Remarks on two symmetric polynomials and some matrices

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Applied Mathematics and Computation 219 (2013) 8770–8778

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Remarks on two symmetric polynomials and some matrices Moawwad El-Mikkawy ⇑, Faiz Atlan Mathematics Department, Faculty of Science, Mansoura University, Mansoura 35516, Egypt

a r t i c l e

i n f o

a b s t r a c t By means of complete symmetric polynomials this paper gives a new proof for the Vandermonde determinant formula. Another alternative proof for this formula is obtained via the collocation matrices. It also gives a generalized relationship between the Vandermonde, the Pascal and the Stirling matrices. A new approach to obtain the explicit inverse of the Vandermonde matrix is investigated. Closed form expressions for some Vandermonde related determinants are obtained. Published by Elsevier Inc.

Keywords: Interpolation Stirling numbers Vandermonde matrix Symmetric polynomials Pascal matrix

1. Introduction and Preliminaries Symmetric polynomials are widely used in many fields such as numerical analysis [6,9,26,29], algebra [2,4,16,17], combinatorics [1,5,8], statistics [21,25], physics [23], discrete mathematics [27], and many others. It is of prime importance to explore new properties of the symmetric polynomials. For the basic properties of symmetric polynomials, see [24]. There are many special types of matrices which are of great importance in many scientific and engineering work. For instance matrices of type Pascal [15], and of type Vandermonde [13,28]. This paper is mainly concerned with matrices of Vandermonde, Pascal and Stirling types. The Vandermonde type matrices, for instance, frequently appear in many applications. Among them curve fitting, interpolation, scattering and in the derivation of explicit Runge–Kutta and Runge–Kutta-Nystrom numerical methods [11,12,14]. The current paper is organized as follows: In the next section, we describe a new proof for the Vandermonde determinant formula using the complete symmetric polynomials. In Section 3, we give a new approach for the explicit inverse of the Vandermonde matrix. A generalized relationship between the Vandermonde, the Pascal and the Stirling matrices is given. Finally, closed form expressions for some Vandermonde related determinants are obtained in Section 4. The rest of this section gives some basic definitions. Throughout this paper x denotes the set fx1 ; x2 ; . . . ; xn g. Definition 1.1 [19]. The Vandermonde matrix VðxÞ is a matrix of the form

 n VðxÞ ¼ xi1 j

i;j¼1

ð1Þ

:

The Vandermonde determinant formula is well known, see for instance [29], in many text books and articles. It is given by

 n   det VðxÞ ¼ xji1 

i;j¼1

¼

Y

ðxi  xj Þ:

16j > < 1 eðnÞ r ðxÞ :¼ > > :

if r > n or n < 0; if r ¼ 0;

X

xi 1 xi 2 . . . xi r

if r ¼ 1; 2; . . . n

ð16Þ

16i1 > :

if r < 0 or n < 0 or ðn ¼ 0 and r > 0Þ; if r ¼ 0;

X

xi1 xi2 . . . xir

16i1 6i2 6...6ir 6n ðnÞ

It should be noticed that each er has written as

X

eðnÞ r ðxÞ :¼

ð17Þ

if r ¼ 1; 2; . . .

    n nþr1 ðnÞ terms and each hr has terms. Moreover these polynomials can be r r

xk11 xk22 . . . xknn

r ¼ 0; 1; . . . ; n

ð18Þ

k1 þ k2 þ    þ kn ¼ r k1 ; k2 ; . . . ; kn 2 f0; 1g and

X

ðnÞ

hr ðxÞ :¼

xd11 xd22 . . . xdnn

r ¼ 0; 1; . . .

ð19Þ

d1 þ d2 þ    þ dn ¼ r d1 ; d2 ; . . . ; dn 2 f0; 1; . . . ; rg ðnÞ

The generating functions for er

EðtÞ ¼

n Y

ð1 þ xi tÞ ¼

ðnÞ

and hr

are given, respectively by

n X

r eðnÞ r t ;

ð20Þ

r¼0

i¼1

and

HðtÞ ¼

n Y

ð1  xi tÞ1 ¼

1 X ðnÞ hr t r :

ð21Þ

r¼0

i¼1

From (20) we see that n n X X r eðnÞ erðn1Þ t r : r t ¼ ð1 þ xn tÞ r¼0

ð22Þ

r¼0

Comparing the coefficients of tr on both sides of (22) yields the following recurrence relation ðn1Þ

