ωnX 

xi

ν X (n,n) 

= 0 (i < n),

ωnX 

xn

=



(xn − xj )

(2.2)

j:j
Definition 2.3. Given two sequences X := (x0 , x1 , x2 , . . .), Y := (y0 , y1 , y2 , . . .) in K, let HnX ,Y



:= p ∈ K[x] : p|νxi

X

(i,i)



= yi (i = 0, . . . , n)

be the set of all polynomials satisfying the Hermite interpolation condition of order n with coefficients from (X , Y ). By the Hermite interpolation sequence associated with (X , Y ) we mean the sequence hnX ,Y (n = 0, 1, . . .) of polynomials in K[x] defined recursively as 1 In the formula below and later on, the symbol pol(x) stands for a suitable polynomial from K[x] which we do not intend to specify any further.

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X ,Y

 X X ,Y ν (n,n) yn − hn−1  xn .

j:j
,Y X ,Y X := y0 x0 , hXn ,Y := hXn− ωn where γnX ,Y := 1 + γn

h0

xj  =xn

Proposition 2.4. For any n, hXn ,Y is the unique polynomial in HnX ,Y with degree  n. X ,Y

X ,Y

: deg(p) = 0}. Provided HnX ,Y = ∅, the difference of two Proof. Clearly {h0 } = {p ∈ H0 X ,Y polynomials from Hn must have root of multiplicity ν X (n, i) at any point xi with i < n + 1, it immediately follows that HnX ,Y

= hXn ,Y + ωnX+1 K[x].

= ∅ for n ∈ Z+ . Trivially y0 x0 + (x − x0 )K[x] = H0X ,Y . Suppose ,Y X = ∅. Then a polynomial p belongs to HnX ,Y if and only if p = hXn− 1 + ωn q for some q ∈ K[x] X X ,Y and p|νxn (n,n) = yn . By choosing q in the form q := γnX ,Y = γnX ,Y x0 we get p = hn−1 + γnX ,Y ωnX with We see by induction that HnX ,Y X ,Y Hn−1

n,n) p|ν( xn

ν(n,n) 

,Y = hnX− 1

xn

ν(n,n) 

+ γnX ,Y ωnX 

xn

Remark 2.5. In the classical case K    (m ) (m ) (0) (0) b1 ,...,b1 1 b2 ,...,b2 2

f a1

a2



ν(n,n) 

,Y = hnX− 1

xn

(xn − xj ) = yn . 

j:j
= R, the interpolation polynomials are of the form (0)

(mr )

... br ...,br ... ar

defined to be the unique polynomial f

(d)

∈ R[x]such that deg(f )  n and f (d) (ak ) = bk (k = (0)

b1

1, . . . , r ; d = 0, . . . , mk ). In our terminology, f a1 Y have the pattern ⎛

⎞

⎜ ⎟ X= ⎝a1 , . . . , a1 , . . . , ar , . . . , ar , . . .⎠ ,       m1 +1

+ γnX ,Y

mr +1

⎛ Y= ⎝

(m1 )

,...,b1

(0)

b1

0!



(0)

(mr )

,..., br ...,br ,..., ar

,...,

(m1 )

b1

m1 !

,...,

= hnX ,Y whenever X and

(0)

br

0!

,...,

(mr )

br

mr !

⎞

, . . .⎠.

 X ,Y Definition 2.6. The Newtonian form of a Hermite sequence is the representation hXn ,Y = nk=0 γk ωkX   (n = 0', 1, . . .). We shall ( write δi for the sequence δ0,i , δ1,i , δ2,i , . . . with the Kronecker symbol δn,i := 1 if n = i, 0 else . We call the members of the double sequences HnX,i

:= hXn ,δi ωXn,i :=

  x − xj

(i, n = 0, 1, . . .)

(2.7)

j:jn, xj  =xi

the basic Hermite interpolation polynomials resp. the complementary Newton factors over  the sequence X. By the Lagrange form of a Hermite sequence we mean the representation hXn ,Y = ni=0 yi HnX,i (n = 0, 1, . . .). Example 2.8. Let K X

= Z2 = {0, 1} and consider the sequences

:= [0, 1, 0, 1, 0, 1, . . . ], Y := [1, 1, 1, . . .].

