Concrete mathematics : a foundation for computer science

also has nice coefficients,
7.3 SOLVING RECURRENCES 325
+ . . . + al P? * (7.27)
We will show that every rational function R(z) such that R(0) # 00 can be
expressed in the form
R(z) = S(z) t T(z), (7.28)
where S(z) has the form (7.26) and T(z) is a polynomial. Therefore there is a
closed form for the coefficients [z”] R(z). Finding S(z) and T(z) is equivalent
to finding the “partial fraction expansion” of R(z).
Notice that S(z) = 00 when z has the values l/p,, . . . , l/pi. Therefore
the numbers pk that we need to find, if we’re going to succeed in expressing
R(z) in the desired form S(z) + T(z), must be the reciprocals of the numbers
&k where Q(ak) = 0. (Recall that R(z) = P(z)/Q(z), where P and Q are
polynomials; we have R(z) = 00 only if Q(z) = 0.)
Suppose Q(z) has the form
Q(z) = qo+q1z+~~~+q,z”‘, where qo # 0 and q,,, # 0.
The “reflected” polynomial
QR(z) = qoP+ q,z"-' +...f q,,,
has an important relation to Q (2):
QR(4 = qo(z - PI 1. . . (2 - P,)
w Q(z) = qo(l -PIZ)...(~ -P~z)
Thus, the roots of QR are the reciprocals of the roots of Q, and vice versa.
We can therefore find the numbers pk we seek by factoring the reflected polynomial
QR(z).
For example, in the Fibonacci case we have
Q(z) = 1 -2-z’; QR(z) = z2-z-l.
The roots of QR ca.n be found by setting (a, b, c) = (1, -1, -1) in the quadratic
formula (-b II: da)/2a; we find that they are
l+ds
+=2
1-d
a n d $ = 2
Therefore QR(z) = (z-+)(2-$) and Q(z) = (1 -+z)(l -i$z).