By Valerie Martin

ISBN-10: 0307833879

ISBN-13: 9780307833877

From the acclaimed writer of the Orange Prize profitable estate comes a clean twist at the vintage Jekyll and Hyde tale, a unique instructed from the point of view of Mary Reilly, Dr. Jekyll's dutiful and clever housemaid.

Faithfully weaving in information from Robert Louis Stevenson's vintage, Martin introduces an unique and desirable personality: Mary is a survivor - scarred yet nonetheless robust - acquainted with evil, but brimming with devotion and love. As a bond grows among Mary and her tortured enterprise, she is distributed on errands to unsavoury districts of London and entrusted with secrets and techniques she could particularly now not know...

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**Example text**

66 for details) oo (27) ZT{s) = Yl J i t 1 ~ N^o)~s~k] for Re(6) > 1; Pell equations and Fuchsian groups 33 the first product being taken over all the conjugacy classes of F, where (28) 7 /N(y) 1 ~ v- 0 1 0 7o being a generator of the class. In brief, the remarkable fact is that the eigenvalues A are directly related to the zeros s of Zr(s) by the simple relation: (29) A = $(1 -s). It is known that Z-p(s) is an entire function of s which satisfies the Riemann Hypothesis in the sense that all its non-real zeros reside on the line Re(s) = ^.

2 in [7]), Gn-z has a torsion-free subgroup H of finite index. Then by Prop. 6 in [21], cd{C00 x H) < 1 + cd(H) (since cd(Coo) = 1), hence vcd((Coo x G n _ 3 ) < 1 + vcd(Gn-z). Therefore, by [21] Prop. 15, vcd(Gn) < mdLx{vcd(Gn-2), 1 + vcd(Gn-3)}. Euler characteristic of graph products PROPOSITION 45 6. For all n > 1, vcd(Gn) = [-^J = w(Gn). PROOF. The proof is by induction on n. If n = 1, then Gn is cyclic of order 2, so vcd(G\) = 0. For n = 2, Gn is the infinite dihedral group, which has an infinite cyclic subgroup of finite index, so vcd(G2) — 1.

A finitely generated Coxeter group Cp is called aspherical by Pride and Stohr [19] if F contains no triangles A such that CA is finite. In this case the formula of Proposition 3 simplifies to 42 Chiswell where the sum is over all edges e (recall that edges are unoriented and are viewed as unordered pairs of vertices, n is the number of vertices of Y and is the labelling function). This formula is valid without the assumption in [19] that Y has no isolated vertices. It can also be obtained from their results.

### Mary Reilly by Valerie Martin

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