Concrete mathematics : a foundation for computer science

284 SPECIAL NUMBERS
We have now boiled down all the information in the Fibonacci sequence
to a simple (although unrecognizable) expression z/( 1 - z - 2’). This, believe
it or not, is progress, because we can factor the denominator and then use
partial fractions to achieve a formula that we can easily expand in power series.
The coefficients in this power series will be a closed form for the Fibonacci
numbers.
The plan of attack just sketched can perhaps be understood better if
we approach it backwards. If we have a simpler generating function, say
l/( 1 - az) where K is a constant, we know the coefficients of all powers of z,
because
1
- =
1 -az
1+az+a2z2+a3z3+~~~.
Similarly, if we have a generating function of the form A/( 1 - az) + B/( 1 - pz),
the coefficients are easily determined, because
A B
- -
1 - a2 +1+3z
= A~(az)"+B~(@)"
1120 ll?O
= xc Aa” + BBn)z” . (6.118)
n>o
Therefore all we have to do is find constants A, B, a, and 6 such that
A B z
1 - a2 t-m= ~~~
and we will have found a closed form Aa” + BP” for the coefficient F, of z”
in F(z). The left-hand side can be rewritten
A B A-A@+B-Baz
- -
1 -az +1-f3z = Il-az)(l-pz) ’
so the four constants we seek are the solutions to two polynomial equations:
(1 -az)(l -f32) = 1 -z-z2; (6.119)
(A-t-B)-(A@+Ba)z = z. (6.120)
We want to factor the denominator of F(z) into the form (1 - az)(l - (3~);
then we will be able to express F(z) as the sum of two fractions in which the
factors (1 - az) and (1 - Bz) are conveniently separated from each other.
Notice that the denominator factors in (6.119) have been written in the
form (1 - az) (1 - (3z), instead of the more usual form c(z - ~1) (z - ~2) where
p1 and pz are the roots. The reason is that (1 - az)( 1 - /3z) leads to nicer
expansions in power series.