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

69

Then, for 1 < n = 2k + r with r := mod2 (n) and k := n/2 we have ωnX = xk+r (x − 1)k = X X xk+r (x + 1)k . From the identity 2 = 1 + 1 = 0 we also get ω2k (t + 1) = (t + 1)k t k = ω2k (t ) and  k k 2 2 X ,Y (1 + x) = 1 + x for any k = 0, 1, . . .. Thus the Newtonian form of hn is simply m:2m −2n ω2Xm −2 M (n) for any index n. Actually also hXn = =0 x with M (n) := max{2m− 2 : 2m− 2  n, m ∈ N}. We obtain X ,Y

(0!,...,k!) (0!,...,(k−1+r )!)

the classical Newtonian form of h2k+r by evaluating f 0 over R with a Newton 1 difference scheme and then taking the coefficients mod2 . The result is a rather sophisticated linear combination from the factors 1, x, . . . , xk+1 , xk+1 (x + 1), . . . , xk+1 (x + 1)k−1+r .

3. Numerical issue: modified Newton difference schemes Classical Hermitian interpolation seems to be well understood in terms of Newtonian difference schemes as done in the nice survey [6]. Working over a field K of general type, can be transferred in a straightforward manner into generalized Newtonian differences schemes, if we consider the classical sequences described in Remark 2.5. However, in a non-classical case as in Example 2.8 we should be more careful when using schemes from classical rearrangements. For the sake of completeness, below we outline a self-contained approach. In order to avoid using any sophisticated notation, throughout this section let the sequences X , Y be fixed arbitrarily and let us write hn , Hn , γn , ν(n, i), ωn instead of hXn ,Y , HnX ,Y , γnX ,Y and ν X (n, i), ωnX ,

respectively. Observe that, for any n, we can rearrange the conditions p|νx0 (0,0) = y0 , . . . , p|νxn (n,n) = yn determining the space Hn into the form X

p|0a1

(0)

m

(n)

(n)

(m1 )

= b1 , . . . , p|a11 = b1

m

(0)

(n)

X

(m

(n)

)

, . . . , p|0ar(n) = br (n) , . . . , p|arr((nn)) = br (nr)(n) .

This can be done in a unique way if we require that a1 , a2 , . . . are the distinct enumeration of x0 , x1 , . . . in order of first appearance, that is for n = 0, 1, . . . we have {x0 , . . . , xn } = {a1 , . . . , ar (n) } where r (n)

(n)

= #{x0 , . . . , xn }, mk = #{j  n : xj = ak } − 1.

(3.1)

As mentioned, to calculate the interpolating polynomials hn we can use Newton difference schemes developed for the setting described in Remark 2.5 with the straightforward modification of replacing the terms

f (d) (ak ) d!

(d)

there with the bk

above. Namely, we can calculate each coefficient

γn as the last

bottom term of a lower triangular matrix (n) whose columns store the column elements of the classical Newton difference scheme. To realize this, let us store the rearranged data sequences in the

(n + 1)-vectors

⎡ a(n) =



⎤

⎢

 ⎢ (n) n := ⎢a1 · · · a1 k=0 ⎢ ⎣  

αk

(n)

1+m1

⎥ ⎥

· · · ar (n) · · · ar (n) ⎥ ⎥ 



(n)

1+mr (n)

and define the columns of the matrix (n)

(n)

if

(n)

αk = as



 (n) n recursively as follows. For the k,d=0

:= lowertr k,d

starting column let

k,0 := b(s0)

(3.2)

⎦

(k = 0, . . . , n).

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L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

Having constructed column d, we set (n)

k,d+1 :=

⎧ (n) (n) ⎪ ⎨ k,d − k−1,d if (n) (n) αk −αk−d−1 ⎪ ⎩ (d+1)

bs

As a final result for any N

hN

=

N  n=0

γ n ωn =

if

(n)

(n)

αk = αk−d−1

(k = d + 1, . . . , n).

αk(n) = αk(n−)d−1 = as

∈ Z+ , we get N  n=0

(nn,n) (x − x0 ) · · · (x − xn−1 ) .

