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Although the following proof is well-known, it makes still sense to quickly go through it because, by extracting the essence of it, we will be able to build a very powerful method out of it (see Section 2.4). If Xi1 = Xi2 with i1 = i2, then the Vandermonde determinant (2.1) certainly vanishes because in that case two rows of the determinant are identical. Hence, (Xi1 − Xi2 ) divides the1determinant as a polynomial in the Xi ’s. But that means that the complete product ≤i<j ≤n (Xj − Xi ) (which is exactly the right-hand side of (2.1)) must divide the determinant. n On the other hand, the determinant is a polynomial in the Xi ’s of degree at most . Combined with the previous observation, this implies that the determinant equals 2 the right-hand side product times, possibly, some constant. To compute the constant, 01 n compare coefficients of X1 X2 · · · Xn −1 on both sides of (2.1). This completes the proof of (2.1). At this point, let us extract the essence of this proof as we will come back to it in Section 2.4. The basic steps are: 1. Identification of factors 2. Determination of degree bound 3. Computation of the multiplicative constant. An immediate generalization of the Vandermonde determinant evaluation is given by the proposition below. It can be proved in just the same way as the above proof of the Vandermonde determinant evaluation itself. Prop osition 1. Let X1 , X2 , . . . , Xn be indeterminates. If p1 , p2 , . . . , pn are polynomials of the form pj (x) = aj xj −1 + lower terms, then 1 det (pj (Xi )) = a1 a2 · · · an (Xj − Xi ). (2.2)
1≤i,j ≤n ≤i<j ≤n...
This unique factorization of the matrix M is known as the L(ower triangular)U(pper triangular)-factorization of M , or as well as the Gauß decomposition of M . Equivalently, for a square matrix M (satisfying these conditions) there exists a unique lower triangular matrix L and a unique upper triangular matrix U , the latter with all entries along the diagonal equal to 1, such that M · U = L. (2.27)...
where the sum is over all perfect matchings π of the complete graph on 2n vertices, where c(π ) is the crossing number of π , and where the product is over all edges (ij ), i < j , in the matching π (see e.g. [169, Sec. 2]). What links Pfaffians so closely to...
The specializations of this determinant evaluation which are of relevance for the enumeration of cyclically symmetric plane partitions and descending plane partitions are the cases µ = 0 and µ = 1, respectively. In these cases, Macdonald, respectively Andrews, actually had conjectures about q -enumeration. These were proved by Mills, Robbins and Rumsey [110]. Their theorem which solves the q -enumeration of cyclical ly symmetric plane partitions is the following. Theorem 33. For nonnegative integers n there holds δ i q = in 1 − q 3i−1 1 1 − q 3(n+i+j −1) +j 3i+1 . det ij + q j 0≤i,j ≤n−1 1 − q 3i−2 ≤i≤j ≤n 1 − q 3(2i+j −1) 3 =1...
For nonnegative integers n there holds = µ +i+j det 0≤i,j ≤n−1 2i − j − i 3 n−1 (µ + i + 1) n−1 i (i+1)/2 ( µ − 3n + i + 2 /2 (−1)χ(n≡3 mod 4) 2( 2 ) , (3.30) i)i =1 where χ(A) = 1 if A is true and χ(A) = 0 otherwise, and µ + =µ +i+j +i+j+2 det 2 0≤i,j ≤n−1 2i − j 2i − j + 1 3 +1 i n (µ + i) i/2 (µ + 3n − i−1 ) 2 2 2n (2i − 1)!! =1
(i+1)/2
...
ADVANCED DETERMINANT CALCULUS
49
This determinant evaluation appears in [193] in the study of certain models in infinite statistics. However, as Varchenko et al. [20, 153, 184] show, this determinant evaluation is in fact just a special instance in a whole series of determinant evaluations. The latter papers give evaluations of determinants corresponding to certain bilinear forms associated to hyperplane arrangements and matroids. Some of these bilinear forms are relevant to the study of hypergeometric functions and the representation theory of quantum groups (see also [185]). In particular, these results contain analogues of (3.63) for all finite Coxeter groups as special cases. For other developments related to Theorem 55 (and different proofs) see [36, 37, 40, 67], tying the sub ject also to the representation theory of the symmetric group, to noncommutative symmetric functions, and to free Lie algebras, and [109]. For more remarkable determinant evaluations related to hyperplane arrangements see [39, 182, 183]. For more determinant evaluations related to hypergeometric functions and quantum groups and algebras, see [175, 176], where determinants arising in the context of Knizhnik-Zamolodchikov equations are computed. The results in [20, 153] may be considered as a generalization of the Shapovalov determinant evaluation [159], associated to the Shapovalov form in Lie theory. The latter has since been extended to Kac–Moody algebras (although not yet in full generality), see [31]. There is a result similar to Theorem 55 for another prominent permutation statistics, MacMahon’s major index. (The ma jor index ma j(π ) is defined as the sum of all positions of descents in the permutation π , see e.g. [46].) Theorem 56. For any positive integer n there holds q = in ma j(σπ−1 ) det (1 − q i )n! (i−1)/i...
where Um (q ) is the Chebyshev polynomial of the necon− kind n s giv.en in Theorem 57, s d a and where a2n,2i = c2n,2i − c2n,2i+2 with cn,h = (n−h)/2 (n−h)/2−1...
However, what if we encounter a sequence where all these nice automatic tools fail? Here are a few hints. First of all, it is not uncommon to encounter a sequence (an)n≥0 which has actually a split definition. For example, it may be the case that the subsequence (a2n )n≥0 of even-numbered terms follows a “nice” formula, and that the subsequence (a2n+1)n≥0 of odd-numbered terms follows as well a “nice,” but different, formula. Then Rate will fail on any number of first terms of (an )n≥0 , while it will give you something for sufficiently many first terms of (a2n)n≥0 , and it will give you something else for sufficiently many first terms of (a2n+1 )n≥0 . Most of the subsequent hints apply to a situation where you encounter a sequence p0 (x), p1(x), p2 (x), . . . of polynomials pn (x) in x for which you want to find (i.e., guess) a formula. This is indeed the situation in which you are generally during the guessing for “identification of factors,” and also usually when you perform a guessing where a parameter is involved. To make things concrete, let us suppose that the first 10 elements of your sequence of polynomials are...
If the above ideas do not help, then I have nothing else to offer than to try some, more or less arbitrary, manipulations. To illustrate what I could possibly mean, let us again consider an example. In the course of working on [90], I had to guess the result of a determinant evaluation (which became Theorem 8 in [90]; it is reproduced here as Theorem 43). Again, the difficult part of guessing was to guess the “ugly” part of the result. As the dimension of the determinant varied, this gave a certain sequence pn (x, y ) of polynomials in two variables, x and y , of which I display p4 (x, y ):
In[1]:= VPol[4] 2 Out[1]= 6 x + 11 x > 6xy 2 +6x 2 y +6x 2 +6y 3 4 2 3 2 + x + 6 y - 10 x y - 6 x y - 4 x y + 11 y 3 -4xy 3 +y 4...
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