Concrete mathematics : a foundation for computer science

9.3 0 MANIPULATION 443
since the right side approaches zero as n t co. OK, we know that p = O(n2);
therefore log p = 0 (log n) and log log p = 0 (log log n). We can now conclude
from (9.45) that
lnp = Inn + O(loglogn) ;
in fact, with this new estimate in hand we can conclude that In In p = In Inn-t
0 (log log n/log n), and (9.45) now yields
lnp = Inn + lnlnn+ O(loglogn/logn)
And we can plug this into the right-hand side of (g.44), obtaining
p = nlnn+nlnlnn+O(n).
This is the approximate size of the nth prime.
We can refine this estimate by using a better approximation of n(n) in
place of (9.42). The next term of (9.31) tells us that
Get out the scratch proceeding as before, we obtain the recurrence
paper again, gang.
p = nlnp (1 i- (lnp) ‘)-‘(1 + O(l/logn)‘) ,
(9.46)
which has a relative error of 0( 1 /logn)2 instead of 0( 1 /logn). Taking logarithms
and retaining proper accuracy (but not too much) now yields
lnp = lnn+lnlnp+0(1/logn)
= Inn l+
(
lnlnp
Ann + O(l/logn)2) ;
lnlnn
lnlnp = lnlnn+ Inn +o(q$y,, .
Finally we substitute these results into (9.47) and our answer finds its way
out:
P, = nlnn+nlnlnn-n+n %+0(C). b@)
For example, when ‘n = lo6 this estimate comes to 15631363.8 + O(n/logn);
the millionth prime is actually 15485863. Exercise 21 shows that a still more
accurate approximation to P, results if we begin with a still more accurate
approximation to n(n) in place of (9.46).