8/18/2019 Mathematical thinking Solution Manual IV
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233 Chapter 16: Differentiation 234
The home run hitters (probability p/4) score
exactly k runs if and only
if there are exactly k home runs before the
third out. This has probabilityak =
k +22
( p/4)k (1− p/4)3. The expectation is
∞k =0
kak = ∞k =0
12
(k + 2)(k + 1)k ( p4
)k (1− p4
)3 = 3 p4− p .
Here we have used
∞k =0
(k + 2)(k + 1)kx k =
x (∞k =0
x k ) = x ( 11−
x )
= 6 x (1− x )−4.
The singles hitters score exactly k >
0 runs if and only if there are
exactly k + 2 singles before the third
out. This has probability bk =k +42
pk +2(1− p)3. The expectation is
∞k =1
kbk =1
2 p2(1− p)3
∞k =1
(k + 4)(k + 3)kpk .
To sum the series, we use
(k +4)(k +3)k =
(k +3)(k +2)(k +1)+ (k +2)(k +1)−
2(k + 1)− 6 to obtain∞k =1
(k +4)(k +3)kx k =∞k =0
(k +4)(k +3)kx k = ( 11− x )
+( 11− x )
−2( 11− x )
−6 11− x
= 6(1− x )4 +
2
(1− x )3 −2
(1− x )2 −6
1− x Setting x =
p and inserting this into the formula for the expectation
yields p2( 3
1− p + 1− (1− p)− 3(1− p)2).
As p approaches 1, the home run hitters score
about .75 runs per in-ning, while the expectation for the
singles hitters grows without bound.
When p is very small, the home run hitters expect
about 3 p/4 runs per
inning, while the singles hitters expect only about 10 p3.
When p > .279
(approximately), the singles hitters do better.
16.66.n
k =0 k x k is a ratio of two polynomials in
x . We compute
n
k =0
kx k = x n
k =0
kx k −1 = x d dx
(
n
k =0
x k )
= x d dx
(1− x n+1
1− x ) = x (nn+1 x − (n
+ 1) x n + 1)
(1− x )2 .
16.67. If q is a polynomial, then∞
k =0 q(k ) x k is the ratio of
two polynomials
in x (for | x | 0 and that the
claim holds
for m − 1. We compute
∞k =0
k m x k = x d dx
∞k =0
k m−1 x k = x
d
dx pm−1( x )(1− x )m
= x mpm−1( x )+ (1−
x ) pm−1( x )(1− x )m+1
=
pm( x )
(1− x )m+1 ,
where pm( x ) =
x [mpm−1( x ) + (1 −
x ) pm−1( x )]. Since pm−1
is a polynomial,also pm is a polynomial.
Proof 2 (generating functions). It suffices to prove that
the statement
holds when q(k ) = k m
, because every polynomial is a finite sum of multi-
ples of such binomial coefficients (see Solution 5.29). By
Theorem 12.35,
∞k =
0 k m x k = x m/(1−
x )m+1.
16.68. The random variable X defined
for nonnegative integer n by
Prob ( X = n) = p(1− p)n ,
where 0 < p < 1.a) The
probability generating function φ for
X is given by φ (t ) = p[1−
(1− p)t ]−1. By definition, φ (t )
=∞n=0 Prob ( X = n)t n. Using the
geometricseries, we compute φ(t ) = p(1 −
p)nt n = p/(1 − (1 − p)t ). Since
φ(1)sums all the probability, we have φ (1) =
p/(1− (1− p)) = 1, as desired.
b) E ( X ) = (1− p)/ p. We
compute
E ( x ) =
nProb ( X = n) = φ(1) = p(1− p)(1−
(1− p)t )−2
t =1= 1− p
p.
c) Prob ( X ≤ 20) = 1− (1− p)21.
We compute
Prob ( X ≤ 20) = p20n=0
(1− p)n = p1− (1− p)21
1− (1− p) = 1− (1− p)21.
16.69. Curvature of y( x ) =
x n , where n ≥ 2.
