Georgian Mathematical Journal 1(94), No. 4, 405-417

ON SOME GENERALIZATIONS OF THE VANDERMONDE MATRIX AND THEIR RELATIONS WITH THE EULER

BETA-FUNCTION

I. LOMIDZE

Abstract. A multiple Vandermonde matrix which, besides the powers of variables, also contains their derivatives is introduced and an explicit expression of its determinant is obtained. For the case of arbitrary real powers, when the variables are positive, it is proved that such generalized multiple Vandermonde matrix is positive definite for appropriate enumerations of rows and columns. As an application of these results, some relations are obtained which in the one-dimensional case give the well-known formula for the Euler betafunction.

It is known from the classical problem of multiple interpolation (cf. [1], pp. 104-106; [2], pp. 13-16) that there exists the unique polynomial P_{N−1} of degree N − 1 which satisfies the conditions

P_{N−1}^{(k−1)}(x_{j}) = y_{j}^{(k−1)}, k = \overline{1, r_{j}}, j = \overline{1, n},  (1)

where x_{j}, y_{j}^{(k−1)} are given elements of an arbitrary field F; x_{j} ≠ x_{l} if

j ≠ l, r_{j} ≥ 1 and N = ∑_{j=1}^{n}r_{j}. Hence the determinant of the N × N

matrix V (x_{1}, r_{1}; …; x_{n}, r_{n}) ∈ M_{N}(F) is nonzero. Here by M_{l}(F) with natural l we denote the set of all l × l matrices with elements from F and by V (x_{1}, r_{1}; …; x_{n}, r_{n}) the N × N block-matrix

V (x_{1}, r_{1}; …; x_{n}, r_{n}) = [w_{1}(x_{1})|w_{2}(x_{2})| ⋯ |w_{n}(x_{n})],  (2)

where w_{j}, j = \overline{1, n}, stand for the rectangular matrices

w_{j}(x) = [(x^{i−1})^{(k−1)}]\begin{matrix}1≤i≤N \\ 1≤k≤r_{j} \end{matrix}.  (3)

405

1991 Mathematics Subject Classification. 15A15, 15A54, 33B15.

1072-947X/94/0700-0405$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

There are other cases when one has to deal with matrices of the form (2). For instance, the determinant of such a matrix can be effectively used when proving the proposition on functional independence of elements from the complete set of invariants of operators (no matter whether they are Hermitian or not) in an N-dimensional unitary space if these operators have r_{j}-dimensional eigenspaces, r_{j} > 1 (this proposition will be published in forthcoming paper of the author). The case r_{j} = 1 has been investigated in [3].

For r_{1} = ⋯ = r_{n} = 1 we have N = n and (2) becomes the Vandermonde matrix V = V (x_{1}, …, x_{n}) ∈ M_{n}(F)

V (x_{1}, 1; …; x_{n}, 1) = [x_{j}^{i−1}]_{1}^{n}

with the determinant

det [x_{j}^{i−1}]_{1}^{n} = ∏_{1≤k<j≤n}(x_{j} − x_{k}).  (4)

1. In what follows ℕ = {1, 2, …}, F stands for the field of real numbers ℝ or that of complex numbers ℂ and we shall call (2) the (r_{1}, …, r_{n})-tiple Vandermonde matrix. For r_{1} = ⋯ = r_{n} = r this matrix is said to be the r-tiple Vandermonde matrix and we denote it by V (x_{1}, …, x_{n}; r).

Theorem 1. For arbitrary r_{1}, r_{2}, …, r_{n} ∈ ℕ the (r_{1}, r_{2}, …, r_{n})-tiple Vandermonde matrix (2) with x_{1}, …, x_{n} ∈ F satisfies the identity

det V (x_{1}, r_{1}; …; x_{n}, r_{n}) = ( ∏_{j=1}^{n} ∏_{k=0}^{r_{j}−1}(k!)) ∏_{1≤i<k≤n}(x_{k} − x_{i})^{r_{k}r_{i}}. (5)

In particular,

det V (x_{1}, …, x_{n}; r) = ( ∏_{k=0}^{r−1}(k!))^{n} ∏_{1≤i<k≤n}(x_{k} − x_{i})^{r^{2}}.  (5_{1})

Before proving Theorem 1 we need some preparation.

