Concrete mathematics : a foundation for computer science

Actually I’m not
into dominance.
9.6 FINAL SUMMATIONS 473
We have to make D, small enough that we can make a good estimate; for
example, we had better not let k get near n, or the term O((n - k)-‘) in
(9.95) will blow up. Yet D, must be large enough that the tail terms (the
terms with k @ Dn) are negligibly small compared with the overall sum. Trial
and error is usually necessary to find an appropriate set D,; in this problem
the calculations we are about to make will show that it’s wise to define things
as follows:
kED, % Ikl < n”‘+‘. bg6)
Here E is a small positive constant that we can choose later, after we get to
know the territory. (Our 0 estimates will depend on the value of e.) Equation
(9.95) now reduces to
lnok(n) = (2n+~)ln2-o-~lnn+O(n~‘)
- (n+k+i) ln(l+k/n) - (n--k+:) 141-k/n). (9.97)
(We have pulled out the large parts of the logarithms, writing
ln(nfk) = lnnfln(1 &k/n),
and this has made a lot of Inn terms cancel out.)
Now we need to expand the terms ln(l f k/n) asymptotically, until we
have an error term that approaches zero as n -+ 00. We are multiplying
ln( 141 k/n) by (n f k+ i), so we should expand the logarithm until we reach
o(n-‘), using the assumption that Ikl 6 n1/2+E:
In l*k = *t-$+O(nP3/2+3C).
( >
Multiplication by n f k + i yields
k2 k2
fk - 2n + ; + O(n-“2+3’) ,
plus other terms that are absorbed in the 0(n-‘/2+3e). So (9.97) becomes
lnok(n) = (2n+f)ln2-o-~lnn-k2/n+O(nP”2+3’).
Taking exponentials, we have
ak(n) = kw8)