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Unformatted text preview: e two ways to prove this.
Verify the guess. Let g (x) = x(N x). Clearly, g (0) = g (N ) = 0. If 0 < x < N
then by considering what happens on the ﬁrst step we have
1
1
g (x) = 1 + g (x + 1) + g (x
2
2
If g (x) = x(N 1) x) then the righthand side is 1
1
= 1 + (x + 1)(N x 1) + (x 1)(N x + 1)
2
2
1
1
= 1 + [x(N x) x + N x 1] + [x(N x) + x
2
2
= 1 + x(N x) 1 = x(N x) (N x + 1)] Derive the answer. (1.25) implies that
g (x) = 1 + (1/2)g (x + 1) + (1/2)g (x 1) Rearranging gives
g (x + 1)
Setting g (1)
general that g (x) = g (0) = c we have g (2)
g (k ) g (k 2 + g (x)
g (1) = c
1) = c 2(k g (x
2, g (3)
1) 1)
g (2) = c 4 and in 53 1.10. INFINITE STATE SPACES*
Using g (0) = 0 and summing we have
0 = g (N ) = N
X 2(k c 1) = cN 2· k=1 N (N
2 1) Pm
since, as one can easily check by induction,
j =1 j = m(m + 1)/2. Solving
gives c = (N 1). Summing again, we see that
g (x) = x
X (N 1) 2(k 1) = x(N 1) x(x + 1) = x(N x) k=1 Example 1.50. Duration of nonfair games. Consider the gambler’s ruin
c...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

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