By Thomas Heath

Quantity 1 of an authoritative two-volume set that covers the necessities of arithmetic and contains each landmark innovation and each vital determine. This quantity gains Euclid, Apollonius, others.

**Read Online or Download A History of Greek Mathematics, Volume 1: From Thales to Euclid PDF**

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**Additional info for A History of Greek Mathematics, Volume 1: From Thales to Euclid**

**Sample text**

Therefore the (V , H )-kernel of f (3,0) contains at most 7 elements. By inspection, one can see that the kernel of f (3,0) contains indeed seven different elements. Since θ (0123) = θ (1133), θ (0213) = θ (1213), and θ (2233) = θ (2323) = θ (3333), the kernel of the Baum–Sweet sequence contains at most three elements. In fact, as one can easily see the kernel of the Baum–Sweet sequence contains three elements. 2. The substitution-graph of the Baum–Sweet substitution. 3. Let A = {0, 1, 2, 3}, = x = Z, H (x j ) = x 2j and V = {x 0 , x 1 } then s : V × A → A deﬁned by 0 1 2 3 induces a (V , H )-substitution S on x0 0 1 2 2 x1 1 2 0 3 (Z, A).

1. The unit e of is in V . 2. For any γ ∈ there exist a unique v ∈ V and a unique γ ∈ such that γ = vH (γ ). For the sake of simplicity, we often call V a residue set (of H ). Remarks. 1. The condition e ∈ V is not really necessary for the deﬁnition of a residue set. However, for all what follows it is very convenient to ensure that e belongs to a residue set. 2. The number of elements in a residue set is equal to the index of the subgroup H ( ). Therefore all residue sets have the same cardinality.

6. t. the generating set E) with expansion ratio C > 1. If the index of the subgroup H ( ) is ﬁnite, then has polynomial growth. o Proof. 4, we have that B C (e) contains no equivalent points. The o n 2 n-th iterate H of H is expanding and B C n (e) contains no H n -equivalent points. 4, we obtain o B C n (e) ≤ d n 2 24 2 Expanding endomorphisms and substitutions for all n ∈ N. For α = log d log C the above inequality becomes Cn 2 o B C n (e) ≤ 2α 2 α . o Since the function r → B r (e) is increasing, we have for all real numbers r such that Cn C n+1 2 ≤ r < 2 the inequality o o B r (e) ≤ B C n+1 (e) ≤ 2α 2 C n+1 2 α = 2α Cn 2 α C α ≤ 2α dr α .