The Combinatorial Identity on the Jacobian Conjecture
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The Combinatorial Identity on the Jacobian Conjecture GEORGY P. EGORYCHEV1 and VITALY A. STEPANENKO2
1 Krasnoyarsk State Technical University, Kirenskogo 26, Krasnoyarsk 660074, Russia. e-mail: [email protected] 2 Krasnoyarsk State University, Svobodny pr. 79, Krasnoyarsk 660041, Russia. e-mail: [email protected]
Abstract. The multidimensional sum with polynomial coefficients is computed with the help of the method of integral representation and computation of combinatorial sums. This sum was obtained in the attempt to prove the well known Jacobian conjecture (two-dimensional case). Mathematics Subject Classifications (2000): combinatorics, algebra. Key words: combinatorial identities, Jacobian problem.
1. Introduction In 1939 Keller [7] postulated the following conjecture: the polynomial mapping f : C m → C m is polynomial invertible, if its Jacobian is a constant and is not equal to zero. A full review of the current state of this problem, including a description of various generalizations and attempts at its solution, was given in 2000 by Arno van den Essen [4]. Using the classical Cayley–Sylvester–Sack formula (see [3, 10, 8] and [1]) for the two polynomials with two variables V. Stepanenko [9] has reduced the proof of Jacobian conjecture to a justification test of a series of the identities of the following type: [ 2 ] k 1 (2n)! − (2n − 2k + 1)!(2k − 2)! (S1(1) + S2(1) ) + (n + 1)!n! (n + 1)! k=1 m=1 n+1
[ 2 ] k 1 + (2n − 2k + 2)!(2k − 3)! (S1(2) + S2(2) ) ≡ 0, (n + 1)! k=1 m=1 n+1
n 2, (1)
Supported in part by the Natural Sciences and Engineering Research Council of Canada on
Grant NSERC-108343 and the Russian Fund of Fundamental Researches.
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GEORGY P. EGORYCHEV AND VITALY A. STEPANENKO
where S1(1)
2k−1
(2k−j −1)/2
j =2m−1
i=0
=∗ ×
(2k−j −3)/2
j =2m−1
i=0
×
j =2m−1
i=0
(2k−j −3)/2
j =2m−1
i=0
×
22k−2i−2m ×
(2k − j ) , (n − 4k + i + j + 2)!(2k − 2i − j )!i!(2k + m − j − 1)!(j − 2m)!(m − 1)! 2k−3
=∗
22k−2i−2m+1 ×
(2k − j − 1) , (n − 4k + i + j + 2)!(2k − 2i − j − 1)!i!(2k + m − j − 2)!(j − 2m + 2)!(m − 1)! (2k−j −1)/2
× S2(2)
2k−1
=∗
22k−2i−2m+1 ×
(2k − j ) , (n − 4k + i + j + 1)! (2k − 2i − j )!i!(2k + m − j − 1)!(j − 2m + 1)!(m − 1)! 2k−3
S2(1) = ∗
S1(2)
22k−2i−2m ×
(2k − j − 1) . (n − 4k + i + j + 3)!(2k − 2i − j − 1)!i!(2k + m − j − 2)!(j − 2m + 1)!(m − 1)!
2k−1 We suppose here by definition ∗ 2k−1 j =2m−1 . . . = j =2m−1,2m−3,... . If the factorials with nonnegative arguments arise under the sign , then we suppose these summands are equal to 0. In this article we postulate the validity (1) by means of the method of integral representation and computation of combinatorial sums [5]. 2. Laurent Formal Power Series over a Field C and Properties of the res Operator Here we shall explore univariate series only, although in further computations the res concept will be also used for multivariate series. Let L be the set of a Laurent formal power series over the field C containing only finitely many terms with negative
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