Lemma 1. Let n, N ∈ ℕ, 1 ≤ n ≤ N and ϕ_{jk}, f_{i} : F → F be such that

ϕ_{jk} ∈ C^{k−1}(F, F), k = \overline{1, r_{j}}, j = \overline{1, n},

f_{i} ∈ C^{r−1}(F, F), r = max\{r_{j}|j = \overline{1, n}\}, i = \overline{1, N}, N = ∑_{j=1}^{n} r_{j},

and let U ∈ M_{N}(F) be the square N × N block-matrix

U = [u_{1}(x_{1})| ⋯ |u_{n}(x_{n})],  (6)

where

u_{j}(x) = [(f_{i}(x)ϕ_{jk}(x))^{(k−1)}]\begin{matrix}1≤i≤N \\ 1≤k≤r_{j} \end{matrix}, x ∈ F.  (7)

Then we have

det U = det [v_{1}(x_{1})| ⋯ |v_{n}(x_{n})] ∏_{j=1}^{n} ∏_{k=1}^{r_{j}}(ϕ_{jk}(x_{j})),  (8)

with

v_{j}(x) = [(f_{i}(x))^{(k−1)}]\begin{matrix}1≤i≤N \\ 1≤k≤r_{j} \end{matrix}.

Proof of Lemma 1. Let us denote by [k]_{j} the column having number k in the matrix u_{j}, k = \overline{1, r_{j}}, j = \overline{1, n}. It is clear that (8) is trivial if ϕ_{j1}(x_{j}) = 0 for some j, 1 ≤ j ≤ n. Now if ϕ_{j1}(x_{j}) ≠ 0 for each j = \overline{1, n}, then it is sufficient to show that by elementary transformations of the columns of (7) it is possible to reduce them to the form

[k]_{j} → [k̃]_{j} = [(f_{i}(x))^{(k−1)}ϕ_{jk}(x)]_{1≤i≤N}.  (9)

This will be proved by induction. The column [1]_{j} in (7) already has the desired form (9) [˜1]_{j} = [1]_{j} = [f_{i}(x)ϕ_{j1}(x)]_{1≤i≤N}, j = \overline{1, n}.

Let us assume that the columns \{[k]_{j}|k = \overline{1, m}\}, 1 ≤ m ≤ r_{j} − 1, also have the desired form (9) for each j, j = \overline{1, n}. If now ϕ_{jk}(x_{j}) = 0 for some k, 1 ≤ k ≤ m, then (8) is proved. Next, if

ϕ_{jk}(x_{j}) ≠ 0,  (10)

for each j, k, j = \overline{1, n}, k = \overline{1, m}, then we add the sum

∑_{q=1}^{m} C_{m}^{q−1}( − ϕ_{j,m+1}^{(m−q+1)}(x)/ϕ_{jq}(x))[q]_{j} = ∑_{q=0}^{m−1} C_{m}^{q}( − ϕ_{j,m+1}^{(m−q)}(x))f_{i}^{(q)}(x)

(which is correctly defined according to (10)) to the column [m+1]_{j} written by virtue of the Leibniz formula as

[m + 1]_{j} =[(f_{i}(x)ϕ_{j,m+1}(x))^{(m)}]_{1≤i≤N} =[ ∑_{q=0}^{m} C_{m}^{q}f_{i}^{(q)}(x)ϕ_{j,m+1}^{(m−q)}(x)]_{1≤i≤N}.

Finally we obtain

[m + 1]_{j} → [{ \atop m + 1}]_{j} = [f_{i}^{(m)}(x)ϕ_{jm+1}(x)]_{1≤i≤N}. ■

Proof of Theorem 1. Denote by \{i\} the row with the number i, i = \overline{1, N}, in the matrix (2). By adding the sum ∑_{q=1}^{r_{1}} c_{q}{q + i − 1} with the coefficients c_{q} ∈ F to the row \{r_{1} + i\}, i = \overline{1, N − r_{1}}, we get

det V (x_{1}, r_{1}; …; x_{n}, r_{n}) = det [ \begin{matrix}w̃_{1}(x_{1}) & | & & | & w̃_{n}(x_{n}) \\ — — & | & ⋯ & | & — — \\ ũ_{1}(x_{1}) & | & & | & ũ_{n}(x_{n}) \end{matrix} ] , (11)

where the matrix w̃_{j}(x), j = \overline{1, n}, is obtained from the matrix (3) after eliminating the last N − r_{1} rows, and the matrix ũ_{j}(x), j = \overline{1, n}, has the form

ũ_{j}(x) = [(x^{i−1}Q(x))^{(k−1)}]\begin{matrix}1≤i≤N−r_{1} \\ 1≤k≤r \end{matrix}.  (12)

Here

Q(x) = x^{r_{1}} + ∑_{q=1}^{r_{1}} c_{q}x^{q−1}

is a polynomial of degree r_{1}.

