FractionalGeometry-Chap2

# Because 718 36 we have hk k0 36 a m d c e b figure

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: /6 0 Taking K0 = {(x, 0) : 0 ≤ x ≤ 1}, the point of K farthest from K0 is a, and the shortest distance from a to K0 is the length of the segment ab, easily seen to be √ 3/6. Thus K ⊆ (K0 )√3/6 . See Fig. 2.18. The point of K0 farthest from K is b, and the shortest distance from K to b is the length of the segment cb for some point c ∈ K. But where is this point c? Is it on the line m perpendicular to the line determined by a and e? This is a plausible guess, but is not obviously correct, due to the wiggly nature of the Koch curve. However, using the law of cosines it is not so diﬃcult to see the distance from d to b is √ (eb2 + ed2 − 2 · eb · ed · cos(π/2))1/2 = 7/18, √ √ √ and so K0 ⊆ K√7/18 . Because 7/18 < 3/6, we have h(K, K0 ) = 3/6. a m d c e b Figure 2.18: Computing the Hausdorﬀ distance between K and K0 . √ Note we do not know if 7/18 is the minimum ǫ for which K0 ⊆ Kǫ , but √ this estimate is suﬃcient for showing h(K, K0 ) = 3/6....
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