Concrete mathematics : a foundation for computer science

442 ASYMPTOTICS
Problem 3: The nth prime number.
Equation (9.31) is an asymptotic formula for n(n), the number of primes
that do not exceed n. If we replace n by p = P,,, the nth prime number, we
have n(p) = n; hence
as n + 00. Let us try to “solve” this equation for p; then we will know the
approximate size of the nth prime.
The first step is to simplify the 0 term. If we divide both sides by p/lnp,
we find that nlnp/p + 1; hence p/lnp = O(n) and
O(&) = o(i&J = “(&I*
(We have (logp))’ < (logn))’ because p 3 n.)
The second step is to transpose the two sides of (g.42), except for the
0 term. This is legal because of the general rule
an= b, +O(f(n)) # b, = a,, +O(f(n)) . (9.43)
(Each of these equations follows from the other if we multiply both sides
by -1 and then add a, + b, to both sides.) Hence
P
- = n+O(&) = n(1 +O(l/logn)) ,
lnp
and we have
p = nlnp(1 + O(l/logn)) . (9.44)
This is an “approximate recurrence” for p = P, in terms of itself. Our goal
is to change it into an “approximate closed form,” and we can do this by
unfolding the recurrence asymptotically. So let’s try to unfold (9.44).
By taking logarithms of both sides we deduce that
lnp = lnn+lnlnp + O(l/logn) , (9.45)
This value can be substituted for lnp in (g.&, but we would like to get rid
of all p’s on the right before making the substitution. Somewhere along the
line, that last p must disappear; we can’t get rid of it in the normal way for
recurrences, because (9.44) doesn’t specify initial conditions for small p.
One way to do the job is to start by proving the weaker result p = O(n2).
This follows if we square (9.44) and divide by pn2,
P (lnp12
= ~ 1 + O(l/logn)) ,
7 P (