FractionalGeometry-Chap2

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 =...
View Full Document

This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

Ask a homework question - tutors are online