FractionalGeometry-Chap2

# B compute the hausdor distance ht n s g for all n

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Unformatted text preview: 1]. The smallest ǫ which works is half the length of this gap, so [0, 1] ⊆ (T ([0, 1]))1/8 . This shows h([0, 1], T ([0, 1])) = 1/8. (b) First, T 2 ([0, 1]) ⊆ T ([0, 1]) = (T ([0, 1]))0 . To ﬁnd the minimum ǫ for which T ([0, 1]) ⊆ (T 2 ([0, 1]))ǫ , we must ﬁnd the largest gap in T 2 ([0, 1]) contained in an interval of T ([0, 1]). The easiest approach, and a good way to discover the pattern, is to note T ([0, 1]) = [0, 1/2] ∪ [3/4, 1] T 2 ([0, 1]) = [0, 1/4] ∪ [3/8, 1/2] ∪ [3/4, 7/8] ∪ [15/16, 1] The largest gap to ﬁll is (1/4, 3/8), so T ([0, 1]) ⊆ (T 2 ([0, 1]))1/16 , Consequently, h(T ([0, 1]), T 2 ([0, 1])) = 1/16. (c) The largest gap of T n+1 ([0, 1]) in T n ([0, 1]) lies in the left-most interval, [0, 1/2n], of T n ([0, 1]). That gap has length 1/2n+2 and is ﬁlled by an ǫ = 1/2n+3-nbhd of T n+1 ([0, 1]). That is, h(T n ([0, 1]), T n+1 ([0, 1])) = 1/2n+3 . (d) The largest agp of T n+m ([0, 1]) in T n ([0, 1]) again is the gap of length 1/2n+2 found in (c). Then for all m ≥ 1, we have h(T n ([0, 1]), T n+m ([0, 1])) = 1/2n+3 . PrxsSol 2.3.2 No: for example, take A =...
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## This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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