To do this we need to calculate all the matrices (n) (n

(3.3)

= 0, . . . , N ). However, large parts in (n+1) and

(n+1) (n) coincide. Clearly, if xn+1  ∈ {x0 , . . . , xn } then (n+1) simply enlarges (n) : in this case k,d = (n) ( n ) k,d for k, d  n. Consider the case xn+1 = as ∈ {x0 , . . . , xn }. Let m∗:= max{j  n : αj = as }+ 1 be the position after the last index where the point xn+1 = as appears in a(n) :

⎡ a(n+1)

⎤

(n)

1+ms

  

⎢

⎥

⎥ =⎢ ⎣· · · as−1 as · · · as as as+1 · · · ar (n)⎦ 







first m∗ terms of a(n)





rest of a(n)

From the recursion rules it follows by induction on d that (n+1)

m∗ ,d = b(sd) (n+1)

k,d

for d

= 0, . . . , m(sn) + 1 = m(sn+1) ,

(n)

= k,d (d  k < m∗ ),

(n+1)

k,d

(n)

= k−1,d (k > max{d, m∗ }). (n)

Thus to obtain (n+1) from (n) , arithmetic operations are required only for the (n − m∗ + 2)(m∗ − ms entries (n+1)

k,d

with



(k, d) ∈ (k, d) : k  m∗ , k − (m∗ − m(sn) )  d  k

)



displayed black in Fig. 1, which form mostly a rather narrow parallelogram. 4. Basic polynomials via formal power series As usual, let Kx denote the field of all formal fractions p/q with p, q ∈ K[x], q  = 0 up to the identification p1 /q1 = p2 /q2 whenever p1 q2 = p2 q1 in K[x]. Given any point a ∈ K along with a fraction (x) := p(x)/q(x) ∈ Kx such that q(a)  = 0, the formal Taylor series

(x) ∼

∞  n=0

|na (x − a)n

(4.1)

 j of ρ around a is well-defined by the requirement p ∼ q ∞ j=0 |a (x n m k i i=0 bi (x − a) then we have the recursion k=0 ak (x − a) and q(x) =

|0a =

p(a) q(a)

;

|ka =

1 q(a)

-

p|ka

−

k −1 j=0

q|ka−j

|ja

.

− a)j . Actually, if p(x) =

(k = 1, 2, . . .)

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

n

ms

m

n

1

n

k ,d

k ,d

n

1

d

m

,d

bs

n

71

1

n

k ,d

k

1 ,d

Fig. 1. Pieces of (n) in (n+1) (gray) and nontrivial new entries (black).

where p|ka = ak (0  k  n), p|ka = 0 for k > n and q|ia = bi (0  i  m), q|ia = 0 for i > m. We p// q with q(a), / q  := / (a) = 0 whenever they have the same Taylor series around a for  := p/q and / ∞ |n (x − a)n and / / p q, because then both T K x i.e. if p q = := represent the same object in a n=0  n n / / / / / / / / | p q  ( x − a ) satisfy q qT ∼ p q resp. qq T ∼ = p q. In the classical case K = R, in terms T := ∞ a n=0 of derivations again we have theorem we have 1

(x − b)m

∼

|na = (n) (a)/n!. In particular, if a = b then by Newton’s binomial

-

1

(a − b)m

1+

. x − a −m a−b

=

∞ 

(−1)k (a − b)m+k k=0



m+k−1 k



(x − a)k

even in the case of generic fields K instead ofR, since the coefficients here satisfy the same recursion  as in the real case. If χ

= char(K) = 0 then

m+k−1 k

= modχ

m+k−1 k

.

For the initial segments of formal Taylor series, we introduce the shorthand notation

Na :=

N  k=0

Lemma 4.3. Let a

|ka (x − a)k

( ∈ Kx ).

∈ K and (x) := p(x)/q(x) with p, q ∈ K[x] and q(a) = 0. Then

Na q(x) = p(x) + (x − a)N +1 pol(x) Proof. Let p(x)

=

(4.2)

N

k=0

ak (x − a)k , q(x)

=

(N = deg(p), deg(p)+ 1, . . .) . m

i=0

bi (x − a)i , (x)

=

∞

j=0 cj (x

− a)j .

