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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...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.
 Fall '14
 AmandaFolsom
 Geometry, Fractal Geometry, Limits, The Land

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