ðn1Þ eðnÞ ðx1 ; x2 ; . . . ; xn1 Þ þ xn er1 ðx1 ; x2 ; . . . ; xn1 Þ; r ðx1 ; x2 ; . . . ; xn Þ ¼ er

ð23Þ

The falling factorial ðxÞn in (7) can be written in the form

ðxÞn ¼

n X k¼0

ð1Þnk enk ð0; 1; . . . ; n  1Þxk : ðnÞ

ð24Þ

M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 219 (2013) 8770–8778

8773

Comparing the coefficients of xk in (5) and (24), yields ðnÞ

ðn1Þ

cðn; kÞ ¼ enk ð0; 1; . . . ; n  1Þ ¼ enk ð1; 2; . . . ; n  1Þ:

ð25Þ

It can also be shown that ðkÞ

Sðn; kÞ ¼ hnk ð1; 2; . . . ; kÞ:

ð26Þ

For any xj 2 fx1 ; x2 ; . . . ; xn g, we have ðnÞ

ðn1Þ

ei ðx1 ; x2 ; . . . ; xn Þ ¼ ei

ðn1Þ

ðx1 ; x2 ; . . . ; xj1 ; xjþ1 ; . . . ; xn Þ þ xj ei1 ðx1 ; x2 ; . . . ; xj1 ; xjþ1 ; . . . ; xn Þ:

ð27Þ

Partial differentiation for both sides of (27) with respect to xj gives

@ ðnÞ ðnÞ ðn1Þ e ¼ ei;j ¼ ei1 ðx1 ; x2 ; . . . ; xj1 ; xjþ1 ; . . . ; xn Þ: @xj i

ð28Þ

Therefore by using (16), we have

ðnÞ

eij ðxÞ ¼

8 0 > > > > > n; if i ¼ 1;

X

> > > 1 6 r 1 < r 2 <    < r i1 6 n > > : r 1 ; r2 . . . ; r i1 –j

xr1 xr2 . . . xri1

if i ¼ 2; 3; . . . n:

ð29Þ

Let G be the n  n matrix defined by G ¼ ðeij Þni;j¼1 . Then (see [19,27]) ðnÞ

det G ¼



det VðxÞ

if n  0 or 1 mod ð4Þ;

ð30Þ

 det VðxÞ if n  2 or 3 mod ð4Þ:

Definition 1.8 [6]. Given a set of n data points relating a dependent variable y ¼ f ðxÞ to an independent variable x as follows

x

x1

x2

….

xn

f(x)

f( x1 )

f( x 2 )

….

f( x n )

The zeroth order divided difference f ½xi  is defined as follows

f ½xi  ¼ f ðxi Þ;

i ¼ 1; 2; . . . ; n:

ð31Þ

Higher order divided differences are given recursively by

f ½x1 ; x2 ; . . . ; xk1 ; xk  ¼

f ½x2 ; x3 ; . . . ; xk   f ½x1 ; x2 ; . . . ; xk1  ; xk  x 1

k ¼ 2; 3; . . . ; n:

ð32Þ

It is known that the divided difference f ½x1 ; x2 ; . . . ; xn  satisfies [26]

f ½x1 ; x2 ; . . . ; xn  ¼

n X i¼1

f ðxi Þ n Y

:

ð33Þ

ðxi  xr Þ

r¼1 r–i Therefore it is a symmetric function of its arguments x1 ; x2 ; . . . ; xn . ðnÞ From [26], for any integer k, the complete symmetric polynomial hk satisfies ðkÞ

hnk ð1; 2; . . . ; kÞ ¼ f ½0; 1; . . . ; k:

ð34Þ

n

where f ðxÞ ¼ x . In conclusion, we have ðkÞ

Sðn; kÞ ¼ hnk ð1; 2; . . . ; kÞ ¼ f ½0; 1; . . . ; k;

ð35Þ

having used (26) and (34), where f ðxÞ ¼ xn . 2. A new proof for the Vandermonde determinant formula This section is mainly devoted to give a new proof for the Vandermonde determinant formula given by (2). The proof is based on using the complete symmetric polynomials.

