By combining the fact that kj index j is the size of

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Unformatted text preview: all (5.9.8), p. 386), and (7.10.43) insures that R (P) = R (B) = R (M) and N (P) = N (C) = N (M). Thus (7.10.39) is proved. If (7.10.38) is used to produce a full-rank factorization ∗ M = U1 (DV1 ), then, because D is nonsingular, ∗ ∗ ∗ ∗ P = (U1 D)(V1 (U1 D))−1 V1 = U1 (V1 U1 )−1 V1 . Equations (7.10.41) and (7.10.42) follow from (5.9.11), p. 386. Formulas (7.10.40) and (7.10.42) are useful because all good matrix computation packages contain numerically stable SVD implementations from which ∗ U1 and V1 can be obtained. But, of course, the singular values are not needed in this application. Copyright c 2000 SIAM Buy online from SIAM Buy from 7.10 Difference Equations, Limits, and Summability 635 Example 7.10.8 Shell Game. As depicted in Figure 7.10.2, a pea is placed under one of four shells, and an agile manipulator quickly rearranges them by a sequence of discrete moves. At the end of each move the shell containing the pea has been shifted either to the left or right by only one position according to the following rules. It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] 1 #1 1/2 1/2 #2 D E #3 1/2 #4 1/2 1 T H Figure 7.10.2 When the pea is under shell #1, it is moved to position #2, and if the pea is under shell #4, it is moved to position #3. When the pea is under shell #2 or #3, it is equally likely to be moved one position to the left or to the right. IG R Problem 1: Given that we know something about where the pea starts, what is the probability of finding the pea in any given position after k moves? Problem 2: In the long run, what proportion of time does the pea occupy each of the four positions? Y P Solution to Problem 1: Let pj (k ) denote the probability that the pea is in position j after the k th move, and translate the given information into four difference equations by writing p2 (k −1) 2 O C p1 (k ) = ⎛ p1 (k ) ⎞⎛ 0 ⎟⎜ ⎜ p3 (k −1) ⎜ p2 (k...
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