Concrete mathematics : a foundation for computer science

The generating function derived earlier is
(qz)
1 1 1 1 1
=- --~
1 AZ 1 F-5 1 pz10 1 pz25-.
1 -z50 '
7.3 SOLVING RECURRENCES 331
this is a rational function of z with a denominator of degree 91. Therefore
we can decompose the denominator into 91 factors and come up with a 91term
“closed form” for C,, the number of ways to give n cents in change.
But that’s too horrible to contemplate. Can’t we do better than the general
method suggests, in this particular case?
One ray of hope suggests itself immediately, when we notice that the
denominator is almost a function of z5. The trick we just used to simplify the
calculations by noting that 1 - 4z2 + z4 is a function of z2 can be applied to
C(z), ifwe replace l/(1 -2) by (1 +z-tz2+z3 +z4)/(1 -z5):
C(z) = - 1 + 2 -t z2 + 23 + z4
-___--
1 1 1 1
1-S 1 M-5 1 vz10 1 yz25 1 pz50
= (1+z+z2+z3+24)c(z5),
11 1 1 1
C(Z) = - .- - - ~
1-21-21-2~1-251-2'0'
The compressed function c(z) has a denominator whose degree is only 19,
so it’s much more tractable than the original. This new expression for C(z)
shows us, incidentally, that Csn = Csn+’ = C5n+2 = Csn+3 = C5,,+4; and
indeed, this set of equations is obvious in retrospect: The number of ways to
leave a 53{ tip is the same as the number of ways to leave a 50# tip, because
the number of pennies is predetermined modulo 5.
Now we’re also But c(z) still doesn’t have a really simple closed form based on the roots
getting compressed
reasoning.
of the denominator. The easiest way to compute its coefficients of c(z) is
probably to recognize that each of the denominator factors is a divisor of
1 - 2”. Hence we can write
A(z)
-c
( z ) = (1 -zlo)5 ' where A(z) =Ao+A’z+...+A3’z3’. (7.39)
The actual value of A(z), for the curious, is
(1 +z+... +z~')~(1+z2+~~~+z~)(l+2~)
= 1 +2z+4z2+6z3+9z4+13z5+18z6+24z7
+ 31z8 $- 39z9 + 452" + 522" + 57~'~ + 63~'~ + 67~'~ + 69~'~
+ 69~'~ t67z" + 63~'~ $57~'~ +52z20 +45z2' + 39~~~ $31~~~
+ 24~~~ t18~~~ + 13~~~ + 9z2' + 6zzs +4z29 +2z30 +z3' .