Math 229: Introduction to Analytic Number Theory Čebyšev (and von Mangoldt and Stirling)

Transcription

1 ath 9: Introduction to Analytic Number Theory Čebyšev (and von angoldt 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 Čebyšev. 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 Čebyšev prove Bertrand s Postulate : every interval (x, x) with x > contains a prime. See [HW 996, p.343 4] for Erdős s simplification of Čebyšev s proof; this simplified proof is also on the Web: problems/kuropatwa html.) For us Čebyšev s method also has the advantage of introducing the von angoldt function and the Stirling approximation to x!, both of which will figure prominently in our future analysis. It is well known that for any prime p and positive integer x the exponent of p in x! (a.k.a. the p-valuation of x!) is c p (x) := x x x + p p + p 3 + = x p k, the sum being finite because eventually p k > x. It was Čebyšev s insight that one could extract information about π( ) from the resulting formula x! = p p cp(x), or equivalently log x! = p c p (x) log(p) = n= x n Λ(n), () where Λ(n) is the von angoldt function { log p, if n = p k for some positive integer k and prime p; Λ(n) := 0, otherwise. To make use of () we need to estimate log x! = x log n n= If only thanks to the perennial problems along the lines of how many zeros end 05!?.

2 for large x. We do this by in effect applying the first few steps of symmetrized Euler-aclaurin summation, to find: Lemma. There exists a constant C such that holds for all positive integers x. log x! = (x + ) log x x + C + O(/x) () Proof : For any C function f we have (by integrating by parts twice) [ / f(y) dy = f(0) + 0 f (y) ( y + ) / dy + f (y) ( ] y ) dy / / 0 = f(0) + / f (y) y + dy, / where z is the distance from z to the nearest integer. Thus N f(k) = N+ / f(y) dy N+ Taking f(y) = log(y) and N = x we thus have The integral is log x! = (x + ) log(x + ) + log x + and the other terms are from which () follows. f (y) y + dy. x+ y + dy y + O(/x), (x + ) log x x + (log + ) + O(/x), y + dy y. [Stirling also determined the value of C (which turns out to be log(π), as we shall soon see), and extended () to an asymptotic series for x!/((x/e) x πx ) in inverse powers of x. But for our purposes log x! = (x + ) log x x + O() is more than enough. In fact, since for the time being we are really dealing with log x! and not log Γ(x + ), the best honest error term we can use is O(log x).] Now let Then from () and () we have ψ(x) := n x Λ(n). ψ(x/k) = (x + ) log x x + O().

3 This certainly suggests that ψ(x) x, and lets us prove upper and lower bounds on ψ(x) proportional to x. For instance, since x + x/m for all x, we have x ψ(x) log x! log m!, which yields ψ(x) [ ] m m log x + O(log x) = ( log )x + O(log x). For a lower bound we can use the inequality ψ(x) ( ) k x! ψ(x/k) = log = (log )x + O(log x) (x/)! for an even integer x = n; This is essentially the same tactic of factoring ( ) n n that Čebyšev used to prove π(x) > π(x). It is true that we are ultimately interested in π(x), not ψ(x). But it is easy to get from one to the other. For one thing, the contribution to ψ(x) of prime powers p k with k > is negligible certainly less than log x k= x/k log x x / log x. (The k = term is O(x / log x); the remaining terms sum to O(x /3 (log x) ), which is negligible in comparison. Once we know that π(x) x/ log x we shall be able to save another factor of log x in the error estimate, reducing it to O(x / ).) The remaining sum, p x log p, can be expressed in terms of π(x) and vice versa using partial summation, and we find: ψ(x) = log(x)π(x) π(x) = ψ(x) log x + x x π(y) dy y + O(x/ log x), dy ψ(y) y log y + O(x/ ). It follows that the Prime Number Theorem π(x) x/ log x holds if and only if ψ(x) x, and good error terms on one side imply good error terms on the other. It turns out that we can more readily get at ψ(x) than at π(x); for instance, ψ(x) is quite well approximated by x, while the right estimate for π(x) is not x/ log x but (x/ log x) + x dy/ log y, i.e., the logarithmic integral x dy/ log y. It is in the form ψ(x) x that we ll actually prove the Prime Number Theorem. Exercises On Čebyšev s method:. How many consecutive 0 s are there at the end of the base- expansion of 05!? Why did I choose rather than any smaller base (including the default 0), and what other bases less than 00 would serve the same purpose? 3