Note that the matrix w̃_{1}(x) = V (x, r_{1}) is the r_{1} × r_{1} Wronski matrix of r_{1} functions \{f_{i}(x) = x^{i−1}|i = \overline{1, r_{1}}\} and therefore

det V (x, r_{1}) = det [w̃_{1}(x)] = ∏_{k=0}^{r_{1}−1} k!.  (13)

Combining the coefficients c_{q} ∈ F, q = \overline{1, N − n}, in such a way that Q has the r_{1}-tiple root at x = x_{1}, we obtain

Q(x) = (x − x_{1})^{r_{1}},  (14)
Q^{(k)}(x_{1}) = 0, k = \overline{0, r_{1} − 1},  (15)

and now the matrix ũ_{1}(x_{1}) vanishes by virtue of the Leibniz formula and (12). Therefore

det V (x_{1}, r_{1}; …; x_{n}, r_{n}) = det [ \begin{matrix}w̃_{1}(x_{1}) & | & w̃_{2}(x_{2}) & | & & | & w̃_{n}(x_{n}) \\ — — & | & — — & | & ⋯ & | & — — \\ 0 & | & ũ_{2}(x_{2}) & | & & | & ũ_{n}(x_{n}) \end{matrix} ] ,

which, taking into account (13) and the Laplace theorem on determinant expansion, gives

det V (x_{1}, r_{1}; …; x_{n}, r_{n}) = ( ∏_{k=0}^{r_{1}−1} k!) det Ũ,  (16)

where the matrix Ũ ∈ M_{N−r_{1}}(F), Ũ = [ũ_{2}(x_{2})| ⋯ |ũ_{n}(x_{n})], satisfies the conditions of Lemma 1. Hence according to (8) and (3)

det Ũ = det [w_{2}(x_{2})| ⋯ |w_{n}(x_{n})] ∏_{j=2}^{n}(Q(x_{j}))^{r_{j}}

and from (16) together with (2) and (14) we obtain a recurrence relation

det V (x_{1}, r_{1}; …; x_{n}, r_{n}) =
( ∏_{k=0}^{r_{1}−1} k!) ∏_{j=2}^{n}(x_{j} − x_{1})^{r_{1}r_{j}} det V (x_{2}, r_{2}; …; x_{n}, r_{n}).  (17)

The induction by n gives (5) from (17) and (13). ■

2. Let the reals α_{i}, x_{j}, i, j = \overline{1, n}, be given. We introduce the notation (α)_{n} = \{α_{i}|α_{i} ≠ α_{j} if i ≠ j, i = \overline{1, n}\}.

It is known that the generalized Vandermonde matrix

[x_{j}^{α_{i}}]_{1}^{n} = V^{(G)}(x_{1}, …, x_{n}; (α)_{n}) ≡ V^{(G)}
(0 < x_{1} < ⋯ < x_{n}; α_{1} < α_{2} < ⋯ < α_{n})

is completely positive (cf.[4], p.372), i.e., all its minors are positive. Let now 1 ≤ n ≤ N and r_{1}, r_{2}, …, r_{n} ∈ ℕ. We shall call the matrix

V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) = [w_{1}^{(G)}(x_{1})| ⋯ |w_{n}^{(G)}(x_{n})]  (18)

the (r_{1}, r_{2}, …, r_{n})-tiple generalized Vandermonde matrix. Here w_{j}^{(G)}, j = \overline{1, n}, is the rectangular matrix

w_{j}^{(G)}(x)=[(x^{α_{i}})^{(k−1)}]\begin{matrix}1≤i≤N \\ 1≤k≤r_{j} \end{matrix}, (r_{j} ≥1, j =\overline{1, n}, ∑_{j=1}^{n} r_{j} =N). (19)

Theorem 2. The (r_{1}, r_{2}, …, r_{n})-tiple generalized Vandermonde matrix

V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N})
(0 < x_{1} < ⋯ < x_{n}; α_{1} < α_{2} < ⋯ < α_{n})

is positively definite.

