Page 294 - HOW TO PROVE IT: A Structured Approach, Second Edition

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280 Mathematical Induction
in this way we can compute f (n) for any natural number n. Thus, the two
equations really do give us a rule that determines a unique value f (n) for each
natural number n, so they define a function f with domain N. Definitions of
this kind are called recursive definitions.
Sometimes we’ll work backwards when using a recursive definition to eval-
uate a function. For example, suppose we want to compute f (6), where f is
the function just defined. According to the second equation in the definition
of f, f (6) = 6 · f (5), so to complete the calculation we must compute f (5).
Using the second equation again, we find that f (5) = 5 · f (4), so we must
compute f (4). Continuing in this way leads to the following calculation:
f (6) = 6 · f (5)
= 6 · 5 · f (4)
= 6 · 5 · 4 · f (3)
= 6 · 5 · 4 · 3 · f (2)
= 6 · 5 · 4 · 3 · 2 · f (1)
= 6 · 5 · 4 · 3 · 2 · 1 · f (0)
= 6 · 5 · 4 · 3 · 2 · 1 · 1
= 720.

Perhaps now you recognize the function f. For any positive integer n,
f (n) = n · (n − 1) · (n − 2) ··· 1, and f (0) = 1. This number is called n facto-
rial, denoted n!. For example, 6! = 720. Often, if a function can be written as
a formula with an ellipsis (...) in it, then the use of the ellipsis can be avoided
by giving a recursive definition for the function. Such a definition is usually
easier to work with.
Many familiar functions are most easily defined using recursive definitions.
n
For example, for any number a, we could define a with the following recursive
definition:
0
a = 1;
n
for every n ∈ N, a n+1 = a · a.
4
Using this definition, we would compute a like this:
3
4
a = a · a
2
= a · a · a
1
= a · a · a · a
0
= a · a · a · a · a
= 1 · a · a · a · a.