Recall that, by definition ak is the coefficient of (x − a) in the formal product q(x)(x) m ∞ i ∞ j k i=0 bi (x − a) j=0 cj (x − a) = k=0 bi cj (x − a) . That is k

=

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L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

ak



=

(k = 0, . . . , N ),

bi cj

0



=

(i,j): i+j=k

bi cj

(k > N ).

(i,j): i+j=k

Hence we conclude that q(x)

N  j=0

cj (x − a)j

= =

m  i=0

bi (x − a)i

N   k=0 i+j=k

N  j=0

cj (x − a)j

bi cj (x − a)k

N +1

= p(x) + (x − a)

+

−1 m 

N +m



k=N +1

i+j=k i m

⎡

bi cj (x − a)k

min{m,N +1+}



⎣

i=0

=0

⎤ ci bN +1+−i ⎦ (x − a) .



Let us introduce some handy notation which make the rest of the formulae more readable and the reasoning shorter. Definition 4.4. Henceforth we write  σ X (n, i) := # j ∈ (i, n] : xj = xi , Proposition 4.6. We have HnX,i HnX,i

(x − xi )ν = ωXn,i

X





μX (n, a) := # j  n : xj = a .

(4.5)

= 0 for n < i. Otherwise, with the terms introduced in (2.1) resp. (2.7),

0ν X (n+1,i)−1

(i,i) 0 0

0 0 0

ωXn,i

(n = i, i + 1, i + 2, . . .).

(4.7)

xi

Proof. In the case n < i the polynomial HnX,i of degree n has roots with total multiplicity n + 1 entailing Hn,i = 0. Assume n  i. Now consider any index j ∈ {0, . . . , n} \ {i} and let a := xj . Observe that the point a must be a root of the polynomial HnX,i of as many algebraic multiplicity as the number of   X appearances of the point a in the sequence x0 , . . . , xn . That is (x − a)μ (n,a) H X (pq standing for p is n,i

a divisor of q). Therefore

ωXn,i =



(x − xj ) =

j:jn xj  =xi



(x − a)μ

a∈{x0 ,...xn }\{xi }

X



(n,a)  X Hn,i

.

The polynomial HnX,i also satisfies the remaining (n + 1)− #{j  n : xj  = xi } constraints d  HnX,i 

xi



ν X (i,i) 



= 0 for d ∈ 0, . . . , n − #{j  n : xj = xi } \ {ν X (i, i)} and HnX,i 

xi

= 1.

Thus HnX,i must be of the form HnX,i

= (x − xi )ν

X

(i,i)

+ (x − xi )n+1−#{jn:xj=xi }−ν

X

(i,i)

pol(x).

Observe here that n + 1 − #{j  n

: xj = xi } = #{j  n : xj = xi } = ν X (n + 1, i). Applying Lemma

4.3 with p := (x − xi )ν(i,i) , N := ν X (n + 1, i)− 1 and q := j:jn (x − xj ) we see the polynomial xj  =xi

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

(x − x)ν h := ωXn,i

X

73

0ν X (n+1,i)−1

(i,i) 0 0

0 ωXn,i has the expansion form h = (x − xi )ν(i,i) + (x − xi )ν(n+1,i)−ν(i,i) pol(x). 0 0  xi  However, also ωXn,i h by definition. Thus h satisfies all the constraints defining HnX,i , which entails h

= HnX,i . 

Next we proceed to the Newtonian form of the sequences HnX,i = hXn ,δi (n ∈ Z+ ). Corollary 4.8. For any fixed index i

HNX ,i

=

N  n=0

∈ Z+ and N = i, i + 1, i + 2, . . . we have σ X (n,i) 

 −1   x − xj   jn,x  =x

nX,i ωnX with nX,i := γnX ,δi =

j

i

(n  i).

xi



< i. According to Definition 2.3, HNX ,i = Nn=0 γnX ,δi ωnX (N = = γnX ,δi ωnX (n = 1, 2, . . .). Thus γnX ,δi is the coefficient of xn in X X X Hn,i − Hn−1,i . Since Hn−1,i is of degree  n − 1, it means that