Lemma 2. For arbitrary simultaneously nonzero reals c_{i} ∈ ℝ, i = \overline{1, N}, α_{i} ≠ α_{k} if i ≠ k, the function f : ℝ → ℝ

f(x) = ∑_{i=1}^{N} c_{i}x^{α_{i}}

has at most N − 1 positive zeros counted according to their multiplicities:

f(x) = ∏_{j=1}^{n}(x − x_{j})^{r_{j}}ϕ(x; (α)_{N}),

∑_{j=1}^{n} r_{j} ≤ N −1; ϕ(x; (α)_{N}) ≠ 0, x>0; x_{j} >0, j = \overline{1, n}, 1≤n≤N.

Proof of Lemma 2. If f has only single zeros, i.e., r_{1} = ⋯ = r_{N} = 1, then n = N, and the assertion of Lemma 2 follows from the inequality det V^{(G)} ≠ 0 ([4], p.372). Assume that the assertion of Lemma 2 is true for 1 ≤ r_{j} ≤ r, j = \overline{1, n}, and prove it for N power-summands also in the case r_{j′} = r + 1 for certain j′, 1 ≤ j′ ≤ n.

Let the opposite be true: say, there exist reals c_{i}, i = \overline{1, N}, at least one of which is nonzero, such that

f(x) = ∑_{i=1}^{N} c_{i}x^{α_{i}} = ∏_{j=1}^{n}(x − x_{j})^{r_{j}}ϕ(x; (α)_{N}),

∑_{j=1}^{n} r_{j} ≥ N; 1 ≤ r_{j} ≤ r + 1, j = \overline{1, n}, 1 ≤ n ≤ N.

Let m, 1 ≤ m ≤ n, be the number of multiple zeros of f and zeros x_{j}, j = \overline{1, n}, be arranged according to the decrease of their multiplicity. Then f satisfies the conditions

f(x_{j}) = 0, j = \overline{1, n},  (20)
f′(x_{j}) = ⋯ = f^{(r_{j}−1)}(x_{j}) = 0, 2 ≤ r_{j} ≤ r + 1, j = \overline{1, m}; (21)

here

n − m + ∑_{j=1}^{m} r_{j} ≥ N.  (22)

The Rolle theorem, together with (20), implies that the function f̃ : ℝ → ℝ

f̃(x) = (f(x)x^{−α_{1}})′ = ∑_{i=1}^{N−1} c̃_{i}x_{α_{i}}  (23)

vanishes at the points ξ_{j}, 0 < x_{j} < ξ_{j} < x_{j+1}, j = \overline{1, n − 1}, i.e.,

f̃(ξ_{j}) = 0, j = \overline{1, n − 1}.

Moreover, from (21) it follows that

f̃(x_{j}) = f̃′(x_{j}) = ⋯ = f̃^{(r_{j}−2)}(x_{j}) = 0, j = \overline{1, m},

i.e., f̃ has the form

f̃(x) = ∏_{k=1}^{n−1}(x − ξ_{k}) ∏_{j=1}^{m}(x − x_{j})^{r_{j}−1}ϕ̃(x; (α̃)_{N}) =

= P_{N}^{∼}(x)ϕ̃(x; (α̃)_{n}),  (24)

all roots of the polynomial P_{N}^{∼} being positive and having the multiplicity ≤ r. From (24) and (22) we find

Ñ = n − 1 + ∑_{j=1}^{m}(r_{j} − 1) = n − 1 + ∑_{j=1}^{m} r_{j} − m ≥ N − 1,

but this contradicts our assumption, since the function (23) is the sum of N − 1 power-summands. ■

Proof of Theorem 2. For α_{i} = i − 1, i = \overline{1, N}, we have

V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) = V (x_{1}, r_{1}; …; x_{n}, r_{n})

and, according to (5),

det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) > 0
(0 < x_{1} < ⋯ < x_{n}; α_{i} = i − 1, i = \overline{1, N}).

It is possible to pass to arbitrary values α_{1} < ⋯ < α_{N} starting from the integers α_{i} = i − 1, i = \overline{1, N}, and changing them continuously, but preserving the inequalities among them. In doing so, the determinant det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) does not vanish according to Lemma 2 and therefore for all 0 < x_{1} < ⋯ < x_{n}; α_{1} < α_{2} < ⋯ < α_{N} we have det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) > 0.

