Page 295 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: Oyk/
0521861241c06 CB996/Velleman October 20, 2005 1:8 0 521 86124 1 Char Count= 0

Recursion 281
1
n
0
2
For another example, consider the sum 2 + 2 + 2 +· · · + 2 , which ap-
peared in the first example of this chapter. The ellipsis suggests that we might
n
2
0
1
be able to use a recursive definition. If we let f (n) = 2 + 2 + 2 + ··· + 2 ,
0 1 2 n
then notice that for every n ∈ N, f (n + 1) = 2 + 2 + 2 +· · · + 2 +
2 n+1 = f (n) + 2 n+1 . Thus, we could define f recursively as follows:
0
f (0) = 2 = 1;
for every n ∈ N, f (n + 1) = f (n) + 2 n+1 .
As a check that this definition is right, let’s try it out in the case n = 3:
f (3) = f (2) + 2 3
2
= f (1) + 2 + 2 3
1 2 3
= f (0) + 2 + 2 + 2
1
0
2
= 2 + 2 + 2 + 2 3
= 15.
Sums such as the one in the last example come up often enough that there
is a special notation for them. If a 0 , a 1 ,..., a n is a list of numbers, then the
n
sum of these numbers is written a i . This is read “the sum as i goes from
i=0
0to n of a i .” For example, we can use this notation to write the sum in the last
example:
n
i 0 1 2 n
2 = 2 + 2 + 2 + ··· + 2 .
i=0
More generally, if n ≥ m, then
n

a i = a m + a m+1 + a m+2 +· · · + a n .
i=m
For example,
6
2 2 2 2 2
i = 3 + 4 + 5 + 6
i=3
= 9 + 16 + 25 + 36 = 86.

The letted i in these formulas is a bound variable and therefore can be replaced
by a new variable without changing the meaning of the formula.
Now let’s try giving a recursive definition for this notation. We let m be an
arbitrary integer, and then proceed by recursion on n. Just as the base case for
an induction proof need not be n = 0, the base for a recursive definition can
also be a number other than 0. In this case we are only interested in n ≥ m,so