Proof. Recall that nX,i = 0 for n 0, 1, . . .) implying HnX,i − HnX−1,i n

n

0

xi

γnX ,δi = HnX,i  = HnX,i  =

 '

(x − xi )ν

X

(i,i)

n  

X (0 0ν (n+1,i)−1

/ωXn,i 0

ωXn,i 

xi

xi

 X 0ν (n+1,i)−1 ν (n+1,i)−1 '  ν X (i,i) X (0  = (x − xi ) /ωn,i 0  xi X

xi

'

ν X (i,i)

= (x − xi )

ν X (n+1,i)−1−ν X (i,i)

( /ωXn,i  x

i

Here we have ν X (n + 1, i) − ν X (i, i) − 1 j  n, xj = xi = σ X (n, i). 

.









= # j  n : xj = xi − 1 − # j < i : xj = xi = # j : i <

5. Closed formulas for the Newtonian representation Inserting the Newtonian form of the basic polynomials into the Lagrange representation of hnX ,Y , on the basis of the observation that Newton’s binomial theorem works even in the case of formal power series, we get the following closed explicit combinatorial form for Hermite interpolation sequences.

Theorem 5.1. For all N

= 0, 1, . . . we have

X nX,i = (−1)σ (n,i)



Kn,i (X )

:= ⎩κ ∈



=

N 

⎛ ⎝

n=0

(xi − a)−[μ

X

n  i=0

⎞ yi nX,i ⎠ ωnX

where

(n,a)+κ(a)]

(5.2)

κ∈Kn,i (X ) a∈An,i (X )

with the notations An,i (X ) ⎧ ⎨



X ,Y hN

:= {x0 , . . . , xn }\{xi } and

An,i (X ) → Z+ functions

:

 a∈An,i (X )

⎫ ⎬

κ(a) = σ (n, i)⎭ . X

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L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

Proof. We have already seen that X ,Y

=

hN

N  i=0

=

yi HnX,i

N 

⎛ ⎝

n=0

n  i=0

⎞ yi nX,i ⎠ ωnX

where ⎤σ (n,i)    ⎢

−1 ⎥  ⎢ ⎥ ( x − x ) j ⎣ ⎦  j:jn  x  =x ⎡

nX,i

=

 σ (n,i)  1/ωXn,i 

=

xi

j

i

⎡

=

⎣



⎤σ (n,i)  −μ(n,i) ⎦

(x − a)

 

a∈Xn \{xi }

.

xi

xi

We know also that in general

(x − a)−m ∼

∞ 

(−1)k (xi − a)m+k k=0



m+k−1



k

(x − xi )k .

Thus in the expression nX,i we get

−μ(n,i)

(x − a)

∼

a∈Xn \{xi }

=

∞  

k=0

 ∞ 

a∈Xn \{xi }



(−1)ka (x − xi )ka (xi − a)μ(n,a)+ka ka =0 

(−1)κ(a)



κ:Xn \{xi }→Z+ a∈Xn \{xi } κ(a) =k a∈Xn \{xi }

(xi − a)μ(n,a)+κ(a)



μ(n, a) + ka − 1 ka

μ(n, a) + κ(a) − 1 κ(a)





(x − a)k .

Hence the theorem follows immediately.  6. Triangular decomposition for Vandermonde matrices of Hermite sequences Throughout this section, let N resp. X = (x0 , x1 , . . .) be a fixed positive integer and a sequence in K. Consider the Hermite–Vandermonde matrix

V

:=

-

ν X (i,i) xn xi



.N i,n=0

=

n



ν X (i, i)

corresponding to the system of equations

n−ν X (i,i) xi

N

n=0 cn x

N (6.1) i,n=0

ν X (i,i)

n  xi

= yi (0  i  N ) established implicitly X ,Y

for the coefficients c0 , . . . , cN ∈ K of the Hermite polynomial hN by Definition 2.3. We can state Rushanan’s hidden idea treating the classical case in [2] in a concise formulation for the setting of Hermite–Vandermonde matrices as V = tx with the formal column resp row vectors 

t

:= ·|νx0 (0,0) ·|νx1 (1,1) . . . ·|νxN (N ,N ) X

X

X

T



, x := 1 x x2 · · · xN



where x is the indeterminate symbol for the polynomial ring K[x] and

·|ka abbreviates the operation

K[x] → K, p → p|ka . Notice that if the points xi are pairwise distinct, for V we get the classical ' (N Vandermonde matrix xin i,n=0 .

