Krylov was made a hero of copyright c 2000 siam buy

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Unformatted text preview: ) ⎟ ⎜ 1 ⎟⎜ ⎜ 2 ⎟=⎜ or ⎜ ⎟⎜ ⎜ p2 (k −1) ⎜ p3 (k ) ⎟ ⎜ 0 p3 (k ) = + p4 (k −1) ⎠⎝ ⎝ 2 p3 (k −1) p4 (k ) 0 p4 (k ) = 2 p2 (k ) = p1 (k −1) + 1/2 0 0 1/2 1/2 0 0 1/2 0 ⎞ ⎛ p1 (k −1) ⎞ ⎟ ⎜ ⎟ ⎜ 0 ⎟ p2 (k −1) ⎟ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟. ⎟ ⎜ ⎟ ⎟ p3 (k −1) ⎟ ⎜ 1⎠ ⎝ ⎠ 0 p4 (k −1) The matrix equation on the right-hand side is a homogeneous difference equation p(k ) = Ap(k − 1) whose solution, from (7.10.4), is p(k ) = Ak p(0), and thus Problem 1 is solved. For example, if you know that the pea is initially under shell #2, then p(0) = e2 , and after six moves the probability that the pea is in the fourth position is p4 (6) = A6 e2 4 = 21/64. If you don’t know exactly where the pea starts, but you assume that it is equally likely to start under any one of the four shells, then p(0) = (1/4, 1/4, 1/4, 1/4)T , and the probabilities Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 636 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 for occupying the four positions after six moves are given by p(6) = A6 p(0), or ⎛ It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu ⎞⎛ p1 (6) 11/32 p2 (6) ⎟ ⎜ 0 ⎜ ⎝ ⎠=⎝ p3 (6) 21/32 0 p4 (6) ⎞⎛ ⎞ ⎛⎞ 0 21/64 0 1/4 43 1 ⎜ 85 ⎟ 43/64 0 21/32 ⎟ ⎜ 1/4 ⎟ ⎠⎝ ⎠= ⎝ ⎠. 0 43/64 0 1/4 256 85 21/64 0 11/32 1/4 43 Solution to Problem 2: There is a straightforward solution when A is a convergent matrix because if Ak → G as k → ∞, then p(k ) → Gp(0) = p, and the components in this limiting (or steady-state) vector p provide the answer. Intuitively, if p(k ) → p, then after awhile p(k ) is practically constant, so the probability that the pea occupies a particular position remains essentially the same move after move. Consequently, the components in limk→∞ p(k ) reveal the proportion of time spent in e...
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