Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 541
7.6 Let the solution to go = LX, gl = fi, gn = g,, I + 29, 2 + (-1)“~ be
4 n= A(n)& + B(n)13 + C(n)y. The function 2” works when 01 = 1, /3 = 2,
y = 0; the function (-1)” works when LX = 1, fi = -1, y = 0; the function
(-1)“n works when 01 = 0, 6 = -1, y = 3. Hence A(n) + 2B(n) = 2”,
A(n) -B(n) = (-l)“, and -B(n) + 3Cln) = (-l)%.
7.7 G(z) = (z/(1 --z)‘)G(z) + 1, hence
I bet that the controversial
“fan of
1 ~ 22 z2 z
G(z) = -- 1 -32+22 + 1 - =l+ 32 + 22 ;
order zero” does
have one spanning
we have gn = Fzn + ‘n=Oj.
tree. 7.8 Differentiate (1 - z) -x-l twice with respect to x, obtaining
Now set x = m.
((H,,, - H,)’ - (H$ ~ HL2’))
7.9 (n + l)(Hi - Hi2)) - 2n(H, ~ 1).
7.10 The identity Hkm,,2-HP,,Z = & +...+ f = 2Hlk-Hk implies that
tk (‘;) (‘; 5 (2H2k - Hk) = 4nH,.
7.11 (a) C(z) = A(z)B(z’)/(l - z). (b) zB’(z) = A(Zz)e’, hence A(z) =
$e ‘l’B’($). (c) A(z) = B(z)/(l -z)‘+‘, hence B(z) = (1 -z)‘+‘A(z) and we
have fk(r) = (‘l’)(-l)k.
7.12 C,. The nunibers in the upper row correspond to the positions of +1’s
in a sequence of +l ‘s and -1 ‘s that defines a “mountain range”; the numbers
in the lower row correspond to the positions of -1’s. For example, the given
array corresponds to
7.13 Extend the sequence periodically (let x,+k = Xk) and define s, =
x1 f...+x,. We have s, = 1, ~2~ = 21, etc. There must be a largest index
ki such that Sk, = j, Sk,+,,, = 1+ j, etc. These indices kl, . . . , kl (modulo m)
specify the cyclic shifts in question.
For example, in the sequence (-2,1, -1 ,O, 1, 1, -1, 1, 1,l) with m = 10
and1=2wehavek, =17,k2=24.
7.14 6 (z) = -2zG(z) + e(z)2 + z (be careful about the final term!) leads
via the quadratic formula to
l+Zz-VTT-Q
G(z) = ~ 2