Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 527
If the harmonic
aa and a bb. Thus each bb behaves like a drone in the bee tree and each aa
behaves like a queen, except that the bee tree goes backward in time while
the rabbits are going forward. There are F,+l pairs of rabbits after n months;
F, of them are adults and F,-, are babies. (This is the context in which
Fibonacci originally introduced his numbers.)
numbers are worm
numbers, the Fi- 6.7 (a) Set k = 1 -- n and apply (6.107). (b) Set m = 1 and k = n- 1 and
bonacci numbers
are rabbit numbers.
apply (6.128).
6.8 55 + 8 + 2 becomes 89 + 13 + 3 = 105; the true value is 104.607361.
6.9 21. (We go from F, to F,+z when the units are squared. The true
answer is about 20.72.)
6.10 The partial quotients a~, al, az, . . . are all equal to 1, because C$ =
1 + 1 /c$. (The Stern-Brocot representation is therefore RLRLRLRLRL.. . .)
6.11 (-1)” = [n=O] - [n=l]; see (6.11).
6.12 This is a consequence of (6.31) and its dual in Table 250
6.13 The two formulas are equivalent, by exercise 12. We can use induction.
Or we can observe that znDn applied to f(z) = zx gives xnzX while 9” applied
to the same function gives xnzX; therefore the sequence (a’, 4’ ,a2,. . . ) must
relate to (z”Do,z’D’, z2D2,. . . ) as (x0, x1,x2,. ) relates to (x”, x1, x2,. . .).
6.14 We have
x(“i”) = (k+l)(~~~) +In-k)(x~~~l),
because (n+l)x= (k+l)(x+k-n)+(n-k)(x+k+l). (It suffices toverify
the latter identity when k = 0, k = -1, and k = n.)
6.15 Since A((‘Ak)) = (iTi), we have the general formula
Set x = 0 and appeal to (6.19).
= A”(x”) = 1y
(-l)mpi(x + j)”
j 0
6.16 An,k = tj>o oj { “i’}; this sum is always finite.
6.17 (a) [;I = [l:T.!,]. (b) /:I = n* = n! [n3 k]/k!. (c) IL/ = k!(z).
6.18 This is equivalent to (6.3) or (6.8). (It follows in particular that
o,(l) = -na,(O) = U&n! when n > 1.)
6.19 Use Table 258.
6’20 xl