Concrete mathematics : a foundation for computer science

448 ASYMPTOTICS
We might as well do Ez now, since it is so easy:
=2 = x(&&T)
k>l
This is the telescoping series (1 -;)+(;-$)+($-&)+... =l.
Finally, X1 gives us the leading term of S,, the coefficient of Inn in (9.53):
=1 = x k($2 - &).
k>l
Thisis (l-i)+(i-$)+(G-&)+... = $+$+$+- =HE’ =7r2/6. (If
we hadn’t applied summation by parts earlier, we would have seen directly
that S, N xk3,(lnn)/k2, because H,t-, -H,tmlP1 N Inn; so summation by
parts didn’t help us to evaluate the leading term, although it did make some
of our other work easier.)
Now we have evaluated each of the E’s in (g.53), so we can put everything
together and get the answer to Golomb’s problem:
S, = glnn+,-&+0(h),
Notice that this grows more slowly than our original hand-wavy estimate of
C(logn)‘. Sometimes a discrete sum fails to obey a continuous intuition.
Problem 6: Big Phi.
Near the end of Chapter 4, we observed that the number of fractions in
the Farey series 3,, is 1 + (#J (n) , where
O(n) = q(l) +(p(2) +...+cP(n);
and we showed in (4.62) that
@(n) = i 1 p(k) ln/k1 11 + n/k1 .
k21
(9.55)
Let us now try to estimate cD(n) when n is large. (It was sums like this that
led Bachmann to invent O-notation in the first place.)
Thinking BIG tells us that Q(n) will probably be proportional to n2.
For if the final factor were just Ln/k] instead of 11 + n/k], we would have
(0(n)( < i xka, [n/k]’ 6 i xk>,(n/k)2 = $n2, because the Mobius function
p(k) is either -1, 0, or +l: The additional ‘1 + ’ in that final factor
adds xka, p(k) Ln/k] ; but this is zero for k > n, so it cannot be more than
nH, = O(nlog n) in absolute value.