Concrete mathematics : a foundation for computer science

452 ASYMPTOTICS
This last estimate follows because, for example,
k>n
(Exercise 54 discusses a more general way to estimate such tails.)
The third sum in (9.60) is
by an argument that’s already familiar. So (9.60) proves that
p%
9 n = 7 + 0 (log n/n)3
Finally, we can feed this formula back into the recurrence, bootstrapping once
more; the result is
en2/b
9 n = 7 + O(logn/n3)
(Exercise 23 peeks inside the remaining 0 term.)
Trick 2: Trading tails.
We derived (9.62) in somewhat the same way we derived the asymptotic
value (9.56) of O(n): In both cases we started with a finite sum but got an
asymptotic value by considering an infinite sum. We couldn’t simply get the
infinite sum by introducing 0 into the summand; we had to be careful to use
one approach when k was small and another when k was large.
Those derivations were special cases of an important three-step asymp- (This importotic
summation method we will now discuss in greater generality. Whenever tant method waS
we want to estimate the value of x k ok (n), we can try the following approach:
pioneered by
Lap/ace [195 ‘1.)
1 First break the sum into two disjoint ranges, D, and T,,. The summation
over D, should be the “dominant” part, in the sense that it includes
enough terms to determine the significant digits of the sum, when n is
large. The summation over the other range T,, should be just the “tail”
end, which contributes little to the overall total.
2 Find an asymptotic estimate
ak(n) = bk(n) + O(ck(n))
that is valid when k E D,. The 0 bound need not hold when k E T,.