FractionalGeometry-Chap2

For the n 10 and k 4 example mentioned above a de

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: of diameter 1. To see that these realize the upper and lower bounds, observe T (AU ) = {(x, 0) : 0 ≤ x ≤ 1/2} has diameter 1/2, and T (AL ) = {(0, y ) : 0 ≤ y ≤ 1/3} has diameter 1/3. Exercises Prob 2.4.1 It is a consequence of the Contraction Mapping Principle that each contraction map T : R2 → R2 has a unique fixed point p. To locate this fixed point, the usual approach is to start with any point q ∈ R2 and begin applying T . The sequence {q, T (q ), T 2(q ), . . . } converges to p. For affine transformations T x a = y c b d x e + y f we can find explicit formulas for the fixed point. (a) Recalling the relation between determinants and invertible matrices, show x that assuming (a − 1)(d − 1) − bc = 0, the fixed point f of T is yf 1 xf −c(d − 1) + bf = yf (a − 2)(d − 1) − bc −f (a − 1) + ce (b) What can be said if this assumption is not valid? 2.5. FASTER CONVERGENCE: DE BRUIJN SEQUENCES 63 Prob 2.4.2 2. From the proof of Lemma 2.4.1, show (x0 , y0...
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