Concrete mathematics : a foundation for computer science

9.3 0 MANIPULATION 449
This preliminary analysis indicates that we’ll find it advantageous to
write
‘(n) = ;fP(k((;) +0(1))2 = ;fp(k)((;)2+o(;))
k=l k=l
= ;&‘i(;)l+fo(;)
k=l k=l
= ifIdk)(E)l + O(nlogn)
This removes the floors; the remaining problem is to evaluate the unfloored
sum 5 x.L, p(k)n2,/k2 with an accuracy of O(nlogn); in other words, we
want to evaluate ,Fi’=, p(k)l/k’ with an accuracy of O(n-’ logn). But that’s
easy; we can simply run the sum all the way up to k = cq because the newly
added terms are
k>n T
= O(g2) = O(&x.&)
k>n
= O (kJA -t,) = o(A).
We proved in (7.88) that tk>, F(k)/k’ = l/, k(k)/k’ =
1 /k2) = 6/7r2, and we have our answer:
‘/(tk>l
CD(n) = $n2 + O(nlogn). (9.56)
9.4 TWO ASYMPTOTIC TRICKS
Now that we have some facility with 0 manipulations, let’s look at
what we’ve done from a slightly higher perspective. Then we’ll have some
important weapons in our asymptotic arsenal, when we need to do battle
with tougher problems.
nick 1: Boots trapping.
When we estimated the nth prime P, in Problem 3 of Section 9.3, we
solved an asymptotic recurrence of the form
P, = nlnP,(l + O(l/logn)) .
We proved that P, = nln n + O(n) by first using the recurrence to show
the weaker result O(n2). This is a special case of a general method called
bootstrapping, in which we solve a recurrence asymptotically by starting with
k=l