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Math 259: Introduction to Analytic Number Theory Cebyˇ^ ˇ sev (and von Mangoldt and Stirling)
Before investigating ζ(s) and L(s, χ) as functions of a complex variable, we give another elementary approach to estimating π(x), due to Cebyˇˇ sev. This method, unlike Euler’s, produces upper and lower bounds on π(x) that remain within a small constant factor as x→∞. These bounds x/ log x π(x) x/ log x are sufficient for many theoretical and practical applications, which thus do not require the more advanced and subtle techniques that enter into the proof of the Prime Number Theorem. (The bounds are also close enough to let Cebyˇˇ sev prove “Bertrand’s Postulate”: every interval (x, 2 x) with x > 1 contains a prime. See [HW 1996, p.343–4] for Erd¨os’s simplification of Cebyˇˇ sev’s proof; this simplified proof is also on the Web: http://forum.swarthmore.edu/dr.math/ problems/kuropatwa.4.3.97.html .) For us Cebyˇˇ sev’s method also has the advan- tage of introducing the von Mangoldt function and the Stirling approximation to x!, both of which will figure prominently in our future analysis.
It is well known^1 that for any prime p and positive integer x the exponent of p in x! (a.k.a. the p-valuation of x!) is
cp(x) :=
x p
x p^2
x p^3
∑^ ∞
k=
x pk
the sum being finite because eventually pk^ > x. It was Cebyˇˇ sev’s insight that one could extract information about π(·) from the resulting formula
x! =
p
pcp(x),
or equivalently
log x! =
p
cp(x) log(p) =
∑^ ∞
n=
⌊ (^) x n
Λ(n), (1)
where Λ(n) is the von Mangoldt function
Λ(n) :=
log p, if n = pk^ for some positive integer k and prime p; 0 , otherwise.
To make use of (1) we need to estimate
log x! =
∑^ x
n=
log n
(^1) If only thanks to the perennial problems along the lines of “how many zeros end 2003! ?”.
for large x. We do this by in effect applying the first few steps of symmetrized Euler-Maclaurin summation, to find:
Lemma. There exists a constant C such that
log x! = (x +
) log x − x + C + O(1/x) (2)
holds for all positive integers x.
Proof : For any C^2 function f we have (by integrating by parts twice)
∫ (^1) / 2
− 1 / 2
f (y) dy = f (0) +
[∫
0
− 1 / 2
f ′′(y)
y +
dy +
0
f ′′(y)
y −
dy
]
= f (0) +
− 1 / 2
f ′′(y)
∥y +^1 2
∥^2 dy,
where ‖z‖ is the distance from z to the nearest integer. Thus
∑^ N
k=
f (k) =
∫ N + 12
1 / 2
f (y) dy +
∫ N + 12
1 2
f ′′(y)
∥y +^1 2
∥^2 dy.
Taking f (y) = log(y) and N = x we thus have
log x! = (x +
) log(x +
log 2 − x −
∫ (^) x+ (^12)
(^12)
∥y +^1 2
∥^2 dy y^2
The integral is
−
(^12)
∥y +^1 2
∥^2 dy y^2
and the other terms are
(x +
) log x − x +
(log 2 + 1) + O(1/x),
from which (2) follows.
[Stirling also determined the value of C (which turns out to be 12 log(2π), as we shall soon see), and extended (2) to an asymptotic series for x!/((x/e)x
2 πx ) in inverse powers of x. But for our purposes log x! = (x + 12 ) log x − x + O(1) is more than enough. In fact, since for the time being we’re really dealing with logbxc! and not log Γ(x + 1), the best honest error term we can use is O(log x).]
Now let ψ(x) :=
1 ≤n≤x
Λ(n).
Then from (1) and (2) we have
∑^ ∞
k=
ψ(x/k) = (x +
) log x − x + O(1).
large x, from which the proof for all x follows by exhibiting a few suitably spaced primes. It turns out that better bounds are available starting from (1). For instance, show that ψ(x) < ( 12 log 12)x + O(log^2 x). Can you obtain Cebyˇˇ sev’s bounds of 0.9 and 1.1? In fact it is known that the upper and lower bounds can be brought arbitrarily close to 1, but alas the only known proof of that fact depends on the Prime Number Theorem!
