Concrete mathematics : a foundation for computer science

326 GENERATING FUNCTIONS
Once we’ve found the p’s, we can proceed to find the partial fraction
expansion. It’s simplest if all the roots are distinct, so let’s consider that
special case first. We might a.s well state and prove the general result formally:
Rational Expansion Theorem for Distinct Roots.
If R(z) = P(z)/Q(z), where Q(z) = qo(l - plz) . . . (1 - pLz) and the
numbers (PI, . . . , PL) are distinct, and if P(z) is a polynomial of degree less
than 1, then
[z”IR(z) = a,p;+..+alp:,
-pkp(l/pk)
where ak = Q,fl,Pkl . (7.29)
Proof: Let al, . , . , a1 be the stated constants. Formula (7.29) holds if R(z) =
P(z)/Q(z) is equal to
S(z) = d!-
1 -P1Z
+...+al.
1 - PLZ
And we can prove that R(z) = S(z) by showing that the function T(z) =
R(z) - S(z) is not infinite as z + 1 /ok. For this will show that the rational Impress your parfunction
T(z) is never infinite; hence T(z) must be a polynomial. We also can ents bY leaving the
book open at this
show that T(z) + 0 as z + co; hence T(z) must be zero.
page.
Let ak = l/pk. To prove that lim,,,, T(z) # oo, it suffices to show that
lim,,., (z - cck)T(z) = 0, because T(z) is a rational function of z. Thus we
want to show that
lim (Z - ak)R(Z) = ;jzk (Z - xk)s(z) .
L’CCI,
The right-hand limit equals l.im,,,, ok(z- c&)/‘(l - pkz) = -ak/pk, because
(1 - pkz) = -pk(z-Kk) and (z-c&)/(1 - PjZ) -+ 0 for j # k. The left-hand
limit is
by L’Hospital’s rule. Thus the theorem is proved.
Returning to the Fibonacci example, we have P(z) = z and Q(z) =
1 - z - z2 = (1 - @z)(l - $2); hence Q’(z) = -1 - 22, and
-PP(l/P) = -1 P
Q/(1/p) -1 -2/p =p+2.
According to (7.2g), the coefficient of +” in [zn] R(z) is therefore @/(c$ + 2) =
l/d; the coefficient of $” is $/($ + 2) = -l/\/5. So the theorem tells us
that F, = (+” - $“)/fi, as in (6.123).