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

75

Proposition 6.2. The matrix V is invertible and we have the UL-decomposition 2 V −1

-

 i .N 

=  where  := ωnX 

Proof. Consider any sequence Y



0 i,n=0

,  := lowertr nX,i

N n,i=0

.

(6.3) ν X (i,i) 

= (y0 , y1 , . . .) in K. We can express the defining relations hXN,Y 

= yi (i = 0, . . . , N ) in terms of the operator vector t as X ,Y

thN



= y where y := y0 , y1 , y2 , . . . yN

T

xi

.

But, by Theorem 5.1 we have 



X ,Y ω0X ω1X · · · ωNX  y = hN .

Observe as well that we can write x 





= ω0X ω1X · · · ωNX . Since each Newton factor ωnX is a polyno-

mial of degree n with unit leading coefficient,  is an upper triangular matrix with terms 1 in the main diagonal. It follows that t [x ]  y = y. Thus, from the arbitrariness of the sequence Y we conclude that IN

= [δin ]Ni,n=0 = t [x ]  = V  .



Corollary 6.4. For the entries with indices (i, n) resp. (n, i) with 0 matrices ,  and their inverses we have

i,n = (−1)n−i



xk1 xk2

0k1 <···
· · · xkn−i , 

n,i = nX,i (defined in 4.8),  −1

 n,i



=

−1

 i,n



=

d

d0 +···+di =n−i d0 ,d1 ,...,di 0

d

x00 x11



0k1 <···
 i  N of the triangular factor


· · · xidi ;

(xn − xk1 ) · · · (xn − xk[i−ν X (n,n)] ). +

+

Proof. Using the results of the previous sections, we can provide closed formulas for all the entries in the matrices  ,  −1 appearing in the LU- resp. UL-decompositions V =  −1 −1 and V −1 =  . The terms nX,i have already been calculated in several ways. Also 



 −1 = V  = t [x ] = t ω0X ω1X · · · ωNX . Since ωiX = (x − x0 ) · · · (x − xi−1 ) 

 −1

ν X (n,n) 

 n,i

= ωiX 

xn

=

= [(x − xn ) + (xn − x0 )] · · · [(x − xn ) + (xn − xi−1 )], 

0k1 <···
(xn − xk1 ) · · · (xn − xk[i−ν X (n,n)] ) (0  i  n  N ). +

+

−1 = [ψ(x , . . . , x )]N The matrices  = [ϕ(x0 , . . . , xN )]N 0 N i,n=0 are well known [4] for the i,n=0 ,  real case with xi  = xj (i  = j). Their entries are polynomials with integer coefficients in the variables  x0 , . . . , xN . By continuity, the identities δij =  ϕi ψj hold for all x0 , . . . , xN ∈ R. Actually, a 2

The terms L and U stand for lower and upper triangular, respectively.

76

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

polynomial expression in several variables that involves only integer coefficients and vanishes for all real substitutions must have vanishing coefficients. Therefore the formulas obtained there admit a straightforward extension to our cases for the entries of  resp. −1 stated.  Remark 6.5. We can express the above results in closed combinatorial formulas as follows. All the below statements follow immediately from the analogous results above, given in terms of x0 , . . . , xN by the aid of interpreting the number of repetitions in terms of standard combinations and combinations with repetitions, respectively. The binomial coefficients have to be taken with modχ if (n)

χ := char(K) = 0. In terms of the rearrangement vector a(N ) and the associated multiplicities rn , ms introduced in (3.1) resp. (3.2), for the non-trivial entries of the LU- resp. UL-decompositions V =  −1 −1 and V −1 =  with 0  i  n  N below we have 

r

(n)

k1 +···+kr (n) =σ X (n,i) kj =0, if ak =xi j

s=1,as  =xi

n,i =



 −1

ν X (n,n) 

 n,i

= ωiX 

xn

n−i

i,n = (−1)



