By Woerle K.

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**Extra resources for 200 Aufgaben aus der Trigonometrie mit Loesungen**

**Sample text**

0]. The usual Poincaré polynomial is obtained by specializing all hi , hij to q and thus is obtained by symmetrizing the “q-Vandermonde”. One could have taken an arbitrary subsum of the expansion of M♥ . Macdonald’s theorem states that the only terms surviving after symmetrization are those having for coefficient the inversion weight of an element of the group. The following theorem shows how to apply this property to generate intervals for the weak order. e. if (w) = (wv −1 ) + (v). In that case I(v) is a factor of I(w).

Instead of enumerating partitions, one can introduce y = {y1 , . . , yn } and test the single function n (x−1 i − yj ) . R(x∨ , y) = i,j=1 Let us first consider R(x∨ , y) πwC0 . The monomials xu in the expansion of R(x∨ , y) which give a non-zero contribution are those such that u + ρ, with ρ = [n, . . , 1], has all its component different in absolute value. Since [0, . . , 1−n] ≤ u + ρ ≤ ρ, the vector u + ρ must be a signed permutation of ρ, in which case w xu πwC0 = ±1. Therefore, the sum w ± xρ R(x∨ , y) (∆C )−1 , which expresses R(x∨ , y) πwC0 , is independent of x.

More precisely, given i, let a = vi , b = vi+1 the corresponding components of the spectral vector. Then, instead of si , ∂i , πi , πi , Ti respectively, one takes si + 1 1 1 t−1 1 , ∂i + , πi + , πi + , Ti + b−a b−a b/a − 1 b/a − 1 b/a − 1 (the careful reader adds “provided b = a”). For n = 3, the Yang-Baxter relations for si , ∂i , πi and Ti , and a spectral vector v are, writing v2 −v1 = a, v3 −v2 = b, v2 /v1 = α, v3 /v2 = β, 7 it only works for πi → πi + 1 = πi . 7 — Yang-Baxter relations 123 s1 + 1 a 123 1 b ∂1 + 132 213 s2 + 213 1 s2 + a+b 1 s1 + a+b 231 s1 + 312 1 b s2 + 1 a 1 a 1 ∂2 + a+b 1 ∂1 + a+b 231 312 ∂1 + 1 b 1 β −1 132 T1 + t−1 α−1 T2 + 213 231 312 1 β −1 π2 + 321 1 α−1 t1 −1 T1 + αβ−1 231 T1 + t−1 β−1 132 t−1 π1 + αβ1−1 T2 + αβ−1 π2 + αβ1−1 π1 + 1 a 123 π2 + 213 ∂2 + 321 123 1 α− 1 1 b 132 321 π1 + ∂2 + 312 t−1 β −1 T2 + t−1 α− 1 321 The fact that each hexagon closes means that the two paths from top to bottom give equal elements when evaluated as products of the labels on the edges.