Since each principal minor of the matrix V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) can be considered as a determinant of a certain (r_{1}, r_{2}, …, r_{n})-tiple generalized Vandermonde matrix, all such minors are positive. ■

Corollary 1. It follows from det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) ≠ 0 that for arbitrary reals α_{i}, α_{i} ≠ α_{k} if i ≠ k, i, k = \overline{1, N}, there exists unique collection of reals c_{i}, i = \overline{1, N − 1}, such that

∑_{i=1}^{N−1} c_{i}x^{α_{i}} + x^{α_{n}} = P_{N−1}(x)ϕ(x; (α)_{N}),

where P_{N−1} is any given polynomial of degree N − 1 having only positive roots, and ϕ(x; (α)_{N}) ≠ 0 for x > 0.

Note that for α_{i} ∈ ℕ the above proposition is nothing but the Viete theorem, since it enables us to determine the coefficients at power-summands provided we know zeros of the sum.

3. Let n, N be integers, 1 ≤ n ≤ N, and 0 < α_{1} < α_{2} < ⋯ < α_{N}, x_{j} > 0, j = \overline{1, n}. Transform the determinant det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}), taking the factor α_{i}, i = \overline{1, N}, out of the i-th row and taking into account that

α_{i}^{−1}(x^{α_{i}})^{(k−1)} = \frac{Γ(α_{i})}{Γ(α_{i} + 2 − k)}x^{α_{i}−k+1} = (x^{α_{i}−1})^{(k−2)}, k = 1, 2, ….

Here we put

(x^{α})^{(−q)} = \underbrace{∫_{0}^{x} dx ⋯ ∫_{0}^{x}}_{q−tiple} dxx^{α} = \frac{1}{Γ(q)} ∫_{0}^{x}(x − t)^{q−1}t^{α} dt,

α ≥ 0, q = 1, 2, …,
(x^{α})^{(−1)} = ∫_{0}^{x} t^{α}dt, α ≥ 0, x > 0,  (25)

and Γ(z) = (z−1)! is the gamma-function. Taking into account the corollary of Theorem 2, after the appropriate transformations we get

det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) =

= (−1)^{σ_{1}} ∏_{i=1}^{N} α_{i} det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α − 1)_{N−n}) det A, (26)

where

σ_{1} =∑_{j=1}^{n} j(r_{j} − 1), (α−1)_{N} =\{α_{i}−1|α_{i} ≠α_{j} if i ≠ j, i, j =\overline{1, N}\},

and A ∈ M_{n}(F) is a square matrix of the form

A = [a_{ij}] = [ ∫_{x_{j−1}}^{x_{j}} ϕ_{i}(t; (α − 1)_{N}) ∏_{p=1}^{n}(t − x_{p})^{r_{p}−1}dt].  (27)

Here we use the notation

Ñ = N − n + 1  (28)

and ϕ_{i}(t; (α − 1)_{N}) is defined by the formula

ϕ_{i}(t; (α − 1)_{N}) ∏_{p=1}^{n}(t − x_{p})^{r_{p}−1} = ∑_{q=0}^{N−n} c_{i+q}t^{α_{i+q}−1}, i = \overline{1, n}, (29)

the coefficients c_{i+q} being defined according to the corollary of Theorem 2.

Rearranging the columns of det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) which contain higher derivatives, we can reduce it to the form

det V^{(G)}(x_{1}, r_{1}; …; x_{n}, r_{n}; (α)_{N}) =
= (−1)^{σ_{2}} det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α)_{N−n}) det Ã, (30)

where σ_{2} = ∑_{j=1}^{n}(j − 1)(r_{j} − 1) and the matrix Ã ∈ M_{n}(F) is written as

Ã = [ã_{ij}]_{1}^{n} = [(ϕ̃_{i}(x; (α)_{N}) ∏_{p=1}^{n}(x − x_{p})^{r_{p}−1})_{|x=x_{j}}^{(r_{j}−1)}]_{1}^{n} =

= [(r_{j} − 1)!ϕ̃_{i}(x_{j}; (α)_{N}) ∏_{\begin{matrix}p=1 \\ p≠j \end{matrix}}^{n}(x_{j} − x_{p})^{r_{p}−1}]_{1}^{n}.  (31)

Here, by definition,

ϕ̃_{i}(x; (α)_{N}) ∏_{p=1}^{n}(x − x_{p})^{r_{p}−1} = ∑_{q=0}^{N−n} c_{i+q}x^{α_{i+q}}, i = \overline{1, n},

from which one can find

ϕ̃_{i}(x; (α)_{N}) = xϕ_{i}(x; (α − 1)_{N}), i = \overline{1, n}.  (32)

Equating the right-hand sides of (26) and (30) and taking into account (31) and (32), we get

\frac{det [ ∫_{x_{j−1}}^{x_{j}} ϕ_{i}(t; (α − 1)_{N}) ∏n ; \begin{matrix}p=1 \\ p≠j \end{matrix} (\frac{t − x_{p}}{x_{j} − x_{p}} )^{r_{p}−1}(x_{j} − t)^{r_{j}−1}dt]_{1}^{n}}{det[ϕ_{i}(x_{j}; (α − 1)_{N})]_{1}^{n} ∏_{p=1}^{n} x_{j}} =

= ∏_{j=1}^{n}(r_{j} − 1)! ∏_{i=1}^{N} α_{i}^{−1} \frac{det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α)_{N})}{det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α − 1)_{N})}.

(x_{p} ≠ x_{j} if p ≠ j; x_{0} = 0; i, j = \overline{1, n}).

For the right-hand side ratio Lemma 1 obviously gives

\frac{det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α)_{N−n})}{det V^{(G)}(x_{1}, r_{1} − 1; …; x_{n}, r_{n} − 1; (α − 1)_{N−n})} = ∏_{j=1}^{n} x_{j}^{r_{j}−1}.

Hence we have the formula

\frac{det [ ∫_{x_{j−1}}^{x_{j}} ϕ_{i}(t; (α − 1)_{N}) ∏n ; \begin{matrix}p=1 \\ p≠j \end{matrix} (\frac{t − x_{p}}{x_{j} − x_{p}} )^{r_{p}−1}(x_{j} − t)^{r_{j}−1}dt]_{1}^{n}}{det[ϕ_{i}(x_{j}; (α − 1)_{N})]_{1}^{n} ∏_{p=1}^{n} x_{j}^{r_{j}}} =

= \frac{∏_{p=1}^{n}(r_{p} − 1)!}{∏_{i=1}^{N} α_{i}}

(x_{p} ≠x_{j} if p≠j; r_{p} ∈ℕ, ∑_{p=1}^{n} r_{p} =N, x_{p} > 0, p, i, j =\overline{1, n}; x_{0} =0),

which can be rewritten as

\frac{det [ ∫_{x_{j−1}/x_{j}}^{1}(1 − u)^{r_{j}−1}ϕ_{i}(ux_{j}; (α − 1)_{N}) ∏n ; \begin{matrix}p=1 \\ p≠j \end{matrix} (\frac{x_{j}u − x_{p}}{x_{j} − x_{p}} )^{r_{p}−1}du]_{1}^{n}}{det[ϕ_{i}(x_{j}; (α − 1)_{N})]_{1}^{n}} =

= \frac{∏_{k=1}^{n} Γ(r_{k})}{∏_{i=1}^{N} α_{i}}  (33)

(x_{p} ≠ x_{j} if p ≠ j; r_{p} ∈ ℕ, ∑_{p=1}^{n} r_{p} = N, x_{p} > 0, p, i, j = \overline{1, n}; x_{0} = 0).

In the particular case where α_{i} = r_{0} − 1 + i, r_{0} > 0, i = \overline{1, N}, we have

ϕ_{i}(t; (α − 1)_{N}) = ( ∏_{p=1}^{n}(t − x_{p})^{r_{p}−1})^{−1} ∑_{q=0}^{N−n} c_{i+q}t^{r_{0}−1+i+q−1} =

= t^{r_{0}+i−2}( ∏_{p=1}^{n}(t − x_{p})^{r_{p}−1})^{−1} ∑_{q=0}^{N−n} c_{i+q}t^{q} = t^{r_{0}+i−2}, i = \overline{1, N}.

The latter equality is valid because, according to the Viete theorem, there exist reals c_{i+q} such that

∑_{q=0}^{N−n} c_{i+q}t^{q} = ( ∏_{p=1}^{n}(t − x_{p})^{r_{p}−1}).

Moreover,

∏_{i=1}^{N}(r_{0} − 1 + i) = ∏_{i=1}^{N} α_{i} = \frac{Γ(r_{0} + N)}{Γ(r_{0})} = \frac{Γ(r_{0} + ∑_{p=1}^{n} r_{p})}{Γ(r_{0})}

and (33) transforms to the equality

\frac{det [x_{j}^{i−1} ∫_{x_{j}−1/x_{j}}^{1} u^{r_{0}+i−2}(1 − u)^{r_{j}−1} ∏n ; \begin{matrix}p=1 \\ p≠j \end{matrix} (\frac{x_{j}u − x_{p}}{x_{j} − x_{p}} )^{r_{p}−1}du]_{1}^{n}}{det[x_{j}^{i−1}]_{1}^{n}} =

= \frac{∏_{p=0}^{n} Γ(r_{p})}{Γ(∑_{p=0}^{n} r_{p})}  (34)

(x_{p} ≠ x_{j} if p ≠ j; r_{p} ∈ ℕ, x_{p} > 0, p, i, j = \overline{1, n}; x_{0} = 0, r_{0} > 0).

For the case n = 1 the formula (34) gives the well-known expression

∫_{0}^{1} u^{r_{0}−1}(1 − u)^{r_{1}−1}du = \frac{Γ(r_{0})Γ(r_{1})}{Γ(r_{0} + r_{1})} = B(r_{0}, r_{1}),  (35)

for the Euler integral of the first kind.

Now, we introduce the notation

B_{n}(r) = B(r_{0}, r_{1}, …, r_{n}) =

=\{ \begin{matrix}1, n = 0, \\ det [x_{j}^{i−1} ∫_{x_{j−1}/x_{j}}^{1}u^{i−1}(1−u)^{r_{j}−1} ∏_{\begin{matrix}k=0 \\ k≠j \end{matrix}}^{n}(\frac{x_{j}u−x_{k}}{x_{j} −x_{k}} )^{r_{k}−1}du]_{1}^{n}/ det[x_{j}^{i−1}]_{1}^{n}, \\ n ≥ 1 (x_{p} ≠ x_{j} if p ≠ j, p, j = \overline{0, n}, x_{0} = 0) \end{matrix}

(36)

for arbitrary \{r_{j}|r_{j} > 0, j = \overline{0, n}\} ≡ r.

Since the Euler formula (35) is valid for arbitrary complex r_{0}, r_{1}, Re r_{0} > 0, Re r_{1} > 0, there arises the problem:

Problem. Is the equality

B_{n}(r) = B(r_{0}, r_{1}, …, r_{n}) = \frac{∏_{j=0}^{n} Γ(r_{j})}{Γ(∑_{j=0}^{n} r_{j})}  (37)

fulfilled for an arbitrary complex r_{j}, j = \overline{0, n}, n ≥ 2? (Note that this case is not covered either by (34) or by (35)).

4. Put 2 ≤ n ≤ N and 1 ≤ m ≤ min\{r_{j}|j = \overline{1, n}\}. Transform the determinant on the left-hand side of (5), taking the factor Γ(i)/Γ(i − m) out of each row having the number i ≥ m + 1 and keeping in mind that

(\frac{Γ(i − m)}{Γ(i)}x^{i−1})^{(k−1)} = \frac{Γ(i − m)}{Γ(i − k + 1)}x^{i−k} = (x^{i−m−1})^{(k−m−1)}

(i = \overline{m + 1, N}, 1 ≤ m ≤ min\{r|j = \overline{1, n}\}, k = 1, 2, …).

After easy but rather long calculations, using (25) and (5) we get the formula

\frac{det[B^{(1)}| … |B^{(n)}]}{det V (x_{1}, …, x_{n}; m)} = ∏_{k=1}^{m}(\frac{Γ(k)}{Γ(N − k + 1)} ∏_{j=0}^{n} Γ(r_{j} − k + 1)), (38)

where B^{(j)}, j = \overline{1, n}, is a rectangular matrix of the form

B^{(j)} = [b_{ik}^{(j)}]\begin{matrix}1≤i≤mn \\ 1≤k≤m \end{matrix} =

= [x_{j}^{i−k} ∫_{x_{j−1}/x_{j}}^{1}(1 − u)^{r_{j}−m}u^{k+i−2} ∏_{\begin{matrix}p=0 \\ p≠j \end{matrix}}^{n} (\frac{x_{j}u − x_{p}}{x_{j} − x_{p}} )^{r_{p}−m}du]\begin{matrix}1≤i≤mn \\ 1≤k≤m \end{matrix}

(x_{p} ≠ x_{j} if p ≠ j; x_{0} > 0; j = \overline{1, n}).  (39)

In particular, for m = 1 (38) reduces to (34).

5. Let us now apply the above results to the case n = 1, N = r, x > 0. It is easy to show that for the determinant of the matrix

V^{(G)}(x, r; (α)_{r}) = [(x^{α_{i}})^{(k−1)}]_{1}^{r},

which is nothing but the Wronski matrix for r functions {f_{i}(x) = x^{α_{i}}|i = \overline{1, r}; x > 0\}, one gets

det V^{(G)}(x, r; (α)_{r}) = x^{β} ∏_{1≤k<j≤r}(α_{j} − α_{k}),  (40)

where β = ∑_{i=1}^{N} α_{i} − r(r − 1)/2.

Obviously, the formula (40) generalizes (13).

Using (40), one can find values of ϕ(x; (α)_{N}) and coefficients c_{i} for which

∑_{i=1}^{N−1} c_{i}x^{α_{i}} + x^{α_{N}} = (x − x_{1})^{N−1}ϕ(x; (α)_{N}).

Namely, the Cramer formulas give

c_{i} = −x_{1}^{α_{N} −α_{i}}L_{i}(α_{N}), i = \overline{1, N − 1},
ϕ(x; (α)_{N})=(x−x_{1})^{−(N−1)}x_{1}^{α_{N}} ((x/x_{1})^{α_{N}} − ∑_{i=1}^{N−1}(x/x_{1})^{α_{i}}L_{i}(α_{N})), (41)

where

L_{i}(α) = ∏_{\begin{matrix}p=1 \\ p≠i \end{matrix}}^{N−1} (\frac{α − α_{p}}{α_{i} − α_{p}} ), i = \overline{1, N − 1},

is the Lagrange elementary interpolation polynomial. (41) gives

ϕ(x_{1}; (α − 1)_{N}) =

= \frac{x_{1}^{α_{N} −N}}{(N − 1)!}(\frac{Γ(α_{N})}{Γ(α_{N} − N + 1)} − ∑_{i=1}^{N−1} \frac{Γ(α_{i})}{Γ(α_{i} − N + 1)}L_{i}(α_{N})), (42)

ϕ(ux_{1}; (α − 1)_{N}) = \frac{x_{1}^{α_{N} −N}}{(u − 1)^{(N−1)}} (u^{α_{N} −1} − ∑_{i=1}^{N−1} u^{α_{i}−1}L_{i}(α_{N})). (43)

Substituting (42) and (43) in (33) for n = 1, N = r_{1} ≥ 2, Ñ = N and performing integration we obtain

\frac{Γ(α_{n})}{Γ(α_{N} − N + 1)} − ∑_{i=1}^{N−1} \frac{Γ(α_{i})}{Γ(α_{i} − N + 1)}L_{i}(α_{N}) =

= (−1)^{N}( ∑_{i=1}^{N−1} α_{i}^{−1}L_{i}(α_{N}) − α_{N}^{−1}) ∏_{k=1}^{N} α_{k}.  (44)

Finally, substitution of (44) in (42) gives

ϕ(x_{1}; (α−1)_{N})= \frac{x_{1}^{α_{N}−N}}{(N −1)!}(−1)^{N}( ∑_{i=1}^{N−1}α_{i}^{−1}L_{i}(α_{N})−α_{N}^{−1}) ∏_{k=1}^{N}α_{k}. (45)

References

1. R. W. Hamming, Numerical methods for scientists and engineers. (Translation into Russian) Mir, Moscow, 1968; English original: McGraw-Hill, 1962.

2. N.A.Lebedev and V.I.Smirnov, The constructive theory of functions of a complex variable. (Russian) Nauka, Moscow, 1964.

3. I.R.Lomidze, Criteria of unitary and orthogonal equivalence of operators. Bull. Acad. Sci. Georgia 141(1991), No. 3, 481-483.

4. F.R.Gantmakher, The theory of matrices. (Russian) Nauka, Moscow, 1988.

(Received 24.11.1992)

Author’s address: Faculty of Physics I. Javakhishvili Tbilisi State University 3, I. Chavchavadze Ave, 380028 Tbilisi Republic of Georgia