4 . Since our upper and lower asymptotic bounds log, log 4 on ψ(x)/x are within a factor of of each other, they do not quite suffice to prove Bertrand s Postulate. But any improvement would prove that π(x) > π(x) for sufficiently 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 (). For instance, show that ψ(x) < ( 4 log 08)x + O(log x). Can you obtain Čebyšev s bounds of 0.9 and.06? In fact it is known [Diamond-Erdős 980] that the upper and lower bounds can be brought arbitrarily close to, 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(x) > π(x) for all x x 0, which reduces Bertrand s Postulate to the finite computation of verifying π(x) > π(x) for each x (, x 0). This can be done by calculating a sequence of O(log x 0) primes, 3, 5, 7, 3, 3,..., 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 ɛ > 0 there exists x 0 such that π(( + ɛ)x) > π(x) for all x x Estimate log (m + n ), where the product extends over all (m, n) Z such that 0 < m + n 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, mod 4) x/ log x? Bernoulli polynomials, Euler-aclaurin summation, and efficient computation of ζ(s) and L(s, χ): 4. The Bernoulli polynomials B n (x) are defined for n = 0,,, 3,... by the generating function te xt e t = B n (x) tn n!. n=0 The Bernoulli numbers B n are the rational numbers B n (0), with generating function t/(e t ) = n=0 B nt n /n!. The first few Bernoulli polynomials are B 0 (x) =, B (x) = x, B (x) = x x + 6, B 3 (x) = x 3 3 x + x, B 4(x) = x 4 x 3 + x 30. Show that in general B n (x) = n ( n k=0 k) Bk x n k ( = (B+x) [n] mnemonically), that B n(x) = nb n (x), and that B n (n =,, 3,...) is the unique polynomial such that B n (x + ) B n (x) = nx n and 0 B n(x) dx = 0. Show that the Bernoulli number B n vanishes for odd n >. What is B n (x) + B n (x + )? The authors report that this result was obtained in 937 by Erdős and László Kálmár, and independently by J. Barkley Rosser at the same time, but apparently was never published before

5 5. Now let f be a C n function on [t, t + ]. Prove that f(t) = t+ t f(x) dx + n + ( ) n+ t+ t ( f (m ) (t + ) f (m ) (t) ) m! f (n) (x) B n(x t) n! Therefore, if f is a C n function on [, N] for some integers, N then N n= f(n) = N f(x) dx + + ( ) n+ N and if f is C n on [, ) then f(n) = n= + ( ) n+ n dx. ( f (m ) (N) f (m ) () ) m! f (n) (x) B n(x x ) n! f(x) dx n f (n) (x) B n(x x ) n! dx; m! f (m ) () dx, (3) provided the integrals converge and each f (m ) (N) 0 as N. This is a rigorous form of the Euler-aclaurin formula f(n) = n= f(x) dx m! f (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 > one can efficiently compute ζ(s) to within exp( N) in time N O() by taking f(x) = x s in (3) and choosing, 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(, χ 4 ) = = 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 > 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( s, χ) and thus achieves the analytic continuation to C in one step. ore about ψ(x): 5

6 6. Show that log p = ψ(x) ψ(x / ) ψ(x /3 ) ψ(x /5 ) + ψ(x /6 ) = µ(k)ψ(x /k ), p x where µ is the öbius function taking the product of r 0 distinct primes to ( ) r and any non-square-free integer to 0. Finally, another elementary approach to estimating π(x) that gets within a constant factor of the Prime Number Theorem: 7. Let P (u) be any nonzero polynomial of degree d with integer coefficients; then Thus 0 P (u) n du /lcm(,,..., dn + ) = exp( ψ(dn + )). ψ(dn + ) > n log min / P (u). 0<u< For instance, taking f(u) = u u we find ψ(4n + ) > 4n log. This is essentially the same (why?) as Čebyšev s trick of factoring ( n n ), but suggests different sources of improvement; try for instance f(u) = (u u )( u), or better yet f(u) = (u u ) ( u). [Unfortunately here the lower bound cannot be brought up to ɛ; see [ontgomery 994, Chapter 0] thanks to adhav Nori for bringing this reference to my attention.] References [Diamond-Erdős 980] Diamond, H.G., Erdős, P.: On sharp elementary prime number estimates, Enseign. ath. () 6 #3 4 (980), [HW 996] Hardy, G.H., Wright, E..: An Introduction to the Theory of Numbers, 5th ed. Oxford: Clarendon Press, 988 [AB / QA4.H37]. [ontgomery 994] ontgomery, H.L.: Ten lectures on the interface between analytic number theory and harmonic analysis. Providence: AS, 994 [AB ]. 6