To recover Bertrand’s Postulate, one needs for once to convert all the O(·)’s to explicit error estimates. One then obtains an explicit x 0 such that π(2x) > π(x) for all x ≥ x 0 , which reduces Bertrand’s Postulate to the finite computation of verifying π(2x) > π(x) for each x ∈ (1, x 0 ). This can be done by calculating a sequence of O(log x 0 ) primes 2 , 3 , 5 , 7 , 13 , 23 ,... , p, each less than twice the previous prime, and with p > x 0. Once we prove the Prime Number Theorem it will follow that for each > 0 there exists x 0 such that π((1 + )x) > π(x) for all x ≥ x 0.
- Estimate log
(m^2 + n^2 ), where the product extends over all (m, n) ∈ Z^2 such that 0 < m^2 + n^2 ≤ x. What is the exponent of a prime p ≤ x in this product? Using this information, how close can you come to the asymptotic formula π(x, 1 mod 4) ∼ 12 x/ log x?
Bernoulli polynomials, Euler-Maclaurin summation, and efficient computation of ζ(s) and L(s, χ):
- The Bernoulli polynomials Bn(x) are defined for n = 0, 1 , 2 , 3 ,... by the generating function text et^ − 1
∑^ ∞
n=
Bn(x)
tn n!
The Bernoulli numbers Bn are the rational numbers Bn(0), with generating function t/(et^ − 1) =
n=0 Bnt
n/n!. The first few Bernoulli polynomials are
B 0 (x) = 1, B 1 (x) = x −
, B 2 (x) = x^2 − x +
B 3 (x) = x^3 −
x^2 +
x, B 4 (x) = x^4 − 2 x^3 + x^2 −
Show that in general Bn(x) =
∑n k=
(n k
Bkxn−k^ ( = “(B+x)[n]” mnemonically), that B′ n(x) = nBn− 1 (x), and that Bn (n = 1, 2 , 3 ,.. .) is the unique polynomial
such that Bn(x + 1) − Bn(x) = nxn−^1 and
0 Bn(x)^ dx^ = 0.^ Show that the Bernoulli number Bn vanishes for odd n > 1. What is Bn(x) + Bn(x + 12 )?
- Now let f be a Cn^ function on [t, t + 1]. Prove that
f (t) =
∫ (^) t+
t
f (x) dx +
∑^ n
m=
Bm m!
f (m−1)(t + 1) − f (m−1)(t)
∫ (^) t+
t
f (n)(x) Bn(x − t) n!
dx.
Therefore, if f is a Cn^ function on [M, N ] for some integers M, N then
N∑ − 1
n=M
f (n) =
∫ N
M
f (x) dx +
∑^ n
m=
Bm m!
f (m−1)(N ) − f (m−1)(M )
∫ N
M
f (n)(x) Bn(x − bxc) n!
dx;
and if f is Cn^ on [M, ∞) then
∑^ ∞
n=M
f (n) =
M
f (x) dx −
∑^ n
m=
Bm m!
f (m−1)(M )
M
f (n)(x) Bn(x − bxc) n!
dx, (3)
provided the integrals converge and each f (m−1)(N )→0 as N →∞. This is a rigorous form of the “Euler-Maclaurin formula”
∑^ ∞
n=M
f (n) =
M
f (x) dx −
∑^ ∞
m=
Bm m!
f (m−1)(M ),
which rarely converges (can you find any nonzero f for which it does converge?), but is often useful as an asymptotic series. For instance, show that for any s > 1 one can efficiently compute ζ(s) to within exp(−N ) in time N O(1)^ by taking f (x) = x−s^ in (3) and choosing M, n appropriately. Do the same for L(s, χ) where χ is any nontrivial Dirichlet character and s > 0. For instance, one can compute Catalan’s constant
G = L(2, χ 4 ) = 1 −
in this way.
We could also use (3) to obtain the analytic continuation of ζ(s) and L(s, χ) to the half-plane σ > 1 − n, and thus to the whole complex plane since n is arbitrary. But this is a less satisfactory approach than using the functional equation which relates L(s, χ) to L(1 − s, χ) and thus achieves the analytic continuation to C in one step.
More about ψ(x):
- Show that
∑
p≤x
log p = ψ(x) − ψ(x^1 /^2 ) − ψ(x^1 /^3 ) − ψ(x^1 /^5 ) + ψ(x^1 /^6 ) · · · =
∑^ ∞
k=
μ(k)ψ(x^1 /k),
where μ is the M¨obius function taking the product of r ≥ 0 distinct primes to (−1)r^ and any non-square-free integer to 0.
Finally, another elementary approach to estimating π(x) that gets within a constant of the Prime Number Theorem:
