Abstract
As extensions of Vandermonde determinant, we establish a general determinant evaluation formula by means of the Laplace expansion formula. Several interesting determinant identities are consequently derived by computing divided differences.
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1 Introduction and motivation
Divided differences (cf. [7, Chapter 1]) are useful tools in mathematics and physics. For example, their applications to symmetric functions and approximation theory can be found respectively in [6, § 1.2] and [8, § 2.4]. For a complex function f(y) and uneven spaced grid points \(\{x_k\}_{k=0}^{n}\), the divided differences with respect to y are defined in succession as follows:
In general, they can be expressed through the Newton formula:
Between the divided differences of monomials and complete symmetric functions, there holds the following well–known formula due to Sylvester (1839), whose proofs can be found in [1,2,3]:
Here and below, we shall denote the elementary and complete symmetric functions (cf. Macdonald [6, § 1.2]) of variables \(X=\{x_0,x_1,\ldots ,x_n\}\), respectively, by
They admit the following generating functions:
Recently, divided differences have been employed by the first author [4, 5] and Tang–Xu [9] to investigate determinant evaluations. In this paper, we shall utilize the Laplace expansion formula to establish an extension of Vandermonde determinant. By computing divided differences, it will further be specialized to several interesting Vandermonde—like determinant identities.
2 Extensions of Vandermonde determinant
For the indeterminates \(X:=\{x_k\}_{k=0}^{n}\), denote the divided differences by
and the Vandermonde determinant by
Theorem 1
(Extension of Vandermonde determinant)
Proof
Consider the extended matrix of order \((n+2)\times (2+n)\) given explicitly by
whose determinant is obviously equal to the determinant stated in the theorem.
Now subtracting \(v_j\) times the first column from each other column, we transform the matrix into the following one:
Applying the Laplace expansion formula to the last matrix with respect to the first column and then to the first row, we have
Let \([x^k]f(x)\) stand for the coefficient of \(x^k\) in the formal power series f(x). Then the determinant displayed in (5) can be evaluated as
Substituting this expression into (5) and then simplifying the result, we obtain the determinant identity stated in the theorem. \(\square \)
When \(u_k=1\), Theorem 1 leads immediately to the following determinant identity.
Corollary 2
Proof
According to (3), there holds the expression
Then we have, from (2), the relation
where \(\chi \) is the logical function defined by \(\chi (\text {true})=1\) and \(\chi (\text {false})=0\). Therefore, the double sum displayed in Theorem 1 with \(u_k=1\) can be rewritten as follows
which reduces to the single term \(v_0\), corresponding to \(k=n\) and \(\jmath =0\). \(\square \)
For \(v_k=v^k\) with \(0\le k\le n\), observing that
we derive from Theorem 1 another determinant identity.
Proposition 3
Suppose further that \(u_k:=u(x_k)\) is a function of \(x_k\) for \(0\le k\le n\), we may state Proposition 3 equivalently in terms of divided differences.
Corollary 4
For \(m\le n\), letting \(u(x_i)=\lambda \prod _{k=1}^{m}(x_i-\gamma _k)\) and appealing to the formula
we derive from Corollary 4 the following determinant identity.
Example 5
Instead, letting \(u(x_i)=\lambda \prod _{k=0}^{n}(x_i-\gamma _k)\) and then applying the formula
we establish from Corollary 4 another determinant identity.
Example 6
Specifying \(\lambda =-1/\prod _{k=0}^{n}(v-\gamma _k)\) further in the last example, we obtain a more symmetric determinant formula:
Finally, if we let in Corollary 4
and applying the formula
we would derive further the following determinant identity.
Example 7
3 Further variants of Vandermonde determinant
For the sake of brevity, this section will use the indeterminates \(\mathbb {X}:=\{x_k\}_{k=1}^n\) and the following notation for the corresponding Vandermonde determinant
Then Theorem 1 can be reformulated equivalently as follows:
For \(u_k=1\), this equality becomes the following determinant identity.
Proposition 8
Proof
Recalling (3), we have the expression
According to (2), it is trivial to check the relations
Then for \(u_k=1\), we may reformulate the double sum displayed in (7) as
Observing that the last double sum vanishes and then evaluating the divided differences in the penultimate line, we confirm the identity stated in the Proposition. \(\square \)
Letting \(v_j=v^j\) in Proposition 8, we get the simplified determinant identity.
Example 9
Notice that the determinant in Corollary 4 can be expressed equivalently as
For \(m\le n\), letting \(u(x_i)=\lambda \prod _{k=1}^{m}(x_i-\gamma _k)\) and appealing to the formula
we obtain from (8) another determinant identity.
Corollary 10
When \(m=n\) and \(\lambda =-1/\prod _{k=1}^{n}(v-\gamma _k)\), the last corollary yields further to a more symmetric determinant identity:
Finally, for \(v_k=h_k(\mathbb {X})\), Proposition 8 reduces to the strange looking identity:
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Acknowledgements
The authors are sincerely grateful to the anonymous referees for their careful reading and valuable comments that have significantly contributed to improve the manuscript during the revision.
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Chu, W., Wang, X. Extensions of Vandermonde determinant by computing divided differences. Afr. Mat. 29, 73–79 (2018). https://doi.org/10.1007/s13370-017-0527-3
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DOI: https://doi.org/10.1007/s13370-017-0527-3