Concrete mathematics : a foundation for computer science

138 NUMBER THEORY
(We must add 1 to O(n) because of the final fraction $.) The sum in (4.59)
looks difficult, but we can determine m(x) indirectly by observing that
(4.60)
for all real x 3 0. Why does this identity hold? Well, it’s a bit awesome yet
not really beyond our ken. There are 5 Lx]11 + x] basic fractions m/n with
0 6 m < n < x, counting both reduced and unreduced fractions; that gives
us the right-hand side. The number of such fractions with gcd(m,n) = d
is @(x/d), because such fractions are m//n’ with 0 < m’ < n’ 6 x/d after
replacing m by m’d and n by n’d. So the left-hand side counts the same
fractions in a different way, and the identity must be true.
Let’s look more closely at the situation, so that equations (4.59) and
(4.60) become clearer. The definition of m(x) implies that m,(x) = @(lx]);
but it turns out to be convenient to define m,(x) for arbitrary real values, not (This extension to
just for integers. At integer values we have the table
real values is a use-
ful trick for many
n 0 12 3 4 5 6 7 8 9 10 11 12
recurrences that
arise in the analysis
v(n) -112 2 4 2 6 4 6 4 10 4 of algorithms.)
o(n)
0 1 2 4 6 10 12 18 22 28 32 42 46
and we can check (4.60) when x = 12:
@,(12) + D,(6) +@(4) f@(3) + O(2) + m,(2) +6.@,(l)
= 46+12+6+4+2+2+6 = 78 = t.12.13.
Amazing.
Identity (4.60) can be regarded as an implicit recurrence for 0(x); for
example, we’ve just seen that we could have used it to calculate CD (12) from
certain values of D(m) with m < 12. And we can solve such recurrences by
using another beautiful property of the Mobius function:
g(x) = x f(x/d)
da1
tr’ (4.61)
This inversion law holds for all functions f such that tk,da, If(x/kd)I < 00;
we can prove it as follows. Suppose g(x) = td3, f(x/d). Then
t Ad)g(x/d) = x Ad) x f(x/kd)
d>l d>l k>l
= x f(x/m) x vL(d)[m=kdl
lTt>l d,kal