−1

 i,n

=

(−1) (xi − as )

 0k1 <···




μX (n − 1, a1 )





1 +···+r (i) =n−i 1 ,...,r (i) 0


+ ks



ks

(xn − xk1 ) · · · (xn − xk[i−ν X (n,n)] ) , +

+





···

m1

m1 +···+mr (n) =n−i m1 ,...,mr (n) 0

=

 (n) (n) −[ms +1+ks ] ms

ks



μX (n − 1, ar (n) ) mr (n)

m

a1 1



m

· · · ar (rn()n) ,

μX (i, a1 ) + 1 −1 μX (i, ar (i) ) + r (i) −1 1 r (i) a1 · · · ar (i) . ··· 1 r (i)

Example 6.6. As in Example 2.8, let X again be the sequence [0, 1, 0, 1, . . .]. We calculate the explicit form of the entries n,i , [ −1 ]n,i , i,n , [−1 ]i,n for the classical case, i.e., over the reals. In terms of Remark 6.5, then we have a1 = 0, a2 = 1, r (n) = min{2, n + 1}, ωnX = xn/2 (x − 1)n/2 . Observe that the product terms in the expressions of n,i , i,n , [−1 ]i,n vanish for all but one case: we get i,n with m1 = 0, m2 = n − i and μX (n − 1, 1) = #{j  n − 1 : xj = 1} =  2n , [−1 ]i,n with 1 = 0, 2 = n−i and the non-vanishing term for n,i appears with the choice of s ∈ {1, 2} such that {as } = {0, 1}\{xi } (n)

and ms = #{j  n : xj = xi } =  respectively. It follows 

n+mod2 (n−i)  2



i − 1 with ks = σ X (n, i) = #{j ∈ (i, n] : xj = xi } =  n− , 2





n/2 i/2 + n − i − 1 i,n = (−1) , [−1 ]i,n = , n−i n−i   k := (n − i)/2 , m+k n,i = (−1)[m if 2|i, k else] with m := n + mod2 (n − i))/2 − 1. k n−i

To calculate [ −1 ]n,i , observe that the products (xn − xk1 ) · · · (xn − xk[i−ν X (n,n)] if xn

= 0 and xk1 = · · · = xk[i−ν X (n,n)] 

products appear

#{k
=



+

) do not vanish either + = 1 or if xn = 1 and xk1 = · · · = xk[i−ν X (n,n)] = 0. Such

(i−mod2 (n−i))/2 i−n/2



[ −1 ]n,i = (−1)[(i−n/2 ) if 2|n, 0 else]

+

times. Thus

(i − mod2 (n − i))/2 . i − n/2

L.L. Stachó, R. Vajda / Linear Algebra and its Applications 439 (2013) 66–77

77

Acknowledgements The authors are indebted to the unknown referee for his valuable suggestions. It is also appropriate to mention the following problem raised by the referee which seems to be extremely worth for future studies: Is there a way to study Birkhoff’s concept [7] of the Hermite type interpolation also in this general algebraic setting? This work was supported by the project TÁMOP-4.2.2.A-11/1/KONV-2012-0073. The research of the second author was partially supported by the HSRF (OTKA), grant number K83219. References [1] [2] [3] [4] [5] [6] [7]

A. Spitzbart, A generalization of Hermite’s interpolation formula, Amer. Math. Monthly 67 (1960) 42–46. J.J. Rushanan, On the Vandermonde matrix, Amer. Math. Monthly 96 (10) (1989) 921–924. J. Stoer, R. Bulirsch, Introduction to Numerical Analysis, Springer, New York, 2002. H. Oruc, G.M. Phillips, Explicit factorization of the Vandermonde matrix, Linear Algebra Appl. 315 (1–3) (2000) 113–123. H. Oruc, LU factorization of the Vandermonde matrix and its applications, Appl. Math. Lett. 20 (9) (2007) 982–987. C. de Boor, Divided differences, Surv. Approx. Theory 1 (2005) 46–69 (electronic). G.D. Birkhoff, General mean value and remainder theorems with applications to mechanical differentiation and quadrature, Trans. Amer. Math. Soc. 7 (1) (1906) 107–136.