8774

M. El-Mikkawy, F. Atlan / Applied Mathematics and Computation 219 (2013) 8770–8778

We begin this section by giving the following result. ðnÞ

Lemma 2.1. The complete symmetric polynomial hr ðx1 ; x2 ; . . . ; xn Þ satisfies ðn1Þ

hr

ðn1Þ

ðx2 ; x3 ; . . . ; xn Þ  hr

ðnÞ

ðx1 ; x2 ; . . . ; xn1 Þ ¼ ðxn  x1 Þhr1 ðx1 ; x2 ; . . . ; xn Þ:

ð36Þ

Proof. The generating function (21) gives 1 1 X X ðnÞ ðn1Þ r hr ð1  xn tÞt r ¼ hr t: r¼0

ð37Þ

r¼0

Comparing the coefficients of t r on both sides of (37), yields the following recurrence relation ðnÞ

ðn1Þ

hr ðx1 ; x2 ; . . . ; xn Þ ¼ hr

ðnÞ

ð38Þ

ðnÞ

ð39Þ

ðx1 ; x2 ; . . . ; xn1 Þ þ xn hr1 ðx1 ; x2 ; . . . ; xn Þ:

Substituting x1 $ xn , in (38) we get ðnÞ

ðn1Þ

hr ðx1 ; x2 ; . . . ; xn Þ ¼ hr

ðx1 ; x2 ; . . . ; xn1 Þ þ x1 hr1 ðx1 ; x2 ; . . . ; xn Þ:

From (38) and (39) by subtraction we obtain the identity (36). h At this point it is convenient to define

PðkÞ r ¼

k Y

ðxiþr  xi Þ;

1 6 r; k 6 n  1

ð40Þ

i¼1

and

nkþ1    ðkÞ W nkþ1 ¼ hi1 ðxj ; xjþ1 ; . . . ; xjþk1 Þ ;

k ¼ 1; 2; . . . ; n:

i;j¼1

ð41Þ

On setting k ¼ 1 in (41), we get

 n  ð1Þ  W n ¼ hi1 ðxj Þ

i;j¼1

 n   ¼ xji1 

i;j¼1

¼ det VðxÞ:

ð42Þ

Putting k ¼ n in (41), we have ðnÞ

W 1 ¼ h0 ðx1 ; x2 ; . . . ; xn Þ ¼ 1

ð43Þ

having used (17). For k ¼ 1; 2; . . . ; n  1, in this order, perform the column operations C r  C r1 ; r ¼ 2; 3; . . . ; n  k þ 1, on W nkþ1 and taking into account the identity (36), we obtain ðnkÞ

W nkþ1 ¼ Pk

W nk ;

k ¼ 1; 2; . . . ; n  1:

ð44Þ

From (44) we see that ðn1Þ ðn2Þ P2

W n ¼ det VðxÞ ¼ P1

ð1Þ

ðn1Þ ðn2Þ P2

. . . Pn1 W 1 ¼ P1

Y

ð1Þ

. . . Pn1 ¼

ðxi  xj Þ:

ð45Þ

16j j. In conclusion

det VðxÞ ¼ det V Q ðxÞ ¼

n Y

Y

qr ðxr Þ ¼

r¼1

ðxi  xj Þ:

16j > < 16j ðxi  xj Þðt i  t j Þ for n  2 or 3 mod ð4Þ > : 16j < xj > :

xij

if i ¼ 1; 2; . . . ; k; k ¼ 0; 1; 2; . . . ; n;

ð69Þ

if i ¼ k þ 1; k þ 2; . . . ; n:

Then, we have ðnÞ

det F n ðxÞ ¼ hr ðx1 ; x2 ; . . . ; xn Þ: det VðxÞ;

ð70Þ

det Gn ðxÞ ¼ f ½x0 ; x1 ; . . . ; xn : det VðxÞ;

ð71Þ

det Hn ðxÞ ¼ ð1Þn1 : det VðxÞ;

ð72Þ

det X n ðxÞ ¼ ð1Þn1 : det VðxÞ; and

ð73Þ

ðnÞ

det V k ðxÞ ¼ enk : det VðxÞ:

ð74Þ

Notice that by setting k ¼ n in (70) we get

V n ðxÞ ¼ VðxÞ:

ð75Þ

Also notice that for the case k ¼ n we see that (75), as expected, yields

det V n ðxÞ ¼ det VðxÞ;

ð76Þ

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