Combinatorics Through Guided Discovery, 2004a

4.3. Generating Functions and Recurrence Relations 85
(ii)
q→1
[n]! q .
(iii)
q→1
[ m + n
n
]
.
q
Why is the limit in Part iii equal to the number of partitions (of any
number) with at most n parts all of size most m? Can you explain
bijectively why this quantity equals the formula you got? (h)
∗ (h)
What happens to [ m+n
n
] if we let q approach −1? (h)
q
4.3 Generating Functions and Recurrence Relations
Recall that a recurrence relation for a sequence a n expresses a n in terms of values
a i for i < n. For example, the equation a i =3a i−1 +2 i is a first order linear constant
coefficient recurrence.
4.3.1 How generating functions are relevant
Algebraic manipulations with generating functions can sometimes reveal the solutions
to a recurrence relation.
• Problem 211. Suppose that a i =3a i−1 +3 i .
(a) Multiply both sides by x i and sum both the left hand side and right
hand side from i =1to infinity. In the left-hand side use the fact that
∞∑
∞∑
a i x i =( x i ) − a 0
i=1
i=0
and in the right hand side, use the fact that
∞∑
∞∑
∞∑
∞∑
a i−1 x i = x a i x i−1 = x a j x j = x a i x i
i=1
i=1
(where we substituted j for i−1 to see explicitly how to change the limits
of summation, a surprisingly useful trick) to rewrite the equation
in terms of the power series ∑ ∞
i=0 a ix i . Solve the resulting equation
for the power series ∑ ∞
i=0 a ix i . You can save a lot of writing by using
a variable like y to stand for the power series.
(b) Use the previous part to get a formula for a i in terms of a 0 .
(c) Now suppose that a i =3a i−1 +2 i . Repeat the previous two steps for
this recurrence relation. (There is a way to do this part using what you
already know. Later on we shall introduce yet another way to deal
with the kind of generating function that arises here.) (h)
j=0
i=0