Concrete mathematics : a foundation for computer science

344 GENERATING FUNCTIONS
Step 3 is also easy. We solve for C(z) by the quadratic formula:
C(z) = 1*di-=G
22
But should we choose the + isign or the - sign? Both choices yield a function
that satisfies C(z) = K(z)’ -1- 1, but only one of the choices is suitable for our
problem. We might choose the + sign on the grounds that positive thinking
is best; but we soon discover that this choice gives C(0) = 00, contrary to
the facts. (The correct function C(z) is supposed to have C(0) = Cc = 1.)
Therefore we conclude that
1-Jl-42
C(z) = 2z *
Finally, Step 4. What is [zn] C(z)? The binomial theorem tells us that
k>O
hence, using (5.37),
(‘f) (-4zjk = 1 + g & (rl/Y) (-4z)k ;
= t (--‘/‘>~ = x (;)A$
nao ll)O
,
The number of ways to parenthesize, C,, is (‘,“) &.
We anticipated this result in Chapter 5, when we introduced the sequence So the convoof
Catalannumbers (1,1,2,5,14,. . . ) = (C,). This sequence arises in dozens luted recurrence
of problems that seem at first to be unrelated to each other [41], because
many situations have a recursive structure that corresponds to the convolution
recurrence (7.65).
For example, let’s consider the following problem: How many sequences
(al,a2.. . , al,,) of +1's and -1's have the property that
al + a2 +. . . + azn = 0
and have all their partial sums
al, al +a2, . . . . al +a2+...+aZn
nonnegative? There must be n occurrences of fl and n occurrences of -1.
We can represent this problem graphically by plotting the sequence of partial
has led us to an
oft-recurring convolution.